R a DeCarlo and P Lin Linear Circuit Analysis s

PROBLEM SOLUTIONS CHAPTER 1. Solution 1.1. (a) Charge on one electron: -1.6019×10-19 C. This means that charge on 1013 electrons is: -1.6019×10-6 C. Net charge on sphere is: 1.6019×10-6 C (POSITIVE). Solution 1.2. (a) 1 atom ≡ -4.646×10-18 C. By proportionality, 64g ≡ NA atoms . 3.1NA 3.1g ≡ ? atoms ⇒ 3.1g ≡ atoms . 64 Total Charge = −4.646 × 10 −18

C 3.1× 6.023 × 10 23 × atoms = − 1.3554 × 10 5 C atom 64

(b) Total charge per atom is -4.646×10-18 C. Total charge per electron is –1.6019×10-19 C. Therefore, there are 29 electrons per atom of copper. ∆Q ∆Q 1.36 × 105 (c) 0.91 A ≡ 0.91 C/s. i = ⇒ ∆t = = = 1.49 × 105 sec . ∆t i 0.91 3.1NA = 2.9174 × 10 22 atoms in the penny. Removing 1 electron from 64 3.1NA 3.1NA 0.05 × atoms means removing 0.05 × electrons. Therefore, 64 64

(d) We know there are

Net charge = 0.05 ×

(

3.1NA × 1.6019 × 10−19 = 234C 64

)

Solution 1.3 (a) 7.573 × 1017 × −1.6019 × 10 −19 = −0.1213C 0.1213 = 121.3A flowing from right to left. 10−3 (c) Again, use proportionality: (b) Current =

x × 1.6019 × 10 −19 10 × 60 10 A = ⇒ x= = 3.75 × 1021 −19 60sec 1.6019 × 10 dq (d) i( t ) = = 1− e −5t A. This is an exponential evolution with an initial value of 0, a final value of 1, dt and a time-constant of 1/5 (signal reaches ~63% of it’s final value in one time-constant).

i(t) 1

0.2

time in sec

(e) Current is the slope of the charge waveform. Therefore, by inspection:

Solution 1.4 (a) 6.023×1023×(-1.6019×10-19) = –9.65×104 C. (b) Current flows from right to left (opposite electrons), and: I=

9.65 × 10 4 = 9.65 × 107 A −3 10

(c) Using proportionality: 5A = (d) i( t ) =

x × 1.6019 × 10−19 60sec

⇒ x=

5 × 60 21 −19 = 1.87 × 10 1.6019 × 10

dq = 1+ 0.5π cos(πt ) ⇒ i (1sec ) = 1− 1.57 = −0.57A . Current flows from left to right. dt

Solution 1.5 (a) i( t ) = 1− 4e −2t + 3e −3t t ≥ 0. Then t

t

(

q(t ) = ∫ i(t )dt = ∫ 1 − 4e 0

−2τ

+ 3e

−3τ

)dτ =

0

τ ]t−∞

t

− 4∫ e 0

[

= t − 4 −0.5e −2τ

−2τ

t

dτ + 3∫ e −3τ dτ 0

]0 + 3[−0.333e−3τ ]0 = t + 2e−2t − e−3t − 1 t

t

(b) By inspection:

(c) q(t ) = 120cos(120πt ) . Hence i( t ) =

dq = −120π × 120sin (120πt ) = − 14400πsin(120πt ) A dt

Solution 1.6. (a) i(t) = 1 − cos(πt) A. Hence t

t

0

0

q(t) = ∫ i(τ)dτ = ∫ (1 − cos(πτ)) dτ = t −

t 1 1 sin(πτ) = t − sin(πt) C π 0 π

(b) Charge is integral of current. Graphically, the charge at time t is the area under the current curve up to time t: (note the quadratic nature between 2 and 4 seconds)

Solution 1.7 Again, Q is the running area under the current curve. Between 0 and 3 seconds, current decreases linearly until zero. So, Qtot = 7.5 C. From 0 to 6: Qtot = 7.5 + Q3_6 = 7.5 -1/1×0.5 + -1/1×0.5 + -1×1 = 5.5 C, where the curve from 3 to 6 was divided into two triangular sections and one rectangular one. Solution 1.8 Charge is the area under the current curve. Thus, Q = 0.1*4 – 0.1*2 = 0.2 C. Solution 1.9 Calculate the change in energy for the electron: ∆E = Q ∆V = 3.218×10-15. Equate this to kinetic energy: 3.218 × 10 −15 =

1 2 mv ⇒ v = 8.4 × 10 7 m / s 2

where the mass of an electron, 9.1066×10-31 has been substituted. 3

Solution 1.10 P = VI. Hence I = P/V = 2×10 /120 = 16.6667 A

PROBLEM Solution 1.11 (a) It is necessary to integrate the i(t) curve to obtain q(t). We do this interval by interval: t

(i) 0 ≤ t ≤ 1 ms, q(t) = 0 + ∫ dτ = t µC 0

t

(ii) 1 ms ≤ t ≤ 2 ms, q(t) = 1 − 2∫ dτ = 3 − 2t µC 1

t

(iii) 2 ms ≤ t ≤ 3 ms, q(t) = −1+ ∫ dτ = −3 + t µC 2

t

(iv) 3 ms ≤ t ≤ 5 ms, q(t) = 0 + ∫ (8 − 2τ )dτ = 8t − t 2 − 15 µC 3

(v) 5 ms ≤ t, q(t) = 0 µC

1 0.8 0.6

Charge in micro C

0.4 0.2 0 -0.2 TextEnd -0.4 -0.6 -0.8 -1

0

1

2

3 Time in ms

4

5

6

(b) Voltage is the ratio of the power and current curve. In this case, the division can be done graphically by inspection. Note that the ratio of a quadratic function and a linear function is a linear function:

2 1.5

Voltage in V

1 0.5 TextEnd

0 -0.5 -1

0

1

2

3 Time in ms

4

5

6

Solution 1.12 (a) VA = P/I = 20/4 = 5 V (b) PB = VI = 2×7 = 14 W (c) VC = P/I = -3W/3A = -1V (d) ID = P/V = -27W/3V = -9A (e) IE = P/V = 2/1 = 2A (f) PF = VI = -4×5 = -20W In all of the above, note that the direction of the current flow relative to the polarity of the voltage across a device determines whether power is delivered or absorbed. Power is absorbed when current flows from the positive terminal of the device to the negative one. Solution 1.13 (a) By inspection: Circuit Element (CE) 1 absorbs –5W, and CE 2 absorbs 6W. (b) Compute power absorbed by all elements including independent sources: I3A : -15 CE1: -5 V3V : -12 CE2: +6 V5V : 10 I2A : 16 ----------

Sum: 0 (Verifies conservation of power.) Solution 1.14 (a) Compute power absorbed: I5A : -85 CE1: 98 V3V : 33 CE2: 16 V7V : -42 I2A : -20 ------------Sum: 0 (b) Add all terms:

( ) V-source: Pabsorbed = −2(3e −t − 1) = −6e −t + 2 watts CE1: Pabsorbed = 3e −t × 3(1− e −t ) = 9e −t − 9e−2t watts CE2: Pabsorbed = (3e −t − 1)(3e −t − 1) = 9e −2t − 6e −t + 1 watts I-source:

Pabsorbed = −3 1− e −t = −3 + 3e −t watts

Simple algebraic manipulation of the the sum of all the above terms reveals that the result is zero. Solution 1.15 (a) When IL = 1, P = VLIL = (16-4)×1 = 12 W. When IL = 2, P = VLIL = (16-16)×1 = 0. (b) P = (16-4I L2 )IL. Differentiate this w.r.t. IL and set to zero: 16 – 12IL2 = 0. Therefore, IL = 1.155A. Solution 1.16 (a) When IL = 2, P = (16-4)×2 = 24W. When IL = 3, P = (16 - 9)×3 = 21 W. (b) Maximum occurs in the interval from 0 to 4: P = (16 - IL2 ) IL Differentiate w.r.t. IL and set to zero: 16 – 3IL2 = 0. Therefore, IL = 2.31 A. Solution 1.17 (a) Power is the product of the current and voltage. We can compute the product graphically:

10

Power in Watts

8 6 4 TextEnd 2 0

0

0.5

1

1.5 Time in s

2

2.5

3

(b) t

t

0

0

(

[

)

]

10 10 −7τ t W (t ) = ∫ p(t )dt = ∫ 10 − 10e −7τ dτ = 10τ ]t0 − − 10 = 10t + e −7t − 7 e 0 7 7

This can be used as an aid to plot the work function:

9 8 7

Energy in J

6 5 4 TextEnd 3 2 1 0

0

0.5

Solution 1.18 (a) Since, i( t ) = 115 − 23t mA ,

1 Time in s

1.5

2

7

23t 2  −3 q( 7) = ∫ i( τ) dτ = 115t −  × 10 = 0.2415 C 2  0 7

0

(b) Energy is the integral of power: 7

7

7

0

0

0

E = ∫ p(t )dt = ∫ v( t ) × i(t )dt = 25∫ i(t )dt = 25 × 0.2415 = 6.0375 C Solution 1.19 (a) ∆t = 100o F, Rate of temp. increase is 2.5 Wh/o F per gallon: Energy = 2.5Wh/o F/gallon×100 o F×30gallons = 7500 Wh = 2.7×107 J. (b) Heater generates P = 120×10 = 1200 W. We want 7500 Wh. Therefore, the total number of hours needed is 7500Wh/1200W = 6.25 h. Solution 1.20 First compute the change in temperature required, in o F: ∆t = 80-25 = 55o C = 55×9/5 o F= 99 o F Next, compute the energy spent every hour, which means on 40 gallons of water: E = 2.5 Wh/o F/gallon×99o F×40gallons = 9900 Wh Since the heater is not 100 % efficient, we spend more energy than is actually needed to heat the water: E_spent = 9900 Wh/0.9 = 11000 Wh So, far, this was the energy spent every hour. Over six hours, the total energy spent is: E6h = 11000×6 = 66,000 Wh Finally, the total energy spent per month is Em = 66,000×30 = 1980 kWh and the bill is 1980 kWh×0.14$/kWh = $277.2

Solution 1.21 Energy = 120 W × 6 h = 720 Wh = 0.72 kWh Therefore, cost per day = 0.72 kWh × 8 = 5.76 cents, and cost per month is 5.76×31 = $1.785.

Solution 1.22

We need to compute the difference between the inner diameter of the tube and the outer one in order to get the cross-sectional area: 2 2 area = πRout − π Rin = π0.0032 − π0.00182 = 1.81 × 10 −3 m2

Then, R = 1.7×10-5×(12/1.81) = 11.3 mΩ.

Solution 1.23 L = 20 m, W = 0.015m, H = 0.001 m. Thus, A = W×H, and R = 5.1×

copper

×L/A

= 0.116 Ω. Solution 1.24. (a) 500 ft, 20 gauge wire: 10.35 Ω/1000 ft from table 1.3. This implies that R = 5.175 Ω. (b) 55 ft, 20 gauge, nickel wire: 10.35 R = 5.1× × 55 = 2.9 Ω 1000 (c) Rtot = 2.9 + 5.175 = 8.08 Ω. Solution 1.25. R(T ) = R(20 )[1 + α (T − 20)] . Substituting at T = –10 yields: 21 = R(20)[1+ 0.0039( −30)] or R( 20) = 23.78 Ω Evaluating at T = +10 yields, R(10) = 23.78 + 23.78 × 0.0039 × ( −10 ) or R(10) = 22.85Ω

Solution 1.26. For tungsten, we know that α = 0.0045. Therefore: R(150) = R (20)[1 + α (T − 20)] = 200[1 + 0.0045(150 − 20)] = 317Ω Rate of change of resistance is (317-200)/(150-20) = 0.9 Ω/o C.

Solution 1.27. Plug numbers directly into the same formula as problem 1.26: 0.0022 = 0.002 + 0.002×0.0039(T-20) Rearrange to obtain: T = 45.64o C. Solution 1.28. (a) Power in a wire: P = I2R. Rearranging, we can express the current as I=

P

R.

Substitute given P and R to obtain I = 0.707 mA. (b) Use the same formula for current obtained above to get 50 A. Solution 1.29. Use formula for power: P = V2/R. Rearranging, R = V2/P = 96 Ω.

Solution 1.30 (a) I = V/R = 12 A, out of the positive terminal of the battery. (b) Up through the resistor. (c) Absorbed power by resistor: P = V2 /R = 14.4 W. Same power is delivered by source. (d) From table 1.2 and 1.3, 1000 feet of 18 AWG aluminum wire has resistance: »R1000ft = 1.6*6.51 R1000ft = 1.0416e+01 By proportionality, 1000 × 0.1 = L × 10.416 . Hence, »L = 100/10.416 L = 9.6006e+00 meters. Solution 1.31 (a) V = 10 V. (b) P = V2 /R, which means that R = V2 /P = 100/25 = 4 Ω. (c) I = V/R = 10/4 = 2.5 A. Current flow is downwards through resistor. (d) Up through resistor. (e) P = V2/R10 = 100/10 = 10 W. Hence, I10 = V/R10 = 1 A. Without applying material from a future chapter, a legitimate way to obtain Isource is to apply conservation of power first and then compute Isource from the power formula. Hence, Psource = 10 + 25 = 35 watts. Using material from a later chapter, in particular KCL, we may conclude that, Isrc = 2.5 + 1 = 3.5 A. Thus, Psource = VIsource = 10×3.5 = 35 W. This approach indicates that power is conserved.

Solution 1.32 (a) From 0 to 1 s, i(t) = 10-3t. Thus, i 2 R = 10−6 t 2 R is the power absorbed during this interval. Integrating this expression for the power from 0 to 1 s gives us the total energy used: 1

10 −6 t 3 R 10 −6 = 5000 = 0.001667 J. 3 3 0 Finally, we need to multiply this by 2 to account for the interval from 1 to 2 seconds. Thus, the total energy spent is 3.33 mJ. (b) The same charge that got transported in one direction during the interval from 0 to 1 is being transported back in the interval from 1 to 2 (by symmetry). Therefore, total charge transfer is zero. Solution 1.33. (a) 60 W + 120 W = 180 W. (b) P = IV þ I = P/V = 180/12 = 15 A. (c) P = Energy/Time þ Time = 1.2 MJ/180 W = 6.67×104 sec = 1.85 h.

Solution 1.34. P = I 2R. Therefore, 325 = 25×(5+4+2R). Solving for R, yields R = 2 Ω.

Solution 1.35. (a) Use definition of power and substitute given power: V2 = P × R = 98 × 2 = 14 V Similarly, I3 =

P 12 = = 2 A, V4 = R 3

P × R = 16 × 4 = 8 V, I5 =

768.8 = 12.4 A, and 5

V6 = 486 × 6 = 54 V. (b) Ptot = P dissipated = 98 + 12 + 16 + 768.8 + 486 = 1380.8 W. V (c) Vin = V2 + V6 = 68 V. Iin + I3 = I5 + I4 and I4 = 4 . Thus, Iin = I5 + I4 – I3 = 12.4 A. 4 Solution 1.36. (a) Sources are the top, right-most, and bottom left. The reason is that current flows out of the positive terminal of the device. (b) The 32/16 element is a 2 Ω resistor. The 54.5/18.167 element is a 3Ω resistor. The 13/2.167 element is a 6 Ω resistor. The 93/2.833 element is a 32.827 Ω resistor. The 24/5 element is a 4.8 Ω resistor.

Solution 1.37. Power: 12 = Ix 2 R, which means that R = 12/Ix 2 . Now, analyze the loop: 16 = Ix (R+4). 12 Substitute the value of R into this expression: 16 = I x 2 + 4Ix . Hence: Ix2 − 4I x + 3 = 0 . Ix This equation has two solutions: one is at Ix = 1 A or R = 12 Ω. The other is at Ix = 3 A or R = 4/3 Ω. Solution 1.38. (a) Conservation of power: 16I x = 4I 2x + 12 + 10 − 6 Hence 2 0 = Ix2 − 4I x + 4 = ( Ix − 2)

Thus, Ix = 2 A. (b) 32Ix = 4I 2x + 28 ⇒ Ix2 − 8Ix + 7 = ( Ix − 7)(I x − 1) = 0 . Hence, Ix = 7A or Ix = 1A . Solution 1.39. (a) (i) AA: I = 36/12 = 3A BB: I = 24/12 = 2A CC: I = 14.4/12 = 1.2A (ii) Sum = 6.2A (iii) P = VI = 6.2×12 = 74.4W. This is equal to the sum of the powers absorbed by the bulbs. (iv) R = V/I AA: R = 12/3 = 4Ω BB: R = 12/2 = 6Ω CC: R = 12/1.2 = 10Ω (b) Each AA bulb draws 3 A. Thus, up to five bulbs can be connected without blowing the fuse (5×3=15). So, 6 or more would blow the fuse. (c) Similar analysis suggests that 13 or more bulbs would blow the fuse. Intuitively, the bulbs draw less current, so more of them can be used. Solution 1.40. (a) p(t) = i 2 (t )R = [ 20cos(2πt )] × 10 = 4000 2

1 + cos(4πt ) = 2000(1 + cos(4πt )) W 2

(b) t

t

0

0

W (t ) = ∫ p(t )dt = 2000t + 2000∫ cos( 4πt )dt = 2000t +

2000sin (4πt ) J 4π

6000 5000

Energy in J

4000 3000 TextEnd 2000

1000 0

0

0.5

1

1.5 Time in s

2

2.5

3

Solution. 1.41. When the switch is closed, a constant current of 5/10000 = 0.5 mA flows through the circuit. When the switch is open, no current flows. So, 50% of the time, a 0.5 mA current flows, and the other 50% no current flows. The average current is therefore 0.25 mA.

Solution. 1.42 When the switch is at A, the current is 5/5000 = 1 mA. When the switch is at B, the current is 5/10000 = 0.5 mA. Now, the switch is at position A 20% of the time (1ms out of a 5ms period, after which the events repeat). So, the average current is 0.2*1 mA + 0.8*0.5 mA = 0.6 mA. 2

Solution. 1.43 The current in the load resistor is 2 A. So, the power is 2 ×RL = 8 W. Solution 1.44. Vin = IinR1 è Iout = µVin /R2 = µIinR1/R2. Solution 1.45 (a) I1 = Vin /R1 . Hence, Vout = αVin R2 /R1 . (b) Vout αR2 100 × 10 = = = 5 ⇒ R1 = 200 Ω Vin R1 R1 (c) Power − gain =

α 2 Vin2 R2 R12 Vin2 R1

=

α 2 R2 = 500 R1

Solution 1.46 (a) V1 = 200 mA × 5 Ω = 1 V implies V2 = 0.8×8 = 6.4 V. Hence Vout = 5×6.4 = 32 V and Iout = 32/64 = 0.5 A. (b) Current Gain = 0.5/0.2 = 2.5. (c) Power values for the 5, 8, and 64 Ω resistors are, respectively, P5 = 0.2 W, P8 = 5.12 W, P64 = 16 W. Solution 1.47 (a) I1 = 5A è I2 = 3×5/3 = 5 A è Iout = 25 A, Vout = 50 V. (b) Voltage Gain = 5. (c) Pin = 5×5×2 = 50 W, P1 = 5×5×3 = 75W, P2 = 25×50 = 1250W.

Solution 1.48 I1 = Vin /10 = 0.1 A, VR = 10×(Vin /10)×R = R; Vout = 5R×10 = 50R =50RVin Vout/Vin = 50R. If we want Vout/Vin to be 150, R has to be 3 Ω.

SOLUTIONS CHAPTER 2 SOLUTION 2.1. Using KCL at the center node of each circuit: (a) I3 = I2 − I1 = −1 − 2 = −3A (b) I3 = I1 + I2 − I4 = 2 − 1 − 0.5 = 0.5A SOLUTION 2.2. KCL at the bottom node gives I1 = −7 − 8 = −15A , and at the right node I4 = − 6 − 8 = −14 . From these, KCL at the top node gives I3 = I 4 − 5 = −19A, and finally at the central node gives I2 = 6 + I3 − 7 = −20 A, SOLUTION 2.3. Use a gaussian surface on the top triangle. Performing KCL around this surface yields 1A − 2A + 3A + 4 A − 5A = I = 1A. SOLUTION 2.4. Use a gaussian surface around the bottom rectangle. KCL yields I1 = 2A + 10A + 3A = 15A . SOLUTION 2.5. Using KVL, V1 = 55V − 15V + 105V − 100V − 30V = 15V . SOLUTION 2.6. Using KVL, Vx = 5V − 1V − 1V − 1V + 1V − 1V = 2V . SOLUTION 2.7. Using KVL once again. v1 = 7 + 6 + 5 = 18V v2 = 6 + 7 − 8 = 5V v3 = −5 − 6 = −11V v 4 = 8 − 7 = 1V SOLUTION 2.8. KVL is used to find the voltage across each current source, and KCL to find the current through each voltage source. I3V = 6A − 7A = −1A I4V = I3V + 8A = 7A I5V = − 8A − 6A = −14A V7 A = 4V + 3V = 7V V8A = −4V + 5V = 1V V6 A = V8 A − 3V = −2V

Chap 2 Probs

P2 - 2

© R. A. DeCarlo, P. M. Lin

SOLUTION 2.9. Using the same method as before, the current and voltages are found through and across each sources. I5V = 9 + 8 − 7 = 10 ⇒ P = 50W I4V = −6 − I5V = −16 ⇒ P = − 64W I2V = 6 − 7 = −1A ⇒ P = −2W I3V = − I2V − 9 = −8A ⇒ P = − 24W V8A = 4 − 5 = −1V ⇒ P = − 8W V9 A = 3 + V8 A = 2V ⇒ P = 18W V7 A = 2 − V9 A = 0 ⇒ P = 0W V6 A = 5 − V7A = 5 ⇒ P = 30W Summing all the power give 0W, hence conservation of power. SOLUTION 2.10. Doing KVL around the right loop does not balance out. Changing 8V to 5V would fix this. SOLUTION 2.11. Using KVL to determine the voltages, and KCL to determine the currents: Vy = 8V Vx = Vy − 4 = 4V Ia = 4 A Iy = 4 − 14 + 2Ia = −2A Ix = Ia − Iy = 6A SOLUTION 2.12. First Vin = I2 ⋅ 8Ω = 24V . Then I1 = Vin / 3Ω = 8A and I3 = 12A − I1 − I2 = 1A. Therefore RL = Vin / I3 = 24Ω ⇒ P = I3 ⋅ Vin = 24W SOLUTION 2.13. (a) First, from current division, get     1/ 3 12 / 3 I1 =   ⋅ (12 − aI1 ) ⇒ I1 =  .  1/ 3 + 1/ 6 + 1/ RL   (1 + a ) / 3 + 1/ 6 + 1/ RL  (b) Using the previous equation and solving for 1/ RL = (12/ 3I1 ) − 1/ 6 − (1+ a ) / 3 = 0.5S or RL = 2Ω .    1/ RL The power P = I32 RL =    ⋅ (12 − aI1 ) ⋅ RL = 18W   1/ 3 + 1/ 6 + 1/ RL   SOLUTION 2.14. For the power delivered by the source to be 60W, the voltage across it should be V = P / 2A = 30V . Therefore the current through the 20Ω must be I20Ω = 30 / 20 = 1.5A , and by KCL the current through IR L = 2 − I20Ω = 0.5A . From this, RL = V / I RL = 60Ω .

Chap 2 Probs

P2 - 3


SOLUTION 2.15. Writing KVL around the loop 25V − 4I − 15V − 5I − I = 0 ⇒ I = 1A, and P5Ω = I 2 R5Ω = 5W SOLUTION 2.16. The total power supplied by the source is P = 50V ⋅ 0.5A = 25W . The power absorbed by the resistor is P60Ω = ( 0.5A ) ⋅ 60Ω = 15W . Therefore by conservation of power, the power absorbed by 2

X is 10W. SOLUTION 2.17. (a) As this loop is open, no current flows through it, so IR is 0A. The output voltage is VOUT = −2V + 3V − 2V = −1V by KVL. (b) Writing out the KVL equation around the loop 3 − 2 − I R R − 2 − I R 2R − I R R = 0 ⇒ −1 = I R 4R. Therefore IR = − 1/ 4R and VOUT = IR R = −1/ 4V . SOLUTION 2.18. Writing out KVL around the loop 60 − 30I − 30 − 20 + 60 − 40I = 0 ⇒ I = 1A . From ohm's law R = V / I = 30Ω. SOLUTION 2.19. (a) Using Ohms law Iin = V2 / ( 20 + 12 ) = 0.75A, and V1 = 12 ⋅ Iin = 9V . To find R, write KCL and get VR = 30 − V2 = 6V . Therefore using Ohms law again, R = VR / IIN = 8Ω . (b) Writing KVL around the loop, 30 = aV1 + Iin R + Iin 20 + V1 , and substituting Iin = V1 / 12, V1 = 30 / [( R + 32) /12 + a] is obtained. Next substitute back V1 = 12Iin and solve for  30  R= − a ⋅12 − 32 = 40Ω  12I IN  SOLUTION 2.20. (a) i. Using R = Vxy / Ibat the value of each resistors starting with the top one are 2.7Ω, 0.6Ω, and 0.25Ω. Using the same relationship, the resistance for the motor is 1.25Ω. ii. Using P = Vxy2 / R the power dissipated by each resistor is 16.875W, 3.75W, 1.5625W, and for the motor 7.8125W. iii. The relative efficiency is = 7.8125/ (12⋅2.5) ⋅100 = 26 % (b) i. Performing voltage division across each resistor VAB = 0 VBC = 12 ⋅ RBC / (RBC + RCD + Rmotor ) = 3.43V VCD = 12⋅ RCD / (RBC + RCD + Rmotor ) = 1.43V Vmotor = 12 ⋅ Rmotor / (RBC + RCD + Rmotor ) = 7.14V

Chap 2 Probs

P2 - 4


ii. Ibat = 12 / (RBC + RCD + Rmotor ) = 5.71A 2 iii. The relative efficiency is = (Vmotor / Rmotor)/ (12 ⋅5.71)⋅100 = 59.5 %

(c) i. Repeating the steps from (b), the voltages across the first two resistance are 0, then across the other and the motor 2V, and 10V ii. Ibat = 12 / (RCD + Rmotor) = 8A 2 iii. And the relative efficiency is = (Vmotor / Rmotor)/ (12 ⋅8)⋅100 = 83.3 %

(d) What is the largest equivalent resistance of the motor that will draw 30A? R = 12/ 30A = 0.4Ω. SOLUTION 2.21. (a) Observe that i = − IO , thus v = ki3 = −kI03 . 3 (b) Using KVL and previous equation, v x = (R1 + R2 )IO + VO + kIO . 4 (c) The power is = IO v x = (R1 + R2 )IO2 + VO IO + kIO

SOLUTION 2.22. I100Ω = I150Ω = 0.02 +

0.04 = 0.02 A. Therefore V300Ω = 0.02 × (100 + 200) = 6 V. By KCL, 100

6 = 0.04 A. Req, seen by the source, is 300 Ω. Therefore Vs = 0.04 × 300 = 12 V. 300

SOLUTION 2.23. Using KCL IR = 5 − 3 = 2A , and KVL VR = 10 + 6 = 16V . Thus R = VR / IR = 16 / 2 = 8Ω. SOLUTION 2.24. Using KCL, KVL, along with Ohm’s law, I5Ω = 6 − 7 = −1A I4V = 8 − I5Ω = 9A I2Ω = 8 + 7 = 15A V6 A = 4 + 5I5Ω = −1V V8A = −4 + 2I2Ω = 26V V7 A = V8 A − 5I5Ω = 31V Now, the power delivered or absorbed by each element is calculated: P6A = I6A ⋅V6 A = −6W P7A = I7A ⋅V7 A = 217W P8A = I8A ⋅ V8A = 208W P4V = I4V ⋅ 4 = 36W 2 P5Ω = I5Ω ⋅5 = 5W 2 P2Ω = I2Ω ⋅ 2 = 450W

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Note that for passive elements, when the power is positive it is absorbed, while for independent sources it is generated when the power is positive. SOLUTION 2.25. Note that I1 = 6A . Thus by KCL I3 = 6 − 0.5I1 = 3A I2 = 2 + 0.2I1 = 3.2A I4 = 8 − 0.3I1 = 6.2A And finally using KVL V2 = 8A ⋅1+ 4I 4 + 3I3 = 41.8V V1 = 2I2 − 3I3 = −2.6V SOLUTION 2.26. (a) Using KCL, I4 = 5 − 4 = 1A I3 = I 4 − 2 = −1A I2 = 3 − 2 = 1A I1 = − I2 − 5 = −6A (b) Using KVL and Ohm’s law, V1 = 3I1 = −18V V2 = 12 I 2 = 12V V3 = 10 I 3 = −10V V4 = 4 I 4 = 4V

(c) P3 A = 3 A(V2 − V1 ) = 90W P2 A = 2(V1 − V2 − V3 ) = −40W P4 A = 4( −V3 − V4 ) = 24W P5 A = 5(V4 + V3 − V1 ) = 60W

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SOLUTION 2.27. Write KVL around the outside loop, 40 = 500Ix + (400 + 200)i . And write KCL equation i = Ix − 2I x . Solving yields Ix = −0.4A . The dependent source delivers 2I x ⋅(−600i) = 192W , and the independent 40I x = −16W . Finally the resistors absorb 500Ix2 + 400i 2 + 200i 2 = 176W verifying the conservation of energy since the source generate 192W-16W=176W. SOLUTION 2.28. By voltage division V2 =

[((90 ||180) + 60)||40] 60 ⋅ ⋅ V = 1/14 ⋅Vs . [((90||180) + 60)||40] + 160 60 + (90||180) s

Therefore Vs = 14V2 = 280V . SOLUTION 2.29. By voltage division v x = 9V ⋅

18 + 3 = 7V (18 + 3) + 6

SOLUTION 2.30. By voltage division we get the following two equations in order to solve for the two unknowns. R1 V2 = V1 ⋅ R1 + R2 R1 + R2 V1 = 100V ⋅ R1 + R2 + 60 Solving yields R1 = 40Ω , and R2 = 100Ω . SOLUTION 2.31. Dividing 1400 in four gives 350. If we only need 1/4 and 2/4, the resistor string can be made of three resistances: 350Ω, 350Ω, and 700Ω.

SOLUTION 2.32. Using voltage division, at t=0 v R = 15⋅ the voltage goes back to 0V.

2R = 10V , and t = 5 s v R = 10V , and at t = 10 3R

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10 9 8 7

Volts

6 5 4 3 2 1 0

0

5

10

15

time (sec)

SOLUTION 2.33. By voltage division Vb =

Rb Rd Vin and Vd = V Ra + Rb Rc + Rd in

By KVL, if Vout = 0, then  Rb Rd  0 = Vout = Vb − Vd =  − V  Ra + Rb Rc + Rd  in

For arbitrary Vin , this requires that

Rb Rd = or equivalently that Rb Rc = Ra Rd . Ra + Rb Rc + Rd

SOLUTION 2.34. First Geq = 1m + 1.5m + 2m + 3m = 7.5mS . By current division I2 = 100mA⋅

1.5m = 20mA, P = 100mA ⋅ I2 /1.5mS = 1.33W . Geq

SOLUTION 2.35. By current division, for I1 to be 2A then 160 + R = 300 ||600 for an even split. Thus R = 40Ω . SOLUTION 2.36. By current division, i1 = 0.4A ⋅

1/10 = 0.32 A . Therefore using KVL 1/10 + 1/ 40

v d = 10i1 − 0.25i1 = 3.12V . SOLUTION 2.37. (a) Req = (8k ||2k) + (9k ||1k) = 2.5kΩ

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(b) Req = 2k ||[(2 k ||2k) + (2k ||2 k)] = 1kΩ SOLUTION 2.38. (a) Req = 2 + 15 + 10 + 10 + 40 + 30 + 20 + 8 = 135Ω . (b) Four of the resistors are shorted, thus Req = 2 + 15 + 10 + 8 = 35Ω . (c) Lumping the series resistance together Req = 8 + [50 ||(50||25)] + 2 = 22.5Ω SOLUTION 2.39. (a) Req = [2R + (4R ||4 R)]||[2 R + (4 R ||4 R)] = 2R (b) Req = 2R ||2 R + (4R ||4 R ||4 R ||4 R) = 2R SOLUTION 2.40. (a) First Req = 150 + [375||(250 + 500)] = 400Ω . Next Iin = 14V / Req = 35mA. The power delivered by the source is then = 14Iin = 0.49W . (b) Req = 150 + [375||(250 + 500)||1k] = 350Ω, and Iin = 14 / Req = 40mA . The power delivered by the source is = 14Iin = 0.56W . As the equivalent resistance decreases, more of it gets dissipated by it.

SOLUTION 2.41. Req1 > Req2. Without going into a detailed analysis using methods of Chapter 3, we present the following intuitive argument. First note that the points a and b represent points on an unbalanced bridge circuit meaning that the voltage between a and b would not be zero. Also note that when two resistors are placed in parallel, the equivalent resistance becomes smaller than either resistance. The addition of the resistor R in circuit 2 essentially creates an internal parallel resistance resulting in an Req2 lower than Req1. SOLUTION 2.42. Req1 = Req2. As was the case in the previous problem, this is a balanced bridge circuit. Hence no voltage appears between a and b making the additional resistor irrelevent. SOLUTION 2.43. (a) Rin = [(20||20) + 10]||(1/ 0.12)||(1/ 0.08) = 4Ω. (b) Rin = 6R ||[( R || R ||0.5 R) + 0.75R + (2R ||2 R)] = 1.5R SOLUTION 2.44. (a) The infinite resistance are essentially open circuits, thus Req = 1 + 2 + 3 + 4 + 2 + 4 + 3 + 2 + 1 = 22Ω (b) 0 resistances are short circuits. Labeling one branch x and the other y, it can be seen that the circuit is a set of 3 resistor strings in parallel to each other between x and y, then added in series to the two 1 Ohm resistor. Thus Req = [(2 + 3)||(4 + 2 + 4)||(2 + 3)]+ 1 + 1 = 4Ω .

Chap 2 Probs

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(c) Writing out Req = 1 + [Rx ||(2 + 3 + 4 + 2 + 4)] + 3 + 2 + 1 = 7 + [Rx ||15] , and solving for Rx = 3.75Ω. (d) No, it requires methods to be covered in the next chapter.

SOLUTION 2.45. Using the formulas for parallel resistances, the circuit of figure 2.45 reduces to

(a) RAC = [((2 + 1) / /6 ) + 8] / /10 = 5 Ω (b) RAB cannot be calculated by series parallel formulas, but RBC can be done. RBC = [((8 + 10) / /6 ) + 2] / /1 = 0.86667 Ω SOLUTION 2.46. (a) Req = 300 + (R ||5.6k), thus R = 800Ω. (b) Req = R + (R ||1.2K) , the following quadratic equation must be solved R2 + 1.4k ⋅ R − 1.2M = 0 . This yields R = 600Ω . (c) Req = 500 + 300 + (800||400 || R). Solving for R yields 800Ω . SOLUTION 2.47. (a) Using the fact that the resistance seen into terminal a-b is the same as that seen in terminal c-d, we can obtain the following relationship. Req = R + R || Req . This produces a quadratic equation whose solution is Req = 1.618R . (b) Using the previous argument Req = 5 + 10 || (5 + Req ) . Solving for Req = 11.18Ω . 1   1/ 18k    6k + (9k ||18k)  SOLUTION 2.48. By current division Ix =  ⋅ 1 1  ⋅36m = 3mA  1/18k + 1/ 9k   +    6k + (9k ||18k) 4k 

Chap 2 Probs

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SOLUTION 2.49. The 500 Ω resistor has no effect on the current entering the circuit to its right. 0.15 =

30 30 30 + = + 0.05 R 600 R

Hence R = 30/0.1 = 300 Ω. SOLUTION 2.50. (a) First, express the total current as I =

120 . Next, find RL2 that 0.5 + (20 ||30||40 || RL2 )

will cause I to be 15A. Thus RL2 = 40Ω or less will cause the fuse to blow as this will cause the current to be 15A or more. (b) Repeating the previous procedure, RL2 = 20Ω . (c) RL2 = 120Ω . SOLUTION 2.51. At time 0, all switches are open and Vout =

260 ⋅ 220 = 190.7V . 260 + 40

Then at t = 5s, switch one closes and Vout =

260||260 ⋅ 220 = 168.2V . (260 ||260) + 40

At t = 10s, Vout =

130 ||260||260 ⋅ 220 = 136.2V . (130 ||260||260) + 40

Finally at t =15 s, Vout =

65||130 ||260||260 ⋅ 220 = 98.5V . (65||130 ||260||260) + 40

200

180

Volts

160

140

120

100

80

0

2

4

6

8

10 12 Time (sec)

14

16

18

20

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SOLUTION 2.52. (a) Lumping the two sources together and the resistors into an equivalent resistor gives 9cos(2t) − 3cos(2t) i1 (t) = = 0.24cos(2t)mA . 7k + 9k + 8k + (2k ||3k ||6 k) 1/ 6k   (b) By current division i2 (t) =   ⋅i (t) = 40cos(2t) A .  1/ 2k + 1/ 3k + 1/ 6k  1

SOLUTION 2.53. (a) Starting with, Req1 = 5||(10 + 10) = 4Ω Req 2 = 10||(6 + Req1 ) = 5Ω Req 3 = 5 + Req2 = 10Ω (b) Using the values R just obtained, R Va = 100⋅ eq 2 = 50V Req3 Vb = Va ⋅

Req1 = 20V 6 + Req1

10 = 10V 10 + 10 (c) Finally, 100 Iin = = 10A Req3 Vc = Vb ⋅

Id =

Va = 5A . 6 + Req1

Ie =

Vb = 1A 10 + 10

SOLUTION 2.54. (a) Circuit a: Using voltage division,  300||(20 + 30 + 50)   30  = 33.75sin(377t)V , and Ohm’s law vout (t) = vin (t) ⋅  ⋅  [300 ||(20 + 30 + 50)] + 5  30 + 20 + 50  iout(t) = vout(t)/ 30 = 1.125sin(377t )A . The instantaneous power is then P(t) = iout (t)⋅ vout (t) = 37.969sin 2 (377t)W . Circuit b: By current division  1/ (20 + 30 + 50)  iout(t) = iin (t) ⋅   = 60sin(377t)A , and from Ohm’s law  1/ (20 + 30 + 50) + 1/ 300 + 1/ (50 + 100)  vout (t) = 50 ⋅ iout(t) = 3000sin(377t)V . The instantaneous power is P(t) = 180sin 2 (377t)kW . (b) No, since the current source forces the amount of current in the circuit.

Chap 2 Probs

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SOLUTION 2.55. (a) Noting that i2 = v1 /10 = 6A, then we can write KCL at the top left node, isource = i2 + v1 / 6 + (v1 − 5i2 )/ 5 = 22A. Thus P = 60⋅ 22 = 1.32kW . (b) First, determine the current through each resistor: i2 = 60 /10 = 6A  60  i2.5Ω =   = 12A  2.5 + (5||5)  i5Ω = 1/ 2 ⋅i2.5Ω = 6A Then calculate the power absorbed by each resistor: P10Ω = 10i22 = 360W 2 P2.5Ω = 2.5i2.5Ω = 360W 2 = 180W P5Ω = 5i5Ω

SOLUTION 2.56. From Ohm’s law I1 = 100m / 200 = 0.5mA. By current division 20k  IRL =  ⋅150I1 = 75.257mA, and PRL = 200I 2RL = 1.103W . 20k + 200  2  SOLUTION 2.57. First, using voltage division, Vx = Vs  = (2 / 3)Vs . Then using KCL and the 2 + 1 previous equation, Is = (Vs / 3) − Vx = −(1/ 3)Vs . Finally using Ohm’s law Req = Vs / Is = − 3Ω. SOLUTION 2.58. Observing the following relationship, V1 = Vin , the following nodal equation can written: Iin = Vin / 3 + Vin / 6 − 2Vin = −1.5Vin .

SOLUTION 2.59. Step 1. From voltage division V1 =

18 22 11 Vs = 0.75Vs and Vin = Vs = Vs 18 + 4 + 2 24 12

Hence V2 11× 11 2 11 2 Pin = in = V = V 22 22 × 144 s 288 s Step 2. For the load, by current division

Chap 2 Probs

P2 - 13

I2Ω =


6 3 3 9A AV1 = A  Vs  = V 6+2 4 4 16 s

Therefore 81A2 2 81A2 2 V = V 256 s 128 s

2 P2Ω = 2 × I2Ω =2

Step 3. P2Ω = 10 × Pin implies that 81A 2 2 11 2 Vs = 10 V 128 288 s Hence A=

SOLUTION 2.60. By voltage division V1 = previous equation I2 =

128 × 110 = 0.7769 81 × 288

6 V = (3/ 4)Vin . By current division, and substituting the 6 + 2 in

3 4V = Vin . Using voltage division and Ohm’s law, and substituting the previous 3+6 1

equation, 10 = 3Vin = 30V 10 + 5 Iout = 4.5I2 / (10 + 5) = 0.3Vin = 3A

Vout = 4.5I2

Finally, from the previous equations | Vout / Vin |= 3 . V − 0.04ibat  SOLUTION 2.61. Writing out KCL when the switch is closed, ibat = 150A +  bat  . Solving 240 gives ibat = 150.02A and Vout ≈ 6V . When the switch is open Vout = Vbat 

240  ≈ 12V . Therefore, 240 + 0.04 

the reason for the radio stopping is insufficient supply voltage.

SOLUTION 2.62. (a) Using the following relationship P = V 2 / R, the resistance of each headlight on low beam is R = V 2 / P = 4.11Ω . (b) Using the same relationship R = 2.22Ω . 240 (c) By voltage division, Vout = 14.7 = 14.698V . 240 + 0.04 240 ||4.11||4.11 (d) Using voltage division, Vout = 14.7 = 14.417V (240||4.11||4.11) + 0.04

Chap 2 Probs

P2 - 14

(e) Using voltage division, Vout = 14.7


240||2.22||2.22 = 14.186V (240 ||2.22||2.22) + 0.04

SOLUTION 2.63. By voltage division 11.96 =

15 180 12 = 15 + R0 15 + R0

Therefore R0 =

180 − 15 × 11.96 = 0.050167 Ω 11.96

SOLUTION 2.64. (a) Using KVL Vt = 102 − 0.05⋅80 = 98 V. (b) Using KVL Vt = 102 + 0.05⋅ 50 = 104.5 V. (c) P = Vt ⋅50 = 5.225. kW

SOLUTION 2.65. Minimum load means the minimum load resistance that the system can handle. »MaxPwr = 0.8*50e6 MaxPwr = 40000000 »Vs = 750e3; »Iline = MaxPwr/Vs Iline = 5.3333e+01 »Rmin = Vs/Iline Rmin = 1.4062e+04, i.e., Rmin = 14.062 kΩ. SOLUTION 2.66. (a) Using the following general form for a non-ideal voltage source: vout = −Rsiout + Vs , one sees that for zero current vout = Vs = 40V . The slope of the line is

−40 = − Rs = −0.04Ω , thus 1000

Rs = 0.04Ω . 60 (b) This curve represents a resistor’s I-v characteristic, thus the slope 1 = R = 45Ω . 1 3 1 (c) The general form for a non-ideal current source is iout = − Vout + I s . When the voltage is zero, Rs iout = Is = 5A . From the slope of the line,

−4000 , Rs = 4000 / 5 = 800 Ω. 5

Chap 2 Probs

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SOLUTION 2.67. Using the following formula:


T  nI    = 1, solve for T, and get 0.625 hrs, or 37.5 min. n  Cn 

T SOLUTION 2.68. Using the same equation as before and solving for Cn = nI   n T=55/60 hrs, the capacity obtained is 20 Ah. SOLUTION 2.69. C20 = 50 Ah (a) In eq. 9, solving for I with n=20, and T=10, I=4.2A (b) Calculate the capacity for n=10 and T=10, this yields 42 Ah. SOLUTION 2.70. (a) Using a sequence of voltage division, 48 = 48mV 50 195 V2 = 50V1 = 2.34V 200 Vload = 2.5V2 = 5.85V V1 = 50mV

2 And the power is RRL = Vload / RL = 2.278W .

(b) Following is the graph, and the script used to generate it.

Power in Watts

5 4 3 2 1 0

0

10

20

0

10

20

30

40

50

60

70

50

60

70

Current in mA

800 600 400 200 0

%Script for Question 70 in chapter 2 RL=8:1:64; V2=2.34; IL=2.5*V2 ./ RL; %Note the use of the ".*" which means that the division %is performed for each value of RL.

30 40 Resistance in Ohms

1/

, with n=10, and

Chap 2 Probs

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PL=RL .* (IL .^ 2); %Plot the Power versus RL subplot(2,1,1); plot(RL,PL); ylabel('Power in Watts'); %Plot the Current versus IL subplot(2,1,2); plot(RL,1000.*IL); xlabel('Resistance in Ohms'); ylabel('Current in mA'); %The use of subplot lets you subdivide the graphing %window in two halfs.

SOLUTION 2.71. (a) Using the following script: %Script for problem 2.71 R1=15; R2=4; R3=9; R4=2; R5=8; R6=18; Ra= R4+R5; Ga= 1/Ra; Gb= Ga+1/R1; Rb= 1/Gb; Rc= 1/(1/R6+1/R3)+Rb; Gc= 1/Rc; Geq= Gc+1/4; Req= 1/Geq; Irc= 20*Gc/Geq; Vrb= Irc*Rb; Vout= Irc*(Ga/Gb)*8; Req Vout

So (a) Req = 3Ω , and (b) Vout = 24V SOLUTION 2.72. Using the following script: %Script for problem 2.72 R1=1e3; R2=2.2e3; R3=2e3; R4=5e3; R5=3e3; R6=R5; R7=3.2e3; R8=1.2e3; R9=1.6e3;


Chap 2 Probs

P2 - 17


Ga=1/R7+1/(R8+R9); Ra=1/Ga; Gb=Ga+1/R6; Rb=1/Gb; Gc=1/R4+1/(R5+Rb); Rc=1/Gc; Gd=1/R2+1/(R3+Rc); Rd=1/Gd; Geq=1/R1+Gd; Req=1/Geq %Going through the same step to find Vout Id=200e-3*(Gd/(Geq)); Ic=Id*((1/(R3+Rc))/Gd); Ib=Ic*((1/(R5+Rb))/Gc); Ia=Ib*Ga/Gb; Iout=Ia*((1/(R8+R9))/Ga) Vout=Iout*R9

The following values are obtained: Req = 5.9121e+02 Iout = 5.5431e-03 Vout = 8.8689e+00

d R1

R2

SOLUTION 2.73. Using the following script: %Script for problem 2.73 R1=20; R2=40; R3=60; R4=30; R5=10; R6=135; R7=150; R8=300; R9=130; R10=200; R11=50; Ga=1/R10+1/R11; Ra=1/Ga; Rb=Ra+R9+(1/(1/R7+1/R8));

c

b

a

R4

R6

R7

R9

Chap 2 Probs

P2 - 18


Gb=1/Rb; Gc=Gb+1/R6; Rc=1/Gc; Rd=Rc+R5+(1/(1/R3+1/R4)); Gd=1/Rd; Ge=Gd+1/R2; Re=1/Ge; Rin=R1+Re Ie=10/Rin; Id=Ie*Gd/Ge; I1=Id*(1/R6)/Gc Ib=Id*Gb/Gc; Vout=Ib*Ra

The following values are obtained: Rin = 50Ω Vout = 0.667V I1 = 33.3mA

R4 e d

R5

R1

c b

R8 R7

R9 R11

a

R3 R2

R6

R10

SOLUTION 2.74. An identical procedure to the one followed in the previous problem will yield the following values: Rin = 50.53Ω Iout = 133.8mA

PROBLEM SOLUTIONS CHAPTER 3. Solution 3.1. Select the bottom node as the reference node, and write a node equation at the positive terminal of the V1 resistor: V1 − V0 V1 V1 − 4V0 + + =0 3R 6R 6R ⇒ 2V1 − 2V0 + V1 + V1 − 4V0 = 0 ⇒ 4V1 = 6V0 ⇒ V1 = 1.5V0

Solution 3.2 Write a node equation at the top node: Vx 2V x V x − − =0 100 100 50 ⇒ −Vx − 2Vx − 2Vx = −60 ⇒ −5Vx = −60 ⇒ V x = 12V 0.6 −

Solution 3.3 Vx 25Vx Vx Vx − 0.2Vx + − − =0 100 100 50 40 3V 25V x 8V x ⇒− x + − = −0.6 100 100 400 0.6 × 400 ⇒ Vx = − 80 ⇒ V x = −3V 0.6 −

Solution 3.4 (a) It is evident from the figure that Vc = 20. We need to write two equations in Va and Vb and put them in matrix form. In this case, we can write the matrix equation by inspection. Note that the resistors are identified by conductance values. 15m −5m Va   0.5 −5m 35m  V  =  0.5   b   

(b) Solve the matrix equation by inverting the left-most matrix: Va   35m 5m  0.5 1 V  = 525 − 25  5m 15m  0.5  b    1  35 5  0.5  = 0.5  5 15  0.5   40  =   20  (c) Vx = Vab = Va − Vb = 20V , Vda = -Va = -40, Vdb = -Vb = -20. (d) Pi = 0.5×40 = 20W, Pv = 20×(20-20) = 0. P diss = 40×40×10m + 20×20×5m + 20×20×5m = 20 W. Power delivered equals dissipated power.

Solution 3.6 Write two nodal equations: Vs1 − V1 V − V2 = Is 3 + 1 3000 6000 V2 V − V2 V1 − V2 = s2 + 30000 12000 6000 Rewrite equations as: 2Vs1 − 2V1 = 6000Is 3 + V1 − V2 2V2 = 5Vs2 − 5V2 + 10V1 − 10V2 Cast into a matrix equation  −3 1   V1   6000I s3 − 2Vs1  −10 17 V  =   5Vs 2   2    Solving the matrix equation yields: V1  181.46  V  = 124.39   2   Power absorbed by the 6k resistor is (V1 -V2 )2 /R = 0.5429W . Similarly, Ps1 = (Vs1 -V1 )/3000×Vs1 = 4.7W, Ps2 = (Vs 2 − V2 ) /12000 × Vs 2 = −0.32W Ps3 = Is 3 × (Vs2 − V1) = −1.21W

Solution 3.7 (a) Again, the matrix equation can be written by inspection:

G1 + G2 + G4  −G4 

−G4   VB   50G1  = G3 + G4 + Gs  VC  50G3 

(b) Substituting the values of conductances and inverting the above matrix equation yields: VB = 34.0132V VC = 33.6842V (c) Power delivered is 80.7566W . Using the Principle of Conservation of Power: Pdel = P1 + P2 + P3 + P4 + P5 or,  V − VB VA − VC  Pdel = 50 ×  A + = 80.7566W 20   20 (d) In this part, we take the above matrix equation and solve it for each value of Gs. If we do this, we can get a feel for the behavior of VB and VC w.r.t. changes in Gs. The following plot is the voltage difference between the two nodes as a function of Gs, and hence as a function of temperature.

As can be seen, in this figure, the voltage difference between B and C does not change linearly with Rs. Since this resistance itself changes linearly with temperature, this means that VB-VC does not change proportionally to temperature.

Solution 3.8 The answer is:

G1 + G2  −G 1   −G2   0

−G1 G1 + G3 + G4 + G7 −G4 −G7

−G2 −G4 G4 + G5 + G6 + G2 −G6

0   V1  − Is1  V   0  −G7  2  =    V3   0  −G6     G6 + G8 + G7  V4   I s2 

Solution 3.9 Write the matrix equation by inspection:  Vs1   4 /100 −1/100 −1/100  V1   100       Vs1  −1/100 4 /100 −1/100  V2  =  100  −1/100 −1/100 4 /100  V3   I s2    Solving the equation in MATLAB, we get: V1 = 7, V2 = 7, V3 = 11, and P = 0.6. 3  V − V V − V2  3 P = Vs1 ×  S1 1 + S1 = 10 × +  10 10 = 0.6W 100 100 

Solution 3.10 (a) Nodal equation for A: VA − Vs1 VA − VB VA − VC + + =0 10 10 10 (b) At node B: VB VB − V A + − Is2 = 0 10 10 (c) At node C: V − VA Is2 − Is 3 + C =0 10 (d) Manipulate algebraically to cast as the following matrix equation:  3 −1 −1 VA   Vs1       −1 2 0  VB  =  10I s2  −1 0 1  VC  10(I s3 − I s2 )   VA   −13 13     1  VB  =  13 3  VC   −43 1   3 

(e) P = (-10-13.333)/10×(-10) = 23.33W.

Solution 3.12 We are required to write the equations in matrix form. First, write a node equation at VA and Vout: V V − Vout VA − 5 + A + A =0 5 10 Vout − V A V − 7.5VA + out = 0 10 10 Now group the coefficients for VA and Vout, and write the matrix equation: 1+ 1 / 5 + 1/10 −1/10   V A  5  =  −1/10 + 7.5 2/10  Vout  0    VA   1  V  = −37  out    where the matrix inversion was performed in MATLAB. The ratio of the output voltage to the input voltage is -37/5.

Solution 3.13 (a) Nodes A and B are already labeled:

(V A − 9)0.1 + (V A − 0.5VB )0.2 + (V A − VB )0.3 = 0 (VB − V A )0.3 + (VB − 9)0.5 + 0.4VB = 0 This can be rearranged Into:

(0.1+ 0.2 + 0.3)VA − 0.3VB − (0.2 × 0.5)VB = 0.9 −0.3VA + (0.3 + 0.5 + 0.4 )VB = 4.5 The matrix equation can now be easily obtained:  0.6 −(0.5)(0.2) − 0.3 V A   0.9 −0.3  V  =  4.5 1.2   B    V A   4.8  V  =  4.95  B   (b) Iin = -(VA-Vs)0.1 - (VB-Vs)0.5 = 2.445A

(c) Ps = VS Iin = 22.005W, Pdep = 0.5 × Vout × (0.5Vout − V A ) × 0.2 = −1.151W . (d) P = V2 /R = (4.95)×(4.95)×0.4 = 9.801W.

Solution 3.14 (a) We write node equations at VA and VB: VA V − VB + gm1VA + A =0 20k 10k VB − VA V − gm2 (V A − VB ) + B + I s2 − gm1V A = 0 10k 2.5k

−I s1 +

Rearranging, we have: 1  V  1 + gm1 + V A − B = I s1  20k 10k  10k 1   −1   1 − gm2 − gm1 V A + + gm2 + V = −I s2  10k   10k 2.5k  B (b) −1/10k 1/20 k + gm1 + 1/10 k  V A   Is1   −1/10k − gm − gm 1/10 k + gm + 1/2.5k   V  = − I   2 1 2  B   s 2  (c) The above matrix is inserted into MATLAB, with all the values substituted, to obtain: V A  9.722  V  = 5.972   B   (d) Vo = VA-VB = 3.75V (e) P1 = VAIs1 = 0.0292W, Pgm1 = -Vo gm1 VA = -0.008W, Pgm2 = VBgm2 Vo = 0.0112W, P2 = -VBIs2 = 0.0119W.

Solution 3.15 I1 = 0.4. Write nodal equations at A and B VA V A − VB + + 0.03(V A − VB ) = 0.4 100 20 V VB − VA VB VB − 80 B 40 + + =0 20 40 40 Rearranging and casting into matrix form:

−1/20 − 0.03  VA   0.4  1/100 + 1/20 + 0.03 =  −1/20 1/20 + 1/40 − 1/40  VB   0   VA   40  V  =  40   B   It is obvious then, that Vx , the voltage between A and B, is zero.

Solution 3.16 V − VB V A − VB VA = 3000i x = 3000 × A = 9000 3 Equation at node B: 1)

2)

VB − VA V V − VD + B + B = 0 is equivalent to: 9000 6000 18000 −2V A + 6VB − VD = 0

Equation at node D:

3)

VD − VB V + D + IS = 0 which can be rewritten as: 18000 9000 3VD − VB = −360 Solving the system formed by equations (1), (2) and (3) we obtain: VA = 9V , VB = −18V , VD = −126V .

Solution 3.17 (a) Choose E as the reference node VA = 2ix At node B 6 = (VB-VC)/3 + (VB-VA)/2 Or 6 = 5/6VB –1/3VC –VA/2 At node C, VC = 2iy At node D, VD = -12V iy = (VD-VA)/2 = VD/2 –ix From here on, the solution involves algebraic manipulations to solve the system of equations. MATLAB or hand analysis can be performed to obtain: VA = 48V, VB = 12V, VC = -60V. (b) P6A = 6(12 = 72W

I12V = 30-8 = 22A ( P12V = 264W. P2i x = 2ix × (−iz − iY ) = 2 × 24 × (+18 + 30) = 2304W v − vD P2i y = 2iy × (−i x + c ) = 2 × (−30)× (−24 − 8) = 1920W 6 (c) P3 Ω = i x × (VB − VC ) = 1728W = 1728W P6 Ω = 6×8×8 = 384W P2 Ω y = 2×iy ×iy = 1800W P2 Ω = (VB-2ix ) (VB-2ix )/2 = 648W. (c) VD = -12V, VA = 2ix , VC = 2iz. Substitute the above VA and VC into the node equation for node B: iz = (VB-2ix )/2 = VB/2 –ix and ix = (VB-2iz)/3 = VB/3 –2/3iz Substitute iz into ix to obtain: ix = 0, VA = 0. Then, VB can be deduced to be 12. Finally, VC = 12V. Now, compute the powers. P6A = 72 W , P3 Ω = 0, P6 Ω = (VC-VD)2 /6 = 96W, P2 Ω y = 72W, P2 Ω z = 72W. V − VD VC − VD P12V = 12 × ( A + ) = 12 × (6 + 4) = 120W, 2 6 P2ix = 0W , P2iy = 12 × 4 = 48W

Solution 3.18 The three node equations at A, C, and D are: −0.8 − 0.3 = 0.015VA + 0.02V A − 0.02VC Vc = 440 −0.8 + 2.5 = −0.005VD + 0.025VC − 0.025VD As can be seen, these really reduce to only two equations in two unknowns. These can be solved rather easily either by hand or by MATLAB to obtain: VA = 220V, VD = 310V . Note that all of these voltages are already referenced to node B (i.e. VA = VAB, etc).

Solution 3.19 (a) Supernode is BC (50 V source). (b) Only one node equation needs to be written:

VB VA VB VC VA VC − + + − + =0 90 90 10 10 10 90 with the constraint that VC − VB = 50. (c) The constraint equation can be substituted into the B node equation to obtain VB = 125V. Thus, VC = 175V, and ix = (VC-VA)/10 = -12.5A. (d) (VA-VB)/90 = 1.94A è P300 = 300×(1.94+12.5) = 4332W. VC/90 = 1.94 è P50 = -528W. V P50 = 50 × (i x + C ) = −528W 90

Solution 3.20 (a) VB = VA – 440 and VC = VA – 460. (b) Supernode is one including A, B, and C. (c) (VC − 40)0.15 + 0.05VB + 0.25V A − 25 + 0.2(V A − 40) = 0 This can easily be rearranged to get VA = 200V. (d) power Ps = V A × I = 200 × 25 = 5000W = 5KW

Solution 3.22 (a) VC = Vs2 = 6V (b) Ix = 0.01VA (c) Supernode at A,B, encompassing the controlled source. So, we have one equation: Is1 = 0.01V A + 0.0125VB + 0.1VB − 0.1VC V (d) Substitute the constraint equations, VA –VB = 20Ix = 0.2VA , (equivalently: VA = B ) into the above 0.8 V equation: VA = B 0.8 V Is1 = 0.01 B + 0.0125VB + 0.1VB − 0.1VC 0.8 ⇒ VB = 6.4V ⇒ V A = 8V (e) Ix = 0.08A. (f) P0.0125Ω = VB2 / R = 0.512W (g) P = I s1 × V A = 0.2 × 8 = 1.6W

Solution 3.23 (a) VC = Vs2 = 50V. (b) ix = VA/100. (c) Supernode A,B: Is1 = 0.01V A + 0.05VB + 0.05VB − 0.05Vs2 + 0.09V A − 0.09Vs2 VA = VB + 300i x = VB + 3V A (d) Solving the above two equations yields: VA = -90V, VB = 180V. V −90 (e) i x = A = = −0.9A 100 100 (f) (VB-VC) (VB-VC)/R = 845W. (g) PS1 = I S1 × VA = 2 × (−90) = −180W

Solution 3.24 (a) VB − VC = 3Vx = 3VB ; VC = −2VB (b) Supernode at B and C, encompassing controlled source. (c) V V V V V V Is2 = ( B − A ) + B + C + ( C − A ); VC = −2VB; 10I s2 = −2(VB + V A ) 10 10 10 10 10 10 equivalently: 2V A + 2VB = −10 (d) (0.1V A − 0.1Vs1 ) + (0.1VA − 0.1VB ) + (0.1V A − 0.1VC ) = 0; 0.3VA + 0.1VB = 0.1VS1 ; equivalently: 3V A − VB = −10 (e) Again, any method can be used to simplify and solve the system of two equations. The solution is: VA = -2.5V, VB = -2.5V.

Solution 3.25 (a) VA = Vs1 = 16V. (b) Supernode at C and D, encompassing controlled voltage source. (c)

Is2 = (0.75mVD − 0.75mVB ) + (1mVC − 1mV A )

(d) (e)

VC = 4VB + V D 0.75mVD + (0.75mVB − 0.75mVD ) + (0.25mVB − 0.25mVA ) = 0 or 1mVB − 0.25mV A = 0

(f) We now have three equations in VB, VC, and VD. These can be solved using any method. By inspection, we can immediately deduce VB from VA using the last equation: VB = 4V. The remaining two equations can be solved to obtain: VC = 20V and VD = 4V.

Solution 3.26 (a) The supernode is the combination of A, C, and the controlled voltage source. (b) Write node equations starting at the supernode: (G2VA − G2Vin ) + (G3VA ) + (G4V A − G4VB ) + (G6VC ) + (G5VC − G5VB ) = 0 ⇒ (G2 + G3 + G4 )V A + (−G4 − G5 )VB + (G6 + G5)VC = G2Vin and (2G6VC ) + (G4VB − G4VA ) + (G5VB − G5VC ) + (G1VB − G1Vin ) = 0 ⇒ (−G4 )VA + (G1 + G4 + G5)VB + (2G6 − G5 )VC = G1Vin and VA − VC = 3V x , VA − VC = 3(Vin − VA ), 4V A − VC = 3Vin In matrix form:  0.8 −0.5 0.2 V A   6        −0.4 0.6 0.1 VB  =  6   4 0 −1  VC  180  (c) The above system of equations can be solved to obtain: VA = 38.75V, VB = 40V, VC = -25V. (d) Iin = (G7Vin )+ (G 2 Vin – G2 VA) + (G1 Vin – G1 VB) = 5A. è Req = 12 Ω and P = 300W. P = Vin × Iin = 60 × 5 = 300W (e) Iout = VCG6 = -2.5A è P = 62.5W.

Solution 3.27 (a) Supernode is A,B encompassing controlled voltage source. (b) (VA – Vs1 ) + 0.4V B + (0.2V B – 0.2VC) = 0 è VA + 0.6V B – 0.2VC = Vs1 (c)

VA – VB = IB = 0.4 VB è VA = 1.4VB.

(d) (e)

Is2 = 0.2VB + (0.2V C – 0.2VB) = 0.2VC. In matrix form:

1 0.6 −0.2 V A  8        1 −1.4 0  VB  = 0  0 0 0.2  VC  2  The solution is: VC = 10V,V A = 7V ,VB = 5V .

i = (VA-Vs1 )1S = -1 è Pccvs = (VA − VB ) × (V A − Vs1 ) × 1S = −2W

(f)

Pvccs = (Vs1 -VC)(0.2VB) = -2W.

Solution 3.28 (a) Supernode at A,C, CCVS. (b) Node equation at supernode: è Constraint:

Is + 0.25mVA = G 1 VA + (G2 VA –G2 VB) + (G5 VC) + (G4 VC – G4VB ) Is = (G 1 + G 2 – 0.25m)VA + (–G2 – G4 )VB + (G4 +G 5 )VC VA – VC = 104 ix = 104 G3 VB è 0 = VA – 104 G3 VB – VC

At node B: G3 VB + (G2 VB – G2 VA) + (G4 VB – G4 VC) = 0 è

–G2 VA + (G3 +G 2 +G 4 ) – G4 VC = 0

è

(c) Matrix equation: G1 + G2 − 0.25m −G2 − G4  1 −10 4 G3   −G2 G3 + G2 + G4

G4 + G5  VA   I s      −1  VB  =  0  ; −G4  VC   0 

 0 −1 1    1000 −1000 −1000  −0.2 1.1 −0.8 

V A  2      VB  = 0  VC  0 

(d) Substitute the values of conductances and solve the above matrix equation in MATLAB to obtain: VA = -38V, VB = -20V, VC = -18V.

[

]

(e) P = V × I = (104 G3VB ) × −I s + VA G1 + G2 (V A − VB) = (−20) × [−2m − 1.9m − 3.6m] = 0.15W

Solution 3.29 Loop equation: è

Vin = 2kI1 + 500(I1 + 20m) Vin = 2500I1 + 10 è I1 = 20mA.

Pvin = 20m×60 = 1.2W. PI = 20m×(500I1 + 500×20m) = 0.4W. P2k = I1 ×I1 ×R = 0.8W. P500 = (I1 + 20m) 2 R = 0.8W. total power absorbed by resistors: PR = 0.8 + 0.8 = 1.6 total power delivered by sources: Ps =1.2 + 0.4 W=1.6W

Conservation of power is verified.

Solution 3.30 Loop equation:

100(I1 − 0.5) + 200I1 + 500 × (I1 + 20m) = 0

è I1 = 0.05A. P0.5A = I × V100Ω; V100Ω = 100 × (0.5 − 0.05); where V100Ω is the voltage on the 100Ω resistor. P0.5A = 0.5(0.5×100 – 0.05×100) = 22.5W. P20m = 20m(I1 +20m)500 = 0.7W.


3.3 = 50I1 + (50m + I1 )100 + (I1 – 30m)40 + (I 1 – 50m)60

è I1 = 0.01A. The power delivered by the independent voltage source: P = I 1 ×3.3 = 0.033W.


50 = 300I1 + (I1 – 0.4I1 )500

50 = (300 + 500 – 200)I1 è I1 = 0.0833A. Power absorbed by the 500Ω resistor. P500 = (I1 – 0.4I1 )2 500 = 1.25W.


1000(I1 − Is) + 4000I1 + 5000(I1 − gmVx ) = 0 →10000I1 − 2Vx = 50

and Vx = 1000(I s − I1) →1000I1 + VX = 50 . Solve the above two equations in I1 and Vx to obtain: I1 = 12.5mA, Vx = 37.5V. Thus, Req = Vx /Is = 750_, P = I vccs × Vvccs = gm × V x × 5000 × (gmVx − I1 )=0.1875W

Solution 3.34 Loop equation: V1 = 2Iin

Vin = 2Iin + 14Iin – 10V1

After replacing V1 in the loop equation we obtain: è Vin = – 4Iin è V R1eg = in = −4Ω Iin


Vs = 500I1 + 100(I1 + 0.5) + 400(I1 − 0.001Vx ) + 100(I1 + 0.005Vy )

Vx = 500I1 ,Vy = 400I1 - 400×0.001Vx = 400I1 – 200I1 = 200I1 After replacing Vx and Vy in the loop equation we obtain: Vs – 50 = 1000I1 è I1 = 0.1A Vy = 200I1 = 20V è P400ohm = Vy2 /400Ω= 1W Req = Vs/I1 = 150/0.1 = 1500Ω.

Solution 3.36 Select clockwise loop current I1 in the left loop. Select anti clockwise loop current I2 in the right loop. The two mesh equations are: 12 = I1 +10(I1 + I2 ) and 10(I2 + I1 ) + 2I2 + 12 = 0 The two simultaneous equations can be solved easily to obtain: I1 = 0.75A, I2 = +0.375A. P10ohm = (I1 + I 2 ) 2 /10 = 0.127W. Battery 1 supplies more current. (I1 > I2 )

Solution 3.37 (a) The equation for the left loop is: 660 = I1 R + 1.296(I1 + I 2 )+ 590 + I1 R The equation for the right loop is: 660 = (0.3 – R) I2 + 1.296 (I1 + I2 ) + 590 + (0.3 – R) I2

Simplifying the two equations: 70 = 1.596I1 + 1.296I2 70 = 1.296I1 + 1.596I2 The solution of these two equations is: I1 = I2 = 24.2A. (b) I1 + I2 = 48.4 , voltage across locomotive = 590 + 48.4×1.296 è power = 31592W. (c) Because the locomotive is 1/3 distance from either station it follows that R = 1/3×0.3 = 0.1 Ω. The two equations become: 70 = I1 (2R + 1.296) + 1.296I2 70 = 1.296I1 + (1.296 + 0.6 – 2R)I2 The solution of these two equations is: I1 = 32.64A, I2 = 16.32A. Current in locomotive motor I1 + I2 = 48.96A. Voltage across locomotive

590 + (I1 + I 2 ) × 1.296

It follows that: è

P = (I 1 + I2 )(590 + 49×1.296) = 31993W.

Solution 3.38 (a)

(b) The three loop equations are: 660 – 590 = 0.1I1 + 1.296 (I1 – I2 ) + 0.1I1 0 = 1.296 ( I2 − I1) + 0.2I2 + 1.296 ( I2 − I3) –70 = 1.296( I3 − I 2) + 0.2I3 These three equations can be solved using any method to obtain:

I1 = 46.8A, I2 = 0, I3 = – 46.8A. (c) Motor currents are 46.8A each. (d) Ps = VI = 660×46.8 = 30.9kW. Each source supplies 30.9kW.

Solution 3.39 (a) Define three meshes with three mesh currents. The first, I1 , is a clockwise current around the first mesh. The second, I2 , is a clockwise current around the middle loop of the circuit (through the 10mS, 5ms, and 5ms conductances). The third, I3 , is a counterclockwise current through the right-most loop containing the voltage source.

*current names shown above. (b)

I1 = 0.5A

I I I I I ( 2 − 1 ) + 2 + ( 2 + 3 ) = 0. 10m 10m 5m 5m 5m I I I ( 3 + 3 ) + 2 = 20 25m 5m 5m These are two equations in two unknown currents. After grouping the terms, it can be verified that: (c)

I2 = 0.1A, I3 = 0. Vx = 20V Vad = (0.5 – I2 )/10m = 40V Vbd = 20V

(d)

P0.5 = Va × 0.5 = 40×0.5 = 20W P20V = 0W Presistors = 2× I2 2 /5m + (0.5 – I2 )2 /10m = 20W

The conservation of power is verified.

Solution 3.40 (a) We can either write down the equations or evaluate the matrix by inspection:

90 ×(I 2 − 4.8m) + 10kI2 + 50 = 0 90kI3 + 10 × (I 3 − 4.8m) = 50 OR 0   I 2   90k × 4.8m − 50 100k =  0 100k   I 3  10k × 4.8m + 50  (b) The solution of the above equation is: I2 = 3.82mA, I3 = 0.98mA. (c) Current source: P = 4.8m×[(4.8m – I2 )90k + (4.8m – I 3 )10k] = 0.61W. Voltage source: P = 50(I3 – I2 ) = – 142mW.

Solution 3.41 (a) By inspection: 112k −90k −10k   I1  180       0   I 2  =  −60  −90k 100k −10k 0 100k   I3   60  (b) Using MATLAB: I1 = 4.4mA, I2 = 3.36mA, I3 = 1.04mA → ix = − I2 = −3.36mA (c) P180 = 180×4.4m = 0.792W, P60 = 60( I3 − I 2 ) = -0.139W.

Solution 3.42 The matrix equation is:  8 −6 3   I1  14       −6 8 −2  I2  =  0   3 −2 4   I 3   6  whose solution is: I1 = 4A, I2 = 3A, I3 = 0 A v = ( I1 + I3 + I2 ) ×2 = 2V

Solution 3.43 (a) First, note that two mesh currents are needed. Two clockwise currents are defined: I1 in the middle loop, and I2 in the right-most loop: Middle loop equation: 100I1 – 100Is1 + 20Ix + 80I1 – 80I2 = 0, where Ix = Is1 – I1 and

Right-most loop equation: 80I2 – 80I1 + 10I2 + V s2 = 0 These can easily be cast into the following matrix equation: 160 −80  I1  16  −80 90   I  = −6   2    (b) The solution of this equation is: I1 = 0.12A, I2 = 0.04A. (c) VA = 100(Is1 – I1 ) = 8V and VB = 80x(I1 − I2 ) = 6.4V . (d) Ps1 = Is1 VA = 1.6W. (e) P0.0125S = (I1 − I2 ) 2 /0.0125 = 0.512W.

Solution 3.44 (a) Create two clockwise mesh currents in the top loop (I 2 ) and the bottom-left loop (I1) . The bottom-right loop has an independent current source. Writing the loop equations: Vs1 = 200 (I1 – I2 ) + 200 (I1 + Is2 ) 200 (I2 – I1 ) + 100I2 + 300Ix + 200 (I2 + Is2 ) = 0, where Ix = I1 – I2 (b) Solving, we get: I1 = -0.1A, I2 = -0.7A, Ix = 0.6A. (c) VB = (I1 + Is2 )200 = 130V. (d) Pvs1 = I1 Vs1 = –25W, Pis2 = (VB + (Is2 + I2 )200)Is2 = 105W, P300ix = (−I 2 )(300Ix ) = 126W.

Solution 3.45 (a) Create two clockwise mesh currents in the top loop (I1) and the middle loop (I 2 ) (all resistor loop): Top loop equation: 0.5vx = 500 (I1 - I2 ,) +500I1 where vx = –500I1 and Middle loop equation: 600 (I2 – Is1 ) + 500 (I2 – I1) + 900 (I2 + Is2 ) = 0 (b) Solving, we get: I1 = 0.015A, I2 = 0.0375A, vx = -7.5V. (c) Pis1 = Is1 [0.5vx + (Is1 – I2 )600] = 109.7W P0.5vx = 0.5vx (I1 – Is1 ) = 1.63W Pis2 = (I2 + Is2 )×900×Is2 = 53.2W

Solution 3.46 Write the mesh equations in terms of R’s and then substitute the values from the matrix: Mesh 1 equation: v1 = R1 (i1 − i3) + R2 (i1 – i2 )-25i2 From this equation, and the first row of the matrix equation, we can deduce that R1 = 5 Ω and R2 + 25 = 40 ⇒ R2 = 15 Ω. Similarly: R1(i3 − i1) + i3R4 + R3 (i3 − i2 ) =0

Mesh 3 equation: From which we can deduce: R3 = 25 Ω

and

R4 = 5 Ω.

Solution 3.47 Modified loop 1 equation: Vs1 = 3MI 1 + v + 2MI1 Constraint equation: – I1 + I2 = Is3 Modified loop 2 equation: v = 2MI2 + V s2 + 8MI2 Or in matrix form: 5M   −1  0 Solving: I1 = -1.1

0 1 −10M

1   I1  Vs1      0   I2  =  Is 3  1   v  Vs2 

A, I2 = -0.95

A.

The power Ps3 = Is3 xv= 7.58 W

Solution 3.50 I2 = 2A, I3 = –7A Loop 1 equation: Vs = 3I1 + 3(I1 − I 2 ) + 6(I1 + I3 − I 2 ) + 2v y + 2(I1 + I 3) = 14I1 + 2v y − 74 ⇒ 14I1 + 2VY = 88 vy = 3(I1 – I2 )= 3I1 − 6

Solving the above system, we obtain: I1 = 5A è vy = 9, v can be found from the loop 3 equation v + (I1 + I3 − I 2 ).6 + 2v y + 2(I3 + I1) = 0 . Solving, we obtain v = 10V Finally, Pvs = Vs×I1 = 70W.

Solution 3.51 I1

I3 I2

Mesh 1 equation: 9kI1 + 3k (I1 – I4 ) + 6k (I1 – I3 ) + 12k (I1 – I2 ) = 0 where we have used the fact that Ix = I1 – I4 (and I4 = – 4mA) Mesh 3 equation: 6k (I3 – I1 ) – v2 +2kI3 + v = 0 Mesh 2 equation: 2.4kI2 + 12k (I2 – I1 ) – v = 0 Constraint equations: I2 – I3 = 0.5mA I3 – I4 = 0.5Ix = 0.5I1 – −0.51I 4 = 0.51I1 + 2mA ; 0.51I1 − I3 = −2mA The above five equations need to be put into matrix form:  30k −12k −6k 0 0   I1   −12   −6k 0 8k 1 −1  I2   0   −12k 14.4k 0 −1 0   I 3  =  0       1 −1 0 0   v   0.5m  0  0.5 0 −1 0 0   v 2   2m   The solution is:

 I1   −0.002A   I   −0.0025A   2    I 3  =  −0.003A       v   −12V  v   −24V   2   I x = I1 − I4 = 2mA Power delivered by 0.5mA current source: P1 = 0.5mxv = −6mW Power delivered by the dependent current source: P2 = 0.5I x ⋅ v 2 = 1m × (−24) = −24mW Power delivered by 4mA current source: P3 = 4m × (3k ⋅ I x − V2 ) = 120mW

Solution 3.52

I5 I6

I1

I4

I2

Write the following equations: Mesh 2 equation: 100 (I2 – I1 ) + 150 (I2 – I4 – I6 ) + v = 0 Mesh 4 equation: 3vx + 30 (I4 – I3 ) + 150 (I4 - I2 + I6 ) = 0 Mesh 5 equation: 800I5 – 3vx + 10( I5 − I1 + I 6) =0 Mesh 6 equation: 250I 6 + 150(I6 + I 4 − I2 ) + 10(I 6 + I5 − I1) = 0 Constraint equation:

I3

Is2 = −I 2 + I3 Substituting the values of I1 = I s1 = 1.15 and I3 = I s3 = 0.95 , and noting that vx = (I1 – I6 – I5 )10 we can write: −150 1   I2   100I1   250 −150 0 −150 180 −30 120 0   I   30I − 30I  1   4   3  0      0 840 40 0 I 5 = 40I1      10 410 0   I6   10I1 −150 150   1     0 0 0 0   v   I3 − I S2    I2  0.65  I  0.45   4    I5  = 0.05       I6   0.1   v   35      Finally, vx = 10V and v is as given above.

Solution 3.53 (a) Replace the voltage source by current sources: At node 1 is1 = (V2 − V1) /12k At node 2 (V2 − V1) /12k + (V2 − V3) /90k + (V2 − V4 ) /10k = 0 At node 3 is2 = (V3 – V2) /90k + V3 /10k At node 4 is2 + (V4 − V2 ) /10k + V 4 /90k = 0 V1 = 180 V4 – V3 = 60 (b) In matrix form:

1/12k 0 0 −1 0   V1   0  −1/12k −1/12k 1/12k + 1/90 k + 1/10k −1/90k −1/10k 0 0  V2   0        0 −1/90 k 1/90k + 1/10 k 0 0 −1 V3   0     =   −1/10k 0 1/90 k + 1/10k 0 1  V4   0   0  1 0 0 0 0 0   is1  180        0 0 −1 1 0 0  is 2   60  The solution from MATLAB is 180.0000 127.2000 33.6000 93.6000 -0.0044 0.00232 (c) Power delivered by S1 is: Ps1 = Vs1 × (−I s1 ) = 0.792W Power delivered by S2 is: Ps2 = Vs 2 × (−I s2 ) = −0.139W

Solution 3.54 (a) Replace the 100 ohm resistor, the controlled voltage source, and vs2 by current source. Then write the node equations: Is1 = ix + I1 + 0.09 (V 1 – V3) I1 = V2 /20 + 0.05 (V2 – V3) – I2 = 0.05 (V3 – V2 ) + 0.09 (V 3 – V1) V3 = Vs2 V1 – V2 = 300ix = 300V1 /100 In matrix form: 0 −0.09 1 0 V1   2   0.1  0 0.1 −0.05 −1 0 V2   0   −0.09 −0.05 0.14 0 1 V3  =  0        0 1 0 0  I1  50  0  2 1 0 0 0  I 2   0   (b) Using MATLAB to solve the above system the solution is: V1 = -90.0000V V2 = 180.0000V

V3 = 50.0000V I1 = 15.5000A I2 = -6.1000A (c) Power delivered by the current source is Ps1 = Is1 × V1 = −180W Power delivered by the voltage source is: Ps2 = Vs 2 × (−I 2 ) = 305W

Solution 3.55 Modify the circuit so that it looks like the following:

The modified node equations are: Is1 = - I y + I x Ib = Iy + 0.2 (V C – VB ) Is2 = 0.2 (VC – VB) + Ix The equations describing the constitutive relationships of elements in the original network are: VA – VB = Ib I x = 0.2Vb Ib = 0.4Vb I y = 8 − VA These can be cast into a matrix equation and solved easily to obtain the same result as previously arrived at. In matrix form:

0 0  0  1 0  0  1

0 −0.2 −0.2 −1 0.2 0.4 0

0 0.2 0.2 0 0 0 0

1 0 1 0 −1 0 0

−1 1 0 0 0 0 1

0 −1 0 −1 0 −1 0

−1 0  0  0 0  0  0 

V A  0 V  0  B   VC  2      I x  = 0  I  0  y    Ib  0      Is1  8

V A   7V  V   5V   B   VC  10V       I x  =  1A   I   1A   y    Ib   2A       Is1   0A 

Solution 3.56 Modify the circuit as follows:

The modified node equations are: At node A:

Is = G 1 VA + G2(VA − VB ) + Ia

At node B:

G2(VA − VB ) = Ix + G4 (VB − VC )

At node C:

0.25mVA + Ia = G4 (VC − VB ) + G 5 VC

The equations describing the constitutive relationships of elements in the original network are: Ix = G 3 VB VA – VC = 104 Ix

These can be cast into a matrix equation that can be solved in MATLAB. In Matrix form: 0 1 0   0.25m −0.2m  −0.2m 1m −0.8m 0 1   −0.25m −0.8m 1m −1 0    −0.1m 0 0 1   0  1 0 −1 0 −104  

V A  2m  V   0   B   VC  =  0       Ia   0  I   0   x  

The solution is: V A   −38V  V   −20V   B   VC  =  −18V       Ia   0.0075A   I  −0.002A   x   We observe that we have obtained the same results as in problem 3.28. Solution 3.57 Replace dependent source by i35 (from 3 to 5). Also, replace voltage source by i10 (from 1 to 0). Now, write the modified noted equations. The reference node is O:VO = 0V : At node 1:

i10 = (V6 – V1 ) + (V 2 – V1)

At node 2:

2 = (V2 – V1 ) + (V 2 – V3)

At node 3:

i35 = (V4 – V3 ) + (V 2 – V3)

At node 4: At node 5: At node 6: Constraints:

2 = 2 + (V 4 – V3 ) + V 4 i35 = (V5 – V6 ) + V 5 2 = (V5 – V6 ) – V6

V3 – V5 = 15vx = 15V4 V1 = 5 The following matrix equation is obtained:

−2 −1  0  0 0  0  0 1 

1 0 0 0 1 −1 0   V1   0 2 −1 0 0 0 0 0   V 2   2 1 −2 1 0 0 0 −1  V 3   0     0 −1 2 0 0 0 0  V 4   0 = 0 0 0 2 −1 0 −1  V 5   0 0 0 0 1 −2 0 0   V 6   2     0 1 −15 −1 0 0 0   i10   0 0 0 0 0 0 0 0   i35  5

The solution of this equation is obtained from MATLAB: V1 = 5.0000V V2 = 3.3571V V3 = 0.2857V V4 = -0.1429V V5 = 1.8571V V6 = -0.0714V i10 = -6.7143A i35 = 3.7857A The power delivered by the dependent voltage source connected between nodes 3 and 5: P35 = 15v x (−i35 ) = 15 × V4 × (−i35) = 8.115W The power delivered by the current source connected between nodes 2 and 4: P24 = (2A) × (V2 − V4 ) = 7W The power delivered by the current source connected between nodes 4 and 6: P46 = (2A) × (V4 − V6 ) = −0.143W The power delivered by the voltage source connected between nodes 1 and 0: P10 = 5V × (I12 + I16 ) = 5x [(V1 − V2 ) + (V1 − V6 )] = 33.57W

Solution 3.59 Using the appropriate element stamps for each element of the circuit, we obtain the following system: −0.15 −0.2 0  V A  −8 − 3 0.15 + 0.2  −0.15 0.15 + 0.05 0 −1  VB   3   =  −0.2 0 0.25 + 0.2 1  VC   25       0 −1 1 0   I x   440  

Solution 3.60

2

1

0 1/20 k + 1/10k + gm1  −1/10k − g − g  m1 m2

−1/10k   V1   Is1  = gm2 + 1/10k + 1/2.5 k  V2  −I s2 

The solution is the same as that of problem 3.14.

Solution 3.62 (a) Because RT (T ) can be approximated by a straight line between (250Ω,0 o C) and (80Ω,50 o C) it follows that: RT (T )= – 3.4T + 250 (b) For T = 25o C , RT = 165 Ω (c) The voltage across the RT + RL series combination can be obtained from voltage division: RT + RL VT ,L = ⋅12 = 4.7857V RT + 2RL + R This is the same as the voltage across Rx because the meter is at zero deflection. Thus, Rx ⋅12 = 4.7857. It follows that Rx = 165.84Ω. Rx + R (d) We first denote the nodes: A - the node common to R,R x and the voltmeter; B - the node common to R,R and the voltage source; C - the node common to RL ,RT and RL ; D - the node common to Rx ,RL and the voltage source. The reference node is D:VD = 0. It follows that VB = 12V . We also have: vout = V A − VC The node equations are:

VA − VC V A − 12 V A + + =0 Rm R Rx Equivalently: VA (RRx + Rm Rx + Rm R) - VC ⋅ RRx = 12Rm R x At node A:

VA × 4199.86 − VC × 41.46 = 19900.8 At node C:

(1)

VC − 12 VC V − VA + + C =0 R + RL RT + RL Rm

Equivalently: (VC − 12) × 0.004 +

VC + (VC − V A ) ⋅10 −4 = 0 RT + 2.5

VC (0.004 ⋅ RT + 0.01 + 1+ 0.00025 + RT × 10−4 ) − −V A (RT × 10 −4 + 0.00025) = 0.48 × (RT + 2.5) The last equation can be rewritten as: VC (0.0041RT + 1.01025) - VA (RT × 10 −4 + 0.00025) = 0.48(RT + 2.5) From (1) and (2), we obtain: (0.0041RT + 1.01025) ×

(2)

−19900.8 + VA × 4199.86 − 41.46

− VA × (RT × 10 −4 + 0.00025) = 0.48(RT + 2.5) Equivalently: VA × (RT × 0.415 + 102.337) = 2.448 × RT + 486.12 It follows that VA =

2.448 × RT + 486.12 0.415 × RT + 102.337

From the equation at node A:  V − 12 VA  2.448 × RT + 486.12 vout = V A − VC = −Rm ×  A + = ×(−100.3) + 480   R Rx  0.415 × RT + 102.337 At T = 0o C:

RT = 250Ω. It follows that vout = − 54.4415V

At T = 50o c:

RT = 80Ω. It follows that vout = 80V

(e): The formula has been derived at part d): T

RT

vout

0o C

250Ω

−54.4415V

5o C

233Ω

−52.4136V

10o C

216Ω

−50.2368V

15o C

199Ω

−47.8938V

o

20 C

182Ω

−45.3650V

25o C

165Ω

−42.6273V

30oC

−39.6537V

148Ω

Solution 3.63 Place a source Vin between C and D, and calculate the current drawn from the source as below:

Loop 1 equation: I1R1 + (I1 − I2 )R2 + (I1 − I 3)R3 = 0 Equivalently: I1(R1 + R2 + R3 ) − I 2R2 − I3R3 = 0 Loop 2 equation: (I 2 − I1)R2 + I2 R4 − 1 = 0 Equivalently: −I1R2 + I2 (R2 + R4 ) = 1 Loop 3 equation: 1 + I 3R5 + (I3 − I1)R3 = 0 Equivalently: −R3I1 + I 3(R3 + R5 ) = −1 We obtain the following system of equations: 30I1 − 4I2 − 6I3 = 0   −4I1 + 6I2 = 1  −6I + 14I = −1  1 3

⇒ I1 = 0.0096A , I2 = 0.1731A , I3 = −0.0673A;

Iin = I 2 − I 3 = 0.2404 A V 1 Reg,CD = in = = 4.16Ω Iin 0.2404

Solution 3.64 The node equation at node A is: VA G1 + (V A − VB )G2 + (VA − VC )G3 = 0 Equivalently: (G1 + G2 + G3)VA − VB G2 − VC G3 = 0 The supernode is identified by a Gaussian surface enclosing the controlled voltage source. The supernode equation is: G2(−VA + VB ) − 6 + G4VC + G3 (VC − VA ) = 0 Equivalently, we have: −V A (G2 + G3) + G2VB + VC (G3 + G4 ) = 6 One way of obtaining the solution to the problem is: We multiply the above two equations by 30. −30(G2 + G3)VA + 30G2VB + 30(G3 + G4 )VC = 180 and 30(G1 + G2 + G3 )V A − 30G2VB − 30G3VC = 0 By equating the coefficient of the above two equations with the coefficients of the first and second given equations, we obtain: 30G2 = 30 ⇒ G2 = 0.1S 30G3 = 2 ⇒ G3 = 0.067S 30(G1 + G2 + G3 ) = 11⇒ G1 = 0.2S 30(G3 + G4 ) = 32 ⇒ G4 = 0.87S can be obtained as follows: VC − VB = V X = (VC − VA ) Equivalently: VA − VB + (1− )VC = 0 By comparing with the third given equation ⇒

Solution of 3.66 (a) At node A: At node B: At node C: At node D:

= 3.

(VA-VC)/2 + (VA-VB)/2 + (VA-VD)/2 = 14 (VB-VA)/2 + (VB-VC)/2 + (VB-VD)/2 = 7 (VC-VA)/2 + (VC-VB)/2 + (VC-VD)/2 + 2VC = 0 (VD-VA)/2 + (VD-VB)/2 + (VD-VC)/2 + 0.5VD = 0

These can be solved in MATLAB to obtain: 22.0000 18.5000 7.5000 12.0000 (b) Mesh analysis would result in the same voltages

The loops and their current loops are: A,C,14A:I1   A,C,D:I2   A,B,D:I3  B,C,D:I 4  C,D,reference node:I 5   B,D,7 A:I6 I1 = 14A,I6 = 7A Loop ACDequation: 2(I 2 − I1) + 2(I 2 − I 3) + 2(I 2 + I4 + I5) = 0 Loop BCD equation: 2(I 3 + I 6 + I4 ) + 2(I2 + I 4 + I 5) + 2I 4 = 0 Loop ABD equation: 2(I 3 + I 4 + I 6 ) + 2(I3 − I 2 ) + 2I3 = 0 Loop CDref node equation: 2(I 5 + I 2 + I4 ) + 2(I5 − I 6 ) + 0.5(I5 + I1) = 0 In matrix form:

6 2  −2   2

−2 2 6 0

2 6 2 2

2  2  0   4.5

 I2   28   I  −14   3=   I 4  −14  ;      I5   7 

 I2   6.75  I  1.75   3=   I 4  −5.5      I5   1 

VA = 2(I1 − I 2 ) + 0.5(I1 + I5 ) = 22V VB = 2(I 4 + I 3 + I 6 ) + 2(I6 − I5 ) = 18.5V VC = 0.5(I1 + I5 ) = 7.5V VD = 2(I 6 − I 5) = 12V (c) Mesh analysis requires more work. (d) The removal of the top resistor will result in more node equations than loop equations. The addition of a resistor between node A and reference node will result in more loop equations than node equations.

Thevenin Probs, 7/11/01

- P4.1 -

@R.A. Decarlo & P. M. Lin

PROBLEM SOLUTIONS CHAPTER 4 SOLUTION 4.1. First, find Vout / Vs for each circuit. Then solve for R knowing Vout =

P ⋅10 = ± 14.142V .

(a) Writing KCL at the inverting terminal, 1 / 1 k(v− − vs ) = 1/ R(Vout − v − ) ⇒ Vout / Vs = − R / 1k , since the inverting terminal is a virtual short. Solving for R = −Vout ⋅1k / Vs = 2.828kΩ . (b) Writing KCL at the inverting terminal, Vs /1.5k = (Vout − Vs )/ R ⇒ Vout / Vs = R / 1.5k + 1, solving for R = 1.5k(Vout / Vs − 1) = 2.743kΩ. (c) From (a) Vout / Vs = −12k / R , thus R = −12k ⋅Vs / Vout = 4.243kΩ. (d) This is the same circuit as (b) except the output voltage is taken across two resistors. Thus Vout =

P (10 + 6) = 22.627V . Using the general form from (b), R = 400(Vout / Vs − 1) = 1.410kΩ 10

SOLUTION 4.2. (a) First, find the voltage at the non-inverting terminal as v + = 1/ 2 ⋅Vs . Then write KCL at the inverting terminal, and make use of the virtual short property, (Vs / 2)/10 k = (Vout − Vs / 2) / 30k ⇒ Vout / Vs = 30k(1/ 20k + 1/ 60k) = 2 . (b) Relating the output of the amplifier to the output of the circuit, Vout = Vamp (500 / 800) . Then writing KCL at the inverting terminal, Vs / 400 = (Vamp − Vs ) /1.2 k ⇒ Vamp / Vs = 1.2k / 400 + 1 = 4. Therefore Vout / Vs = (Vamp / Vs )⋅ (Vout / Vamp ) = 2.5. (c) Note that since no current goes into the non-inverting terminal of the op-amp, the voltage at that node is –Vs. KCL at the inverting terminal, −Vs / 4k = (Vout + Vs )/ 20k ⇒ Vout / Vs = −6. SOLUTION 4.3. Write KCL for both terminals, (V− − Vi ) / 1 k = (Vo − V− ) / 2k V− /1k = (Vo − V− )/ 3k Solving and doing the appropriate substitutions, Vo / Vi = −8 . SOLUTION 4.4. This is essentially the basic inverting configuration, which is defined as Vo / Vi = −2k / 1k = −2 .


SOLUTION 4.5. (a) By voltage division VL = 1V ⋅

- P4.2 -


100 = 0.5V . Using Ohm’s law 200

1 Is = I L = = 5mA . 100 + 100 (b) No current flows in the input terminal of an ideal op-amp, thus Is = 0A and VL = 1V . From Ohm’s law Ia = I L = VL /100 = 10mA. SOLUTION 4.6. (a) Using voltage division,  32||(8 + 24)  2 V1 = Vs  = V  [32||(8 + 24)] + 8 3 s 24  Vout = V1  = 0.5Vs  24 + 8 (b) By voltage division, 32  V1 = Vs  = 0.8Vs  32 + 40  24  Vout = V1  = 0.6Vs 24 + 8 32  (c) Using voltage division, V1 = Vs  = 0.8Vs , as no current enters the non-inverting terminal of the 32 + 8  24  op-amp. Due to the virtual short property, Vout = V1  = 0.6V1 . This is indeed the same results as 24 + 8 (b), which should be expected because of the isolation provided by the ideal buffers. SOLUTION 4.7. Write KCL at the inverting terminal, −Vs1 /1k − Vs 2 / 2k = Vout / 4k ⇒ Vout = −4Vs1 − 2Vs2 = 40mV . SOLUTION 4.8. (a) The voltage at the non-inverting terminal is V+ = 3 / 2V , KCL at the inverting terminal gives (1.5− 2.5) /10k = (Vout − 1.5)/ 30k ⇒ Vout = −1.5V . The power is 2 P = Vout / 500 = 4.5mW .

(b) The voltage at the non-inverting terminal is 3V this time, thus KCL 2 (3 − 2.5)/ 10k = (Vout − 3) / 30k ⇒ Vout = 4.5V . The power is P = Vout / 500 = 40.5mW .

SOLUTION 4.9. (a) Define the point between the two op-amp as Vint . Observe that the first op-amp is in the basic inverting configuration, and the second the non-inverting configuration. By inspection,


- P4.3 -


Vint / Vs = − R1 / 1k Vout / Vint = (1+ R2 /1k ) Cascading the two stages, (Vint / Vs )(Vout / Vint ) = Vout / Vs = − R1 /1k(1+ R2 /1k) . Solving for R1 = 20 ⋅1k / (1+ R2 /1k) = 5kΩ. The power absorbed is P = (20 ⋅0.5)2 / 8 = 12.5W . (b) Using the same equations as (a), solve for R2 = (20 ⋅1k / 2k − 1)1k = 9kΩ . (c) Rewriting the equation obtained in (a), R12 + 1kR1 − 20M = 0, and solving the quadratic equation yields R1 = R2 = 4kΩ . SOLUTION 4.10. This is a cascade of two non-inverting configuration op-amp of the form Vo / Vs = (1+ 10k /10k ) for each. Therefore 2 ⋅ 2 = 4. SOLUTION 4.11. This system is made up of a non-inverting stage with a gain of 1+10k/10k, a voltage divider of gain 8k/(2k+8k), and a second non-inverting stage of gain 1+10k/10k. The product of all three yields Vout / Vin = (2)(0.8)(2) = 3.2. SOLUTION 4.12. (a) By inspection, the gain of the first stage is –1. Then write KCL for the second stage Vs1 / 2R − Vs2 / R = Vout / 2R ⇒ Vout = Vs1 − 2Vs2 = 10V . (b) The first stage gain is –0.5, thus Vout = 2R(0.5Vs1) / 2R − 2R(Vs 2 )/ 0.5R = − 7.5V , using the same procedure as in (a). SOLUTION 4.13. (a) This is a cascade of a summing amplifier with the following transfer characteristic, Vo = −4Vs1 − 2Vs2 , and an inverting stage of gain –1.5. Thus Vout = 1.5(4Vs1 + 2Vs2 ) = 2.25V . (b) Notice that the only difference is the gain of the inverting stage, which is now –2. Therefore Vout = 2(4Vs1 + 2Vs2 ) = 3V . SOLUTION 4.14. This circuit is a cascade of two summing amplifier where the output of the first is an input of the second stage. The transfer function of the first stage is Vo = −2RVs1 / 2R − 2RVs2 / R , which is substituted in the transfer function of the second stage to obtain Vout = − R[−2RVs1 / 2R − 2RVs2 / R]/ R − RVs3 / R = Vs1 + 2Vs2 − Vs3 = − 2V . SOLUTION 4.15. Writing KCL at the inverting node, −V1 / R1 − V2 / R2 − V3 / R3 = Vout / R f , and Rf Rf   Rf solving for Vout = −  V1 + V2 + V . R2 R3 3   R1


- P4.4 -


SOLUTION 4.16. Referring figure P4.15, the value of the resistance must satisfy the following R1 = R2 = R3 = 3R constraints: Rf = R These will yield the inverted average. If polarity is a concern, a second inverting stage should be added with a unity gain, i.e. both R’s equal. SOLUTION 4.17. Using the topology of 4.12 the following parameters are chosen, Ga1 = 3, Ga2 = 5, Gb1 = 2, Gb2 = 4 For the time being assume G f = 1. Now we calculate δ = (1+ 3 + 5) − (2 + 4) = 3, this sets Gg = 3. (a) The requirement for G f = 10µS sets the scaling factor K = 10µ / 1 = 10µ . This then yields the following set of parameters, Ga1 = 30µS, Ga2 = 50 µS, Gb1 = 20µS, Gb2 = 40µS, G f = 10µS, Gg = 30µS (b) The requirement for G f = 2µS , sets the scaling constant to 2uS. So the following parameters are obtained: Ga1 = 6µS, Ga2 = 10 µS, Gb1 = 4 µS, Gb2 = 8µS Furthermore for Gg = 12µS , ∆G = 6µS in order to make the incident conductance equal at both terminal. (c) Using the starting values from (a), one could choose a scaling constant of 5 µS. This will yield the following resistances: Ra1 = 66.67kΩ, Ra2 = 40kΩ, Rb1 = 100kΩ, Rb2 = 50kΩ, R f = 200kΩ, Rg = 66.67kΩ These are all reasonable values for circuit implementation. SOLUTION 4.18. (a) Choosing the following initial values: Ga1 = 3S, Ga2 = 5S, Gb1 = 11S, Gb2 = 4S, G f = 1S


- P4.5 -


then calculate δ = (1+ 3 + 5) − (11+ 4) = − 6 . Thus Gg = 1S, and ∆G = 1 + 6 = 7S . Scaling everything by 1µS, yield this final set of parameters, which meet the requirements. Ga1 = 3µS, Ga2 = 5µS, Gb1 = 11µS, Gb2 = 4µS, G f = 1µS, Gg = 1µS, ∆G = 7uS (b) The set of parameters remains unchanged, except for ∆G which now becomes 6uS in order to maintain the equal termination conductance requirement due to Gg = 0S. (c) Scale the initial parameters of (a) by 5uS, and get the following set of resistances: Ra1 = 66.67kΩ, Ra2 = 40kΩ, Rb1 = 18.18kΩ, Rb2 = 50kΩ, R f = 200kΩ, Rg = 200kΩ, ∆R = 28.57kΩ

SOLUTION 4.19. (a) Choosing the following initial set of parameters: Ra1 = 1/ (4S) = 0.25Ω, Ra2 = 1/ (2S) = 0.5Ω, Rb1 = 1/ (5S) = 1/ 5Ω, Rb2 = 1/ (4S) = 0.25Ω, R f = 1Ω and δ = (1+ 4 + 2) − (5 + 4) = −2 , thus choose Rg = 1/ (1S) = 1Ω and ∆R = 1/ (1+ 2) = 1/ 3Ω . To meet the R f = 50kΩ requirement, all the parameters must be scaled by 50k, which gives Ra1 = 12.5kΩ, Ra2 = 25kΩ, Rb1 = 10kΩ, Rb2 = 12.5kΩ, R f = 50kΩ, Rg = 50kΩ, ∆R = 16.67kΩ (b) Same as (a) with a 100k scaling constant: Ra1 = 25kΩ, Ra2 = 50kΩ, Rb1 = 20kΩ, Rb2 = 25kΩ, R f = 100kΩ, Rg = 100kΩ, ∆R = 33.33kΩ SOLUTION 4.20. (a) When the op-amp is in its active region vout / vs = −5 . Thus it will operate in its active region when −3 ≤ vs ≤ 3, and will saturate at 15V when vs ≤ −3 , and at –15V when vs ≥ 3. SPICE yield the following plot:


(b) Using SPICE the following plot is obtained:

- P4.6 -



- P4.7 -


SOLUTION 4.21. The first stage is in a summing configuration, thus its output is, assuming it's in the active region of operation, -15 V which means it is just about to saturate. The second stage is in the inverting configuration with a gain of -1.5, which means that the overall output will be saturated at 15V. v + 1.5  SOLUTION 4.22. When vin − 80k  in > 0, or vin > 6 the output of the comparator saturates at 100k  –15 V, when it is vin < 6 , it will saturate at 15 V. The following plot is obtained from SPICE.

v + 20  SOLUTION 4.23. When vin − 10k  in > 0 , or vin > 2 the output of the comparator will be 110k  saturated at -15V. Otherwise when it is < 2V the output saturates at 15V. In SPICE:


- P4.8 -


SOLUTION 4.24. Based on the same reasoning as the previous questions,  R + R2  R1 The output will be +Vsat, when vin < vref  1 − 1 v , and –Vsat for =−  R2  R2 ref  R + R2  R1 vin > vref  1 − 1 v . =−  R2  R2 ref SOLUTION 4.25. Using the previously derived relationship, and the topology of figure P4.24, set vref = −1.5V , and R1 = 2kΩ and R2 = 3kΩ . Set the power supplies to the Op-amp to +/– 10V to satisfy the Vsat requirement. Also the input to the inverting and non-inverting terminal are reversed for fig. P4.24. Verifying in SPICE we obtain the following,


- P4.9 -


SOLUTION 4.26. The design that fulfills the requirement is the same as for P4.25, with the input to the op amp reversed. The following is obtained from SPICE,


- P4.10 -


SOLUTION 4.27. First, for the comparator to give +Vsat for the lower voltages, the inputs to the op amp in the topology of P4.24 must be interchanged. Then the components are chosen to satisfy the following  R + R2  relationship, vswitch = vref  1− 1  =−  R2  Verifying in SPICE,

R1 v . Choose vref = −1.5V , and R1 = 2k and R2 = 1k. R2 ref


- P4.11 -


SOLUTION 4.28. Write KCL at the inverting terminal, noting that the no current flows into it: (V− − vin )/ R = (vout − V− )/ R . Use the following relationship vout = A(V+ − V− ) = −AV− . Solving A using the previous two equations yields vout / vin = − . A+2 SOLUTION 4.29. (a) By inspection the voltage gain for the ideal case is –1. When A=1000, the gain becomes –0.998, thus 0.2%. ARf (b) Repeating the method of P4.28, and setting vout / vin = − to –1 and solving for R f + R1 + AR1 R f = 10.417kΩ. (c) Solving the previous equation when the gain is –1, R f / R1 = (A + 1)/ ( A − 1). SOLUTION 4.30. (a)The first part was obtained in P4.29. Rearranging the equation yields  Rf  1  . vout / vin = − R1  1 + 1+ R f / R1 / A   

(

)


- P4.12 -


(b) The error is caused by (1+ R f / R1) / A in the denominator, and may be defined, in percent, as   1   ⋅100. Thus for the conditions listed in the problem, it will always be less than 100 −  1+ 1+ R f / R1 / A    2.05%. With A = 10000 it will be less than 0.21%.

(

)

SOLUTION 4.31. (a) Substituting the non-ideal model, and writing KCL at the inverting terminal, (V− − vin )/ R1 + V− / Rin = (vout − V− ) / R f is obtained. Now observe the following dependencies, iout = vout / RL , and vout = − AV− − (iout + (vout − V− )/ R f )Rout . Using these three equations, substitute the second into the third and then solve for vout / vin using the last two. This yields  −1 − Rout − Rout  Rf Rf RL Rf  Rf  V− = vout + vin 1/  + + 1 = vout  R R1  R1 Rin   A − out  Rf  and

Rf vout / vin = − R1

            1        R R    1 + Rout + Rout   1+ f + f      Rf RL   R1 Rin     1+    R   A − out    Rf    

A gain of –9.988 (b) For an ideal op-amp the gain is − R f / R1 =-10. (c) The error is about 0.1175%. SOLUTION 4.32. The gain is –9.883, and the error 1.16% SOLUTION 4.33. This derivation was performed in P4.31.

     


- P4.13 -


SOLUTION 4.34. Assume that the sliding contact is at the bottom of Rp . Then, writing KCL at the inverting terminal yields vin / Ro = (vout − vin )/ R p . This implies vout / vin = 1 + Rp / Ro . When the Rp slider is at the top, it is evident that vout = vin . Therefore 1 ≤ vout / vin ≤ 1 + . Ro SOLUTION 4.35. Writing KCL at inverting input, and making use of voltage division,

[

]

Rf v −vin / R1 = βvout / R f where β is the fraction of vout that appears across Rf. Hence, out = − . vin βR1 Rf v When the slider is at the top β = 1 and out = − . When the slider is at the bottom, the fraction of vout vin R1 appearing across Rf is β =

R f / / R0 R f R0 R f R0 1 = × R R = . Hence R f / / R0 + Rp R f + R0 f 0 R R + R (R + R ) f 0 p f 0 + Rp R f + R0

1 R f R0 + R p (R f + R0 ) = =1 + β R f R0

Rp R0

+

Rp Rf

. It follows that

Rf Rf  Rp Rp  vout =− =− 1 + + . vin βR1 R1  R0 R f 

Therefore the range of achievable voltage gain is

−

Rf

Rf v ≥ out ≥ − R1 vin R1

 Rp Rp  1 + R + R   0 f

SOLUTION 4.36. Using the basic non-inverting configuration of figure 4.10 characterized by Rf  Rf    vout / vin =  1+  , i.e., µ =  1+ .  R1   R1  SOLUTION 4.37. At first glance, one might use two inverting configurations, figure 4.5, in cascade. However, such would not have infinite input resistance. To circumvent this problem we add a buffer amplifier as per figure 4.7 at the front end of a cascade of two inverting configurations. The resulting  R f1   R f 2  overall gain is µ =    . Indeed, such a configuration can achieve theoretically any gain greater  R11   R12  than zero.


- P4.14 -


SOLUTION 4.38. Using a single inverting amplifier configuration, figure 4.5, preceded by a buffer stage Rf of figure 4.7. The gain is µ = − . R1 SOLUTION 4.39. By KVL for figure P4.39a, Vo = −i1R f . Thus to achieve Vo = −i1rm in figure P4.39b, we set R f = rm . SOLUTION 4.40. Writing KCL at the inverting node of the ideal op amp yields IL = Vi / Ra , which is indeed independent of the load resistor which has no effect on the load current. R SOLUTION 4.41. The current through the LED is IL = 10 1  / 3.8k , so for (a) it is 1.32mA and for 10k (b) 2.11mA. SOLUTION 4.42. Applying KCL at the inverting terminal, IL = vin / R1 . Again, ideally, RL does not affect IL. SOLUTION 4.43. (a) Defining a temporary voltage Vo at the output of the op-amp, we can write KCL at the inverting and non-inverting terminal: (V− − 2)/ 1k = (Vo − V− ) / 2k V− /100 + (V− − Vo ) / 200 = Iout Substituting the first equation into the second and simplifying causes Vo to drop out and Iout = 20mA . (b) The answer remains the same as the value of the load resistance was not used for finding the load current. SOLUTION 4.44. Using the same approach as for the previous question, but with resistor labels instead, the following equations are obtained from KCL: R V + R1Vo V− = 2 s R2 + R1  R + R1  Iout = V−  2 −  R1R2  Substituting the first into the second yields Iout = Vs / R1 .

Vo R2


- P4.15 -


SOLUTION 4.45. (a) Vs = 5 V, (b) Iout = 10mA sets R1 = Vs / Iout = 500 Ω. (c) From KVL and Ohm's law, Is = (Vs + RL Iout ) / αR1 . We require Is < 0.5 mA. This means that in the worst case, RL = 500 Ω, (Vs + RL Iout ) 5 + 500 × 0.01 = = 40 < α . 0.25 0.5 × 10−3 R1 (d) From KVL and Ohm's, Vo = − RL Iout − (Iout + (RL Iout )/ R1)R2 ≥ −20 V Hence R2 ≤

20 − RL Iout 20 − 5 = = 750 (Iout + (RL Iout ) / R1 ) 0.01 + 5/ 500

Hence one design is to pick R2 = 750 Ω and α = 40 which impliesαR2 = 30 kΩ. SOLUTION 4.46. The exact same design as P4.45 can be used with the isolation buffer of figure 4.7 placed at the input of it in order to provide the infinite input resistance needed by P4.46b. SOLUTION 4.47. The general expression for this summing circuit is

Vout = −

Rf Ro

Vo −

Rf R1

V1 −

Rf R2

V2 −

Rf R3

.

(a) Using the expression above | Vout |=| −1 − 0 − 0 − 8 | E = 9E . (b) | Vout |=| −0 − 2 − 4 − 0 | E = 6E . (c) It has to be a linear combination of 8, 4, 2, 1, thus [1 1 0 1] would yield 13E. (d) With the same approach, [0 1 1 1]. SOLUTION 4.48. For this implementation we add an extra R-2R branch along with an extra summing input to the op amp. From the theory developed in Example 4.9 the total resistance seen by the source is E2 2R. For the total power supplied by the source to be less than 0.02 W, we require R ≥ = 2.5 kΩ. 2( 0.02)


- P4.16 -


SOLUTION 4.49. The same steps as in the previous questions are repeated. Because the resistance seen by the source is unchanged no matter how many branches are added to the R-2R network, R≥

E2 = 5 kΩ. 2( 0.01)

SOLUTION 4.50. (a) If the input is 3vmax / 8, then the first comparator will give –Vsat, keeping S2 down. The next comparator will output +Vsat, causing S1 to go up. After subtraction, the input to the last comparator is vmax / 8 yielding +Vsat at its output since its input is slightly above the reference input level. Thus the logic output values are [0 1 1]. (b) Putting in 6v max / 8, will cause +Vsat and S2 to go up. The input to the second comparator will be 2vmax / 8, which will cause +Vsat and S1 to go up. The input to the last comparator will be 0, thus it will output –Vsat. The corresponding logic output is [1 1 0]. SOLUTION 4.51. Simply add a subtractor and switch to the last comparator, followed by an additional − comparator. The reference level to the new (additional) comparator will be vmax /16 , and its output will be

the new least significant bit. SOLUTION 4.52. (a) Writing the node equation for figure P4.52c, Vout (Vout − V1) A(0 − V1) − Vout + = 10k Rout R'L which implies that

Vout / V1 = 

 1 A   10k − R  out

1 1 1   R' + 10k + R  out L

For figure P4.52a, the corresponding node equation is Vout A(0 − Vin ) − Vout = RL Rout which leads to


- P4.17 -


 A   − R  out

Vout / Vin =  1 1   R + R  L out Note that

1 1 1 = 1 + , which when substituted into the later equation make both of them RL RL 10k

approximately the same since the 1/10k term in the numerator of Vout / V1 has a negligible contribution. (b) Writing the node equation for figure P4.52d, yields Vout (Vout − V2 ) A(0 − V1 ) − Vout + = 10k Rout R'L Hence   V2 100|| Rin V1 = V2  ≈  (100 || Rin ) + 10k  101  1 A   10k − 101R  out

Solving produces Vout / V2 = . Note that as in (a) the 1/10k term in the numerator is  1 1 1   R1 + 10k + R  out L V V negligible; after eliminating this negligible term, one sees that out is 101 time smaller than out . V2 Vin SOLUTION 4.53. (a) Using the equation just derived, after substituting in the values, the gain is –980.392 (b) From the previous equation, Vout / V1 = −980.382 ; write KCL at the non-inverting terminal to obtain, Vin − V1 V V − Vout = 1 + 1 ; substitute Vout = −980.382V1 ; solve for V1 / Vin , and then multiply both gains 10k Rin 10k to obtain (Vout / V1)(V1 / Vin ) = Vout / Vin = −0.9979 . (c) They only differ by about 0.01%, thus they are very similar. R R SOLUTION 4.54. Writing out the transfer equation, Vout = 2 Vs2 − 2 Vs1 , thus R2 / R1 = 4 . Using R1 R1 R2 = 100kΩ , R1 = 25kΩ . As expected SPICE shows to noticeable difference in outputs when the source resistances are varied. SOLUTION 4.55. Due to the ideal nature of the op-amp, the voltage VRb = Vs2 − Vs1 . By KVL


- P4.18 -


V2 = Vs2 + Ra (Vs2 − Vs1 )/ Rb V1 = Vs1 − Ra (Vs 2 − Vs1) / Rb Next, V1 − V2 = (Vs1 − Vs 2 )(1+ 2Ra / Rb ) . R R SOLUTION 4.56. (a) Noticing that the final stage is a summing op-amp in which Vout = 2 V1 − 2 V2 . R1 R1 R From the previous question, Vout = 2 (V1 − V2 ) = R1

R2 (1+ 2Ra / Rb )(Vs1 − Vs2 ) . Thus R1

R α = 2 (1+ 2Ra / Rb ). The gain α can be varied by adjusting the single resistance Rb . R1 (b) Picking the set of values below will satisfy the requirement: R2 = 100kΩ, R1 = 100kΩ, Ra = 20kΩ, Rb = 10kΩ . (c) Doing the SPICE simulation using the parameters from (b) yield 5 V at the output for Vs1 − Vs2 = 2 − 1 V. Setting Rb arbitrarily to 20 kΩ, the output now becomes 3 V, which agrees with the relationships developed earlier.

PROBLEM SOLUTIONS CHAPTER 5.

Solution 5.1. (a) Vs = 10 V, P = 20 W and P = Vs ×Is implies Is = 2 A. (b) Rin = Vs /Is = 10/2 = 5 Ω

(c) By the linearity/proportionality property:

Vsnew Isnew 2 Isnew = which implies = implies 10 2 Vsold Isold

Isnew = 0.4 A. (d) P new = Vsnew × Inew = 2 × 0.4 = 0.8 watts. Observe that s P new 0.8 Vsnew 2 = ≠ = 20 Vsold 10 P old It follows that the proportionality property does not hold for power calculations.

Solution 5.2 First note that the ratio IR/VS is constant. With the given values of voltage and current, this ratio is: IR/VS = 0.25/25 = 0.01 Power dissipated in the resistor is P = I R2 R = 2.5 è IR2 = 2.5/R = 0.25 è IR = 0.5 Since IR is always 0.01×VS, it follows that VS = 50V.

Solution 5.3 Label the resistances R1 , R2 , and so on in the manner shown in Example 5.11. In this problem, we have R1 to R10 (the last being the 2 Ohm resistance at the voltage source). First, assume that V1 (the voltage across R1 ) is 1V. Then evaluate the rest of the currents and voltages until you deduce the

resulting VS. It should be noted that the equivalent resistance looking into R3 , R5 , R7 , and R9 is always 2Ω. V V1 = 1 ⇒ I1 = 1 = 0.25 ⇒ I 2 = 0.25 ⇒ V2 = I2 × 2 = 0.5 V 4 V V3 = V1 + V2 = 1.5 è I3 = 3 = 0.5 è I4 = I3 + I 2 = 0.75 è V4 = I4 × 4 = 3 3 V5 = V3 + V4 = 4.5 è I5 =

V5 = 1.5 è I6 = I5 + I 4 = 2.25 è V6 = I5 × 4 = 9 3

V V7 = V6 + V5 = 13.5 è I7 = 7 = 4.5 è I8 = I7 + I6 = 6.75 è V8 = I8 × 4 = 27 3 V9 = V8 + V7 = 40.5 è I9 =

V9 = 13.5 è I10 = I9 + I8 = 20.25 è V10 = I10 × 2 = 40.5 3

VS = V9 + V10 = 40.5 + 40.5 = 81 V Thus, an 81 V input produces a 1 V output è Vout = (1/81)×VS = 2 V.

Solution 5.4 Label the resistances R1 to R10 progressively from right to left just like in the previous problem. Then, assume Iout = 1 and proceed as follows: Iout = 1 è V1 = Iout × 4 = 4 è I2 =

V1 =1 4

V I3 = I1 + I 2 = 2 è V3 = I3 × 4 = 8 è V4 = V3 + V1 = 12 è I4 = 4 = 4 3 I5 = I4 + I 3 = 6 è V5 = I5 × 4 = 24 è V6 = V5 + V4 = 36 è I6 =

V6 = 12 3

V I7 = I6 + I5 = 18 è V7 = I7 × 4 = 72 è V8 = V7 + V6 = 108 è I8 = 8 = 36 3 V I9 = I8 + I 7 = 54 è V9 = I9 × 4 = 216 è V10 = V9 + V8 = 324 è I10 = 10 = 108 3

IS = I10 + I9 = 162 è Iout/IS = 1/162 è Iout = (1/162)×40.5 = 0.25A.

Solution 5.5 (a) MATLAB code given in problem. (b) Subsitute to obtain V1 = 10V. (c) Req = VS/IS = 11.6667Ω. (d) First, define r1 = 1:0.25:10; then create an outermost loop around the code of part (a) as: for j=1:length(r1) then, in the statement defining R, do R = [R1(j), R2, R3, R4, R5, R6, R7, R8]’; Finally, replace the last statement with Vs(j) = V(n) + V(n-1); end; The following is the resulting plot:

Solution 5.6 (a) The following code can be used: n = 9;

v = zeros(n,1); i = zeros(n,1); r = [r1 r2 r3 r4 r5 r6 r7 r8 r9]’; i(1) = 1; v(1) = i(1)*r(1); i(2) = i(1); for k=2:2:n-2 v(k) = r(k)*i(k); v(k+1) = v(k)+v(k-1); i(k+1) = v(k+1)/r(k+1); i(k+2) = i(k+1) + i(k); end; v(8) = i(8)*r(8); v(9) = v(8) + v(7); i(9) = v(9)/r(9); Is = i(9) + i(8); It follows that Is = 16.9877A . (b) By the proportionality property: (c) Req = v(9)/Is = 38.15Ω.

Solution 5.7 Va = 12V, iB = 60m By inspection: Vout_a = 300/900×12 = 4V Vout_b = (300||600)×60m = 12

I1new Isnew 1 = old → I1new = × 200mA = 11.77mA old 16.9877 I1 Is

_

Vout = 4 + 12 = 16V.

*S OLUTION 5.8. Part 1: Set the 3 A current source to zero. This generates an open circuit in place of the current source eliminating the effect of the series 0.1 S resistor. The equivalent circuit is:

By voltage division, 1 2 0.25 + 0.2 + 0.05 VL12V = × 12 = × 12 = 2 V 1 1 2 + 10 + 0.25 + 0.2 + 0.05 0.1

Part 2: Set the 12 V source to zero. This generates a short circuit in place of the voltage source which shorts out the effect of the 0.5 S resistor. The equivalent circuit is:

Note that the 0.1 S resistor in series with the 3 A source is redundant to the calculation of VL. Hence, by Ohm's law, VL3A =

1 3 ×3= =5 V 0.25 + 0.2 + 0.05 + 0.1 0.6

Therefore by superposition, VL = VL12V + VL3A = 2 + 5 = 7 V

Solution 5.9 Replace the dependent source by an independent voltage source VS:

VS

In the following analysis, we have to always compute Va because It defines the constraint on Vs. So, when only the 88V source Is active, VA Is the result of voltage division between the 60||30 Ω resistor and the 120||30 Ω resistor. So, Va_1 = 40V And, since deactivated VS, Vout_1 = 0. Now, due to the 55V source, we have

Now, the 120 and 60 Ω resistors are in parallel, and the same can be said about the 30 and 30 Ω resistors. Thus, another voltage divider gives: Va_2 = – 15V and Vout_2 = 0V. Finally, when VS1 is active, the left part of the circuit consists only of resistances, so Va_3 = 0. Vout is given by another divider formula: Vout_3 = 90/100×VS Now add all contributions: Va = 40 –15 + 0 = 25V Vout = 0 + 0 + 0.9VS, where VS = 2Va. Vout = 0.9×2×25 = 45V. Finally, P = V2 /R = 22.5W.

Solution 5.10 Due to 3A source:

iout = 1A by current division between the two paths. So, vout_1 = 2 V. Due to 1A source:

iout = 2/3A again by current division. So, vout_2 = 4/3V. Due to 1vV source:

iout = 14/6 = 7/3A (by Ohm’s Law). So, vout_3 = 14/3V. Finally, vout = 6/3 + 4/3 + 14/3 = 8V, and the power delivered by the source is 8×1 = 8 W.

Solution 5.11 Due to 22 V source:

Req = 900||225 = 180. Now, by voltage divider:

Vout_1 = 0.5×22 = 11 V. Due to the 20 V source:

Req = 180||225 = 100 Ω. So, again, by voltage division: Vout_2 = 900/(900+100)×20 = 18 V. Finally, due to current source:

We have three resistances in parallel with a resistance equal to 90 Ω. So, Vout_3 = 0.1×90 = 9 V.

Vout = 11 + 18 + 9 = 38V and P = 38×38/900 = 1.6 W.

Solution 5.12 Find contribution to Vout :

First, note that no current flows through R1 - R3 because of the virtual ground property of the op-amp. Thus, this circuit is identical to the inverting amplifier studied in Chapter 4. So, Vout _ 0 = −

Rf R0

V0

Similarly, when each of the other sources is activated, the circuit will be an inverting amplifier. So, Vout _1 = −

Rf Rf Rf V1, Vout _2 = − V2, Vout _3 = − V R1 R2 R3 3 V V V V  Vout = −R f  0 + 1 + 2 + 3   R0 R1 R2 R3 

Solution 5.13 Due to the 4 V source, the circuit looks like an inverting amplifier:

So, Vout_1 = -30/10×(-4) = 12 (from the results of Chapter 4). Again, note here that no current flows through the two resistances connected to the + terminal of the op amp. Since no current flows through them, then no voltage develops across them. So, the + terminal can be assumed to be connected to ground, and this is why we say that the circuit looks like that of the inverting amplifier.

Now, due to the 6 V source:

The voltage at the + terminal is established by a resistive divider between the two 10K resistors. So, this voltage is 3V. Thus, the voltage at the negative terminal is also 3V. We can now use KVL on the inverting side of the op-amp to get: Vout_2 = 3 + 30k×0.3m = 12V

So, Vout = 12 + 12 = 24V, and P = (24)×(24)/500 = 1.15W.

Solution 5.14 (a) When VS2 is deactivated. The circuit looks like two inverting amplifiers in cascade. Thus, by inspection, V1 = –2VS1 and Vout_1 = –3V1 = 6VS1 = 3V. (b) Similarly, when VS1 is zero, V1 is zero because the first inverting amplifier has zero input. Thus, the circuit consists just of the second inverting amplifier:

Vout_2 = – 3/2VS2 = – 4.5 (c)

Vout = 3 – 4.5 = – 1.5V.

*S OLUTION 5.15. For Vs1 and Vs2 , the situation reduces to the analysis of two inverting amplifiers in cascade. For Vs3 , the situation is simply a single stage inverting amplifier. Note that because of the virtual ground at the inverting terminal of the op amp, when Vs1 and Vs2 are zero, they have no contribution to the input of the second stage. (a) With Vs2 and Vs3 set to zero, s1  − R   −2R  Vout = V = Vs1 = 5 V R   2R  s1

(b) With Vs1 and Vs3 set to zero, s2  − R   −2R  Vout = V = 2Vs 2 = 2 × (−2.5) = −5 V R   R  s2 (c) With Vs1 and Vs2 set to zero,

 −R  s3 Vout =   Vs3 = − Vs3 = −2 V R (d) By superposition, s1 s2 s2 Vout = Vout + Vout + Vout = −2 V

Solution 5.16 If the op-amp were ideal, we would get: Vout = – 4VS1 – 2VS2 = – 26 This is clearly beyond the linear range of operation of the op-amp. In other words, the amplifier responds in a non-linear manner to this level of input. Hence, superposition, which relies on linearity, cannot be used.

Solution 5.17 We know by linearity that Vout = aIs1 + bV s2 Substitute the first measurement to obtain: 5 = 0 + b×10 è b = 0.5 Now, take the second measurement: 1 = a×10 + 0 è a = 0.1 So, Vout = 0.1Is1 + 0.5V s2 At 20A, 20V: Vout = 12V

Solution 5.18 Again

Vout = aIs1 + bV s2

Substitute the two measurements to obtain: 5a + 10b = 15 2a + 5b = 10 These two simultaneous equations in a and b can easily be solved to obtain: a=–5

and

Therefore, at 1A and 5V, Vout = – 5 + 20 = 15V

Solution 5.19 (a) Again, Vout = aIs1 + bV s2

b=4

Substituting the result of the first measurement into this equation yields: a×4× cos(2t) + 0 = –2cos(2t) è a = – 0.5 Now, substitute the second measurement: 0 + 10b = 55 è b = 5.5 Therefore, for the given input current and voltage: Vout = –cos(2t) –55cos(2t) = –56cos(2t) V (b)

Vout = 2cos(5t) + 110cos(5t) = 112cos(5t) V

Solution 5.20 (a) First of all, Iload = aVa + bIb Now, substitute the two measurements into this equation: 7a + 3b = 1 9a + b = 3 Solving these two equations for the unknowns a and b, we get a = 0.4 (b)

and

b = – 0.6

Iload = 0.4×15 –0.6×9 = 0.6A

Solution 5.21 (a) We know that the output is going to be a linear combination of the three inputs: Vout = aIs1 + bV s2 + cVs3 Now, substitute the three measurements into this relationship: 50ma – 2b + 5c = – 13 0 + 3b + 5c = 2

0 + 2b + 4c = 0

These equations can be written in matrix form and solved as follows: 0.05 −2 5   a  −13  a  100          3 5   b =  2  ⇒  b  =  4   0  0  c   −2  2 4   c   0  (b) Substitute the given values of the input sources to obtain: Vout = 100×1 – 4×40 + 2×10 = – 40V

Solution 5.22 Vout = AIs1 + BV s2 + CV s3 Substituting the measurements into this equation results in a system of three equations and three unknowns. This system can be written in matrix form by inspection:  30 ⋅10 −3  −3 −20⋅10 −10⋅10−3 

2 −1  A  11.5  A  150         4 2   B  =  27  ⇒  B  =  5.5   C   4  −3 1  C   −14 

This can be solved to obtain:

Solution 5.23 (a) The coefficient matrix is inverted, and both sides of the nodal equation are multiplied by it to obtain: 0.233   0.02Vs1 − 0.00125Vs2  Va   43.0108 V  =  43.0108 −1.4337   Vs 2  b    Expanding the first row of the above equation gives: Va = 0.8602Vs1 – 0.0538Vs2 + 0.233V s2 This is exactly in the form required, where A = 0.8602 and B = 0.1792

(b) For this part, we expand the second row of the equation: Vb = 0.8602Vs1 – 0.0538Vs2 – 1.4337Vs2 Again, this is in the desired form, where A = 0.8602 and B = – 1.4875 (c)

Vab = Va – Vb = 0 + 1.667 V s2

Thus, A = 0 and B = 1.667

Solution 5.24 (a) Again, we invert the coefficient matrix to obtain

0.0022 0.6296 −0.0741  Vs1   I1   0.0022 0.0022  I   0.0022 0.0022 0.0022 −0.7037 0.2593   0   2   I 3  =  0.0022 0.0022 0.0022 −0.3704 −0.0741   0        v1   −0.2222 0.7778 −0.2222 137.037 −92.5926  I s2  v   −0.3333 −0.3333 0.6667 −11.1111 77.7778   0   2   

Expanding the first row of this equation: I1 = 0.0022Vs1 + 0.6296Is2 So, A = 0.0022, B = 0.6296 (b) Similarly, expanding the third row: I3 = 0.0022Vs1 + -0.3704Is2 So, A = 0.0022, B = -0.3704 (c) By the same procedure, A = -0.3333 and B = -11.111

Solution 5.25 Invert the coefficient matrix and mujltiply both sides of the equation in the problem by this inverse matrix to obtain

−3 2 −1 −4   I s1  V1  0 0 0 V  0 0 0 −3 1 −1 −4   0   2  V3  0 0 0 12 −3 3 16   0       −4 1 −1 −5   0   Ia  = 0 0 0  I  0 −1 0 −16.6 9.2 −5.2 −21.8  Vs2   b   Ic  1 1 0 −42.1 −16.2 11.7 53.8   0        Id  1 1 1 26.1 −12.2 7.7 32.8   0 

The second to last row can be expanded to get Ic = Is1 – 16.2Vs2 è A = 1, B = –16.2

Solution 3.26 (a) First compute the response due to V1 :

By voltage divider: Vout_1 = 6/(2+6+1)×V1 = 2/3V1

Then, due to I2 :

I2 flows through the 20 ohm resistor in series with 6||3. Thus, Vout_2 = –2I2 Therefore,

Vout = 2/3V1 – 2I2

(b) Vout = 8cos(10t) – 4

*S OLUTION 5.27. (a) By linearity Vout = A Vs1 + B I s2 . To find A, let Is2 = 0. The circuit becomes a ladder network as follows.

A Let Vout = 1 V. Then

»I1 = 1/420 + 1/70 I1 = 1.6667e-02 »V400 = I1*40 + 1 V400 = 1.6667e+00 »I400 = V400/400 I400 = 4.1667e-03 »I2 = I1 + I400 I2 = 2.0833e-02

»Vs1 = I2*20 + V400 Vs1 = 2.0833e+00 »A = 1/Vs1 A = 4.8000e-01 To find B, let Vs1 = 0. The circuit becomes a ladder network as follows.

B Again assume that Vout = 1 V. Then

»I1 = 1/420 + 1/70 I1 = 1.6667e-02 »V400 = I1*40 + 1 V400 = 1.6667e+00 »I400 = V400/400 I400 = 4.1667e-03 »I20 = V400/20 I20 = 8.3333e-02 »Is2 = I20 + I400 + I1 Is2 = 1.0417e-01 »B = 1/Is2 B = 9.6000e+00 Hence by linearity Vout = 0.48 Vs1 + 9.6 Is 2 . (b) »Vout = A*20 + B* 0.5 Vout = 1.4400e+01 (c) Doubling resistances does not change voltage ratios hence A is the same. However, the doubling also doubles the voltage to current ratio. Hence, B is doubled. It follows that if all resistances are doubled, then Vout = 0.48 Vs1 + 19.2 I s2

Solution 5.28 (a) (b) We solve parts (a) and (b) at the same time. First, we find the responses to Vs1 :

Equivalent resistance across Vo2 : 60||30 = 20 Now, by voltage divider:

Vo2_1 = 20/140Vs1

This voltage now divides between Vo1 and the 10 ohm resistance: Vo1_1 = 20/30Vo2_1 = 2/3×1/7Vs1 = 2/21Vs1 Now, compute the responses due to Is2 :

Equivalent resistance across Vo2 : 60||120 = 40Ω By voltage division:

Vo2_2 = 40/50Vo1_2

Where

Vo1_2 = Is2 ×(50||20) = 14.286Is2

Now, add the contributions: Vo1 = 0.0952Vs1 + 14.286Is2 Vo2 = 0.1429Vs1 + 11.4288Is2

Solution 5.29 (a)

Find contribution due to v1 :

The parallel combination results in a 12 ohm resistance in series with the remaining two. Thus, by voltage division: vout1 = 24/48v1 = 0.5v1 Now, due to v2

The equivalent resistance across v in this figure is 36||(24+12) = 18, which means that by voltage divider: v = 0.5v2 Similarly, by another voltage division application vout2 = 24 / 36 × v= 0.66×0.5v2 = 0.333v2

Now, due to i3

Define Req1 = 24 + 36||18 = 36 Ω. This resistance is in parallel with the 12 ohm resistance to introduce an equivalent of 9 Ω. The total voltage that develops across this 9 ohm resistance is –9is3 . This voltage divides between vout and the 36||18 resistance: vout3 = 24/36v1 = -9×24/36is3 = -6is3 Finally, due to i4:

A similar analysis of this resistive network can reveal that vout4 = 6is4 . Thus

vout = 0.5v1 + 0.333v2 – 6is3 + 6i s4

(b) For this part, note that scaling resistance values does not affect voltage ratios. This can be evident from the application of any voltage divider formula. On the other hand, scaling resistances does affect current-

to-voltage or voltage-to-current ratios. This is by definition of a resistance! So, in the above equation, doubling the resistances does not affect the first two terms, but doubles the second two terms.

Solution 5.30 (a) Define two clockwise mesh currents: I1 in the bottom left loop and I2 in the top loop. The bottom right loop has a current source, so it will not be considered:

Va – (I1 -I2 )3 – I2 – (I1 – ib ) = 0 and 6I2 – I2 + (I2 – I1 )3 = 0 Solving these two equations for the two currents gives: I1 = 8 A

and

I2 = 3 A

The power delivered by the dependent source is: P = i x × I ix where I ix is the current leaving the ' +' terminal of the dependent voltage: P = i x × (I2 − I1) = I 2 × (I2 − I1) = −15W è vout = I1 + i b = 8 + 26 = 34V (b) Now, we express vout = Ava + Bi b Turning off ib , we still have two loops, in which we can define the same mesh currents as above to obtain: va – 3(I1 – I2 ) – I2 – I1 = 0 I2 – 3(I2 – I1 ) – 6I2 = 0 Solving for the two currents, we get I1 = 4/13va which sets up vout across the 1 ohm resistor: Vout1 = 4/13va Now, turn off the voltage source:

Write a node equation at vout: vout vout vout − vout 6 + + − ib = 0 6 1 3 9 ? vout 2 = ib 13

So,

vout = 4/13va + 9/13i b

(c) Substitute into the above equation: vout = 35V

Solution 5.31 By inspection:

where the leftmost current is 0.25 with a resistance of 32Ω. Similarly, the downward current is 0.75 and its resistance is 32Ω. This reduces to:

Solution 5.32 The circuit can be transformed as follows:

Write two nodal equations at 1 and 2: v1 v −v − 9m + 1 2 = 0 2k 2k v2 − v1 v v + 9m + 2 + 2 − 5m = 0 2k 9.6k 16k

− 1m +

These two equations can be solved using any method to obtain: v1 = 11.2V

and

v2 = 2.4V

The power absorbed by the 9.6kΩ resistor is: P=

Solution 5.33

Then

V22 = 0.6mW 9.6 × 10 3

Output voltage is 5V (b) P = 1.25W (c) For a given resistance, doubling the voltage increases the current by two times. So, the current is new old doubled. It follows that Vout = 2 × Vout = 10V

Solution 5.34

This circuit is easy enough to solve by inspection. Vs = 28V.

Solution 5.35

Now, write two node equations: v1 v v − 2 + 1 = 12m 500 500 500 v2 v v − 1 + 0.032v1 + 2 = 80m 500 500 125

These two equations can be solved to get v1 = 2.8V, v2 = -0.4V.

Solution 5.36

Now, we can write the node equations: v1 v2 v1 100 v − + − + 0.2 + 0.4 + 1 − 0.95 = 0 50 50 25 25 20 v2 100 v2 v1 v − + − + 2 + 1 − 0.1 − 0.2 = 0 100 100 50 50 100

Solving these two equations yields: v1 = 25V

and

v2 = 20V

*S OLUTION 5.37. After a source transformation on the 30 V independent source and one on the 9Vx dependent source we obtain the circuit below.

Transforming the two dependent voltage sources and combining yields the following circuit.

Writing a single node equation for Vx yields V V V 7.5 = x + x + x = 1.25Vx 4 2 2 Hence, Vx = 6 V.

Solution 5.38 Replace the dependent source with a temporary independent source. When doing the analysis, always compute Va in order to keep track of the constraint on the dependent source.

When Vt is not active, vout is obtained from a voltage divider between the 60||30 combination and the 15 ohm resistor: vout1 = 4/7Vs Similarly

va1 = 3/7Vs

Now, compute the response due to the temporary source. Straightforward voltage division also applies here to get:

va2 = -2/7Vt and

vout2 = -5/7Vt

So,

vout = 4/7Vs – 5/7Vt va = 3/7Vs – 2/7Vt,

where Vt = µva è va = 3/7Vs – 2µ/7va

Rearranging, va =

3 Vs 7+2

Then, 4 5µ Vs − v 7 7 a 4−µ vout = V 7 + 2µ s vout =

Solution 5.39 Replace the dependent source by a temporary independent source:

When Vt is shorted, the result is a ladder network. The input resistance looking each of the vertical branches is R. Label these vertical branches V1 , Va, V2 from left to right. It follows that V1 = R/3RVs _ Also,

Va_1 = R/2R × R/3RVs = 1/6Vs Vout_1 = 0

Now, short VS and turn on Vt.

Again, the result is a ladder network (note the symmetry in the above figure). Thus, we can write by inspection: Va_2 = 1/6Vt Vout_2 = Vt

Adding the contributions: Vout = 0 + Vt = Vt = µVa and

where we have substituted the constraint on Vt

Va = 1/6Vs + 1/6Vt = 1/6Vs + 1/6 µVa è Va = [1/(6 – µ)]Vs

Substitute this Va into the expression for Vout: Vout =

6−

Vs

Solution 5.40 The first step is to replace the dependent source with a temporary independent source. Then, superposition can proceed as usual.

Now, let’s short the temporary source, Vt. Again, this network is a ladder network, like the one in the previous problem. However, now, the equivalent resistance looking into each of the vertical branches (from left to right) is different. Now, it is Req = 20R||(R+4R) = 20R×5R/25R = 4R

Now, again, define the voltages across these three vertical branches (from left to right) as V1 , Va, V2 . It follows by voltage division that V1 = 4/9Vs è

Va_1 = 4/5 × 4/9 × Vs

è

Va_1 = 16/45 Vs

It should be noted that Vout = 0. Now, short the input source and find the response due to Vt. Again, in this case, the circuit is identical to the case when Vs was active (note the symmetry in the above figure). Therefore, Va_2 = 16/45 Vt Vout_2 = Vt

Adding the two contributions: 16 16 Vs + Va 45 45 1 V ? Va = (4516 − ) s

Va =

Vout = Vt = Va

and

Vout =

16 Vs 45 − 16

Solution 5.41 First, replace the controlled source by a temporary independent source:

Now, do a source transformation on the Vt source:

The equivalent resistance seen by the current source is 99Ω. Therefore: Vout = -99/100Vt = -0.99Vt Vt = µVg

Now, è

and

IL = -0.99Vt/990k where IL is the current through R2

VR2 = ILR2 = (100k/990k)×(-0.99Vt) = -0.1Vt

By KVL Vg = VR2 + V s = -0.1Vt + Vs = -0.1µVg + V s Rearranging Vg =

1 0.1 + 1

Vs

Finally, and substituting: Vout 1 = −0.99 Vs 1 + 0.1 Vout = −8.91 Vs

Solution 5.42 Vout b0 + b1 x b0 + b1R = = Vin a0 + a1 x 1+ a1R

At R = 0 Vout b0 = = 20 Vin 1 ? b0 = 20

At infinite R Vout b1 = =8 Vin a1 ? b1 = 8a1

Finally, at R = 10 20 + 80a1 = 10 1 + 10a1 ? a1 = 0.5 ? b1 = 4

Substituting these values for R = 2 yields Vout 20 + 4 ↔2 = = 14 Vin 1 + 0.5 ↔2

SOLUTION 5.43. From the chapter

Vout b0 + b1µ b0 + b1µ = = Vin a0 + a1µ 1+ a1µ Equivalently, V V −µ out a1 + b0 + µb1 = out Vin Vin Plugging in the data yields three equations in three unknowns which in matrix form are: 1 0  a1  264   0 −154 1 1  b  = 154     0    −168 1 2  b1   84  »A=[0 A= 0 -154 -168

1 0;-154 1 1;-168 1 2] 1

0 1 1

1 2

»y = [264 154 84]' y= 264 154 84 »Coefs = A\y Coefs = 2.8571e-01 2.6400e+02 -6.6000e+01 Therefore, a1 = 0.28571, b0 = 264, and b1 = –66. Making a1 = 1, will yield a different set of answers. When m = ∞, Vout /Vin = b1 /a1 = –231.


- P6.1 -


PROBLEM SOLUTIONS CHAPTER 6 SOLUTION 6.1. (a) Voc is found by removing RL and doing voltage division. VOC = 63V

(600 || 700) 600 × = 28V (600 || 700) + 300 600 + 100

RTH is found by setting the source to zero and by calculating the equivalent resistance seen looking back between the A and B terminal. RTH = [(300 || 600) + 100] || 600 = 200Ω

(b) Using P = RL ⋅ I L

the power for each resistance may be found by substituting the appropriate RL in the following equation. 2

 VOC  P =   ⋅ RL  RTH + RL 

For 50 Ω, 200 Ω, and 800 Ω , the power obtained is 627.2 mW, 980.0 mW, and 627.2 mW respectively. The use of Thevenin equivalent does reduce the effort in obtaining the answer. SOLUTION 6.2. (a) To find RTH, open circuit the current source and short-circuit the voltage source. The resulting resistance seen from terminal A-B is 1 kΩ. Using superposition, the contribution of the current and voltage source at the open circuit output may be summed as 30 V (2 k/4 k) + 10 mA (2 k/4 k) (2 kΩ). VOC is then 25 V and ISC = VOC/RTH is 25 mA. (b) Following is a plot of 2

 VOC  P =   ⋅ RL  RTH + RL 

for RL from 100 Ω to 4 k Ω.


- P6.2 -


SOLUTION 6.3. (a) Turning off the two sources RTH = (60 + 60)||40 = 30 Ω, and using superposition VOC = 6V 

40 60  + 0.1A  40 = 3 V.   40 + 60 + 60 40 + 60 + 60 

ISC is the obtained as 100 mA. (b) Using 2

 VOC  P =   ⋅ RL  RTH + RL 

a load of 90 Ω will absorb 56.25 mW. (c) It absorbs 75 mW; hence the 30 Ω resistor absorbs more power. SOLUTION 6.4. As both resistor divider ration are the same (3/6), the voltage at A and B is the same resulting in a VOC of 0 V. RTH = (3K ||6 K) + (9K ||18 K) = 8 kΩ The relation VOC/ISC cannot be used in this situation. SOLUTION 6.5. Using superposition     20 × 103 5 × 103 VOC = 20 − 10    + 20 ⋅ 20sin(50t) = 14 + 400sin(50t) V,  20 × 103 + 5 × 103   20 × 103 + 5 × 103  and


- P6.3 -


RTH = (20k ||5k) + 20k = 24 kΩ. ISC can then be found using the VOC/RTH relationship as 0.58+20sin(50t) mA. SOLUTION 6.6. Once again using superposition 6K 4K     VOC = 72V   + 54mA  ⋅ 6 K + 54mA ⋅ 2 K = 204V ,  6 K + 8K + 4 K   4 K + 8 K + 5K 

and RTH = [( 4 K + 8 K ) || 6 K ] + 4 K + 2 K = 10 KΩ

SOLUTION 6.7. Using source transformation, (a) is obtained from the original circuit. Then combining in series the resistors and voltage sources, and retransforming them (b) is obtained. Finally adding the two currents and transforming back the circuit to its Thevenin form (c) is obtained.

From (c), Voc = 60V RTH = 6kΩ I L = 60 /(6k + 6k ) = 5mA P = I L RL = 0.15W 2


- P6.4 -


SOLUTION 6.8. First, each source in series with 2R, can be replaced by an up going current source of V x / 2 R in parallel with 2R. Then starting from the left, the two 2R in parallel are combined and then

retransformed to a voltage source of Vo / 2 in series with 2R once added to the series resistance. Repeating the previous steps, Vo / 2 + 2 R ? Vo / 4 R|| 2 R|| 2 R|| V1 / 2 R Vo / 4 R + V1 / 2 R|| R ? Vo / 4 + V1 / 2 + R + R Vo / 4 + V1 / 2 +2 R ? Vo / 8 R + V1 / 4 R|| 2 R|| 2 R|| V2 / 2 R Vo / 8 R + V1 / 4 R + V2 / 2 R|| R ? Vo / 8 + V1 / 4 + V2 / 2 + R + R Vo / 8 + V1 / 4 + V2 / 2 +2 R ? Vo / 16 R + V1 / 8 R + V2 / 4 R|| 2 R|| 2 R|| V3 / 2 R Vo / 16 R + V1 / 8 R + V2 / 4 R + V3 / 2 R|| R ? Vo / 16 + V1 / 8 + V2 / 4 + V3 / 2 + R + R

Thus Voc = Vo / 16 + V1 / 8 + V2 / 4 + V3 / 2 and RTH = 2 R . SOLUTION 6.9. First we find RTH RTH = 2K + [(6K ||3K ) + (15K ||10 K )]||24 K = 8KΩ Use nodal analysis to solve for VOC. At node a 1mA = 1/ 6K(Va − Vc ) + 1/ 3K(Va − Vd ) + 1/ 24K(Va ) or (1/ 6K + 1/ 3K + 1/ 24K)Va − 1/ 6K(Vc ) − 1/ 3K(Vd ) = 1m doing the same for the supernode, and the equation inside the supernode yields the following two equations: (1/ 6K + 1/ 3K)Va − (1/ 6K + 1/15 K)Vc − (1/ 3K + 1/10 K)Vd = 0, and Vc − Vd = 30 using these three equations, it is now possible to solve for the three unknowns VA, VC, and VD. Using MATLAB they are respectively 4.5V, 22.8750V, and -7.1250V. VOC being VA, 4.5V.

a

c

SOLUTION 6.10. RTH = [(18K ||9K) + 66K]||36K = 24K

d


- P6.5 -


By superposition, noting that all resistances are in k-ohms, VOC =

66+(9||18) (66 + 36)||9 36 36 ⋅36 ⋅ 2.5 + ⋅ ⋅18 − 30 = 52 V 66+(9||18)+36 [(66 + 36)||9] + 18 36 + 66 36 + (9||18) + 66

When a 2 kΩ is connected, the current IL becomes VOC/(RTH+2k) = 2mA; thus the power absorbed is 8 mW. SOLUTION 6.11. (a) Introduce a test voltage source at the output, and write the nodal equations in matrix form: −1/ 50 −1/100 (1/100 + 1/ 50)   Vtest   itest       −1/ 100 −1/100 (1/ 100 + 1/ 200 + 1/100)   VC  =  0      VD  Vs / 100 −1/ 50 (1/ 50 + 1/100 + 1/100) −1/100 Solving we obtain, Vtest = 100itest + 2Vs / 3. From eq. 6.10 RTH = 100 Ω, voc = 2Vs / 3.

C

D

2

 voc  (b) To obtained the power the following equation is used, P =   ⋅ RL .  RTH + RL 


- P6.6 -


1.2 1.1 1

Power/Vs in mW

0.9 0.8 0.7 0.6 0.5 0.4 0.3

0

20

40

60

80 100 120 Load resistance in Ohms

140

160

180

200

SOLUTION 6.12. VOC = 0, as no independent sources are present. Writing the following nodal equation where vx is the voltage across both ports, 2.5 + ix = ((v x − iz ) − ix ) + (v x − i z )/ 2, RTH can be found as v x / ix = , 1.5 SOLUTION 6.13. Defining vi and ii as the voltage across and current into the input ports, writing the nodal equation at the input: ii + gmv x = 1/ 200K(vi − vx ) . We can also get the following equation v x = 200K(gmv x + ii ) . Using   200K ⋅ gm 1 the previous two equations we can solve for RTH = vi / ii = − 200K  1+ + .  1 + 200K ⋅ gm 1 + 200K ⋅ gm  This yields a gm of 10 µS. SOLUTION 6.14. ISC is null as no independent source are present. To find RTH vi and ii are defined as the voltage across and current into the input ports. Writing the nodal equation we get: ii = Vx / 1.8K + (1/ 200)(Vx − 3 / 4Vx ) , and vi = Vx − 300 ⋅ ii . Solving RTH = vi / ii = − 600Ω . SOLUTION 6.15. First, write out the equation around loop 1:


- P6.7 -


Vx = i1 ⋅1200 − i2 ⋅100 − i3 ⋅300 . Then substituting the following relationships, i1 = Ix , i2 = 2I x , and i3 = 0.01Vx + i2 = 0.01Vx + 2Ix , and solving for Vx / Ix = RTH = 100Ω . VOC = 0 as no independent sources are present.

2

3

1

SOLUTION 6.16. Introduce a test source at the output terminals, and write out the nodal equations in matrix for the top node, and the supernode comprised of the current controlled voltage source (ccvs). vb is the node left of ccvs, and vc the node to the right. −1/ 20 −1/ 30  vtest   itest   1/ 30 + 1/ 20 −1/ 20 − 1/ 30 + 0.4 1/ 20 − 0.4 2 / 30   v  =  0    b         1/ 6 −1 5/ 6   vc   0  Solving, we obtain vtest = − 90itest , thus RTH = − 90 Ω.

b

c

SOLUTION 6.17. (a) Turn off independent source. Introduce a test source and write loop equation: v x = 6ix − 4i1 + 10ix . Note that i1 = ix . Now solve for v x / ix = RTH = 12Ω .


- P6.8 -


Short the input and write the loop and nodal equations: 3A = i1 + ISC 10i1 = 4i1 + 6ISC Solving yields ISC = 1.5A, and VOC = ISC RTH = 18V . 2

 VOC  (b) In MATLAB the following plot is generated: P =   ⋅ RL for 1 ≤ RL ≤ 24Ω.  RTH + RL  7

6

Power in Watts

5

4

3

2

1

0

5

10 15 Resistance in Ohms

20

25

Maximum power is absorbed by 12 Ω load.

SOLUTION 6.18. To find thevenin resistance, introduce a test source and write the following equations: v s = 20i1 + 40i1 + 40i1 = 100i1 , and is = vs /100 + i1 . Solving for v s / is = RTH = 50Ω . Next, use the following nodal equation; 0.2A = i1 + ISC , and loop equation 20i1 + 40i1 = 40ISC . Solving using these two equations yields ISC = 0.12A , and consequently VOC = RTH ISC = 6V . SOLUTION 6.19. (a) Introduce a test source, vs, and get the following two equations: v s − 15is = − Vx , and 1   kVx + Vx / 3 + Vx / 5 + is = 0. Solving yields v s / is = RTH = 15 +   , or 95/6 Ω for k = 2/3.  k + 8 /15  Next, write the following nodal equation kVx + Vx / 3 + (Vx − 1) / 5 = 0 and observe that VOC = 1V − Vx .

Thus solving yields VOC

  1/ 5   = 1−  8  = 5/ 6V for k = 2/3, and ISC = VOC / RTH = 1/19 A . k +   15 


- P6.9 -


(b) Solving the previously obtained equation for k when VOC = 0, yields k = -1/3, and consequently RTH = 20 Ω. SOLUTION 6.20. Introduce a test source, vt, and get the following equations: it + ix = ix ⋅

Vs − 300ix ⇒ Vs = 300it + 300ix , 300

and vt = Vs − 300ix + 100it . Solving yields vt / it = RTH = 400Ω . Observe how no current flows through the 300 resistor in parallel with the dependent source. Thus VOC will always be 0 V and is independent of VS . SOLUTION 6.21. (a) For this part, consider the modified circuit below.

Step 1: Applying KCL to node A, we have 1 V (VA − 100) + A + 0.1 − Is = 0 400 800 Multiplying through by 800 yields 3VA = 800Is + 120 Step 2. Computing Vs , we have 2000 Vs = VA − 2000i x = VA − (100 − VA ) = 6VA − 500 400 Hence 800 Vs = 6 Is + 40 − 500 = 1600Is − 260 = RthIs + Voc 3 Therefore Rth = 1.6 kΩ and Voc = −260 V. (b) By linearity, Voc = −130 V.


- P6.10 -


SOLUTION 6.22. Introduce a test source, vt,, and get the following two equations: vt = (30m − V1 / 100)400 − V1 = 12 − 5V1 , and it = 0.06V1 + 30m − V1 / 100 = 30m + 0.05V1 . Solving obtain vt = −100it + 15. Thus RTH = −100Ω , and ISC = VOC / RTH = − 150mA . SOLUTION 6.23. Insert Itest as per text. Hence  1 + 1  V − 16 = I test  40 120  C 40 Solving, we obtain VC = 30Itest + 12 By KVL, 60 Vtest = VC + 30 × ( 2i1 + Itest ) = VC + (16 − VC ) + 30Itest = − 0.5VC + 30Itest + 24 40 Substituting for VC yields Vtest = 15Itest + 18 Hence, Voc = 18 V and Rth = 15 Ω. Thus i =

18 = 0.2 A. Further, P75 = 75(0.2)2 = 3 W. 75 + 15

C

SOLUTION 6.24. (a) and (b)


- P6.11 -


+ 2

v

+ vs

1

(c) Writing KVL around loop 1, loop 2, and finally relating v to v x , vs = 10(0.02 − I1) + v − 20I1 v = − 5I2 − 15(I2 − 0.02) 3vx = I1 − I2 = 3[10(0.02 − I1)] = 0.6 − 30I1 (d) Rewriting these in matrix form, −30 0 1  I1  1   −0.2   0 20 1  I  = 0  v +  0.3    2    s    31 −1 0  v  0   0.6  (e), (f), and (g). Using MATLAB, »A=[-30 0 1;0 20 1; 31 -1 0]; »b1 = [1 0 0]'; »b2 = [-.2 .3 .6]'; »I1 = inv(A)*b1 I1 = -1.5385e-03 -4.7692e-02 9.5385e-01 »I2 = inv(A)*b2 I2 = 1.9231e-02 -3.8462e-03 3.7692e-01 »Rth=-1/I1(1) Rth = 650 Note: Isc = I2(1) = 0.01923 A »Voc= -Rth*I2(1)

-


- P6.12 -


Voc = 1.2500e+01 (h)

SOLUTION 6.25. (a) (b) Write four nodal equations, is = (VA − VC ) / 2k is + 1m = (VC − VD )/ 6k + (VC − VE )/ 3k VE / 2k = (VC − VD ) / 6k − VD /15k VE / 2k = VE /10k + (VE − VC )/ 3k (c) 1/ 2k −1/ 2k  0 1/ 2k   0 1/ 6k  −1/ 3k  0

0 0   VA   1 0      1m −1/ 6k −1/ 3k   VC   1 = i +  −(1/ 6k + 1/ 15k) −1/ 2k   VD   0 s  0        0 −1/ 15k   VE   0 0

(d) Solving in MATLAB VA = 5.818kis + 3.8182 , thus RTH = 5.818kΩ VOC = 3.8182V (e) This only changes VOC = 3818⋅8m = 30.54V . SOLUTION 6.26. For this circuit, no current flows though the 20 Ω resistor. Therefore VAB = VCB . Further, from the examples in the chapter, VCB = 4Vs . Hence, Voc = VAB = VCB = 4Vs . Also, shorting V 4V V V terminals A and B, yields Isc = CB = s = s . It follows that Rth = oc = 20 Ω. Note that the 20 20 5 Isc Thevenin equivalent to the left of C-B is a voltage source of value 4Vs . Therefore the Thevenin equivalent to the left of A-B is 20 Ω in series with 4Vs . SOLUTION 6.27. (a) From the previous problem VCB = 4Vs . Thus by voltage division, VOC = VAB = VCB 

180  = 3.6Vs . Next find ISC = VCB / 20 = Vs / 5, and then RTH = VOC / ISC = 18Ω . 180 + 20 


- P6.13 -


 180||162  (b) This changes the voltage division at the output, thus VAB / Vs = 4  = 3.24 .  (180||162) + 20  SOLUTION 6.28. (a) Writing the following two KCL equations, vtest / 4k = (Vo − vtest ) /12 k Itest = (vtest − Vo ) /15k where Vo is the voltage at the output of the op-amp. Doing the appropriate substitution get vtest = −15k / 3⋅ Itest , thus RTH = −5kΩ VOC = 0V Since no independent source exist right of A-B (b) Applying Ohm’s law Is = 1/ (10k − 5k )Vs = 0.2mVs SOLUTION 6.29. (a) Adding a test source at terminal A-B, and noting that the voltage at the output of the op-amp is Vo = −5 / 2Vs . Write KCL at terminal A, vtest / 900 = (Vo − v test )/ 100 + itest vtest = 90itest − 2.25Vs Where one sees by inspection that RTH = 90Ω VOC = −2.25Vs ISC = VOC / RTH = −0.025Vs (b) Noticing the virtual short to ground provided by the ideal op-amp, RTH = 20kΩ , and VOC = 0V since no independent sources are present right of the input terminal. SOLUTION 6.30. The output voltage of the ideal op amp is −2.5Vs1 − 2Vs2 which drives a voltage divider circuit. Hence Voc = VAB = 0.9(−2.5Vs1 − 2Vs2 ) = − 2.25Vs1 − 1.8Vs 2 Further, Isc =

−2.5Vs1 − 2Vs 2 = −0.0225Vs1 − 0.018Vs 2 100

Finally, V 0.9(−2.5V − 2V ) Rth = oc = (−2.5V s1− 2V s)2 = 90 Ω Isc s1 s2 100


- P6.14 -


Equivalently if one sets Vs1 and Vs2 to zero, then the ouput terminal of the op amp goes to ground. Hence Rth = 100 / /900 = 90 Ω SOLUTION 6.31. Define the node at the output of the op-amp as Vo . Note how the circuit left of this node is a general summing circuit as per text. Thus, Vo = 2Vs2 − 4Vs1 . Hence we simply replace the op amp circuit to the left of the 20 Ω resistor by an ideal voltage source of value V0 . Hence

Voc =

80 −V V0 + 2I = 0.8V0 + 2 0  = 0.78V0 = 1.56Vs2 − 3.12Vs1 80 + 20 100

Alternately, one can introduce a test source at the output terminal and write out a set of equations using KVL, Vo = 2Vs2 − 4Vs1 = −20I − 80(I − itest ) vtest = 2I + 80(itest − I) Solving yields vtest = 17.6itest + 0.78(2Vs2 − 4Vs1 ) , and by inspection, RTH = 17.6Ω VOC = 0.78(2Vs2 − 4Vs1 ) = 56Vs2 − 3.12Vs1

SOLUTION 6.32. Upon inspection when the op-amp is in active range, for inputs from –3V to 3V, RTH = 1kΩ VOC = 0V However when the input exceeds 3V, the output of the op-amp will saturate at –15V, and Vs = −15V + 6kI s . Therefore from eq. 6.10, RTH = 6kΩ VOC = −15V When the input is less than –3V, Vs = 15 + 6kI s , thus RTH = 6kΩ VOC = 15V SOLUTION 6.33. (a) The op-amp configuration has a gain of –4. So when the input is between –3V and 3V it is operating in its active region, thus


- P6.15 -


RTH = 4kΩ VOC = 0V When the input is greater than 3V the output saturates at –12V and Vs = 20kI s − 12 RTH = 20kΩ VOC = −12V When the input is less than –3V the output saturates at 12V, Vs = 20kI s + 12 RTH = 20kΩ VOC = 12V (b) When the input is in the active range, RTH = 16kΩ VOC = −4Vs When it is greater than 3V, RTH = 0kΩ VOC = −12V And when less than –3V, RTH = 0kΩ VOC = 12V The last two obtained using figure 6.28. The maximum power is when the output is in saturation, P = (VOC ) / 28k = 6mW . 2

SOLUTION 6.34. From the table the following two equations can be written: 6 = 2RTH + v oc 12 = 8RTH + voc Putting in matrix form and solving, 1 1  RTH   6  8 1  v  = 12    oc    −1/ 6 1/ 6   6   RTH   4 / 3 −1/ 3 12 =  v       oc  Thus RTH = 1 kΩ since current was in mA, and voc = 4 V. SOLUTION 6.35. (a) From Ohm’s law, IAB = VAB / RL . Thus 0.2uA and 0.1uA. (b) Note how using this topology VAB = voc − I AB RTH , thus


- P6.16 -


1 −0.2u   voc   0.4  1 −0.1u   R  =  1     TH    2   0.4  voc   −1 −10M 10M   1  =  R       TH  Thus, voc = 1.6V , and RTH = 6MΩ . SOLUTION 6.36. (a) Although the text describes finding Rth from a measurement or calculation of both Voc and Isc , measurement of Isc is often impractical. Hence the procedure outlined in this problem provides a more practical means of determining the Thevenin equivalent. Since the internal meter resistance is 10 MΩ, V V IAB (µA) = AB + AB 10 RL Hence, the completed table is: RL(M Ω)

vAB (V)

IAB (µA)

2

0.4

0.24

10

1

0.2

(b) The terminal relationship assuming a Thevenin equivalent is given by VAB = Voc − RthI AB In matrix form with the data from the table 1 −0.24   Voc   0.4 1 −0.2   R  =  1     th    Hence Voc = 4 V and Rth = 15 MΩ. Note that since we used µA and V, Rth is in MΩ. SOLUTION 6.37. (a) For this scenario, the circuit is essentially a voltage source with a resistance R in series with the circuit under test, in parallel with a voltmeter measuring the voltage division between the


- P6.17 -


later two. Therefore if replacing R = 0 by R = R2 causes the voltage measured by the voltage meter to  RTH  drop by half, then by voltage division, Eo / 2 = Eo   , and RTH = R2 .  R2 + RTH  (b) Using the same reasoning and voltage division, RTH Eo = Vs RTH + Rs RTH Eo / 2 = Vs RTH + Rs + R2 Therefore RTH + Rs = R2 or equivalently RTH = R2 − Rs . (c) Again by voltage division, RTH || Rm Eo = Vs (RTH || Rm ) + Rs RTH || Rm Eo / 2 = Vs (RTH || Rm ) + Rs + R2 RTH Rm R R − Rs Rm Therefore (RTH || Rm ) + Rs = R2 or = R2 − Rs . Solving for RTH = 2 m . RTH + Rm Rm + Rs − R2 SOLUTION TO 6.38. (a) The voltmeter measures the voltage division between the two resistors involved thus, Eo = Voc Eo / 2 = Voc

R2 RTH + R2

and RTH = R2 . (b) Now Rm Rm + RTH Rm || R2 Eo / 2 = Voc (Rm || R2 ) + RTH Eo = Voc

From the division of the former by the later R2 = RTH || Rm . And from the former Voc = (Rm + RTH )Eo / Rm = (1+ RTH / Rm )Eo . SOLUTION TO 6.39. Using the relation developed in problem 6.37 RTH =

SOLUTION TO 6.40. Using the equation developed in problem 6.38,

R2 Rm − Rs Rm = 5kΩ Rm + Rs − R2


RTH =

- P6.18 -


R2 Rm = 4MΩ Rm − R2

Voc = (1+ RTH / Rm )E o = 20V SOLUTION TO 6.41. Writing the line equation in the following general form, i = vab / RTH − isc i = 2v ab − 4 Thus RTH = 0.5Ω , and isc = 4A . SOLUTION TO 6.42. (a) In this range the curve appears to be varying linearly between (0V,0mA) and (0.2V, 0.1mA) pair, thus writing the line equation i = (0.1m / 0.2)v yields RTH = 2kΩ ⇒ Voc = 0V . (b) Writing the line equation of the following form, i = v / RTH − isc , into a matrix equation, 0.2 0.7 

−1 1/ RTH   0.1m  = −1  isc  10.1m

and solving gives RTH = 50Ω isc = 3.9mA voc = 0.195V (c) First make a guess as to which region of the curve, N will operate in. Assuming that it will operate in the A-B region, then by KVL i(t) = (vs (t) + vb − v oc )/ (R + RTH ) = (50sin(1000t) − 0.095) / 550. It can be seen that this guess is wrong as the range of i(t) is not in the appropriate region. Assuming the 0-A region, by KVL i(t) = (vs (t) + vb − v oc )/ (R + RTH ) = (50sin(1000t)m + 0.1)/ 2500 = 0.02sin(1000t) + 0.04mA. Note that the highest current is 0.06mA, thus still in the appropriate region of operation. (d) Repeating the procedure above and guessing the region A-B, by KVL i(t) = (vs (t) + vb − v oc )/ (R + RTH ) = (200sin(1000t)m + 305m) /100 = 2sin(1000t) + 3.05mA. The maximum and minimum current are 5.05 and 1.05 mA respectively, thus the assumption made was correct. SOLUTION TO 6.43. (a) By the power transfer theorem RL = RTH . For circuit (a) RL = 80 ||240 = 60Ω, and circuit (b) RL = (900||180) + 50 = 200Ω. (b) Finding isc for each circuit: (a) isc = 36 / (80 + 240||60) ⋅ 240 / 300 = 225mA and (b)  900 ||180  2 isc = 60mA  = 22.5mA. The power is now obtained from P = (isc / 2 ) RL . Thus  (900||180) + 50 + 200  759mW for (a) and 25.3mW for (b).


- P6.19 -


SOLUTION TO 6.44. Finding the Thevenin equivalent, RTH = [(30||15 ) + 10 ] ||80 = 16Ω Voc = −32V 2

V The value for the load resistance is 16Ω, and the power is P =  oc  / RL = 16W . 2 SOLUTION TO 6.45. (a) Find the Thevenin equivalent, RTH = 12k ||6k + 8k = 12kΩ = RL Voc = 8k(2mA) + 24

6  = 24V 6 + 12 

The power is P = (Voc / 2 ) / RL = 12mW . 2

(b) The maximum power will be transferred to the load, when the value of its load is closest to 12kΩ. Thus 2

  Voc the power is P =   ⋅10k = 11.9mW .  RTH + 10k  2

 Voc  (c) Same reasoning as (b) the power is P =   ⋅15k = 11.9mW .  RTH + 15k  SOLUTION TO 6.46. (a) The Thevenin equivalent to the left of RL has Rth = 12 + 20 / /180 = 30 Ω and Voc = 1 × 18 +

180 40 = 54 V. Therefore, for maximum power transfer 180 + 20 GL + 2GL + 3GL = 6GL =

6 1 = RL 30

Hence RL = 180 Ω. (b) For this part, let VL denote the voltage (top to bottom) across the load. With RL = 180 Ω, then the parallel combination equals Rth and hence VL = 27 V. It follows that (27)2 = 4.05 watt. 180 Since the terminal voltages are the same, the absorbed power is inversely proportional to the resistance. P180 =

Hence P90 = 8.1 watt, and P60 = 12.15 watt.


- P6.20 -


SOLUTION 6.47. To find the Thevenin equivalent introduce a test voltage source at the output, and write KCL at the two node in the circuit. By inspection the following matrix expression is obtained: −1/ 400 (1/ 200 + 1/ 400)   VA   0.1  =  −1/ 400 0.0015 + 1 / 1 k + 1/ 400   Vtest   Itest   Solving Vtest = 240Itest + 8. And RTH = 240 Ω, Voc = 8 V. By voltage division, the voltage across the load resistor is 4V, and the power delivered to it is 66.7mW.

VA

Vtest

SOLUTION 6.48. Using KCL get ix = 1mA, which independent of what is connected to the output. Thus isc = 10ix = 10mA RTH = 3kΩ The power is then P = (10mA / 2) ⋅3k = 75mW . 2

SOLUTION 6.49. Performing a source transformation on (a), and combining the elements will simplify to one current source going up of 2/3A in parallel with a 10Ω resistor. This is essentially the Norton equivalent of the circuit, isc = 2/ 3A RTH = 10Ω For (b), combine the voltage sources and resistor in series. The circuit obtained is one voltage source of 5V in series with a 45Ω resistor. This is the Thevenin equivalent, voc = 5V RTH = 45Ω (a) The value of the load resistor is simply the thevenin resistance obtained above. (b) Using Ohm’s law for (a) VL = isc (10 ||10) = 10/ 3V , and voltage division for (b) VL = v oc (1/ 2) = 2.5V (c) Using the following formula, P = VL2 / RL , (a) absorbs 1.1W and (b) 139mW, thus (a) absorbs more power.


- P6.21 -


SOLUTION 6.50. (a) Note that the circuit left of the terminal is already in its Thevenin form. The load RL = R ||( R + 300) R2 + R(300 − 2RL ) − 300RL = 0 Solving, R = 71.6Ω . By voltage division, the voltage across the load is 5V. The power absorbed is 2 P = (Voc / 2 ) / RL = 416.7mW . (b) The following script can be used to plot the power absorbed by the load versus R: %Script for problem 6.50b R=0:2:400;

%Calculate Load resistance RL= 1./((1./R)+1./(R+300));

%Calculate the power P=(10.*(RL./(RL+60))).^2./RL;

%Plot the power versus R plot(R,P); ylabel('Power in Watts'); xlabel('Resistance in Ohms')

SOLUTION 6.51. First, find the Thevenin equivalent by writing out the transfer equation v ab = 200i + 40. Thus RTH = 200Ω⇒ Voc = 40V . The maximum power will then be P = (Voc / 2) 2 / 200 = 2W . SOLUTION 6.52. The assumption that all controlling voltages or currents for dependent sources within Ni are assumed to be in Ni , implies that the nodal equation matrix of figure P6.52a has the partitioned form:  G11  G21  0

G12 G22 G32

0   VN1   IN1      G23   Vm  =  Im  G33  VN 2   IN 2 

(*)

where VN1 is the vector of UNKNOWN and INDEPENDENT node voltages internal to N1 and VN2 is the vector of UNKNOWN and INDEPENDENT node voltages internal to N2 . The right side of the


- P6.22 -


equation consists of (effective) currents injected into the appropriate node. However, IN1 depends only on sources in N1 and IN2 depends only on sources in N2 . At this point we must presume that the matrix equation (*) has a unique solution, i.e., the determinant of the coefficient matrix is non-zero. Hence we can calculate VN1, Vm, and VN2 uniquely. As such, by considering the first row of (*), we can assert that VN1 satisfies G11VN1 = I N1 − G12 Vm

(**)

Note that we are not claiming that we can solve for VN1 from this equation. Replacing N2 by a voltage source of value Vm results in the network of figure P6.52b. For this network, the nodal equations are G11VN1 = I N1 − G12 Vm

(***)

where Gij is the same as in (*). Clearly, this is the same as equation (**). Again we presume there is a unique solution to this equation, i.e., the determinant of G11 is non-zero. If so, we can solve for VN1 uniquely and the result is the same as that obtained by solving (*). This theorem can be extended to RLCM networks (to be studied in later chapters) or even nonlinear networks under appropriate conditions. To emphasize the subtlety of this result and the need for unique solvability in each network, consider the following circuit.

The resulting nodal equations are:


- P6.23 -


−1 0 0   V1   −2 2 −1 0.5 −0.5 0   V   0     2 =    0 −0.5 2.5 −1  V3   0       0 −1 2   V4   4  0 There exists a unique solution and from MATLAB, we find Vm = V3 = 2 V To apply the voltage source substitution, we replace N2 by a voltage source of 2 V and obtain the following circuit.

The nodal equations here are  2 −1  V1   −2 −1 0.5 V  =  1    2   Observe that the coefficient matrix has a zero determinant. Thus there is either no solution or many solutions, i.e., no unique solution. This demonstrates that unique solvability of the larger network does not imply the unique solvability of the smaller derived network.

SOLUTION 6.53. The assumption that all controlling voltages or currents for dependent sources within Ni are assumed to be in Ni , implies that the loop equation matrix of figure P6.53a has the partitioned form:  R11   R21  0

R12 R22 R32

0   I N1   EN1      R23   Im  =  Em  (*) R33   I N 2   EN 2 


- P6.24 -


where IN1 is the vector of unknown and independent loop currents internal to N1 and IN2 is the vector of unknown and independent loop currents internal to N2 and Im a single independent loop current common to N1 and N2 . The right side of the equation represents the net contribution of voltage sources present in the appropriate loop. However, EN1 depends only on N1 and EN2 depends only on N2 . At this point we must presume that the matrix equation (*) has a unique solution, i.e., the determinant of the coefficient matrix is non-zero. Hence we can calculate IN1, Im, and IN2 uniquely. As such, by considering the first row of (*), we can assert that IN1 satisfies R11IN1 = EN1 − R12 Im

(**)

Note that we are not claiming that we can solve for IN1 from this equation. Replacing N2 by a current source of value Im results in the network of figure P6.53b. For this network, the loop equations are R11IN1 = EN1 − R12 Im

(***)

where Rij is the same as in (*). Clearly, this is the same as equation (**). Again we presume there is a unique solution to this equation, i.e., the determinant of R11 is non-zero. If so, we can solve for IN1 uniquely and the result is the same as that obtained by solving (*). For some subtlety in the proof refer to the solution of 6.52. This theorem can be extended to RLCM networks (to be studied in later chapters) or even nonlinear networks under appropriate conditions. SOLUTION 6.54. (a) The thevenin equivalent to the left of terminal A-B is RTH = [(30||60) + 20]||10 = 8Ω , and using Ohm’s law along with voltage division 15V   30  VOC = 15 − 10 = 14V 15 + 60   30 + 30  (b) Doing the same for the circuit right of terminal A-B. RTH = [(30||15) + 10]||20 = 10Ω and 7.5V   30  VOC = 7.5 − 20 = 5V . 15 + 15   30 + 30  10 8 (c) Using superposition, VAB = 14  + 5  = 10V . 18 18 (d) (e) (vCB − 15) / 30 + vCB / 60 = (10 − vCB )/ 20 Hence, vCB = 10V .


- P6.25 -


SOLUTION TO P6.55. The proof is based on superposition. Let us consider the figure below where N1 and N2 are differently named but identical networks.

We first compute the contribution to ia from the independent sources in N1 with those of N2 deactivated. Let this current be i1a . The contribution to ia from the independent sources in N2 with those of N1 deactivated is ia2 . But because N1 and N2 are identical, i1a = −ia2. Hence by superposition ia = ia1 + ia2 = 0. By the current source substitution theorem we can replace the lines by current sources of value 0-amp. This defines an open circuit and the connecting line can be replaced by an open circuit. From the given network we also note by KVL that Vx + Vy = 0 which implies that Vx = –Vy . On the other hand, since the networks are identical, Vx = Vy . Thus we conclude that Vx = Vy = 0. Thus we can replace Vx and Vy by a voltage source of 0-volt (voltage source substitution theorem) which is the definition of a short circuit. SOLUTION 6.56. Label the potential between each line starting from the top as Vx1 , Vx2 on the left, and Vy1 and Vy2 on the right. Now by superposition and linearity notice that Vx1 = −Vy1 Vx2 = −Vy2 because the independent source is negative on the right side. Additionally, from KVL, Vx1 = Vy1 Vx2 = Vy2 The only way all these condition can be met, is if all the voltages are 0 V, or short circuited.


- P6.26 -


SOLUTION 6.57. (a) Using the results from P6.55, no current flows between the two halves. So the right hand side circuit may be analyzed as if it was stand alone. By voltage division then, 3+2  v a =  18 = 15V . 3 + 2 + 1 (b) From the results of P6.56, all the lines crossing the symmetry line are shorted together. Consequently,  6 ||3  by voltage division, v a =   (−18) = −12V .  (6 ||3) + 1 SOLUTION 6.58. Note how this circuit is the same as in P6.57: it is just redrawn with the neighboring resistors added in parallel or in series. Using superposition, we can solve for va when the sources [vs1 vs2 ] are [18 18], and then [18 –18]. By linearity adding the two contribution will be equivalent to solving for [36 0] directly, since adding the source contributions [18 18]+[18 –18] =[36 0]. The contributions of 15V–12V were obtained in P5.57; thus v a = 3V . SOLUTION 6.59. Yes since

[ 45

27] = [36 36] + [ 9 −9] = 2[18 18] + 1/ 2[18 −18] = 2(15) + (12) / 2 = 24V .

SOLUTION 6.60. For this proof we attach an arbitrary network N to each of the networks N1 and N2 in figure P6.60 as shown below.

N may have internal independent sources, but we consider N1 and N2 external excitations to N and we assume no violation of KVL in the attachment. Choose node 3 as a reference node. Then V13 = Va − Vc and V23 = Vb − Vc for both figures. Hence N1 and N2 provide the identical external


- P6.27 -


excitations to N and hence all currents and voltages in N remain unchanged. The extension of this result to 4 external nodes is shown in the figure below. The verification is the same as above. The extension of course to n-terminals is clear.

SOLUTION 6.61. Using the E-shift theorem, remove the 9 V source from each branch and add it to the 4V source, and notice by inspection that VOC = −4 + 9 = 5 V. SOLUTION 6.62. (a) Writing a KCL equation for each node in N2: I1 = Ia I2 = Ia − Ia = 0 I3 = I a − I a = 0 I4 = − Ia Do the same for N1: I1 = Ia I2 = 0 I3 = 0 I4 = − Ia This shows that the two have identical outcomes. (b)


- P6.28 -


SOLUTION 6.63. Using the I-shift theorem, this circuit is essentially a –2A source in parallel with a series combination of resistors, and a 5A source. Thus isc = 5 − 2 = 3A . SOLUTION 6.64. (a) This can be done by inspection. An equal source is connected between A-C and CB; thus by the I-shift theorem, it is equivalent to the same source just connected between A-B. (b) In figure 6.64c the VCCS is replaced by a resistor using the Ohm’s law relationship 100Ω = V1 / (0.01V1 ).  100||900  (c) (d) By voltage division V1 = Vs   , and by Ohm’s law  (100 ||900) + 10   100||900  Vout = Vs   0.01( 20k ||5k ) = 36Vs  (100 ||900) + 10  Vout / Vs = 36

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PROBLEM SOLUTIONS CHAPTER 7 SOLUTION 7.1. Given the coil has 48 turns and 12 turns/cm, we know that the length of the coil is 4 cm. Since the length of the coil is greater than 0.4 times its diameter, the formula given in the question can be used: L=

4 × 10−5 × (0.02)2 × ( 48)2 = 18.81 H 18 × (0.02) + 40 × (0.04)

SOLUTION 7.2. Part 1. Applying (7.1) The voltage vL(t) can be computed using the inductor v-i relationship: v L (t) = L

diL (t) dt

The calculations for vL(t) for t = 0s to 5s are summarized in the following table: Time Interval 0s< t ≤1s 1s< t ≤3s 3s< t ≤4s 4s< t ≤5s

d/dt (iin (t)) 2 As-1 -2 As-1 2 As-1 0 As-1

vL(t) 1V -1V 1V 0V

Below is the plot of VL vs time.

Part 2. Applying (7.4) wL (t) = In the time interval 0s < t ≤ 1s, iin (t) = 2t . Thus wL (t) =

1 (0.5 iin2 (t)) 2

(

)

1 0.5iin2 (t) = t 2 2

In the time interval 1s < t ≤ 3s, iin (t) = 4 − 2t . Thus

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P:7-2

wL (t) =

(


)

1 1 0.5iin2 (t) = (4 − 2t) 2 = t 2 − 4t + 4 2 4

In a similar way, wL (t) can be computed in the remaining intervals. The calculations for wL(t) for t = 0s to 5s are summarized in the following table: Time Interval 0s< t ≤1s 1s< t ≤3s 3s< t ≤4s 4s< t ≤5s

iin (t) 2t 4-2t -8+2t 0

wL(t) t2 2 t -4t+4 t2 -8t+16 0

Below is the plot of wL vs t.

SOLUTION 7.3. Applying (7.2) 1 t v ( )d L ∫0 L It is assumed that iL(0) = 0 A. Using the preceding formula and the fact that vin (t) = t in the time interval 0s < t ≤ 2s, the current is iL (t) = iL (0) +

iL (t) =

1 t d = t 2 in the time interval 0s < t ≤ 2s ∫ 0.5 0

In the time interval 2s < t ≤ 3s vin (t) = 2 . Hence iL (t) =

1 t 1 t v L ( )d = iL (2) + 2dt = 4 + 4 t2 = 4t − 4 ∫ 0.5 0 0.5 ∫2

In a similar way iL (t) can be computed in the remaining intervals. The calculations for iL(t) for t = 0s to 6s are summarized in the following table:

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Time Interval 0s< t ≤2s 2s< t ≤3s 3s< t ≤5s 5s< t ≤6s


vin (t), V t 2 -2 0

iL(t), A t2 4t-4 20-4t 0

Below is the plot of iL vs t.

SOLUTION 7.4. Part 1. Using (7.1), we have d vin (t) = 0.2 × 10 −3 [is (t)] = 0.2 × 10−3 × 1000cos(1000t) = 0.2cos(1000t)mV dt For the 2mH inductor t

t

1 1 iout (t) = iout (0) + ∫ 10v in ( )d = −3 ∫ 2cos(1000 )d = sin(1000t)mA L 2 × 10 0 0 Below is a sketch of iout vs t.

Part 2. Instantaneous power delivered by the dependent source is given by pL (t) = v L (t) × iL (t) = 10vin (t) × iout (t) = 2cos(1000t) × sin(1000t) = sin(2000t) W

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Part 3. The energy stored in the 2mH inductor is given by 1 1 WL (t) = LiL 2 (t) = Liout 2 (t) = sin2 (1000t)nJ 2 2 Below is a sketch of WL vs t

SOLUTION 7.5. Part 1 For the excitation in Figure P7.5b, t

1 i1 (t) = i1(0) + v ( )d , 0.5 ∫ in 0

t

1 i2 (t) = i2 (0) + v ( )d = 2i1(t) 0.25 ∫ in 0

It is assumed that i1 (0) = i 2 (0) = 0 A . In the interval 0s < t ≤ 1s, vin (t) = −10V . Hence, in this interval i1 (t) = i2 (t) =

1 t 1 t v in ( )d = (−10)d = −20t ∫ 0.5 0 0.5 ∫0

1 t 1 t v in ( )d = (−10)d = −40t ∫ 0.25 0 0.25 ∫0

In the interval 1s < t ≤ 2s, vin (t) = −5V . Hence, in this interval, i1 (t) = i1(1) +

i2 (t) = i2 (1) +

1 t 1 t v in( )d = −20 + (−5)d = −10t − 10 ∫ 0.5 1 0.5 ∫1

1 t 1 t v in( )d = −40 + (−5)d = −20t − 20 ∫ 0.25 1 0.25 ∫1

In a similar fashion i1 (t) and i2 (t) can be computed in other intervals. The calculations for i1 (t) and i2 (t) for t = 0s to 5s are summarized in the following table:

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Time Interval 0s< t ≤1s 1s< t ≤2s 2s< t ≤3s 3s< t ≤4s 4s< t ≤5s

vin (t), V -10 -5 0 5 10


i1 (t), A -20t -10-10t -30 -60+10t 100+20t

i2 (t), A -40t -20-20t -60 -120+20t 200+40t

Below are the plots of i1 vs t and i2 vs t.

Part 2 For the excitation in Figure P7.5c, t

1 i1 (t) = i1(0) + v ( )d , 0.5 ∫ in 0

t

1 i2 (t) = i2 (0) + v ( )d = 2i1(t) 0.25 ∫ in 0

It is assumed that i1 (0) = i2 (0) = 0 A. In the interval 0s < t ≤ 1s, vin (t) = 10t . Hence, in this interval i1 (t) = i1(0) +

1 t 1 t v in( )d = (10 )d = 10t 2 ∫ ∫ 0.5 0 0.5 0

and i2 (t) = i2 (0) +

1 t 1 t v in( )d = (10 )d = 20t 2 ∫ ∫ 0.25 0 0.25 0

In the interval 1s < t ≤ 3s, vin (t) = 10t − 20 . Hence, in this interval i1 (t) = i1(1) + i2 (t) = i2 (1) +

1 t 1 t v in( )d = 10 + (10 − 20)d = 10t 2 − 40t + 40 ∫ 0.5 1 0.5 ∫1

1 t 1 t v in( )d = 20 + (10 − 20)d = 20t 2 − 80t + 80 ∫ 0.25 1 0.25 ∫1

The calculations for i1 (t) and i2 (t) for t = 0s to 3s are summarized in the following table:

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Time Interval 0s< t ≤1s 1s< t ≤3s


vin (t), V 10t 10t-20

i1 (t), A 10t2 10t2 -40t+40

i2 (t), A 20t2 20t2 -80t+80

Below are the plots of i1 vs t and i2 vs t.

SOLUTION 7.6. For the circuit in Figure P7.6, the parallel combination of the 0.75mH and 1.5mH inductors can be replaced by an inductor with the inductance of (0.75||1.5)mH. The v-i relationship for this inductor is 1

t

1 iin (t) = iin (0) + v ( )d = ∫ s ∫ v s ( )d A 0.5 × 10 −3 0 [(0.75) (1.5)] × 10−3 0 t

The series combination of the 0.8mH and 0.6mH can be replaced by an (0.8mH + 0.6mH) inductor. The v-i relationship for this inductor gives: t  0.8 + 0.6 d  −3 d vout (t) = (0.8 + 0.6) × 10 [6i (t)] = 0.5 × 6 dt ∫ v s ( )d  = 16.8v s (t) V dt in 0  Below is a plot of vout(t) vs t.

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Assume iL(0) =0 for all inductors.


t

t

6 iout (t) = 6iin (t) = vs ( )d = 12 × 10 3 ∫ vs ( )d A ∫ −3 0.5 × 10 0 0 In the interval 0s < t ≤ 1s, v s (t) = 1mV . Hence, iin (t) =

t 1 v ( )d = 2t(A), iout (t) = 6iin (t) = 12t(V ) ∫ −3 0 s 0.5 × 10

The power p(t) on the same interval is computed as: p(t) = 6iin (t) × vout (t) = (12t) × (16.8× 10 −3 ) = 0.202t(W ) In the interval 1s < t ≤ 3s, v s (t) = 2mV . Hence, iin (t) = iin (1) +

and

t 1 v s ( )d = 2 + 2 × 2 1t = 4t − 2(A) ∫ −3 1 0.5 × 10

iout (t) = 6iin (t) = 24t − 12(A) p(t) = 6iin (t) × vout (t) = (24t − 12) × (16.8× 2 × 10−3) = 0.806t− 0.403(W )

In a similar fashion, iout(t) and p(t) can be computed for the remaining intervals. The calculations for iout(t) and the instantaneous power delivered by the dependent source, p(t), for t = 0s to 6s are summarized in the following table: Time Interval 0s< t ≤1s

vs(t), V 1×10-3

vout(t), V 16.8×10-3

iout(t), A 12t

p(t), W 0.202t

1s< t ≤3s

2×10-3

33.6×10-3

-12+24t

0.806t-0.403

3s< t ≤5s 5s< t ≤6s

-2×10-3 0

-33.6×10-3 0

132-24t

0.806t-4.435

12

0

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SOLUTION 7.7. For 0 ≤ t < 2 s, using the inductor v-i relationship, we have t

1 2 8 iin (t) = v in ( )d = [− cos(0.25π )]t0 = [1− cos(0.25πt)] A ∫ 0.5 0.25π π 0

The associated energy stored as a function of t for this time interval is W0.5 (t) =

16 [1− cos(0.25πt)]2 J 2 π

The energy for the second inductor remains zero over this interval. For 2 s ≤ t, we have t

1 iin (t) = i0.5 (t) + i0.25 (t) = i0.5 (t) + v ( )d 0.25 ∫ in 2

8 4 8 16 [1− cos(0.25πt)] + [− cos(0.25π )]t2 = [1− cos(0.25πt)] − cos(0.25πt) π 0.25π π π 8 24 = − cos(0.25πt) A π π Here the current 8 i0.5 (t) = [1− cos(0.25πt)] A π =

in which case the energy stored over the interval [2,t] is W0.5 (2,t) =

(

)

1 1  2 64  2 2 0.5 i0.5 (t) − i0.5 (2) = 0.5 i0.5 (t) − 2  J 2 2  π 

Further i0.25 (t) = −

16 cos(0.25πt) A π

in which case the energy stored over the interval [2,t] is W0.25 (2,t) =

(

)

1 1 2 2 0.25 i0.25 (t) − i0.25 (2) = 0.25i 20.25(t) J 2 2

SOLUTION 7.8. Let the 5mH inductor be L1 and the 20mH inductor be L2 . For 0 ≤ t < 3ms, t

t

1 1 iL1(t) = iL1 (0) + ∫ v s ( )d = −3 ∫ 12cos (500 )d = 4.8sin(500t) mA L1 5 × 10 0 0

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P:7-9

t

For t ≥ 3ms,


t

1 1 iL 2 (t) = i L2 (0) + v s ( )d = ∫ −3 ∫ 12cos (500 )d = 1.2sin(500t) mA L2 20 × 10 0 0

iL1(t) = 4.8sin(500t) mA as L1 is still subjected to the same voltage. on the other hand, iL 2 (t) = i L2 (3− ms) +

1 L2

t

∫ +

3 ms

1 v L2 ( )d = iL 2 (3− ms) + L

2

t

∫ +

0d

3 ms

−

= iL 2 (3 ms) = 1.2sin(500 × 3 × 10−3) = 1.197mA For 0≤ t < 3ms, the energies stored in the inductors are given as follows: 1 2 L1[iL1 (t)] = 57.6sin 2 (500t) nJ, 2 1 2 WL2 (t) = L2 [iL2 (t)] = 14.4sin 2 (500t) nJ 2 WL1 (t) =

For t ≥ 3ms, the energies stored in the inductors are given as follows: 1 2 L1[iL1 (t)] = 57.6sin 2 (500t) nJ, 2 1 2 WL2 (t) = L2 [iL2 (t)] = 14.328 nJ 2 WL1 (t) =

SOLUTION 7.9. Given the dielectric parameters and the dimensions of the capacitor, the capacitance of the paper capacitor is given by A 0.04 × 0.8 C = r 0 = 3× 8.854 × 10−12 × = 8.5nF d 10−4 SOLUTION 7.10. . Part 1. Applying the capacitor v-i relationship: iC (t) = C

d (v (t)) = 1 × 100 × 1000 × (− sin(1000t)) = −0.1sin(1000t) A dt C

Part 2. Applying the capacitor v-i relationship:

L&C Probs, 11/15/01

P:7-10


d (v (t)) dt C d 10 × 10 −3 cos(1000t) = C (sin(1000t)) dt iC (t) = C

10 × 10 −3 cos(1000t) = 1000C cos(1000t) Therefore, C=10µF.

SOLUTION 7.11. Applying the capacitor v-i relationship for C1 and C2 : iC1(t) = (2mF)

d (v (t)), dt in

iC 2 (t) = (6mF)

d (v (t)) = 3iC1(t), dt in

In the interval –1s < t ≤ 0s, vin (t) = 5t + 5. Hence, in this time interval iC1(t) = 2 × 10−3 ×

d(v in (t)) = 10−2 A dt

and iC 2 (t) = 6 × 10 −3 ×

d(vin (t)) = 3× 10 −2 A dt

Using KCL, iin (t) can be computed as iin (t) = iC1 (t) + iC 2 (t) = 4 × 10 −2 A In the interval 0s < t ≤ 1s, vin (t) = 5. Thus iC1(t) = iC2 (t) = 0 A Using KCL, iin (t) = iC1 (t) + iC 2 (t) = 0 A . In a similar fashion iC1 (t), iC2 (t) and iin (t) can be computed for the remaining intervals.

The calculations for iC1 (t), iC2 (t) and iin (t) for t = -1s to 6s are summarized in the following table: Time Interval -1s< t ≤0s 0s< t ≤1s 1s< t ≤3s 3s< t ≤4s 4s< t ≤5s 5s< t ≤6s

d/dt (vin (t)), Vs-1 5 0 -5 20 0 -15

Below are the plots of iC1 (t), iC2 (t) and iin (t).

iC1 (t), mA 10 0 -10 40 0 -30

iC2 (t), mA 30 0 -30 120 0 -90

iin (t), mA 40 0 -40 160 0 -120

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P:7-11


SOLUTION 7.12. Applying (7.6) 2

2

1 1 vC (2) = vC (0) + ∫ iC ( )d = 4V + ∫ d = 5V C C 0

vC (3) = vC (2) +

0

3

3

2

2

1 1 iC ( )d = vC (2) + ∫ 2d = 5V + 1V = 6V ∫ C C

Applying (7.11), the energies stored in the capacitor over the intervals [0,2] and [2,3] are given by

[

]

[

]

WC[0,2] =

1 C vC2 (2) − v C2 (0) = 9 J 2

WC[2,3] =

1 C vC2 (3) − v C2 (2) = 11 J 2

SOLUTION 7.13. Part 1 Applying (7.6)

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P:7-12


t

1 1 vC1 (t) = vC1(0) + iin ( )d = ∫ C1 0.25 0

t

vC 2 (t) = v C2 (0) +

1 1 iin ( )d = ∫ C2 0.1 0

t

∫ iin (

)d

∫ iin (

)d

0 t

0

In the interval 0s < t ≤ 1s, iin (t) = 2t . Hence, in this interval vC1 (t) =

1 0.25

t

∫o

iin ( )d =

t 1 ( 2 ) = 4 × 106 t 2 (V ) 0.25 0

1 0.1

t

iin ( )d =

t 1 ( 2 ) = 10 × 10 6 t 2 (V ) 0.1 0

and vC 2 (t) =

∫o

In the interval 1s < t ≤ 3s, iin (t) = 2 × 10−3 (A). Thus vC1 (t) = vC1(1) +

t t 1 1 × ∫ iin ( )d = 4 + × (2 × 10−3 ) = 8 × 10 3 t − 4(V ) 1 0.25 0.25 1

vC 2 (t) = v C2 (1) +

t t 1 1 × ∫ iin ( )d =10 + × (2 × 10−3 ) = 20 × 10 3 t − 10(V ) 1 0.1 0.1 1

and

In a similar fashion, vC1 (t) and vC2 (t) can be computed for the remaining intervals. The calculations for vC1 (t) and vC2 (t) for t = 0s to 8ms are summarized in the following table: Time Interval 0s< t ≤1ms

iin (t), A 2t

vC1 (t), V 4×106 t2

vC2 (t), V 10×106 t2

1ms< t ≤3ms

2×10-3

-4 + 8×103 t

-10 + 20×103 t

3ms< t ≤5ms 5ms< t ≤8ms

-2×10-3 0

44 - 8×103 t 4

110 - 20×103 t 10

Below are the plots of vC1 (t) and vC2 (t).

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P:7-13


Part 2 Applying (7.12), WC1 (t) =

1 C v 2 (t) , 2 1 C1

WC 2 (t) =

1 C v 2 (t) 2 2 C2

The expressions for WC1 (t) and WC2 (t) for t = 0ms to 8ms are listed in the following table: Time Interval

W C1 (t), µJ

W C2 (t), µJ

0s< t ≤1ms

0.125 × (4 ×10 t )

0.05 × (10×106 t2 )2

1ms< t ≤3ms

0.125 × (-4 + 8×103 t)2

0.05 × (-10 + 20×103 t)2

3ms< t ≤5ms

0.125 × (44 – 8×103 t)2

0.05 × (110 – 20×103 t)2

5ms< t ≤8ms

0.125 × (4)2

0.05 × (10)2

6 2 2

Below are the plots of WC1 (t) and WC2 (t).

Part 3 Since the current iC1 (t) stays constant at 0A after t = 5ms, vC1 (∞) = v C1 (5) = 4V vC 2 (∞) = vC 2 (5) = 10V SOLUTION 7.14. Part 1 Applying (7.6),

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P:7-14

t


t

1 1 vin (t) = v in (0) + ∫ iin ( )d = −3 ∫ iin ( )d C 0.5 × 10 0 0 In this part, we use the current excitation signal described in Figure P7.14b. In the interval 0s < t ≤ 1s, iin (t) = −10(mA). Thus, in this interval vin (t) =

t 1 (−10 × 10−3)d = −20t(V ) ∫ −3 o 0.5 × 10

In the interval 1s < t ≤ 2s, iin (t) = −5(mA). Hence, in this interval vin (t) = v in (1) +

t 1 (−5 × 10−3)d = −20 + (−10t + 10) = −10t − 10(V ) ∫ −3 1 0.5 × 10

In a similar fashion, vin (t) can be computed for the remaining intervals. The calculations for vin (t) for t = 0s to 5s are summarized in the following table: Time Interval 0s< t ≤1s 1s< t ≤2s 2s< t ≤3s 3s< t ≤4s 4s< t ≤5s

iin (t), mA -10 -5 0 5 10

vin (t), V -20t -10 -10t -30 -60+10t -100+20t

Below is the plot of vin (t) vs. time.

Part 2. In this part, we use the current excitation signal described in Figure P7.14c. In the interval 0s < t ≤ 1s, iin (t) = 10(mA) . Thus, in this interval vin (t) =

t 1 (10× 10 −3 × )d = 10t 2 (V) ∫ −3 o 0.5 × 10

In the interval 1s < t ≤ 3s, iin (t) = 10t − 20(mA) . Hence, in this interval vin (t) = v in (1) +

t −3 1 10 × (10 − 20)d = 10 + 10(10 2 − 40 ) 1t = 10t 2 − 40t + 40 (V) ∫ −3 o 0.5 × 10

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P:7-15


The calculations for vin (t) for t = 0s to 3s are summarized in the following table: Time Interval 0s< t ≤1s 1s< t ≤3s

iin (t), mA 10t 10t-20

vin (t), V 10t2 10t2 - 40t + 40

Below is the plot of vin (t) vs. time.

SOLUTION 7.15. Part 1 Using the capacitor v-i relationship, iin (t) = C1

d (v (t)) = 20 × 6 × 1500 × cos(1500t) = 0.18cos(1500t) A dt s

Then, we can find vout by applying (7.6), t

vout (t) = vout (0) +

1 2i ( )d 0.5m ∫ in 0

= 10 +

2 × 0.18 sin(1500t) 0.5m × 1500

= 10 + 0.48sin(1500t)

V

Part 2 The instantaneous power delivered by the independent source is given by p(t) = 2iin (t) × vout (t) = 0.36cos(1500t) × [10+ 0.48sin(1500t)]

W

Part 3 Applying (7.11), the energy stored in the capacitor over the interval [0,t] is given by WC1[0,t] =

[

]

1 C1 vs 2 (t) − v s2 (0) = 360sin2 (1500t) J 2

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P:7-16


SOLUTION 7.16. Part 1 From the solution of problem 7.15, t

1 vout (t) = vout (0) + 2i ( )d 0.5m ∫ in 0 t

1 d = vout (0) + 2 × 20 (v s ( ))d ∫ 0.5m d 0

t

= vout (0) + 0.08∫

0

d (v s ( ))d d

= vout (0) + 0.08v s( ) If we assume vout(0) = 0V, then vout (t) = 0.08 v s ( ) The following is a plot of vout(t) vs time.

Part 2 The instantaneous power delivered by the dependent source is given by p(t) = v out (t) × 2iin (t) = ( vout (0) + 0.08 v s(t)) × 2 × 20 = 40( vout (0) + 0.08 vs (t)) If we assume vout(0) = 0V, then p(t) = 3.2 v s (t)

d (v (t)) dt s d (v (t)) dt s

d (v (t)) dt s

W

W

The calculations for d/dt (vS(t)) and p(t) for t = 0s to 6ms are summarized in the following table:

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P:7-17


Time Interval 0ms< t ≤1ms

d/dt (vS(t)), Vs-1 3 2 × 10

1ms< t ≤3ms

-2 × 10

3

12.8 × (t-2)

3ms< t ≤4ms 4ms< t ≤6ms

3

12.8 × (t-4) 0

p(t), W 12.8 × t

2 × 10 0

The following is a plot of p(t) vs time.

Part 3 The energy stored in the 0.5-mF capacitor is given by WC (t) =

1 × 0.5m × v out 2 (t) = 0.25( vout (0) + 0.08 v s (t)) 2 2

If we assume vout(0)=0V, then WC (t) = 1.6(v s (t)) 2

J

In the interval 0s < t ≤ 1ms, v s (t) = 2 × 103 t(V ). Thus, in this interval WC (t) = 6.4t 2J In the interval 1s < t ≤ 3ms, v s (t) = 2 × 103 × (2 − t)(V ). Hence, in this interval WC (t) = 6.4 × (2 − t) 2J In a similar way WC(t) can be computed for the interval 3ms < t ≤ 4ms. The following is a plot of W C(t) vs time.

mJ

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P:7-18


SOLUTION 7.17. Part 1 From the solution of problem 7.16, vout (t) = v out (0) + 0.08 v s ( ) If we assume vout(0)=0V, then vout (t) = 0.08 v s ( ) The following is a plot of vout(t) vs time.

Part 2 The instantaneous power delivered by the dependent source is given by. d p(t) = 40( v out (0) + 0.08 v s (t)) (v s (t)) W dt If we assume vout(0)=0V, then d p(t) = 3.2 v s (t) (v (t)) W dt s The calculations for d/dt (vS(t)) and p(t) for t = 0s to 6ms are summarized in the following table:

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P:7-19

Time Interval 0ms< t ≤1ms 1ms< t ≤2ms 2ms< t ≤4ms 4ms< t ≤5ms 5ms< t ≤6ms


d/dt (vS(t)), Vs-1 2 × 103 0 -2 × 103 0 2 × 103

p(t), W 12.8 × t 0 12.8 × (t-3) 0 12.8 × (t-6)

The following is a plot of p(t) vs time.

Part 3 The energy stored in the 0.5-mF capacitor is given by WC (t) = 0.25( v out (0) + 0.08 v s (t))2 If we assume vout(0) = 0V, then

WC (t) = 1.6(v s (t)) 2

J

In the interval 0ms < t ≤ 1ms, v s (t) = 2 × 103 t(V ). In this interval WC (t) = 6.4t 2J In the interval 1ms < t ≤ 2ms, v s (t) = 2(V ). In this interval WC (t) = 6.4( J ) In a similar fashion WC(t) can be computed for the remaining intervals. The following is a plot of W C(t) vs time.

mJ

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P:7-20


SOLUTION 7.18. Part 1 Applying the capacitor v-i relationship for the equivalent capacitor of the series combination of 0.3 mF and 0.6 mF capacitors vin (t) = v in (0) +

= (0) +

t

1

∫ is (

0.3 × 10−3 0.6 × 10−3 0 t

1 0.3× 10 −3

)d

0.6 × 10 −3

∫ 60 × 10 −3 sin(100 )d 0

= 3 − 3cos(100t) V

Therefore, iout (t) = (0.2 × 10−3 + 0.8 × 10−3)

d (10v in (t)) = 3sin(100t) dt

Below is a sketch of iout(t) vs time.

Part 2 The instantaneous power delivered by the dependent source is given by p(t) = iout (t) × 10v in (t) = 90sin(100t) × [1− cos(100t)] W

V

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P:7-21


Part 3 The instantaneous energy stored in the 0.8mF capacitor is given by WC (t) =

1 1 2 CvC 2 (t) = (0.8mF)[10vin (t)] = 0.361 [ − cos(100t)]2 J 2 2

SOLUTION 7.19. Part 1 Part 2 Part 3 Part 4

Since Q = CV, the charge that resides on each plate of the capacitor = 10µF × 100V = 1mC Since V = Q/C, the required voltage = 1mC/5µF = 200V Since V = Q/C, the required voltage = 50µC/1µF = 50V The energy required = 0.5 × 10µF × (100V)2 = 0.05J

SOLUTION 7.20. When 0s≤ t <2s, vC(t) = 25V. Conservation of charge requires that q1(2 − ) + q2 (2− ) = q1 (2+ ) + q2 (2 + ). Since q2 (2− ) = 0 C it follows that

25 V × 150 mF = vC(2+)(150 mF + 100 mF) Hence vC (2+ ) =

25V × 150mF + 0V × 100mF = 15 V 150mF + 100mF

Thus, for t > 2s, vC(t) = 15 V. SOLUTION 7.21. For this solution consider the figure below in which C1 and C2 are labeled.

There are two cases to consider: (i) t < 2 ms and (ii) t > 2 ms. Case 1. t < 2 ms. Here, since the current source is zero for t < 0, and C1 is uncharged at t = 0, 1 vC (t) = C1 =

t

∫

−∞

t

t

0

0

1 1 is ( )d = vC (0) + is ( )d = i ( )d ∫ C1 C1 ∫ s

[

]

(

)

−12 −500 t e = 1.6 1− e −500t V −3 0 15 × 10 × 500

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P:7-22


Note that vC (2− ms) = 1.0114 V. Hence the energy stored over [0, 2 ms] is

(

WC1 (0 < t < 2ms) = 0.0192 1− e −500t and

)

2

J

WC2 (0 < t < 2ms) = 0

Case 2. 2 ms < t. At t = 2 ms the switch closes, forcing a discontinuity in the capacitor voltages. To + calculate the capacitor voltages at 2 ms, we use conservation of charge. Here, the relevant equation is: qC1 (2+ ms) + qC 2(2 + ms) = qC1(2 − ms) + qC2 (2 − ms) +

2+ ms

∫

is ( )d

−

2 ms

Note that since vC 2 (2 − ms) = 0, qC 2 (2 − ms) = C2vC 2 (2− ms) = 0 and the integral of the bounded continuous function is (t) over an infinitesimal interval is zero, this equation reduces to qC1 (2+ ms) + qC 2(2 + ms) = qC1(2 − ms) or equivalently, since for t > 2 ms, vC1 (t) = vC 2 (t) = vC (t) , C1vC (2 + ms) + C2vC (2+ ms) = C1vC (2− ms)

Therefore

vC (2+ ms) =

C1 15 vC (2− ms) = × 1.0114 = 0.37927 V C1 + C2 15 + 25

and it follows that 1 vC (t) = vC (2 ms) + Ceq +

= 0.379 +

t

∫

2ms

[

1 is ( ) d = 0.379 + C1 + C2

]

t

∫

is ( )d

2ms

(

)

−12 −500 t e = 0.379 + 0.6 0.36788 − e −500t V −3 0.002 40 × 10 × 500

Hence the energy stored in the two capacitors over the interval [2+ ms, t] is WCeq (2+ ms < t) =

1 1 Ceq vC2 (t) − Ceqv C2 (2 + ms) J 2 2

whereas the instantaneous stored energy, i.e., the energy stored over (-∞,t > 2 ms] is given by WCeq (t) = -

+

1 Ceq vC2 (t) 2

What happens between 2 ms and 2 ms is beyond the scope of the material in this chapter. Please refer to problem 51 in chapter 8 for an explanation.

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SOLUTION 7.22.

P:7-23


Part 1 Applying the inductor v-i relationship, v L (t) = L

d i (t) dt in

(

= 2.5mH × −200te−10t + 20e −10t

)

= 0.05e −10t (−10t + 1) Applying the capacitor v-i relationship, we have: t

1 vC (t) = vC (0) + ∫ iin ( )d C 0

t

=

20 te−10 1m ∫ 0

[

[

]0 t

= 20 × 103 −0.1 e −10 − 0.01e −10

]

= 2 × 103 − te−10t − 0.1e −10t + 0.1 V and vin (t) = v L (t) + vC (t)

[

]

= 0.05e −10t (−10t + 1)+ 2000 − te−10t − 0.1e −10t + 0.1 = −2000.5 × te−10t − 199.95 × e −10t + 200(V) The sketches of vL(t), vC(t) and vin (t) are shown below.

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P:7-24


Part 2 The energy stored in the inductor is given by WL (t) =

1 L[i (t)]2 = 0.5t 2e −20t 2 in

J

The sketch of WL(t) vs. time is shown below. Part 3 The energy stored in the capacitor is given by WC (t) =

[

]

2 1 2 C[vC (t)] = 2 −te−10t = 0.1e −10t + 0.1 kJ 2

The sketch of WC(t) vs time is shown below.

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SOLUTION 7.23. Applying the inductor v-i relationship we have that v L (t) = 0.25

d[is (t)] = 4cos(4 t) (V ) dt

Applying the capacitor v-i relationship it follows that iC (t) = 0.25

d[2v 2 (t)] = 0.25 × 2 × 4 × 4 × [− sin(4t)] = −8sin(4 t) A dt

SOLUTION 7.24. Part 1. By KCL is (t) = iC1(t) + iC 2 (t). Applying the v-i relationship for capacitor C1 and C2 we have: iS (t) = (10mF + 20mF)

d v (t) dt s

= 30 × 10−3 × (12te −5t − 30t 2e −5t ) = 0.36te −5t − 0.9t 2e −5t

A

Part 2. Applying the v-i relationship for the 20mH inductor, it follows that vout (t) = (20mH )

(

d [12is (t)] dt

= 0.24 × 4.5t 2e −5t − 3.6te−5t + 0.36e −5t

)

= 1.08t 2e −5t − 0.864te−5t + 0.0864e −5t (V ) Part 3 The energy stored in the 20mH inductor for t > 0 is given by WL (t) =

1 (20mH )[12iS (t)]2 2

(

= 1.44 0.36te−5t − 0.9t 2e −5t

)

2

J

= 1.44 × t 2e −10t (0.36 − 0.9t)2 J SOLUTION 7.25. We denote by vC(t) the voltage across the capacitors C1 and C2 and by iL(t) the current through the inductors L1 and L2 . The equivalent capacitance of the parallel combination of C1 and C2 is (C1 + C2 ) and thus: iS (t) = (C1 + C2 )

dvC (t) dt

Using the v-i relationship for the capacitor C2 it follows that iC 2 (t) = C2 By replacing

dvC (t) , it follows that dt

dvC (t) dt

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P:7-26

iC 2 (t) =


C2 2 × i S (t) = is (t) C1 + C2 3

The equivalent inductance of the series combination of L1 and L2 is (L1 + L 2 ) and thus: 9iC 2(t) = (L1 + L2 )

diL (t) dt

Using the v-I relationship for the inductor L2 it follows that vout (t) = v L2 (t) = L2 By replacing

di2 (t) dt

diL (t) , it follows that dt vout (t) =

L2 2 × 9iC 2 (t) = 9iC 2(t) = 6iC 2 (t) = 4is (t) L1 + L2 3

SOLUTION 7.26. Using the v-i relationship for the capacitor we can write: 1 vC 2 (t) = C2 1 vC1 (t) = C1

t

∫

−∞

iC ( )d

t

∫

−∞

iC ( )d

where iC(t) is the current through the capacitors C1 and C2 . Since C1 and C2 are connected in series: 1 1  vin (t) = v C1(t) + vC 2 (t) =  +  C1 C2  t

It follows that

∫ iC ( )d

−∞

t

∫ iC (

)d

−∞

 CC  = v in (t) 1 2  . Hence  C1 + C2 

1 vC 2 (t) = C2

t

∫

−∞

iC ( )d =

1 CC C1 × 1 2 v in (t) = v (t) C2 C1 + C2 C1 + C2 in

Since L1 and L2 are combined in parallel it follows that 1 1 vout (t) =  +   L1 L2  =

−1

d(AvC 2(t)) dt

L1L2 dv (t) LL C1 dvin (t) × A× C2 = 1 2 × A× L1 + L2 dt L1 + L2 C1 + C2 dt

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P:7-27

where the last equality follows by replacing vC2 (t) with


C1 v (t) C1 + C2 in

SOLUTION 7.27. Observe first that the 0.3H and 0.6H parallel inductances combine to make a 0.2H inductance. Also the 0.4H and 1.2H parallel inductances combine to make a 0.3H inductance. Finally, the 0.2H and 0.3H inductances are combined in series and the equivalent inductance is 0.5H. Shortly, all the above steps can be written as: Leq = (0.3H || 0.6H) + (0.4H || 1.2H) = 0.5H. SOLUTION 7.28. Observe first that the 1mH and 5mH inductors combine to make a 6mH inductance. This inductance combines in parallel with the 3mH inductance to make a 2mH inductance. The next step is to combine in series the 2mH inductance with the 10mH inductance. The equivalent inductance is 12mH. This inductance is combined in parallel with the 36mH inductance and the result is 9mH. Finally, the 9mH inductance is combined in series with the 4mH inductance and the result is 13mH. Shortly, all the above steps cam be written as: Leq = [(5mH + 1mH) || 3mH + 10mH] || 36mH + 4mH = 13mH. SOLUTION 7.29. Step 1. The parallel combination of the 0.6mH and 1.2mH inductors is equivalent to a 0.4mH inductor. Step 2. The series combination of the 2.4mH and 0.4mH inductors is equivalent to a 2.8mH inductor. Step 3. The parallel combination of the 2.8mH and 7mH inductors is equivalent to a 2mH inductor. SOLUTION 7.30. The three inductors can be arranged in the seven fashions as shown below.

(a)

(b)

(c)

(d)

(e)

(f) Laeq Lbeq Lceq Ldeq Leeq f Leq Lgeq

= 1mH = 1mH + 1mH = 2mH = 1mH || 1mH = 0.5mH = 1mH + 1mH + 1mH = 3mH = 1mH + (1mH || 1mH) = 1.5mH = 1mH || (1mH + 1mH) = 0.667mH = 1mH || 1mH || 1mH) = 0.333mH

(g)

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SOLUTION 7.31. For Fig. P7.31a, Leq = 2 + 15 + 10 + 10 + 40 +30 + 20 + 8 = 135mH The circuit in Fig. P7.31b is equivalent to the following circuit.

Therefore, Leq = 2 + 15 + 10 + 8 = 35mH The circuit in Fig. P7.31c is equivalent to the following circuit.

Therefore, Leq = 2 + [(15+10) || (10+40) || (20+30)] + 8 = 22.5mH

SOLUTION 7.32. When the switch is open, the circuit in Fig. P7.32 can be rearranged as the following.

Therefore, Leq = 8L || 8L = 4L When the switch is closed, the circuit in Fig. P7.32 can be rearranged as the following.

Therefore, Leq = 4L || 4L + 4L || 4L = 4L

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SOLUTION 7.33. L eq1 =Leq2 Without going into a detailed analysis, we present the following intuitive argument. Note that the points a and b represent points on a balanced bridge circuit meaning that the voltage between a and b would be zero. Therefore, no current will flow through the additional inductance L. Therefore the presence of L does not affect the equivalent inductance value.

SOLUTION 7.34. L eq1 >Leq2 Without going into a detailed analysis, we present the following intuitive argument. Note that the points a and b represent points on an unbalanced bridge circuit, meaning that the voltage between a and b would not be zero. Also note that when two inductors are placed in parallel, the equivalent inductance becomes smaller than either inductance. The addition of the inductor L in circuit 2 essentially creates an internal parallel inductance resulting in an Leq2 lower than Leq1 .

SOLUTION 7.35. First we add an iin label to the circuit as shown below.

From KVL and the derivative definition of the capacitor di (t) di (t) di (t) vin (t) = v L1(t) + vL 2 (t) = L1 in + L2 in = ( L1 + L2 ) in dt dt dt Equivalently, diin (t) 1 = v (t) dt ( L1 + L2 ) in It follows that v L k (t) = Lk

diin (t) Lk = v (t) dt L1 + L2 in

which is the required voltage division formula.

SOLUTION 7.36. Leq = (11mH) || (19.25mH) + 3mH = 10mH. Applying the voltage division formula, v L1 (t) = vin (t) × and

3mH = 60te−t mV Leq

v L2 (t) = v in (t) − vL1(t) = 140te−t mV

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SOLUTION 7.37. First consider the circuit below which contains the additional label of vin (t).

From KCL and the integral definition of the inductor, t

t

0

0

1 1 iin (t) = iL1 (t) + iL 2 (t) = ∫ v in ( )d + v ( )d L1 L2 ∫ in Equivalently     1 ∫ vin ( )d =  1 1  iin (t) + 0  L1 L2  t

Therefore 1    L1L2   L   Lk  1 iL k (t) = v in ( )d =  1 k 1  iin (t) =   iin (t) for k = 1,2 ∫ Lk  +   L1 + L2  0  L1 L2    t

SOLUTION 7.38. In the circuit illustrated in Fig. 7.38, Leq = (12mH + 27mH) || (130mH) = 30mH Applying the current division formula, 2 12 + 27 iin (t) = 0.231e −t mA 12 + 27 + 130 2 130 iL 2 (t) = iin (t) = 0.769e −t mA 12 + 27 + 130

iL1(t) =

Also, vin (t) = Leq

2 d iin (t) = −60te−t dt

V

The instantaneous energy stored in the 130-mH is given by WL1 (t) =

2 1 2 L1[iL1 (t)] = 0.8e −2t nJ 2

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SOLUTION 7.39. Using the inductor v-i relationship, vin (t) = Leq

d d iin (t) = 30mH × iin (t) dt dt

The calculations for vin (t) for t = -2s to 7s are summarized in the following table: Time Interval -2s< t <=0s 0s< t <=2s 2s< t <=3s 3s< t <=4s 4s< t <=6s 6s< t <=7s

d/dt (iin (t)), mAs-1 100 -200 0 200 100 -200

From the solutions to problem 7.38, iL1(t) = 0.231 iin (t) iL 2 (t) = 0.769 iin (t) Below are the plots for vin (t), iL1(t) and iL2(t).

vin (t), mV 3 -6 0 6 3 -6

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SOLUTION 7.40. Part 1 Using the current division formula, iL 2 (t) =

18 i (t) = 90cos(300 t) mA 18 + 6 in

Part 2 Leq =1.5mH + 6mH || 18mH = 6mH vin (t) = Leq

d i (t) = -0.679sin(300 t) V dt in

Part 3 Instantaneous power delivered by the source is given by p(t) = v in (t) × i in(t) = [-0.679sin(300 t)]× [120cos(300 t)] mW = -40.7sin(600 t) mW Below is a plot of p(t)

SOLUTION 7.41. Part 1 Leq = 0.3 || 0.9 + 0.4 || 0.4 = 0.425 H For computing iin (t) we need to apply 1 iin (t) = iin (t0 ) + Leq

t

∫ v in (

)d

t0

to each interval [0, 1], [1, 2], …, [n, n+1], … The initial condition for the interval [n, n+1] for n even is: n

1 iin (n) = iin (0) + v ( )d Leq ∫ in 0

We assume that iin (0) = 0 and it follows that iin (n) = 0 for n even, since

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n

∫ vin (

)d = 0

0

for n even. The initial condition for the interval [n, n+1] for n odd is: 1 iin (t) = iin (n − 1) + Leq

n

∫

n−1

vin ( )d =

1 (16 ) nn−1 = 37.6 A 0.425

For the interval [n, n+1] with n even: t

1 1 iin (t) = iin (n) + v in ( )d = 0 + 16 tn = 37.6( t − n) A ∫ Leq 0.425 n

For the interval [n, n+1] with n odd: t

1 1 iin (t) = iin (n) + v in ( )d = 37.6 + (−16 ) tn = 37.6( n + 1− t) A ∫ Leq 0.425 n

For the interval [n, n+1] with n even: t

1 1 iin (t) = iin (n) + v in ( )d = 0 + (16 ) tn = 37.6(t − n) A ∫ Leq 0.425 n

Hence,

37.6(t − 2n), 2n < t ≤ 2n + 1 iin (t) =  37.6(2n + 2 − t), 2n + 1< t ≤ 2n + 2

where n is a non-negative integer. Below is a sketch for the input current vs. time.

Part 2 The total energy stored in the set of four inductors is given by

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1 2 Leq [iin (t)] 2 300(t − 2n)2 J = 300(2n + 2 − t) 2 J


W (t) =

for 2n < t ≤ 2n + 1 for 2n + 1< t ≤ 2n + 2

where n is a non-negative integer. Below is a sketch of W(t) vs. t.

Part 3 By the current division formula, the current through the 0.9H inductor is given by 0.3 i (t) 0.3 + 0.9 in 9.4(t − 2n) A = 9.4(2n + 2 − t) A

iL 0.9H (t) =

where n is a non-negative integer. Below is a sketch of the iL0.9H vs. time.

for 2n < t ≤ 2n + 1 for 2n + 1< t ≤ 2n + 2

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SOLUTION 7.42.. a) Step 1. The parallel combination of the 1µF and 2µF capacitances is equivalent to a 3µF capacitance. Step 2. The series capacitors 1.5µF and 3µF combine to make a 1µF capacitance. Step 3. Finally, the 2µF capacitance is combined in parallel with the 1µF capacitance that was the result of Step 2 to make a 3µF capacitance. Shortly, the above steps can be written as: Ceq = [(1µF + 2µF) || 1.5µF] + 2µF] = 3µF b) Proceeding in a similar fashion as in part a: Ceq = 30mF + [9mF || (9mF + 18mF) || 5.4mF] = 33mF SOLUTION 7.43. a) Step 1. Combine the parallel capacitances of 1µF and 2µF to make a 3µF capacitance. Step 2. Combine the series capacitances of 1.5µF and 3µF to make 1µF capacitance. Step 3. Combine the parallel capacitances of 2.5µF and 1µF (the result of Step 2) to make a 3.5µF capacitance. Shortly, the above steps can be written in a condensed form as: Ceq = 2.5µF + [1.5µF || (1µF + 2µF) = 3.5µF b) Proceeding in a similar fashion as in part a: Ceq = (1mF || 2 mF) + (1.2mF || 2mF || 1.5mF) + (4mF || 2.6667mF || 3.2mF || 1.6mF) = 2.1667mF

SOLUTION 7.44. Part 1 Ceq = ((( 30 || 60 + 40 ) || 30 + 40 ) || 30 + 40 ) || 30 + 40 = 60mF Part 2 The value of vin (t) is given by t

t

0

0

1 1 vin (t) = v in (0) + ∫ iin ( )d = 0 + 200sin(20 )d = 166.7 − 166.7cos(20t) mV C 0.06 ∫

SOLUTION 7.45. When the switch is open, the circuit in figure P7.45 can be represented by the following.

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Therefore, Ceq = 6.4C When the switch is closed, the circuit in figure P7.45 can be represented by the following.

Therefore, Ceq = ( 4 + 4 ) || ( 16+ 16 ) = 6.4C

SOLUTION 7.46. C eq1
SOLUTION 7.47. The three capacitors can be arranged in the seven fashions as shown below.

For configuration (a), Ceq = 1 || 1 || 1 = 0.3333µF For configuration (b), Ceq = 1 || (1 + 1) = 0.6667µF For configuration (c), Ceq = 1 || 1 + 1 = 1.5µF For configuration (d), Ceq = 1 + 1 + 1 = 3µF For configuration (e), Ceq = 0 + 1 + 1 = 2µF For configuration (f), Ceq = 1 || 1 + 0 = 0.5µF For configuration (g), Ceq = 1 + 0 + 0 = 1µF

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SOLUTION 7.48. As Q = CV for any capacitor, C=Q/V. The equivalent capacitance of the network is given by 1C 1C 1C Ceq = || || = 0.1 F 2V 3V 5V

SOLUTION 7.49. First consider the circuit below which contains the additional label of iin (t).

From KVL and the integral definition of the capacitor, t

t

0

0

1 1 vin (t) = v C1(t) + vC 2 (t) = iin ( )d + i ( )d ∫ C1 C2 ∫ in Equivalently, t

∫ iin ( 0

1 )d = 1 1 v in (t) + C1 C2

Therefore, 1 Ck

C1C2 1 Ck vCk (t) = iin ( )d = 1 v (t) = v in (t) ∫ in 1 Ck C + C 1 2 + 0 C1 C2 t

Hence, C2 v (t) C1 + C2 in C1 vC 2 (t) = v (t) C1 + C2 in which is the required voltage division formula. vC1 (t) =

SOLUTION 7.50. By the voltage division formula, vC 3 (t) =

6 ||6 vin (t) = 21(1− e −20t ) V 6|| 6 + 9

SOLUTION 7.51. Part 1 By the voltage division formula,

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0.05 + 0.15 v (t) = 6.667sin(120 t) V 0.05 + 0.15 + 0.1 in vC 2 (t) = v in (t) − vC1(t) = 3.333sin(120 t) V vC1 (t) =

Part 2 Let the 0.1µF capacitor be C1 and the 0.15µF capacitor be C2 . The energies stored in the 0.1µF and the 0.15µF capacitors are given by WC1[0,t] =

[

]

[

]

1 C v 2 (t) − vC12 (0) = 2.222sin 2 (120 t) J 2 1 C1

1 WC2 [0,t] = C2 vC 22 (t) − v C 22 (0) = 0.8333sin 2 (120 t) J 2 SOLUTION 7.52. Part 1 As both the terminals of 0.08F are tied to the voltage source vin (t), vC1 (t) = vin (t) = 100e −2t V By the voltage division formula, vC 2 (t) =

0.06 v in (t) = 66.67e −2t V 0.03 + 0.06

Part 2 Let the 0.03-F capacitor be C2 . The energy stored in the 0.03-F capacitor over the interval [0,t] is given by

[

]

[

]

1 WC2 [0,t] = C2 vC 22 (t) − v C 22 (0) = 66.67 e −4t − 1 J 2

SOLUTION 7.53. First consider the circuit below which contains the additional label of vin (t).

From KCL and the derivative definition of the capacitor iin (t) = iC1 (t) + iC 2 (t) = C1

dv in (t) dv (t) dv (t) + C2 in = (C1 + C2 ) in dt dt dt

Equivalently, dvin (t) 1 = i (t) dt (C1 + C2 ) in It follows that iCk (t) = Ck which is the required current division formula.

dv in (t) Ck = i (t) dt C1 + C2 in

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SOLUTION 7.54. Part 1 By the current division formula, iC1(t) =

0.08 i (t) = 80e −2t A 0.08 + 0.03 0.06 in

iC 2 (t) =

0.03 0.06 i (t) = 20e −2t A 0.08 + 0.03 0.06 in

Part 2 Let the 0.03-F capacitor be C2.

t

1 vC 2 (t) = v C2 (0) + iC 2 ( )d = 333.3(1− e −2t ) V ∫ C2 0

The energy stored in the 0.03-F capacitor over the interval [0,t] is given by

[

]

1 WC2 (0,t) = C2 vC 22 (t) − v C 22 (0) = 1666.5(1− e −2t ) 2 J 2 SOLUTION 7.55. Part 1 By the current division formula, 2 iC 2 (t) = i (t) = 20sin(250t) mA 1+2 s Part 2 By the voltage division formula, vout (t) =

2 × 9ic2 (t) = 0.12sin(250t) V 1+ 2

Part 3 The current in the 2mH inductor is given by t

1 iL 2 (t) = i L2 (0) + v ( )d = −0.24[cos(250t) − 1] A L2 ∫ out 0 The energy stored in the 2-mH inductor is given by 1 WL2 (t) = L2iL 22 (t) = 57.6[cos(250t) − 1]2 J 2

SOLUTION 7.56. Part 1 The dual network N* is shown below.

Part 2 Since the equivalent conductance seen by Is* is 500S, the equivalent resistance seen by Is* is equal to 0.002Ω.

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SOLUTION 7.57. Part 1 The dual network N* is shown below.

Part 2 S*eq = 350S I*out = 8A.

SOLUTION 7.58. First we draw the graph, given below, of the circuit in figure P7.58. We construct the graph associated with dual network graph for N* from the graph of N. The dual network graph is given by the dashed lines.

This circuit and its dual have the branch characteristics given in the following table: ORIGINAL NETWORK V1 = 40 V V2 = 500 I2 I3 = 2 I2 V4 = 400I4 V5 = 200 I5

DUAL NETWORK * I1 = 40 A * * I2 = 500 V2 * * V3 = 2 V2 * * I4 = 400V4 * * I5 = 200 V5

Finally replace the branches by the elements given in the table above. This produces the dual network below.

SOLUTION 7.59. The dual for the circuit in figure P7.59 is shown below.

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SOLUTION 7.60. First we draw the graph, given below, of the circuit in figure P7.60.

This circuit and its dual have the branch characteristics given in the following table: ORIGINAL NETWORK V1 = 5 I1

DUAL NETWORK I1* = 5V1*

I2 = 8 A

V2* = 8 V

I3 = 6 A

V3* = 6 V

V4 = 4 V

I4* = 4 A

V5 = 2 I5

I5* = 2V5*

I6 = 7 A

V6* = 7 V

Now we construct the graph associated with dual network graph for N* from the graph of N. The dual network graph is given by the dashed lines.

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The graph of the dual network then is pulled out and flipped vertically to produce the graph topology of the dual network, N*.

Finally replace the branches by the elements given in the table above. This produces the dual network below.

SOLUTION 7.61. Part 1 We first redraw the circuit to eliminate branch crossing. schematic is shown below.

The resultant

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Then we draw the graph, given below, of the above circuit. We construct the graph associated with dual network graph for N* from the graph of N. The dual network graph is given by the dashed lines.

The resultant dual network given by the above graph is shown below.

Part 2 The equivalent resistance Req * seen by the current source in the dual circuit is equal to 1/Req = 1/α Ω.

SOLUTION 7.62. Part 1 We first redraw the circuit to eliminate branch crossing. schematic is shown below.

The resultant

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Then we draw the graph, given below, of the above circuit. We construct the graph associated with dual network graph for N* from the graph of N. The dual network graph is given by the dashed lines.

The resultant dual network given by the above graph is shown below.

Part 2 If we label the conductances in the dual network by their corresponding resistance values, we have the circuit below.

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By comparing to the original network, we conclude that Req * = Req . Part 3 Since the general relationship between Req and Req * is given by Req = 1/Req *. We can write Req = 1/Req * =1/R eq Solving the above equation, we know that Req = 1 Ω. Part 4 To retain the special properties of parts (b) and (c), we require one resistance value must be a reciprocal of the other one.

SOLUTION 7.63. The dual network is shown below.


From the answer of problem 7.55, iC2 (t) = 20 sin (250t) mA, and vout(t) = 0.12 sin (250t) V * Therefore, vC2 *(t) = 20 sin (250t) mV and iout (t) = 0.12 sin (250t) A

The energy stored in the 2mF capacitor is given by

[

]

2 1 1 2 * WC2 (t) = C2* v *out (t) = L2[iL 2 (t)] = 57.6[cos(250t) − 1]2 J 2 2


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The equivalent inductance of the dual circuit (Leq *) = (1+2) || 1.5 + 2.5 = 3.5µH.

SOLUTION 7.66. By voltage division 1 0.4

0.2 1 v1(t) = 1 1 v s (t) = 0.2 + 0.4 v s (t) = 3 vs (t) + 0.4 0.2 Note that the 0.4 F resistor in parallel with the dependent voltage source has no effect on i1 (t) and hence is redundant. By the definition of a capacitor, i1 (t) = 0.2

dv1(t) 1 dv s(t) = × dt 15 dt

Finally, since the last capacitor is initially uncharged, t

t

0

0

1 2 dv ( ) 2 vout (t) = i1 ( )d = ∫ s d = v s (t) ∫ 0.5 15 d 15 A sketch is given below.

SOLUTION 7.67. Following the method outlined in page 269 of the text, we require 1 vC (t0 ) − C

1/200

∫

i 0( )d ≥ 14

0

With i0 (τ) = 2A, vC(0) = 20 V, it follows that 1 1 × 2A × s C 200 C ≥ 1.667mF Therefore, the standard capacitor value of 1.8mF should be chosen for this application. 20 − 14 ≥

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SOLUTION 7.68. From the relationships given in Problem 7.67, we have  1 d d t  v L (t) = [ L(t)i L (t)] = I 0 [ L(t)] = I 0L1  4 − 2   dt dt   T1 T1   Below is a plot of vL(t) vs t.

From the plot, we notice that when t=T1 , the value of vL(t) changes from positive to negative. This means that vL(t) will change sign when a car is inside the loop. Therefore, we can make a circuit to monitor the voltage vL(t) and whenever a negative voltage is detected, the left turn signal should be initiated during the next traffic light change.

1st Order Circuit Probs 11/26/01

P8-1


SOLUTIONS PROBLEMS CHAPTER 8 SOLUTION 8.1. (a) By KCL, C

dvC (t) v (t) dv (t) v (t) =− C or C + C = 0 . Using 8.12 dt R dt RC

vC (t) = vC (0)e −t / = 10e −t V where

= RC = 1s . Plotting this from 0 to 5 sec

»t = 0:.05:5; »vc = 10*exp(-t); »plot(t,vc) »grid »xlabel('time in s') »ylabel('vc(t) in V') 10

vc(t) in V

8

6

4

TextEnd

2

0

0

1

2

3

4

5

time in s

(b) The solution has the same general form as for (a),iC (t) = iC (0+)e −t / = − −vC (t) . R

10 e −t = −2e −t mA = 3 5 × 10


P8-2


0

ic(t) in mA

-0.5

-1 TextEnd -1.5

-2

0

1

2

3

4

5

time in s

(c) By linearity, if vC(0) is cut in half, all resulting responses are cut in half. If vC(0) is doubled, then all resulting responses are doubled. Alternately, one view this as a simple change of the initial condition with the same conclusion reached from linearity.

SOLUTION 8.2. (a) From inspection of the general form, 8.12, 0.1/ = 1= 0.1/ RC ⇒ C = 0.1/R = 5 µF. (b) Since τ = RC = 0.1, vC (t) = 10e −10t V. SOLUTION 8.3. (a) The general solution form is vC (t) = vC (0)e −t / = v C (0)e

−t /Req C

. Using the

v (0.001) 18.394 v (0)e −(0.001)/ given data, take the following ratio, C = = 2.7183 = C = e 0.001/ . Hence, −(0.002)/ vC (0.002) 6.7668 v C (0)e »K = 18.394/6.7668 K =2.7183 »tau = 1e-3/log(K) tau =0.0010 »C = 5e-6; »Req = tau/C Req =200.0008 »% Req = R*4e3/(R+4e3)


P8-3


»R = Req*4e3/(4e3-Req) R =210.5272 »vC0 = 6.7668/exp(-0.002/tau) vC0= 49.9999 V (b) » % vC (t) = 50e −1000t V »t = 0:tau/100:5*tau; »vc = vC0*exp(-t/tau); »plot(t,vc) »grid »xlabel('Time in ms') »ylabel('vC(t) in V')

SOLUTION 8.4. After one time constant the stored voltage, 8 V, decays to 8/e = 2.943 V. From the graph, the time at which the output voltage is 2.94 V is approximately 0.19 s. Thus

= 0.19 s, and R =

τ/C = 190 Ω. SOLUTION 8.5. (a) The general form of the inductor current is iL (t) = iL (0)e −t / = 0.15e −t / A where = L / R = 2 × 10 −3 s. Plotting iL (t) = 0.15e −500t A from 0 to 10 msec yields: »t = 0:.01e-3:10e-3; »iL = 0.15*exp(-t/2e-3); »plot(t,iL) »grid


P8-4


»xlabel('Time in s') »ylabel('Inductor Current') 0.16 0.14 Inductor Current

0.12 0.1 0.08 0.06 TextEnd

0.04 0.02 0

0

0.001 0.002

0.003 0.004

0.005 0.006 0.007 Time in s

0.008 0.009

0.01

0.005 0.006 0.007 Time in s

0.008 0.009

0.01

(b) Here v L (t) = −RiL (t) = −22.5e −500t V. 0

Inductor Voltage in V

-5 -10 -15 TextEnd

-20 -25

0

0.001 0.002

0.003 0.004

(c) Using linearity, for iL(0) = 50 mA, then v L (t) = mA, then v L (t) = −22.5

SOLUTION 8.6. Since

−22.5 −500t e = −7.5e −500t V and for iL(0) = 250 3

250 −500t e = −37.5e −500t V. 150

= L / R, we can solve for L = 5 mH. Then solving iL (t) = iL (0)e −t / for iL(0)

at t = 4 µsec yields iL(0) =15 mA.


P8-5

SOLUTION 8.7. (a) We desire to solve iL (t) = iL (0)e −t / i (0.05 ms) 9.197 Using the following ratio, L = = iL (0.15 ms) 1.2447


for iL(0) and R in

= 0.08 /(R + 103 ).

iL (0)e −(0.05m)/ 0.1×10 −3 / = e = 7.3889. Hence iL (0)e −(0.15m)/

»K = 9.197/1.2447 K =7.3889 »tau = 0.1e-3/log(K) tau =5.0000e-05 »L = 0.08; »Req = L/tau Req =1.6000e+03 »R = Req - 1000 R = 599.9862 »iL0 = 9.197e-3/exp(-0.05e-3/tau) iL0 = 0.0250 A (b)

= 80m /(R + 1000) = 50 sec , iL (t) = 0.025e −t /50×10

−6

A.

SOLUTION 8.8. By Ohm’s law, v R (0+) = −(4k ||16k)iL (0+) = −32 V. The time constant = L /(4 k ||6 k) = 25 sec , i.e., »Req = (4e3*16e3/(4e3+16e3)) Req =3200 »L = 0.08;


P8-6


»tau = L/Req tau =2.5000e-05 Using the general equation, v R (t) = v R (0+)e −t / = −32e −t /25 V. Equivalently, 4 v R (t) = − Req iL (t) = −Req 0.01e −4×10 t V.

SOLUTION 8.9. (a) Note that the Thevenin resistance seen by the capacitor is Rth: »R1 = 360+60*120/(60+120) R1 =400 »Rth = 400*1200/1600 Rth =300 Hence, vC (t) = vC (0)e −t / = 80e −t /0.15 V where

= RTH C = 300 × 0.5 × 10−3 = 0.15 s.

80

Capacitor Voltage (V)

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(b) Here iC (0+) = −v C (0 + ) / RTH = −0.2667 . Therefore iC (t) = iC (0+)e −t / = −0.2667e −t /0.15 A. −v (t) Equivalently, iC (t) = C = −0.2667e −t /0.15 A. Rth  60||120  (c) By voltage division, v R (0+) = vC (0) = 8 V; for t > 0 v R (t) = 8e −t /0.15 V.   (60||120) + 160 + 200 


P8-7


SOLUTION 8.10. First, find the Thevenin equivalent seen at the left of the inductor. Introducing a test source in place of the inductor we obtain the following KCL equation at that node.  v   itest = v test /1k +  vtest − 200 test   200 . Let vtest = 1 V. Then 1k   »itest = 1e-3 + (1 - 200*1e-3)/200 itest =0.0050 »Rth = 1/itest Rth =200 Thus Rth = 200 Ω,

= L / Rth = 0.25 ms, and iL (t) = 0.025e −4000t A. Next from 8.13b find

v L (t) = −(RTH × iL (0))e −4000t = −5e −4000t V, and from Ohm’s law i x (t) = −5e −4000t mA.

SOLUTION 8.11. For all parts it is necessary to find the Thevenin equivalent resistance seen by the capacitor. To this end we apply an external test current to the remainder of the circuit to obtain: v test = R1i test + R2 (itest + itest ) Thus Rth = v test / itest = R1 + R2 (1+ ) = 120 + 70(1+ ) (a) With α = 4, Rth = 470 Ω,

= RthC = 0.1175 s, and vC (t) = vC (0)e −t / = 50e −8.51t V.

50


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(b) With α = -4, Rth = −90 Ω,

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= RthC = −0.0225 s, and vC (t) = 50e 44.44t V.

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Note how this is not a stable design as V increases exponentially without bound. (c) From the general equation developed at the beginning, Rth = 120 + 70(1+ ) > 0 requires that > −2.7143.

SOLUTION 8.12. Find the Thevenin resistance left of the inductor. Forcing a test current source into the output node, v test = R1i test + R2 (itest − R1itest ) = (100 + 50(1− 100))itest = (150 − 5000 )i test and Rth = v test / itest = 150 − 5000 (a) Using the above equation, Rth = −350 Ω and

Ω

= L / Rth = −7.1429 × 10-5 s. Hence,

iL (t) = 0.1e14000t A, an unbounded response due to the presence of the negative equivalent resistance. = L / Rth = 2.5 × 10−4 s, v R (0+) = −R2 (1− R1)i L (0) = 0 , but more importantly v R2 (t) = −R2 (1− R1)i L (t) = 0 × iL (t) = 0 .

(b) Rth = 100 Ω,

(c) R1 + R2 (1− R1) = (150 − 5000) > 0 implies < 0.03.

SOLUTION 8.13. Over a long period of time the inductor L, is seen as a short circuit. Thus at time 0-, the current through the inductor is, by current division, Is /2. As such, iL (t) = 0.5I s e −Rt /2L A. A sketch will reveal an exponentially decreasing current from an initial 0.5Is A.


P8-9


SOLUTION 8.14. This is similar to problem 8.13. Here the current turns off at time zero instead of a switch opening. By current division iL (0 + ) = Is /2 . The difference between this problem and problem 8.13 is that the Thevenin resistance seen by the inductor is different. Here, RTH = 2R ||0.5 R = 0.4R . So for t > 0, iL = (Is /2)e −R th t /L = 0.5I se −Rt /2.5L A. A sketch of this function plotted with respect to this new time constant will be identical to the one in problem 8.13. Define the time constant of problem 8.13 as τold . The slower decay in the plot below represents the fall of the inductor current for problem 8.14 relative to that of problem 8.13 which is the faster decaying curve in the plot below.


P8-10


SOLUTION 8.15. Over a long period of constant applied voltage, a capacitor looks like an open circuit. By voltage division, vC (0+ ) =

3 V and τ = 3RC. Hence 4 s vC (t) = vC (0+)e −t / = 0.75Vse −t /(3RC ) V

SOLUTION 8.16. Same problem as 8.15 for t < 0. For t > 0 the effective resistance changes. Rth = 3R / /R = 0.75R and τ = 0.75RC. Thus, vC (t) = 0.75Vse −t /(0.75RC) V. Same behavior as in the previous problem, except for a faster decay than in problem 8.15 due to a smaller effective resistance. Note how the decreased resistance affects the RC and RL circuit differently.

SOLUTION 8.17. For t < 0 the inductor looks like a short circuit. Let R1 = 1333/ /800 = 500 Ω. The 12 current supplied by the source is Is = = 0.02 A. By current division, 100 + 500 iL (0 + ) = 0.02 ⋅

1333 = 0.0125 A 800 + 1333

For t > 0, the switch is opened and the inductor sees only the 800 Ω resistor. Hence, and

= L / R = 25 µsec


P8-11


iL (t) = 12.5e −40000t mA 14

Inductor current in mA

12 10 8 6 4 TextEnd 2 0

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20

40 60 Time in micro seconds

80

100

SOLUTION 8.18. (a) For t < 0 the applied voltage is constant and at t = 0, the capacitor is like an open circuit. By voltage division, vC (0+ ) = 30

25k ||6.25k = 25 V. For 0 < t < 1 ms, the source is off (25k ||6.25k) + 1k

and the capacitor discharges through three resistors in parallel; thus = RthC = 833.33 × 0.6 × 10−6 = 0.5 ms and vC (t) = 25e −2000t V. (b) From continuity vC (0.001)= 25e −2000(0.001) = 3.383. For t > 1ms, the capacitor keeps on discharging through only one resistance, the 25 kΩ resistor; thus the new time constant is τnew = 15 ms, and vC (t) = 3.383e −(t −0.001)/0.015 V. (c) PlotvC (t) = 25e −2000t[ u(t) − u(t − 0.001)] + 3.383e −(t −0.001)/0.015 u(t − 0.001) V »t = 0:.01:12; »vc = 25*exp(-2*t) .*(u(t)-u(t-1))+3.3834*exp(-(t-1)/15) .*u(t-1); »plot(t,vc) »grid »xlabel('Time in milli-secs') »ylabel('Capacitor Voltage in V')


P8-12


25

Capacitor Voltage in V

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SOLUTION 8.19. (a) From Ohm’s law iL (0 + ) =

8

10

12

54 = 0.6 A. For t > 0, the Thevenin resistance 60 + 30

seen by the inductor is Rth = (60 + 30) || 720 = 80 Ω and τ = L/Rth = 1/160 s. Thus iL (t) = iL (0+ )e −t / = 0.6e −160t A. From Ohm’s law and current division v(t) = −60 ×

720 × i L (t) = −32e −160t V 90 + 720

(b) From continuity property, iL (0.01±) = 0.6 −160(0.01) = 121.14 mA. For t > 10 ms, the Thevenin resistance seen by the inductor is Rth = (690 + 30) || 720 = 360 Ω and τnew = L/Rth = 1/720 s. Hence, iL (t) = 121.14e −720(t −0.01) mA for t > 10 ms. From Ohm’s law and current division v(t) = −690 ×

720 × iL (t) = −41.793e −720(t −0.01)u(t − 0.01) V 720 + 720

Therefore iL (t) = 0.6e −160t [u(t) − u(t − 0.01)] + 0.12114e −720(t −0.01)u(t − 0.01) A


P8-13


0.7

Inductor Current in A

0.6 0.5 0.4 0.3 0.2

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Time in milli-seconds

SOLUTION 8.20. For both circuits we first compute the Thevenin resistance seen to the right of the capacitor for 0 ≤ t ≤ 60 ms. If we excite the circuit to the right of the capacitor over this time interval, v (1− 0.25)v test 1 then itest = test + = 0.0125v test . Let Rth1 = = 80 Ω. 200 100 0.0125 (a) For t < 0, the capacitor acts as an open circuit. Using voltage division, vC (0+ ) =

80 80 = 30 80 + 133.3

V. For 0 ≤ t ≤ 60 ms, the time constant is 1 = Rth1C = 40 ms, and vC (t) = vC (0+ )e −t

= 30e −t (Rth1C ) = 30e −25t V

From continuity, vC (60+ ms) = 6.694 V. The new Thevenin resistance is Rth2 = 200 Ω. Thus for t > 60 ms, the time constant is 2 = 200C = 100 ms, and vC (t) = 6.694e −10(t −0.06) V. The resulting capacitor voltage is plotted below.


P8-14


30


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250

time in ms

(b) It is the same circuit as above for t < 0; thus vC (0+ ) = 30 V. However the Thevenin resistances seen by the capacitor are different because there is no switch to disconnect the independent voltage source and its series resistance. First for 0 ≤ t ≤ 60 ms, the Thevenin resistance to the right remains as 1 = 80 Ω. However, for 0 ≤ t ≤ 60 ms, the Thevenin resistance seen by the capacitor 0.0125 changes to Rth3 = Rth1 //133.3 = 50 Ω. Then new time constant is 3 = Rth 3C = 25 ms and for 0 ≤ t ≤ Rth1 =

60 ms vC (t) = 30e −40t V From continuity, vC (60+ ms) = 2.72 V. The new Thevenin resistance is Rth4 = 200/ /133.3 = 80 Ω. Thus for t > 60 ms, the time constant is 4 = 80C = 40 ms, and vC (t) = 2.72e −25(t −0.06) V. The resulting capacitor voltage is plotted below.


P8-15


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time in ms

(c) For t < 60 ms, the voltage decays faster in (b) due to the smaller time constant. Similarly, for t > 60 ms. SOLUTION 8.21. Following, are the switching times with the time constants associated with them. t = 0 ⇒ Rth = 20 kΩ ⇒ = 20 ms t = 5 ms ⇒ Rth = 4 k Ω ⇒ = 4 ms t = 7.5 ms ⇒ Rth = 800 Ω ⇒ = 0.8 ms It follows that with t in ms, vC (t) = 10e −0.050t [u(t) − u(t − 5)] + 7.788e −0.250( t −5)[ u(t − 5) − u(t − 7.5)] + 4.169e −1.25(t −7.5)u(t − 7.5) V


P8-16


10 9


8 7 6 5 4 TextEnd

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6 time in ms

8

*S OLUTION TO 8.22. This solution is done in MATLAB. % Define switching times, inductance, and Thevenin equivalent resistances. t1= 12e-6; t2=18e-6; t3=21e-6; L= 0.1; rth1= 800; rth2= 8000; rth3=1600; rth4= 32000; % Define time constants for each of the four time intervals. tau1= L/rth1 tau2= L/rth2 tau3= L/rth3 tau4= L/rth4 tau1 = tau2 = tau3 = tau4 =

1.2500e-04 1.2500e-05 6.2500e-05 3.1250e-06

% Compute initial inductor currents for each of the four time intervals. il1= 100e-3; il2=il1*exp(-t1/tau1)

10

12


P8-17


il3=il2*exp(-(t2-t1)/tau2) il4=il3*exp(-(t3-t2)/tau3) il2 = 9.0846e-02 il3 = 5.6214e-02 il4 = 5.3580e-02 % Determine inductor currents for each of the four time intervals. Plot. t = 0:0.5e-7:36e-6; seg1= il1*exp(-t/tau1) .*(ustep(t)-ustep(t-t1)); seg2 =il2*exp(-(t-t1)/tau2) .*(ustep(t-t1)-ustep(t-t2)); seg3=il3*exp(-(t-t2)/tau3) .*(ustep(t-t2)-ustep(t-t3)); seg4= il4*exp(-(t-t3)/tau4) .* ustep(t-t3); iL=seg1 + seg2 +seg3 + seg4; plot(t,iL) grid

SOLUTION 8.23. For circuits with a forced voltage, equation 8.19c is used as a general solution, vC (t) = vC (∞) +

[ ( ) − v (∞)]e vC to+

C

 t  −   RTH C 

.

(a) At time zero the voltage is 0 V. As time approaches infinity, the capacitor looks like an open with voltage 10 V. The Thevenin resistance is 10 kΩ. Thus for t > 0.


P8-18

vC (t) = 10 + [−10]e +

(b) With vin(t) = 0 and vC (0 ) = 5 V, vC (t) = 5e


 t −   2

 t −   2

= 10(1− e −0.5t ) V. = 5e −0.5t V.

Part (a)

Part (b)

(c) From linearity, vC (t) = 10(1− e −0.5t ) + 5e −0.5t = 10 − 5e −0.5t V. Using Ohm’s law, v (t) v (t) iC (t) = in 4 − C 4 . Thus iC (t) = 1− 1− 0.5e −0.5t = 0.5e −0.5t mA. 10 10

(

)

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Capacitor Current (mA)

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time in s

(d) This is the same as (a), under the condition that the input is 1.5 times larger. Hence by linearity, vC (t) = 1.5 × 10(1− e −0.5t ) = 15(1− e −0.5t ) V. (e) By linearity, ANSWER = –2x(ANSWER to (b)) + 3x(ANSWER to (a)):


P8-19


vC (t) = −2 × 5e −0.5t + 3× 10(1− e −0.5t ) = 15(1− e −0.5t ) = 30 − 40e −0.5t V SOLUTION 8.24. (a) At t = 0-, the capacitor looks like an open circuit; therefore, by voltage division and the continuity 3R 3R property, vC (0− ) = vC (0+ ) = Vs1 = 0.75Vs1 . Similarly, at t = ∞, vC (∞) = V = 0.75Vs2 . The 4R 4R s2 circuit time constant is = (3R / /R)C = 0.75RC . Hence vC (t) = 0.75Vs 2 + 0.75[Vs1 − Vs 2 ]e

t   −   0.75RC 

.

(b) A sketch will show an exponentially varying voltage from 0.75Vs1 converging to 0.75Vs2 with the computed time constant. (c) The response to the initial condition when the inputs are set to zero, zero-input response, is vC (t) = Vs1e

t   −   0.75RC 

. The zero order response, the response with 0V initial condition to a forced

voltage, is vC (t) = Vs 2 − Vs 2e

t   −   0.75RC 

.

SOLUTION 8.25. For RL circuits with a forced current, equation 8.19b is used as a general solution: iL (t) = iL (∞) +

[ ( ) − i (∞)]e iL to+

Since Rth = R = 100 Ω and L = 0.4 H, we have

L

−

R th (t −t o) L .

= 4 ms and

[( )

]

iL (t) = iL (∞) + iL to+ − iL (∞) e −250(t −to ) (a) Here, iL (0) = 0 and as time approach infinity, the inductor becomes a short and iL (∞) = 10 100 = 0.1 A. Thus iL (t) = 0.11 − e −250t A.

(

)

(b) Here iL (0) = −50 mA and because the input is zero, iL (∞) = 0 . Thus, iL (t) = −0.05e −250t A. Plots for parts (a) and (b) appear below.


P8-20


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(

10 Time in ms

15

20

)

(c) By linearity iL (t) = 0.11 − e −250t − 0.05e −250t = 0.1− 0.15e −250t A. Further, by KVL and Ohm's law, v L (t) = vin (t) − 100iL (t) implies v L (t) = 10 − 10 + 15e −250t = 15e −250t V.

Inductor Voltage (V)

15

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10 Time in ms

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(d) Observe that the new initial condition is –0.5 times the old one and that the new input voltage is 1.5 times the old one. Therefore, by linearity,

(

)

(

)

iL (t) = 1.5 × 0.11 − e −250t + (−0.5) × −0.05e −250t = 0.15 − 0.125e −250t and thus

v L (t) = 15 − 15 + 12.5e −250t = 12.5e −250t V

The plot is similar to part (c) with initial point 12.5 instead of 15. SOLUTION 8.26. For this problem Rth = 2R / /0.5R = 0.4R in which case

= L Rth = L 0.4R .


P8-21


(a) At t = 0, the inductor looks like a short circuit. Hence by current division, iL (0 − ) = iL (0+ ) = 0.5Is1 . A similar argument yields iL (∞) = 0.5Is 2 . Using the general form of the solution, −

0.4Rt L

iL (t) = 0.5I s2 + 0.5[ Is1 − Is 2 ]e . (b) A sketch will show an exponentially varying current from 0.5Is1 A converging to 0.5Is2 . (c) The response to the initial condition when the inputs are set to zero, zero-input response, is iL (t) = 0.5I s1e

−

0.4Rt L

. The zero state response, the response with no initial condition, to the input 0.4Rt   − Is2 u(t), is iL (t) = I s2  1− e L  .    

(

SOLUTION 8.27. For this problem, vC (0) = 20 − 10e −0.4t

(

)

)t =0 = 10 V and

vC (∞) = 10Is = 20 = lim 20 − 10e −0.4t . Hence Is = 2 A. Further, t→∞ −0.4t

dvC (t) = 4Ce = 0.4e −0.4t which implies that C = 0.1 F. Since dt = 1/0.4 = 2.5 = (10+ R)C = 0.1(10 + R) , it follows that R = 15 Ω.

iC (t) = C

SOLUTION 8.28. (a) The Thevenin resistance for this configuration is Rth = 1000/ /1000 = 500 Ω and Hence vC (t) = vC (0+ )e −t = 15e −4t V is the zero-input response.

= RthC = 0.25 s.

(

)

(b) Using a source transformation and voltage division, vC (∞) = 3 V. Thus vC (t) = 3 1− e −4t V.

(

)

(c) Here vC (∞) = 4 V, thus vC (t) = 4 1− e −4t V.

(

(d) This is the superposition of parts (b) and (c), i.e., vC (t) = 7 1− e −4t

)

(e) The complete response is the superposition of parts (d) and (a), i.e., vC (t) = 7 + 8e −4t V. (f) From linearity, vC (t) = 0.5 × 7 1− e −4t + 2 × 15e −4t = 3.5 + 26.5e −4t

(

)

SOLUTION 8.29. Solution done in MATLAB %Problem 8.29 %RTH= (60||120)+120 %tau=L/RTH RTH=1/(1/60+1/120)+120; tau=0.2/RTH; %Using superposition at t<0 il01=36/120/2; il02=(72/180)*(60/180); il0= il01+il02;


P8-22


% At t>0 36volts is off thus ilinf=il02; t=0:5*tau/1000:5*tau; ilt=ilinf+(il0-ilinf)*exp(-t/tau); plot(t,1000*ilt); xlabel('Time in seconds'); ylabel('Current in mA'); 300 280 260

Current in mA

240 220

200 180 160 140 120

0

1

2

3 4 Time in seconds

*S OLUTION 8.30. » % Rth = 120 + 120\\60 = 160 kohm » % tau = Rth*C »tau = 160e3*0.5e-6 tau = 0.0800 » % Initial capacitor voltage is computed by » % voltage division and superposition »vc0 = 24+24 vc0 = 48 » % At t = ∞, capacitor looks like an open circuit. Hence »vcinf=24; »t = 0:1e-3:5*tau; »vct = vcinf+(vc0-vcinf)*exp(-t/tau); »plot(t*1000,vct) »grid »xlabel('Time in msec') »ylabel('Capacitor voltage (V)')

5

6

7 x 10

-3


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Capacitor voltage (V)

45 40 35 30 25 20

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100

150

200 250 Time in msec

300

*S OLUTION 8.31A »vc0 =0; »% Consider 0 ≤ t ≤ 0.5 »vcinf = 50; »% Rth = 600\\300 = 200 ohms »% tau1 = Rth*C »tau1 = 300e-6*200 »vc0 = 0; »t = 0:5e-3:1; »vct = (vcinf+(vc0-vcinf)*exp(-t/tau1)) .* (ustep(t)-ustep(t-.5)); » »% Consider 0.5 ≤ t ≤ 1 »tau2 = tau1 »vc5 = (vcinf+(vc0-vcinf)*exp(-.5/tau1)) »vcinf2 = 80; »vct2 = (vcinf2+(vc5-vcinf2)*exp(-(t-0.5)/tau1)) .* ustep(t-0.5); »vca = vct+vct2; »plot(t,vca) »grid »xlabel('Time in sec') »ylabel('Capacitor Voltage (V)')

350

400


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80


70 60 50 40 30 20 10 0

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Time in sec

0.6

*S OLUTION 8.31B »vc0 =0; »% for 0 ≤ t ≤ 0.5 »vcinf = 75; »tau1 = 300e-6*300 »vc0 = 0; »t = 0:5e-3:1; »vct = (vcinf+(vc0-vcinf)*exp(-t/tau1)) .* (ustep(t)-ustep(t-.5)); »% for 0.5 ≤ t ≤ 1 »tau2 = 300e-6*200 »vc5 = (vcinf+(vc0-vcinf)*exp(-.5/tau1)) »vcinf2 = 80; »vct2 = (vcinf2+(vc5-vcinf2)*exp(-(t-0.5)/tau1)) .* ustep(t-0.5); »vcb = vct+vct2; »plot(t,vcb) »grid »xlabel('Time in sec') »ylabel('Capacitor Voltage (V)') »pause »plot(t,vca,t,vcb,'b') »grid

0.8

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80


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1

0.4

0.8

1

Comparison of the two responses. 80 70 60 50 40 30 20 10 0

0

SOLUTION 8.32. %Problem 8.32 %Consider t < 0 %vin=-20V, thus vc0=(8/10)*(-20); %For 0
0.2

0.6


P8-26


vcinf=(8/10)*20; t=20e-3; vc20ms=vcinf+(vc0-vcinf)*exp(-t/tau1); %For 20ms
15

10

Volts

5

0

-5

-10

-15

-20

0

0.005

0.01

SOLUTION 8.33. (a) %Problem 8.33 %(a) %at t<0 only one source is contributing thus, il0=24/60*0.5; %For t>0 RTH=60+1/(1/30+1/60); tau=16e-3/RTH; %As t goes to infinity, by superposition, ilinf=24/(60)*0.5+24/80*1/3; t=0:5*tau/1000:5*tau; ilt=ilinf+(il0-ilinf).*exp(-t/tau); figure(1); plot(t,1000.*ilt); grid; xlabel('time in seconds'); ylabel('Current in mA');

0.015

0.02 0.025 time in seconds

0.03

0.035

0.04


P8-27


300 290 280

Current in mA

270 260 250 240 230 220 210 200

0

0.1

0.2

0.3

0.4 0.5 0.6 time in seconds

0.7

0.8

0.9

1 x 10-3

(b) Using 8.23, The time constant remains the same as in (a). In order to find the inductor voltage, we must do so indirectly by solving for v L (t) = V A − 60IL . At t=0+, iL (0 + ) = 200mA 60 30 VA = 24 + 24 − iL (0+ )(60|| 30) = 20V 90 90 v L (0+ ) = x(t0+ ) = V A − 60iL (0+ ) = 8V

For t > 0, X e = v L (∞) = 0V , Thus v L (t) = 8e −5000tV . (c) By linearity, iL (t) = 600 − 200e −5000t mA v L (t) = 16e −5000tV VA

SOLUTION 8.34. (a) Since the voltage has been constant for a long time, the capacitor acts as an open circuit. Thus by voltage division and continuity, vC (0− ) = vC (0+ ) = 0.75Vo . (b) RTH = 6R ||18R || 3R = 1.8R . (c) For that period of time the switch is closed, vC (t) = 0.9Vo − 0.15Voe −t /1.8RC . (d) Using the previous equation, and by continuity, vC (T − ) = vC (T + ) = 0.9Vo − 0.15Voe −T /1.8RC . (e) The time constant remains the same as the only difference is the source turning off. (f) For t > T, vC (t) = vC (T + )e −(t −T )/1.8RC . (g)


P8-28


0.9 0.8

Normalized Volts in terms of Vo

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

0

2

4

6 8 10 12 Normalized time in terms of tau

14

16

18

SOLUTION 8.35. (a) For t < 0, The switch is closed, the current source off, and the voltage source has been providing a constant voltage for a prolonged period of time. Thus, vC (0− ) = −50uV . (b) Since voltage in continuous across a capacitor, vC (0+ ) = v C (0− ) = −50uV . (c) The thevenin resistance seen by the capacitor is RTH = 200||200 = 100Ω , thus = RTH C = 0.2s . (d) As t goes to infinity the capacitor voltage goes to 16V, thus vC (t) = 16 + (−50u − 16)e −5tV . (e) Again using the continuous property of a capacitor, vC (0.5+ ) = v C (0.5− ) = 14.687V . (f) For t > 0.5s, the switch is open, thus RTH = 200Ω . = RTH C = 0.4s . (g) As t goes to infinity, the capacitor voltage goes to 32V. vC (t) = 32 + (14.687− 32)e −2.5(t −0.5)V . (h) 35

30

25

Volts

20

15

10

5

0

-5

SOLUTION 8.36. (a) vC (0+ ) = −5V (b) Doing so in matlab. %Problem 8.36b %Initial condition vc0=-5; %From 0
0

0.5

1 1.5 Time in seconds

2

2.5


P8-29


%So the initial condition on Vout vout0=(10+5)/300e3*60e3-5; %-2V voutinf1=10; t=80e-6; vout80us= 10+(vout0-10)*exp(-t/tau); %For t > 80us %tau stays the same voutinf2=-5; t=0:160e-6/1000:160e-6; vout= (ustep(t)-ustep(t-80e-6)).*(10+(vout0-10).*exp(-t./tau)) ... +ustep(t-80e-6).*(-5+(vout80us+5).*exp(-(t-80e-6)./tau)); plot(t,vout); xlabel('time in seconds'); ylabel('Volts'); grid; %Notice that Vout is vin-vct times a constant 60/300 plus vct %Thus Vout=vin(60/300)+vct(1-60/300) 5

4

3

Volts

2

1

0

-1

-2

(c)

0

0.2

0.4

0.6

0.8 1 time in seconds

1.2

1.4

1.6 -4 x 10


(d)

SOLUTION 8.37.

P8-30



P8-31


If the source voltage has been –10 V for a long time then the switch is open and vC (0+ ) = −10V . The time constant with the switch open is = 5us . At t > 0, the input voltage changes to 20V. It then follows that vC (∞) = 20V , and vC (t) = 20 − 30e −t /5uV . Using the elapsed time formula, 8.24, we wish to find when the switch closes.  −10 − 20  ta − 0 = 5us ⋅ln = 2.03us . At that time the input voltage is still 20V, and the switch closes. 0 − 20  The time constant is now = 98ns , and vC (∞) = 0.39V . Note that because the voltage converges to a value greater than zero, this time interval will be from 2.03us to 5us when the input changes back to –10V, thus vC (t) = 0.39 − 0.39e −t /98nV . vC (5us) = 0.39V . At t > 5us, the voltage changes to –10 V, so vC (t) = −0.2 + (0.39 + 0.2)e −t /98nV . Using the elapsed formula, we get tb = 0.1us for the voltage to go down to 0V again and cause the switch to open again. At this point the time constant becomes the original value again and vC (t) = −10 + 10e −t /5uV for t > (0.1+5) us. SOLUTION 8.38. (a) Introduce a test current source at the output and write KVL,  3  itest = v test /200 + (v test − 6V ) /200 + (6V − v test ) / 400. Solving for itest = v test  − 0.015 . This 400 implies the following, RTH = 400 /3Ω iSC = 15mA vOC = RTH iSC = 2V (b) Using the general form, vC (t) = 2 − 8e −15000t . SOLUTION 8.39. (a) Introduce a test voltage and solve for KVL, v test = 5kitest + 101(v test + 1− 40itest ) − 1 + 40itest v test = −10itest − 1 vOC = −1V RTH = −10Ω (b) The complete response is vC (t) = −1 + e t V. Note that the voltage goes to infinity as t goes to infinity because of the negative time constant. SOLUTION 8.40. Compute the thevenin equivalent seen by the inductor at t > 0. Using KCL write, itest = ⋅100⋅ itest + (v test − 100itest + 25) /50 . Then one obtains the following,


P8-32


itest = v test /125 + 1 / 5 A RTH = 125Ω iSC = −1 / 5 A vOC = −25V A t < 0, the applied voltage has been –25V for a long time. Using the previously obtained thevenin equivalent and linearity, iL (0 − ) = 1/5 A . iL (∞) = i SC = −1/5 A , So iL (t) = 0.2 + 0.4e −6250t A . SOLUTION 8.41. (a) Introducing a test source and using KCL, v − 1.5v s v test itest = test + 40 100 v 1.5v s − v test 0.2 = s + 40 40 v Solving for the test source current in terms of the test voltage, itest = test − 0.12 A. Thus the thevenin 50 equivalent is, Rth = 50Ω voc = i sc * Rth = 6V (b) vC (t) = 6 − 6e −200t V. (c) From linearity all the currents increase by the same ratio, thus vC (t) = 15 − 15e −200t V. (d)

(e)


P8-33


SOLUTION 8.42. %Problem 8.42 C= 1e-6; vc0=0; %For 0 < t < 5ms Rth= 20e3; tau1= Rth*C; vcinf1= 50e-3*Rth; vc5ms= vcinf1+(vc0-vcinf1)*exp(-5e-3/tau1); %For 5ms < t < 7.5ms Rth= 4e3; tau2= Rth*C; vcinf2= 50e-3*Rth; vc75ms= vcinf2+(vc5ms-vcinf2)*exp(-(7.5e-3-5e-3)/tau2); %For t > 7.5ms Rth=800; tau3= Rth*C; vcinf3= 50e-3*Rth; t=0:12e-3/1000:12e-3; vct= (ustep(t)-ustep(t-5e-3)).*(vcinf1+(vc0-vcinf1).*exp(-t/tau1)) ... + (ustep(t-5e-3)-ustep(t-7.5e-3)).*(vcinf2+(vc5ms-vcinf2).*exp(-(t-5e-3)/tau2)) ... + (ustep(t-7.5e-3)).*(vcinf3+(vc75ms-vcinf3).*exp(-(t-7.5e-3)/tau3)); plot(t,vct); grid;


P8-34


xlabel('time in seconds'); ylabel('Volts'); 250

200

Volts

150

100

50

0

0

0.002

0.004

0.006 time in seconds

0.008

0.01

0.012

SOLUTION 8.43. (a) From the thevenin resistance (b) From the thevenin resistance (c) From the thevenin resistance (d) From the thevenin resistance (e) 0 mA. (f) In MATLAB: %Problem 8.43f

RTH = 800Ω , 1 = 125 s . RTH = 8kΩ , 2 = 12.5 s . RTH = 1.6kΩ , 3 = 62.5 s . RTH = 32kΩ, 4 = 3.125 s .

tau1= 125e-6; tau2= 12.5e-6; tau3= 62.5e-6; tau4= 3.125e-6; vs= 100; il0= 0; ilinf1= vs/800; ilinf2= vs/8e3; ilinf3= vs/1.6e3; ilinf4= vs/32e3; il12us= ilinf1+(il0-ilinf1)*exp(-12e-6/tau1); il18us= ilinf2+(il12us-ilinf2)*exp(-(18e-6-12e-6)/tau2); il21us= ilinf3+(il18us-ilinf3)*exp(-(21e-6-18e-6)/tau3); t= 0:36e-6/1000:36e-6; ilt= (ustep(t)-ustep(t-12e-6)).*(ilinf1+(il0-ilinf1).*exp(-t/tau1))+ ... (ustep(t-12e-6)-ustep(t-18e-6)).*(ilinf2+(il12us-ilinf2).*exp(-(t-12e-6)/tau2))+ ... (ustep(t-18e-6)-ustep(t-21e-6)).*(ilinf3+(il18us-ilinf3).*exp(-(t-18e-6)/tau3))+ ... (ustep(t-21e-6)).*(ilinf4+(il21us-ilinf4).*exp(-(t-21e-6)/tau4)); plot(t,1000*ilt); grid; xlabel('time in seconds'); ylabel('Current in mA');


P8-35


15

Current in mA

10

5

0

0

0.5

1

1.5

2 2.5 time in seconds

3

3.5

4 x 10-5

SOLUTION 8.44. The first stage is a differentiator, and from 8.25, the output of the first op-amp is dv (t) = −RC in = 0.25RCe−0.25t . The second stage is an integrator and using 8.26, dt t −1 t vout (t) = (0.25RCe−0.25 d ) = 0.5 e −0.25 = 0.5e −0.25t ∫ 2RC 0 0

[

]

SOLUTION 8.45. (a) First note the following relationships, 1 t vout (t) = vC 2 (t) = i ( )d C2 ∫0 C 2 vin (t) = v C1(t) dv (t) v (t) dv (t) iC 2 (t) = −v in (t) / R − C1 C1 = − in − C1 in dt R dt Doing the appropriate substitution, and solving, 1 t C C vout (t) = − v in ( )d − 1 vin (t) + 1 vin (0). ∫ C2R 0 C2 C2 4 C (b) e −0.25t − 1 + 1 1− e −0.25t V. C2 R C2 −1 C (c) sin( t) + 1 (1− cos( t)) V. C2R C2

(

)

(

)

SOLUTION 8.46. These are two integrator in cascade. Using 8.26, the output of the first stage is 1 t − vin ( )d = 2[cos(50 )]t0 mV = 2cos(50t) − 2 mV. Using the same equation again, ∫ 0 RC t 1 t  2  vout (t) = − (2cos(50 )d = −10 − sin(50 )  mV = 2cos(50t)mV RC ∫0  50 0


P8-36

*S OLUTION 8.47. The following solution is done in MATLAB c= 1e-6; rf= 10e6; rs=1e6; tau=c*rf; vgain= -10/1; vofinal= -3*vgain; % Part (a) voinit= 0; t= 0: tau/100: tau; vout= vofinal + (voinit -vofinal).*exp(-t/tau); % Time at which output voltage reaches saturation is tsat tsat= tau*log((0 - vofinal)/(15- vofinal)) plot (t,vout) grid xlabel('Time in secs') ylabel('Output Voltage in volts') tsat = 6.9315e+00

% Part (b) voinit=-5; vout= vofinal + (voinit -vofinal).*exp(-t/tau); tsat= tau*log((-5 - vofinal)/(15- vofinal)) plot (t,vout) grid xlabel('Time in secs')



P8-37


ylabel('Output Voltage in volts') tsat = 8.4730e+00

-

(c) vC(0 ) = 0. Observe saturation at about 6.4 seconds. Leaky Integrator-Transient-7 +0.000e+000

(V)

+2.000

+4.000

+6.000

+15.000

+10.000

+5.000

+0.000e+000 V(IVM)

-

vC(0 ) = -5 V. Observe that the time of saturation is 8.198 seconds.

Time (s) +8.000

+10.000


P8-38


Prb Sol 8.47-Transient-16 +0.000e+000

(V)

+2.000

+4.000

Time (s) +6.000

+8.000

+10.000

+15.000

+10.000

+5.000

+0.000e+000

-5.000

V(IVM)

Note that in both cases the time of saturation is much lower than in the MATLAB computations which assume an ideal op amp. In the Burr Brown model used by the SPICE simulation of this circuit, the input resistance is 2 MΩ which is comparable with the external input resistance. Hence the assumption of an infinite input resistance is not valid for the SPICE simulation and causes the discrepancy in the time of saturation. However, if the external input resistance is changed to 10 kΩ and the feedback resistance to 100 kΩ with a corresponding change in the capacitor to 100 µF, one obtains results comparable to the MATLAB computations.

SOLUTION 8.48. Since the op-amps do not load the first stage of the circuit, we can find its transfer function for the opv (t) amp stage as out = (1+ K) . v + (t) (a) From the problem statement, we know that the overall function, is a scaled integrator. As the op-amp stage only provides gain it is logical to assume that the R-C stage will perform the integration of the input times some constant, G. With this in mind we have t

vout (t) = G(1+ K)∫ 10sin ( 0

)d =

−10G(1+ K)

(b) Using the same reasoning, vout (t) = G(1+ K)

(cos( t) − 1) , where G must be negative. d(10sin( t)) = 10G(1+ K) cos( t)where G is dt

positive. For low frequency (a) yields a big output, while (b) a small one. For high frequency the reverse happens.


P8-39


SOLUTION 8.49. For (b) the integral i-v relationship is v(t) = vCeq (0 + ) +

1 t C + C2 t i( )d = vC1 (0+ ) + v C 2 (0+ ) + 1 i( )d . Repeating the same for (a), ∫ Ceq 0 C1C2 ∫0

1 t i( )d C1 ∫0 1 t vC 2 (t) = v C2 (0 + ) + i( )d C2 ∫0 vC1 (t) = vC1(0 + ) +

By KVL the two capacitor voltage can be added together, thus give the same relationship as for (b). SOLUTION 8.50. First calculate Ceq = C1 ||C2 = 0.2F . Then find the initial voltage vCeq (0 + ) = vC1 (0+ ) + v C 2 (0+ ) = 30V . The final voltage will be 12V, and the time constant is = Req ⋅Ceq = 0.4s . Thus vout (t) = 12 + (30 − 12)e −2.5tV . *S OLUTION 8.51. (a) After the switch closes, we have the circuit shown below.

From Chapter 7, Ceq = 0.5 F and vC (0+ ) = v C1(0+ ) − v C 2 (0+ ) = 2 − 0 = 2 V. Hence, iR (t) = i R (0+ )e −t / = iR (0+ )e

−t /RC eq

v (0 + ) −t /RCeq = C e = 4e −4t u(t) A R

(b) For this part we apply the integral definition of the capacitor. Specifically,


P8-40


t

t

t

0

0

0

1 vC1 (t) = vC1(0 ) + iC1( ) d = 2 − ∫ iR ( ) d = 2 − 4 ∫ e −4 d = 1+ e −4t V ∫ C1 +

and t

t

t

0

0

0

1 vC 2 (t) = v C2 (0 ) + i ( ) d = 0 + ∫ iR ( )d = 4 ∫ e −4 d = 1− e −4t V C2 ∫ C 2 +

+

(c) The energy stored at t = 0 for each capacitor is: 2 WC1 (0+ ) = 0.5C1vC1 (0 + ) = 2 J

and WC2 (0 + ) = 0.5C2vC2 2 (0+ ) = 0 J Further at t = ∞, 2 WC1 (∞) = 0.5C1vC1 (∞) = 0.5 J

and WC2 (∞) = 0.5C2vC2 2 (∞) = 0.5 J Computing total instantaneous stored energies, we have WCtot (0 + ) = 2 J and WCtot (∞) = 1 J +

Hence the decrease in stored energy from 0 to ∞ is 1 J. +

(d) Computing the energy dissipated in the resistor over [0 , ∞) is ∞

WR (0,∞) = R ∫

0

iR2 (

 vC2 (0+ )  ∞ −2t /RCeq vC2 (0+ ) RCeq −2t /RCeq ∞ )d = R  e d = × × e ∫ R −2 0  R2  0

]

Ceqv C2 (0 + ) = =1J 2 (e) From the expressions developed in part (d), the dissipated energy is independent of the value of R. R only affects the rate at which energy is dissipated. Clearly, the energy stored at 0 is 2 J while the energy dissipated over [0,∞) is 1 J and the remaining energy at t = ∞ is 1 J. Hence conservation of energy is verified.


P8-41


SOLUTION 8.52. (a) Using the relations developed in P8.49, Ceq = 0.2F vCeq = 1− 0.5 = 0.5V vCeq (0+ ) −t /RC eq iR (t) = i R (0 )e = iR (0 )e = e = e −10t u(t) R (b) For this part we apply the integral definition of the capacitor. Specifically, +

−t /

−t /RC eq

+

t

t

t

0

0

0

1 1 vC1 (t) = vC1(0 ) + iC1( ) d = 1− ∫ iR ( ) d = 1− ∫ e −10 d = (9 + e −10t ) V ∫ C1 10 +

and t

t

t

0

0

0

1 1 vC 2 (t) = v C2 (0 ) + iC 2 ( ) d = 0.5 + 4 ∫ iR ( )d = 0.5 + 4∫ e −10 d = (9 − 4e −10t ) V ∫ C2 10 +

+

(c) The energy stored at t = 0 for each capacitor is: 2 WC1 (0+ ) = 0.5C1vC1 (0 + ) = 0.5 J

and WC2 (0 + ) = 0.5C2vC2 2 (0+ ) = 31.25mJ Further at t = ∞, 2 WC1 (∞) = 0.5C1vC1 (∞) = 405mJ

and WC2 (∞) = 0.5C2vC2 2 (∞) = 101.25mJ Computing total instantaneous stored energies, we have WCtot (0 + ) = 531.25mJ

and WCtot (∞) = 506.25mJ

+

Hence the decrease in stored energy from 0 to ∞ is 25 mJ. +

(d) Computing the energy dissipated in the resistor over [0 , ∞) is


∞

WR (0,∞) = R ∫

iR2 (

0

P8-42


2 2  vCeq (0 + ) ∞ −2t /RCeq v Ceq (0+ ) RCeq −2t /RCeq ∞ )d = R  e d = × × e  2  ∫ R −2 0  R 0

]

2 Ceqv Ceq (0+ ) = = 25mJ 2 (e) From the expressions developed in part (d), the dissipated energy is independent of the value of R. R only affects the rate at which energy is dissipated. Clearly, the energy stored at 0 is 531.25 mJ while the energy dissipated over [0,∞) is 506.25 mJ and the remaining energy at t = ∞ is 25 mJ. Hence conservation of energy is verified.

SOLUTION 8.53. As all the switches are open initially, the initial current through the inductors is 0A. For 0 < t < 50 ms, iL (t) = 54.54 − 54.54e −20000t mV. At t > 50 ms, the equivalent inductance is 10 mH, the initial current through the 110 mH inductance is 54.54 mA, and through the 11 mH inductance 0 A. So assuming the current splits equally between the two branches in steady state, iL1 = 27.27 + (54.54 − 27.27)e −220000t iL 2 = 27.27 − 27.27e −220000t SOLUTION 8.54. (a) Charges will distribute in order to achieve equal voltage by KVL. Since q=CV, v R (0 − ) = 0V , due to equal capacitance the charges will distribute half and half, v R (0 + ) = 0.5V . (b) The equivalent capacitance is 2 F, thus v R (t) = 0.5e −0.5t . *S OLUTION 8.55. (a) Writing a node equation at v we have for all t,

4

dv v d (v − v s ) = 0 + + 4 (v − v s ) + dt 4 dt 2

(*)

Equivalently, 8

dv 3v dv v = − +4 s + s dt 4 dt 2

(**)

Grouping terms and dividing by 8 yields when t > 0, dv 3 1 =− v+ dt 32 16

(***)


P8-43

Notice that vs = 1 for t > 0 and for t > 0,


dvs (t) du(t) = = 0. dt dt

-

-

(b) By inspection v(0 ) = 0; both capacitors are uncharged at 0 . Recall from part (a) that KCL at the node for v yields (*) which is equivalent to (**). Since conservation of charge follows by integrating (*) or equivalently integrating (**) we have 0+

8∫ 0

−

dv 3 d =− d 4

0+

0+

∫− v d

+4

∫

0−

0

0+

dv s d + 0.5 ∫ v s d d − 0

Since the integral of a finite integrand over an infinitesimal interval is zero, we have equivalently, 0+

8∫

0−

dv d = 0+ 4 d

0+

∫−

0

dv s d +0 d

Evaluating these integrals we obtain

(

) (

)

8 v(0+ ) − v(0 − ) = 4 v s (0+ ) − v s (0− ) = 4u(0+ ) = 4 -

+

(c) Since v(0 ) = 0, v(0 ) = 0.5 V. Since v satisfies (***), i.e., dv 3 1 1 = − v + ≡ − v + F (***) dt 32 16 from equation 8.17,

[

]

2 1  v(t) = F + v(0+ ) − F e −t / =  − e −3t /32 u(t) V 3 6 -

Using v(0 ) would have led to an incorrect answer. SOLUTION 8.56. (a) x(t) = K1e −t / + K 2 dx(t) K = − 1 e −t / dt K 1 doing the substitution, − 1 e −t / = − (K1e −t / + K 2 ) + F . In order to satisfy the equality, K 2 = F .


P8-44

[ ] + K 2 = [ x(t0+ ) − F ]e −(t −t )/ + F


(b) x(t +0 ) = K1e −t 0 / + F , and K1 = x(t0+ ) − F e t 0 / . (c) x(t) = K1e −t /

0

.

SOLUTION 8.57. (a) From the graph, the initial and final values are 0 and 80 V respectively. That sets the following constraint, 100

R2 = 80V . From vC ( ) = 80 − 80e −1 = 50.57V . Looking at the Graph R1 + R2

= 5ms . Thus R1 ||R2 ⋅C = 5ms . Solving, R1 = 6250Ω and R2 = 25kΩ . (b) Using the same equalities, R1 = 0.25R2 = 2kΩ , and C =

2k || 8K

= 3.125uF .

SOLUTION 8.58. (a) From the graph, the initial and final values are 0 and 100 mA respectively. Thus R1 = 200 /100m = 2kΩ. From iL ( ) = 100 − 100e −1mA = 63.21mA , the graph shows a L /(R1 || R2 ) = 20ms, and R2 = 0.25mΩ. (b) R1 stays the same, L = 20ms(R1 || R2 ) = 20H . SOLUTION 8.59. This question is done in matlab %Problem 8.59 tau1= 20*1; tau2= 1; %For 0 < t < ta vo= 0; vinfa= 10; %Using the elapsed time formula, ta=tau1*log((0-10)/(9-10)); %For ta < t < tb vinfb=0; tb=tau2*log((9-0)/(1-0)); %For tb < t < tc vinfc=vinfa; tc=tau1*log((1-10)/(9-10));

= 20ms . Thus


P8-45


%Next switching is just a repeat of ta < t < tb t1=ta; t2=ta+tb; t3=t2+tc; t4=t3+tb; t5=t4+tc; t=0:t5/1000:t5-1/1000; vt= (ustep(t)-ustep(t-t1)).*(10-10.*exp(-t/tau1))+ ... (ustep(t-t1)-ustep(t-t2)).*(9.*exp(-(t-t1)/tau2))+ ... (ustep(t-t2)-ustep(t-t3)).*(10-9.*exp(-(t-t2)/tau1))+ ... (ustep(t-t3)-ustep(t-t4)).*(9.*exp(-(t-t3)/tau2))+ ... (ustep(t-t4)-ustep(t-t5)).*(10-9.*exp(-(t-t4)/tau1)); Frequency= 1/(tb+tc) plot(t,vt); grid; xlabel('time in seconds'); ylabel('Volts'); (b) The frequency is 0.0217 Hz. 9 8 7

Volts

6 5 4 3 2 1 0

0

20

40

60 80 time in seconds

100

120

140

SOLUTION 8.60. When the switch is in position A, = 18.18ms . In position B it is the elapsed time formula, find ta, when the output voltage reaches 90 V.

= 99.5us . Using


P8-46


 60 − 136.36  ta = 18.18msln = 9.07ms . At this point the switch goes to B, and the elapsed time until the 90 − 136.36   90 − 59.45 voltage reaches 60 V is tb = 99.5usln = 0.4ms . Adding both time, F = 105.6Hz . 60 − 59.45 SOLUTION 8.61. (a) The circuit can be rearranged in a series of one Vsolar V voltage source, one Lstore inductor, and one Rstore +R solar resistor. (b) iL (t) =

[

]

Vsolar 1− e −(Rstore+Rsolar )t /L A . Rsolar + Rstore

(c) In this time period the circuit reduces to an Lsotre inductor in series with a Rstore +R 1 resistor. (d) Vsolar Rsolar + Rstore

iL (T1− ) =

iL (t) = iT − e

−(Rstore+ R1)(t −T 1)/L

1

. A

(e) The two elements in series are an Lstore inductor and a resistor Req = Rstore + (R1 || R2 ). (f) iL (T2− ) = iT − e −(Rstore+ R1)(T 2 −T 1)/Lsotre A 1

iL (t) = iT − e −(Re q)(t −T 2)/Lstore A 2

(g)

[

]

V 2 solar PVsolar = Vsolar iL (t) = 1− e −(Rstore+Rsolar)t /L W Rsolar + Rstore PRsolar = Rsolar iL2 (t)W PRstore = RstoreiL2 (t)W di (t) PLstore = LstoreiL (t) L W dt 1 (h) WL (0,t) = LiL2 (t)J 2 SOLUTION 8.62. The light turns off when the current through it goes down to 0.5 mA. This corresponds to ib = 10uA , and a voltage across the capacitor of vC = ib(R1 + 2k) + 0.5 = 0.7V . The time constant of this circuit is = (R1 + 2k) || 5k ⋅1000uF = 4s. The final voltage across the capacitor is by  1.5 − 0.1 voltage division, 0.1V. Thus using the elapsed time formula t1 = 4ln  = 3.39s . 0.7 − 0.1


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-3

SOLUTION 8.63. (a) Since RC = 10 s, from equation 8.25, v a (t) = − RC


dv s (t) = cos(1000t) V. dt

Rf dv a (t) == − sin(1000t) V, and vout (t) = − v (t) = sin(1000t) = v s(t) V. dt R b (b) With the switch moved to position B, there is no source in the circuit. But the output at the switching instant is sin(1000t) V which coincides with vs(t). Hence, the input to the first amp remains Hence vb (t) = − RC

the same and the circuit continues to produce vout (t) = sin(1000t) V, i.e., the circuit becomes an oscillator.

*S OLUTION 8.64. Before attacking the problem proper, consider driving an ideal unity gain integrator with the square wave of figure P8.64b. If we start the integration when the square wave goes positive, then we have a triangular waveform as follows:

On the other hand, if we start the integration when the square wave goes negative, we get the following waveform

One concludes that without some further physical assumptions, there is no unique solution to this problem. Physically speaking all capacitors have a leakage resistance. Hence, in modeling the capacitor we put a very large resistance in parallel with an ideal C, producing a nearly ideal leaky integrator circuit. The leaky integrator circuit has a first order response. Hence over time, when the circuit reaches steady state, the dc level of the resulting output waveform will be proportional to the average level of


P8-48


the square wave which is zero in this case. Comments: Actually the proportionality constant in the above statement is the overall dc gain of the integrator-inverter. See the formulas given in P22.16. Adding up the two formulas, we have Output(t)max + Output(t) min = K (Vmax + V min ) and K = H(0), i.e., the dc gain of the first order low pass system. This leads to Average of output = (dc gain) (average of input levels) See the analysis in example 8.7 and later an exact analysis is given in problems 22.15 and 22.16. In other words, one would expect that the output of our (leaky) integrator in steady state to be given by the waveform below.

Now we can start to solve the problem. The first part is to design a (leaky) integrator circuit to produce a triangular waveform of value 2 V peak-to-peak. For this we consider the following figure which consists of the leaky integrator followed by an inverter.


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To handle this analysis, recall that i ≅ ∆q/∆t in which case ∆q = C∆v. Hence, to have a peak-to-peak voltage at v2 (t) and v3 (t) of 2 V, we require that i ∆t 9 50 × 10 −6 ∆v = in = × =2 C R1 C Hence R1 C = 2.25×10-4. If we choose R1 = 10 kΩ, then C = 22.5 nF. At this point the waveform of v3 (t) is given below.

In order to complete the design, we must raise the portion of the curve with positive slope by 1 V and lower the portion with negative slope by –1 V. This can be done by adding one-ninth of vin (t) to v3 (t). This can be done by using the following circuit. In this circuit, there is a voltage-divider at the non-


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inverting terminal of the second op amp. Here V+ equals one-eighteenth of vin (t). However the gain of the non-inverting portion is 2; therefore one-ninth of the input is added to v3 (t) as desired.

CHAPTER 9 PROBLEM SOLUTIONS SOLUTION TO PROBLEM 9.1. If we can compute expressions for K and q that are real, then these quantities exist by construction. Consider that A, B, K and q must satisfy the following relationship: K cos(ωt + θ) = ( K cos(θ)) cos(ωt) + (−K sin(θ))sin(ωt) ≡ Acos(ωt) + Bsin(ωt) Therefore K cos(θ) = A and − K sin(θ) = B . Consequently,

( K cos(θ))2 + (− Ksin(θ))2 = K 2 = A2 + B2 in which case K =

A 2 + B2 . Further, K sin(θ) −B = tan(θ) = K cos(θ) A

in which case θ = tan−1

− B A

with due regard to quadrant.

SOLUTION TO PROBLEM 9.2. For the inductor, 2 1  C  1  CV02  1  2 1 WL (t) = 2 L ⋅ iL (t) = LVo sin t  = sin2  t    LC  2  L LC  2 and for the capacitor, 2  CV 2  1   1  WC (t) = 12 C ⋅vC2 (t) = 12 C Vo cos t  = 0 cos2  t .  LC    LC  2  Hence, CV02  2  1    CV02 2 2 2 1 1 1 WC + WL = 2 C ⋅ vC (t) + 2 L ⋅ iL (t) = sin  t + cos  t =  LC   LC   2  2 SOLUTION TO PROBLEM 9.3. Since x(t) = (K1 + K2 t)e −αt , x' (t) = − α ⋅K1e −αt + K2e −αt − α t⋅ ⋅ K2e −αt and x' ' t)( = α2 ⋅ K1e −αt − α K ⋅ 2e −αt − α K ⋅ 2e −αt + α2 ⋅ t ⋅ K2 e −αt Substituting into the differential equation, we have α 2 ⋅ K1e −αt − 2α ⋅K2e −αt + α2 ⋅ t ⋅ K 2e −αt

[

]

[

]

+ 2α − α ⋅K1e −αt + K2 e−α t − α t⋅ ⋅ K2 e −αt + α2 K1e −αt + K2te−αt = 0

This means that the solution form satisfies the differential equation.

SOLUTION TO PROBLEM 9.4. (a) Suppose x(T ) = 0 at some T. Then K1e s1T = −K2e s2 T . Since e siT > 0 whenever si is real and T is finite, K1 & K2 must have opposite signs. (b) For this we solve for T and show there can only be one solution. Since K1e s1T = −K2e s2 T and e siT > 0 ,  es2 T   K  K1 es2 T = s T implies ln  1  = ln s T  = ( s2 − s1 )T − K2 e 1  −K2  e 1  Hence the unique solution is given by  K  T = ln 1   −K2 

(s2 − s1 )

provided s2 ≠ s1 which is the case for distinct roots. SOLUTION TO PROBLEM 9.5. Suppose x(T ) = 0 at some T >0. This is true if and only if (*) K1e s1T = −K2Te s1T s1T Since e > 0 and T > 0, (*) is true if and only if K1 = − K2T which is true if and only if K1 & K2 have opposite signs. SOLUTION 9.6. (a) Denote one period of oscillation by T. Then by definition 9950T = 2π. Hence, T = 0.63148 ms. The time constant of decay is 1 ms. Therefore, NT = N × 0 . 6 3 1 4 8 =. 1Hence N = 1.5836 cycles. 2π 1 (b) Here observe that the time constant of decay is 1/σ s. Hence NT = N = . One ωd σ ω concludes that N = d . 2πσ SOLUTION 9.7. The differential equation for the capacitor voltage is d 2 vC (t) 1 1 + vC (t) = V 2 LC LC s dt 1 For t > 0, the characteristic equation is s2 + = 0 . Hence from table 9.2, the solution LC for either the inductor current or capacitor voltage has the general form

vC (t) = Acos(ω d t) + Bsin(ω d t) + X F 1 V 1  . From table 9.2, X F =  s   = Vs . Further,    LC LC LC vC (0 + ) = A + X F = A + Vs = 0 Hence A = −Vs . Also, where ω d =

v'C (0 + ) = ωd B = Hence B = 0. Therefore

1 i (0 + ) B= L =0 LC C

  t  vC (t) = −Vs cos(ω d t) + Vs = Vs 1 − cos  V  LC   

To obtain the expression for iL(t) (= iC(t)), we can either repeat the above derivation or differentiate and multiply by C. We choose the latter. Therefore iL (t) =

CVs V  t   t  sin   = s sin  A  LC  L LC  LC  C

SOLUTION TO PROBLEM 9.8. Essentially this is example 9.7, case 1, with literals and R = ∞. Clearly the circuit is a driven parallel LC circuit having characteristic equation 1 j  j   s2 + = s + s −  = (s + jω d )(s − jω d ) = 0 LC  LC   LC  Thus we obtain  1   1  iL (t) = Acos(ω d t) + Bsin(ω d t) + X F = Acos t  +Bsin t  +X F  LC   LC  Here X F = I S is the value of the current when the inductor is shorted and the capacitor is open. Applying the initial conditions, iL (0+) = A + Is = 0 ⇒ A = − Is Further v (0+) vC (0+) i'L (0+) = L = =0= L L

1 B LC

because capacitor voltage is continuous and because i'L (0+) =

d    1   1  Acos t + B sin t + XF  =      dt  LC LC  t =0

Hence B = 0. Therefore,

B LC

1   iL (t) = Is 1− cos( t)  LC   Rather than repeat the above derivation, vC (t) = vL (t) = L

diL (t) d = LIs dt dt

 L  1   1  1 − cos LC t   = Is C sin LC t   

SOLUTION TO PROBLEM 9.9. Observe that the circuits of figures (a) and (b) are dual circuits. Hence the numerical value of vout(t) and iout(t) are the same for the same excitation. Since the circuit is linear, when the excitation is doubled, the response is doubled (given zero initial conditions) by linearity. Therefore, iout (t) = 2g(t) .

SOLUTION TO PROBLEM 9.10. (a) The initial conditions are:

vc (0−) = 0 = vc ( 0+) iL (0−) = 10 / 0.5 = 20 = i L ( 0+)

(b) 1 L ⋅i 2L ( 0 ) = 0.5 J 2 WC (0) = 0

WL ( 0 )=

(c) Maximum value of vC when all energy is in capacitor must satisfy

1 2 C ⋅ vC,max = 0.5 2

or equivalently vC,max = 1000 V. (d) From the text development, the parallel LC circuit has the differential equation, d 2 vC 1 2 + LC vC = 0 dt The solution form is:  1   1  vC (t) = A ⋅cos t  +B ⋅sin t  LC   LC  The initial conditions are:

vC ( 0 ) = A = 0

and vC ' ( 0+) =

1 i (0+) i ( 0+) B= C =− L LC C C

Hence B = −1000 Thus

 1  vC (t) = −1000 ⋅ sin t V  LC 

SOLUTION TO PROBLEM 9.11. The switch has been closed for a long time which means that the inductor acts like a short and the capacitor like an open. Hence, at t = 0 , vL = 0 and iC = 0. Hence Is divides equally between the resistors, i.e., iL (0−) = i L ( 0+) = Is / 2 and vC (0−) = vC ( 0+) = 0 For t ≥ 0, the differential equation is d 2 vC 1 2 + LC vC = 0 dt with corresponding response  1   1  vC (t) = A ⋅cos t  +B ⋅sin t  LC   LC  Evaluating at the initial conditions, vC ( 0 ) = A = 0 and vC ' ( 0 )= B ⋅

1 i (0+) −iL (0+) = C = LC C C

in which case I B=− S 2

L C

Therefore, I v C (t) = − s 2

L  1  sin t C  LC 

SOLUTION TO PROBLEM 9.12. Natural frequency is

1 = 5000 ⋅ 2π . Hence, LC

»C = 100e-9; »L = 1/((10e3 *pi)^2 *C) L= 1.0132e-02

By voltage divider,

vC (0−) = vC ( 0+) = 20 mV

Current through L,

iL (0−) = i L ( 0+) = 0

Voltage across capacitor satisfies,  1   1  vC (t) = Acos t + B sin t  LC   LC  Using the ICs, vC (0+) = A = 20 mV and 1 i (0+) i ( 0+) = C =− L =0 LC C C

vC ' ( 0 )= B ⋅ Hence

vC (t) = 20 ⋅cos(10,000πt ) mV

SOLUTION TO PROBLEM 9.13. By definition ω=

1 = 2π ⋅40 LC

in which case »C = 0.1e-3; »w = 2*pi*40; »L = 1/(w^2 *C) L= 1.5831e-01 (rad/s). Observe that

iL (0−) = i L ( 0+) = 1A

From the given circuit, the capacitor is never connected to a source. Therefore, vC (0−) = vC ( 0+) = 0. Also, since d 2 iL 1 2 + LC iL = 0 dt it follows that From the initial conditions and

iL (t) = Acos(ωt) + Bsin(ωt) iL ( 0 )= A = 1

v (0 + ) iL ' (0 + ) = Bω = C =0 L iL (t) = cos(80πt) A

SOLUTION TO PROBLEM 9.14. (a) From the continuity property, the capacitor voltage and inductor current remain the same, vC (0−) = vC ( 0+) = 5 V iL (0−) = i L ( 0+) = 1 However, iC (0−) = 0 but iC (0+ ) = −iL (0+) = −1 A and v L (0−) = 0 but v L (0+) = 5 V. These values change to maintain satisfaction of KVL and KCL. (b)  1   1  vC (t) = Acos t + B sin t  LC   LC  where

vC (0+) = 5 = A 1 i (0+) 1 π ⇒ B=− vC ' ( 0+) = B = C =− LC C C 2 vC (t) = 5cos(2πt) −

π sin(2πt) V 2

Alternately, from equation 9.4, vC (t) = K cos(2πt + θ) and from equation 9.17b K = »A = 5; B = -pi/2; »K = sqrt(A^2 + B^2) K= 5.2409e+00 »theta = atan2(-B,A) theta = 3.0440e-01 »thetadeg = theta*180/pi thetadeg = 1.7441e+01

A 2 + B2 and θ = tan−1

− B in which case A

Hence, K = 5.24 and θ = 17o. (c) 6


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SOLUTION TO PROBLEM 9.15. As before, ω =

1.4

1.6

1.8

2

1 = 2π ⋅40 in which case L = 0.158 LC

H. Now, vC (0−) = vc (0+) = and

100 ⋅ 25 = 20 mV 125

iL (0−) = i L ( 0+) = 10 mA

The solution form is:

iL (t) = Acos(80πt) + Bsin(80πt)

where

iL (0+) = A = 10 mA

and v ( 0+) vC (0+) iL ' (0+) = L = = 80πB ⇒ B = 0.50265 mA L L

SOLUTION TO PROBLEM 9.16. (a) Note: vC (0−) = vC ( 0+) = −1 V and iL ( 0 )= 5 A. The characteristic equation is:

s2 +

1 1 s+ = s2 + 5s + 4 = (s + 1)(s + 4 )= 0 RC LC

As stated, the circuit is overdamped. Hence, vC (t) = K1e s1t + K2 es2 t = K1e −t + K2e −4t vC (0+) = K1 + K2 = −1 i (0+) vC ' ( 0+) = s1K1 + s2 K2 = −K1 − 4K2 = C =0 C ⇒ K1 = Finally,

−s2 s1 = −1.3333 and K2 = = 0.3333 . s2 − s1 s2 − s1 vC (t) = −1.3333e −t + 0.3333e−4t V

Using MATLAB to plot: »t = 0:.02:4.5; »vc = -1.33333*exp(-t) + 0.33333*exp(-4*t); »plot(t,vc) »grid »ylabel('Capacitor Voltage in V') »xlabel('Time in s') 0 -0.1


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K1 + K2 = 1

3

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i (0+) −10 vC ' ( 0+) = s1K1 + s2 K2 = −K1 − 4K2 = C = = −10 C C Hence, »A = [1 1;-1 -4] A= 1 1 -1 -4 »b = [1;-10] b= 1 -10 »K = A\b K= -2 3 yielding

vC (t) = −2e −t + 3e −4t V 1

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Obviously, there is only one zero crossing.

SOLUTION TO PROBLEM 9.17. (a) First, 100 30 = 10 V 300 iL (0−) = i L ( 0+) = 0.1 A

vC (0−) = vC ( 0+) =

3.5

4

4.5

Clearly, iC (0−) = 0 and v L (0−) = 0 . However, v (0+) iL (0+) + iC (0+) + C = 0.1 + iC (0+) + 0.15 = 0 66.667 Hence, iC (0+ ) = −0.25 A. Further, v L (0+) = 10 V. (b) The characteristic equation is: s2 +

1 1 s+ =0 RC LC

From MATLAB, »R = 66.667;C = 25e-6; L = 0.5; »b = 1/(R*C) b = 6.0000e+02 »c = 1/(L*C) c= 80000 »si = roots([1 b c]) si = -3.9999e+02 -2.0000e+02 We take the roots to be: s1 = −400 and s2 = −200 . (c) Overdamped response implies, vC (t) = K1e s1t + K2 es2 t = K1e −400t + K2e −200t (d)

vC (0+) = K1 + K2 = 10

i (0+) vC ' ( 0+) = s1K1 + s2 K2 = −400K1 − 200K2 = C = −10 4 C »A = [1 1;-400 -200]; »b = [10;-1e4]; »K = A\b K= 40 -30 Finally,

vC (t) = 40e −400t − 30e−200t V

(e) »t=0:0.01e-3:25e-3; »vc = 40*exp(-400*t) - 30*exp(-200*t); »plot(t,vc) »grid »ylabel('Capacitor Voltage in V') »xlabel('Time in s')

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SOLUTION TO PROBLEM 9.18. (a)

iL (0−) = i L ( 0+) = 0 vC (0−) = vC (0+) = 5 V

At t = 0+, the circuit is a series RLC with R = 12.5 Ω, L = 2.5 H, and C=0.1 F. The resulting characteristic polynomial is:

s2 +

R 1 s+ = s2 + 5s + 4 = (s + 4)(s + 1) = 0 L LC

Hence, s1,s2 = −4,−1 and the form of the response is: vC (t) = K1e −4t + K 2e −t At t = 0+,

vC (0+) = K1 + K2 = 5

and i (0+) iL (0+) vC '(0) = −4K1 − K2 = C = =0 C C Solve for K1 and K2 we obtain: »A = [1 1; -4 -1]; »b = [5;0];

»K = A\b K= -1.6667e+00 6.6667e+00 Hence,

vC (t) = −1.66667e −4t + 6.66667e −t V

(b)

iL (0+) = i L (0−) = 0 Since, the stable (passive) circuit contains no source for t > 0, all initial energy is absorbed by the resistor. Hence limt →∞ iL (t) = 0 , i.e., iL (∞) = 0 . iL (t) = iC (t) = C

dvC 2 −4t 2 −t = e − e A dt 3 3

SOLUTION TO PROBLEM 9.19. This circuit is a series RLC in which case the R 1 characteristic equation is always: s2 + s + =0 L LC (a) For this time period, R = 2 kΩ in which case the characteristic equation is found and solved in MATLAB as follows: »R = 2e3; L = 0.1; C = 0.1e-6; »b = R/L b= 20000 »c = 1/(L*C) c= 100000000 »s12 = roots([1 b c]) s12 = -10000 -10000

The roots are repeated and: s1, s2 = –10,000. The form of the response is: iL (t) = (K1 + K2t)e −10000t To find K1 and K2: iL (0+) = K1 = 2.5 and

v (0+) vC (0+) − RiL (0+) 6 − 2000 × 2.5 iL '(0+) = −25000 + K 2 = L = = = −49940 L L 0.1 ⇒ K2 = −24940 Thus

iL (t) = (2.5− 24940t)e −10000t

0 ≤ t ≤ 0.1m sec

(b) After the switch closes, R = 1 kΩ, iL (0.1× 10 −3) = (2.5 − 24940 × 0.1× 10 −3 )e −1 = 0.0022073 and

iL '(10 −3) = −25000e −1 − 24940e −1 + 24940e −1 = −9197

The new characteristic equation is computed as follows: »R = 1e3; L = 0.1; C = 0.1e-6;b = R/L b= 10000 »c = 1/(L*C) c = 100000000 »s12new = roots([1 b c]) s12new = -5.0000e+03 + 8.6603e+03i -5.0000e+03 - 8.6603e+03i Hence s1,s2 = −5000 ± j8660.25. The form of the new solution is: iL (t') = e −5000t' [ Acos( t') + Bsin( t') ] A where t'= t − 0.1× 10 −3 and

= 8660.25 rad/sec. Observe that iL (t'= 0) = 0.0022073 = A

and

iL '( t'= 0) = −5000A + 8660.25B = −9197

From MATLAB, »w = imag(s12new(1)); »sig = real(s12new(1)); »B = (-9197 -sig*2.2073e-3)/w B= -1.0607e+00 Hence for t > 0.1 ms, iL (t) = e −5000(t −0.1ms) [0.0022073cos( (t − 0.1ms)) − 1.0607sin ( (t − 0.1ms))] A (c)

2.5 2

iL(t), in A

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3

3.5

K1 = 2.5; K2 = -24940; t = 0:0.01e-3:4e-3; A = 0.0022073; B = (-9197 -sig*2.2073e-3)/w B = -1.0607e+00 iL = (K1 + K2*t) .* exp(-10000*t) .* (u(t)-u(t-1e-4)) ... + exp(-5000*(t - 1e-4)) .* (A*cos(w*(t - 1e-4))+B*sin(w*(t - 1e-4))) ... .* u(t -1e-4); plot(t,iL) grid iL = (K1 + K2*t) .* exp(-10000*t) .* (u(t)-u(t-1e-4)) (d) »K1 = 2.5; K2 = -24940; »t = 0.1e-3; »iL = (K1 + K2*t) .* exp(-10000*t) iL = 2.2073e-03 » % The energy stored in the inductor over [0,0.1ms] is in J: »WL = 0.5*0.1*(iL^2 - 2.5^2) WL = -3.1250e-01 » % The energy stored in the capacitor over [0,0.1ms] first » % requires computation of vL and then vC. »vL = 0.1*(K2*exp(-10000*t) - 10000*(K1 + K2*t) .* exp(-10000*t)) vL = -9.1970e+02 »vC = vL + 2000*iL vC =

4 x 10

-3

-9.1528e+02 » % The energy stored in the capacitor over [0,0.1ms] is in J: »WC = 0.5*0.1e-6*(vC^2 - 6^2) WC = 4.1885e-02 » % To compute energy dissipated in resistor, we make » % use of conservation of energy: WR + WC + WL = 0 »WR = -WL - WC WR = 2.7061e-01

SOLUTION TO PROBLEM 9.20. For this problem, iL (0) = 8 A, vC (0) = 20 V. For the 1 1 parallel RLC, the characteristic equation s2 + s+ = 0 is solved as follows: RC LC »R = 20; C = 0.01e-3; L = 25e-3; »si = roots([1 1/(R*C) 1/(L*C)]) si = -4.0000e+03 -1.0000e+03 »s1 = si(1); s2 = si(2); Hence

vC (t) = K1e −4000t + K2e −1000t

To compute the constants, vC (0) = K1 + K 2 = 20 and iL (0+) + i (0+) vC '(0) = −4000K1 − 1000K 2 = C =− C C »iL0 = 8; vC0 = 20; »vCprime = -(iL0 +vC0/R)/C vCprime = -9.0000e+05 »A = [1 1;-4000 -1000]; »b = [20;vCprime]; »K = A\b K= 2.9333e+02 -2.7333e+02

v C (0+) R

Hence,

vC (t) = 293.33e −4000t − 273.33e −1000t V

Also, v (t) iL (t) = − C − CvC' (t) = −2.933e −4000t + 10.933e −1000t A R where »K1 = K(1); K2 = K(2); »KR1 = K1/R; KR2 = K2/R; »KCp1 = -4000*K1*C; KCp2 = -1000*K2*C; »KL1 = -KR1-KCp1 KL1 = -2.9333e+00 »KL2 = -KR2 - KCp2 KL2 = 1.0933e+01 »t = 0:0.01e-3:4e-3; »iL = KL1*exp(-4000*t) + KL2*exp(-1000*t); »plot(t,iL) »grid »xlabel('Time in s') »ylabel('Inductor Current in A') » 9 8

Inductor Current in A

7 6 5 4 3

TextEnd

2 1 0

0

0.5

1

1.5

2 Time in s

2.5

3

3.5

SOLUTION TO PROBLEM 9.21. (a) The circuit is critically damped and the characteristic polynomial is:

4 x 10

-3

s2 +

1 1 s+ = (s + 20)2 = s2 + 40s + 400 = 0 RC LC

»R = 2; C = 1/80; »L = 1/(400*C) L= 2.0000e-01 i (0+ ) −iL (0 + ) − vC (0 + ) / 2 (b) vC (0) = 10 V and vC '(0 + ) = −800 = C . = C C »vC0 = 10; vCp0 = -20*10 -600 vCp0 = -800 »iL0 = C*800-vC0/2 iL0 = 5 Hence: iL(0-) = iL(0+) = 5 A. (c) By simple KCL, iL (t) =

−1 dv C v C − = 5e −20t + 150 ⋅ t ⋅e −20t A 80 dt 2

SOLUTION TO PROBLEM 9.22. (a) The characteristic equation for the series RLC s2 +

Req L

s+

1 =0 LC

For critically damped response, want (Req / L)2 − 4 / (LC ) = 0 . Solving yields Req = 4 Ω. Hence, 5R Req = 5+R implies that R = 20 Ω. (b) Solving for the resulting roots implies that »si = roots([1 Req/L 1/(L*C)]) si = -50 -50 »s1 = 50; Hence

vC (t) = (K1 + K 2t)e −50t From the initial conditions, vC (0) = K1 = 5 and i (0+) iL (0+) −5 vC '(0) = −250 + K2 = C = = = −500 C C 0.01 Hence, K2 = –250 and vC (t) = (5 − 250t)e −50t V Set to zero and solve for t = 0.02. See MATLAB plot. 5

4

3

2

1

0

-1

0

0.005

0.01

0.015

0.02

0.025

0.03

0.035

SOLUTION TO PROBLEM 9.23. (a) The circuit is a series RLC. Hence »R = 40; C = 0.25e-3; L = 0.1; si = roots([1 R/L 1/(L*C)]) si = -200 -200 »s1 = si(1); Since the circuit is critically damped, iL (t) = (K1 + K 2t)e −200t A Using initial conditions,

0.04

0.045

0.05

iL (0) = K1 = 1 v (0+) −vC (0+) − 40i L (0+) iL '(0) = s1K1 + K2 = −200K1 + K 2 = L = = −350 L L Hence

K 2 = −150

Thus

iL (t) = (1− 150t)e −200t A

Now using MATLAB, »R/L ans = 400 »1/(L*C) ans = 40000 »y = dsolve('D2y + 400*Dy + 40000*y = 0, y(0) = 1,Dy(0) = -350') y= exp(-200*t)-150*exp(-200*t)*t This answer coincides with the analytical solution. (b) As above, the form of the solution is vC (t) = (K1 + K 2t)e −50t Applying initial conditions, vC (0) = K1 = 5 i (0+) iL (0+) vC '(0) = s1K1 + K2 = −200K1 + K 2 = C = = 4000 C C Hence, K 2 = 5000 and

vC (t) = (5 + 5000t)e −200t V

In MATLAB, »y = dsolve('D2y + 400*Dy + 40000*y = 0, y(0) = 5,Dy(0) = 4000') y= 5*exp(-200*t)+5000*exp(-200*t)*t which verifies the analytical solution. Plotting we obtain,

12

10

8

6

4

2

0

0

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

Observe, there is no zero crossing as per problem 9.5.

SOLUTION TO PROBLEM 9.24. Source has been on for a long time and turns off at t = 0. Hence iL (0−) = iL (0+) = 0 vC (0−) = vC (0+) = 0.1× 40 = 4 V At t ≥ 0, we have a series RLC circuit: »R = 40; C = 4e-3; L = 2; si = roots([1 R/L 1/(L*C)]) si = -1.0000e+01 + 5.0000e+00i -1.0000e+01 - 5.0000e+00i »wd = imag(si(1)) wd = 5 »sig = -real(si(1)) sig = 10 Hence

0.045

0.05

vC (t) = e −10t ( Acos(5t) + B sin(5t)) Applying initial conditions, vC (0) = A = 4 i (0+) iL (0+) vC '(0) = −10A + 5B = C = =0 C C Hence, B = 8. The final form is: vC (t) = e −10t (4 c o s ( 5t) + 8sin(5 t)) V Similarly,

iL (t) = e −10t ( A cos(5t) + Bsin(5t)) A Applying initial conditions, iL (0) = A = 0 v (0+) −v C (0+) − 40iL (0+) iL '(0) = −10A + 5B = L = = −2 L L Thus

iL (t) = −0.4e −10t sin(5t) A

SOLUTION TO PROBLEM 9.25. -

At t = 0 , the capacitor is an open circuit and the inductor is a short circuit. The resulting circuit is a simple resistive network. The first step in the solution is to solve this network for the initial conditions on the capacitor and inductor. Specifically, solve for the capacitor voltage (i.e. the voltage across the series connection of the 6 Ω resistor and the independent voltage source) and the inductor current (i.e. the current flowing through the 4 Ω resistor). Verify that iL (0−) = iL (0+) = 1 A and vC (0−) =v C (0+) = 12 V. When the switch opens, the branch containing the independent voltage source is eliminated. So, we end up with a series RLC circuit. The equivalent resistance is Req = 4 + 24 / / 4 8= 20 Ω. »R = 20; C = 0.01; L = 2;si = roots([1 R/L 1/(L*C)]) si = -5.0000e+00 + 5.0000e+00i -5.0000e+00 - 5.0000e+00i

»sig = -real(si(1)) sig = 5 »wd = imag(si(1)) wd = 5

iL (t) = e −5t [ A cos(5t) + Bsin(5t)] A

Applying ICs, iL (0) = A = 1 v (0+) −8 iL '(0) = −5A + 5B = L = = −4 L 2 Hence B = 0.2. It follows that 1 iL (t) = e−5t cos(5t) + sin(5t) A  5 

SOLUTION TO PROBLEM 9.26. The series RLC circuit has characteristic equation s2 +

R 1 10 1 s+ = s2 + s + = s2 + 2 s + 2 + 2d = 0 L LC L LC

From the given response,

= 10 =

10 which implies that L = 0.5 H. Further, 2L

(

)

2 1 2 = = 2 + d2 = 10 2 + 10 3 = 400 LC C

Hence, C = 5 mF. v (0) Now, from given response, iL (0) = 0 and iL '(0) = 500 3 = L = 2v L (0) . L Hence v L (0) = 250 3 V. In addition, vC (0) = −10iL (0) − v L (0) = −250 3 V and vC '(0) = iL (0) / C = 0 To find the capacitor voltage we have

(

vC (t) = e −10t Acos(10 3t) + B sin(10 3t)

)

It follows that vC (0) = −250 3 = A and vC '(0) = 0 = 2500 3 + 10 3 B B = –250. Therefore

(

)

vC (t) = −e −10t 250 3 cos(10 3t) + 250sin(10 3t)

SOLUTION TO PROBLEM 9.27. (a) From the given information, vC (t) = Ae− t cos( t + ) V 2 where = 2 f = 2 / T = = 4000π rad/s. f = 2 kHz. 0.5 × 10 −3 The time constant of the exponential term is 1/σ which is 2 ms from the figure. Hence σ = 500 . The characteristic equation of the parallel RLC is: s2 + Since

1 1 1 1 s+ = s2 + 4 s + = s2 + 2 s + 2 + 2d = 0 RC LC LC 10 C

1 = 2 = 1000, C = 0.1 µF. Further, 10 4 C 1 107 = = 2 + d2 = 10 6 + 16π 2106 LC L

implies that »L = 1e7/(1e6 + 16*pi^2 *1e6) or L = 63 mH. (b) From the figure and the above calculations, vC(0) = 10 V and vC (t) = 10e −500t cos(4000π t) V, Hence

vC '(t) = −5000e −500t cos(4000π t) − 40000π e −500t sin(4000π t) implying that vC' (0) = −5000 . Thus iL (0) = −vC (0) /10 4 − CvC' (0) = −0.001 + 5000 × 0.1× 10−6 = −0.5 × 10 −3 A

SOLUTION TO PROBLEM 9.28. As given vC (t) = Ae−3t cos(4t + ) and R = 10 Ω. The characteristic polynomial of the parallel RLC circuit is: 1 1 1 1 s2 + s+ = s2 + s+ = s2 + 2 s + 2 + 2d = 0 RC LC 10C LC

Hence 1/(RC) = 6 implies C = 1/60 F. Further

d

=

4 × 60 − 36 L = 4 . Hence L = 2.4 2

H. To change R to obtain a cricially damped circuit, 2 4  60  = = 100  R LC Hence R2 = 36 or R = 6 Ω. It follows that 2σ = 10 or σ = 5. The form of the response is:

vC (t) = (K1 + K 2t)e −5t

SOLUTION TO PROBLEM 9.29. For all cases, vC(0-)= vC(0+) = 0 and iL(0-)= iL(0+) = 20/20 = 1 A. Further for all cases the circuit is a parallel RLC with characteristic equation: s2 +

1 1 s+ = s2 + 2 s + 2 + d2 = 0 RC LC

(a) »L = 2e-3; C = 5e-6; »c = 1/(L*C); »R = 10; »b = 1/(R*C); »si = roots([1 b c]) si = -10000 -10000 »% Solution is cricitally damped. Thus 4 vC (t) = (K1 + K 2t)e s1t = (K1 + K2 t)e −10 t V

From ICs,

vC (0) = K1 = 0 i (0+) −iL (0+) vC '(0) = s1K1 + K2 = K 2 = C = = −2 × 105 C C

Hence 4 vC (t) = −2 × 10 5 te−10 t V

(b) »R = 100; »si = roots([1 1/(R*C) 1/(L*C)])

si = -1.0000e+03 + 9.9499e+03i -1.0000e+03 - 9.9499e+03i vC (t) = e −1000t ( Acos(9950t) + B sin(9950t)) V From ICs.

vC (0+) = A = 0

i (0+) −i L (0+) vC '(0) = −1000A + 9950B = C = = −2 × 10 5 C C in which case B = –20.1. Thus vC (t) = −20.1e −1000t sin(9950t) V (c) »R = 87/17;si = roots([1 1/(R*C) 1/(L*C)]) si = -3.6328e+04 -2.7527e+03 We define s1,s2 = –2753, –36327. Thus vC (t) = K1e s1t + K 2e s2t = K1e −2753t + K 2e −36327t V From the IC's

vC (0) = K1 + K 2 = 0 vC '(0) = s1K1 + s2K 2 = −2 × 105

»A = [1 1;si(2) si(1)]; »b = [0; -2e5]; »K = A\b K= -5.9568e+00 5.9568e+00 Therefore

(

)

vC (t) = −5.9568 e −2753t − e −36327t V

SOLUTION TO PROBLEM 9.30. For all cases, vC(0-) = vC(0+) = 0 and iL(0-) = iL(0+) = 10/100 = 0.1 A. Further for all cases the circuit is a parallel RLC with characteristic equation: 1 1 s2 + s+ = s2 + 2 s + 2 + d2 = 0 RC LC

(a) »R = 50; C = 0.04e-3; L = 0.625; »si = roots([1 1/(R*C) 1/(L*C)]) si = -400 -100 Define the two roots as: s1 = –100 s2 = –400 Hence,

vC (t) = K1e −100t + K2e −400t V

From, the initial conditions vC(0+) = 0 = K1 + K2 and vC '(0) = s1K1 + s2K 2 = −100K1 − 400K2 =

−iL (0+) = −2500 C

»A = [1 1;si(2) si(1)]; »b = [0;-2500]; »K = A\b K= -8.3333e+00 8.3333e+00 Therefore

(

)

vC (t) = −8.3333 e −100t − e −400t V

(b) »L = 0.4; »si = roots([1 1/(R*C) 1/(L*C)]) si = -2.5000e+02 -2.5000e+02 Thus From IC's,

vC (t) = (K1 + K 2t)e −250t V vC (0) = K1 = 0 i (0+) −iL (0+) vC '(0) = s1K1 + K2 = K 2 = C = = −2500 C C

Therefore

vC (t) = −2500te−250t V

(c) »L = 0.2; »si = roots([1 1/(R*C) 1/(L*C)]) si = -2.5000e+02 + 2.5000e+02i -2.5000e+02 - 2.5000e+02i vC (t) = e −250t [A cos(250t) + Bsin(250t)] V From ICs.

vC (0+) = A = 0

i (0+) −i L (0+) vC '(0) = −250A + 250B = 250B = C = = −2500 C C in which case B = –10. Thus vC (t) = −10e −250t sin(250t) V

SOLUTION TO PROBLEM 9.31. (a) The indicated behavior occurs when the resistance causes the circuit to be critically damped, i.e., 2 4  4 2  1  − = − 16 = 0  RC  LC  R  Thus R = 1 Ω and R = 1 = R0 + e t −5 = 0.8 + e t −5 So »t = 5+log(1-0.8) t= 3.3906e+00 (years) (b) Here, we have a series case: the indicated behavior occurs when the resistance causes the circuit to be critically damped, i.e., 2 4  R 2  L  − LC = R − 144 = 0 Thus R = 12 Ω and R0 15 R = 12 = = t −5 1+e 1 + e t −5 »t = 5 + log((15-12)/12) t= 3.6137e+00 (years)

SOLUTION TO PROBLEM 9.32. Step 1: Since the step functions are 0 from t = ∞ up to t = 0-, vC(0-) = vC(0+) = 0, iL(0-) = iL(0+) = 0 Step 2: At t = 0+, we have v (0+) iC (0+) = 2 − C − iL (0+) = 2 A 20 and

v L (0+) = v C (0+) − 5iL (0+) − 50 = −50 V

SOLUTION TO PROBLEM 9.33. Since the step function is 0 from t = ∞ up to t = 0-, vC(0-) = vC(0+) = 0, iL(0-) = iL(0+) =0 Since the circuit is a parallel RLC s2 +

1 1 s+ = s2 + 100s + 1600 = 0 RC LC

»R = 4; L = 0.25; C = 2.5e-3; »si = roots([1 1/(R*C) 1/(L*C)]) si = -80 -20 Hence s1,s2 = -20, -80. The general form is: vC (t) = K1e −80t + K2e −20t + X f V When the capacitor is open and the inductor is a short, Xf = 0. Thus, vC (t) = K1e −80t + K2e −20t V From the ICs vC(0+) = 0 = K1 + K2 and

i (0+) −iL (0+) − v C (0+) 8 + (20 − vC (0+)) 8 vC '(0) = −80K1 − 20K 2 = C = C C 2.5 = = 1000 2.5 × 10 −3

»A = [1 1;-80 -20]; »b = [0; 1000]; »K = A\b K=

-1.6667e+01 1.6667e+01 Hence,

[

]

vC (t) = −16.667 e −80t − e −20t V

SOLUTION TO PROBLEM 9.34. From the continuity property and the fact that at t = 0-, the capacitor looks like an open and the inductor looks like a short at t = 0-, iL(0-) = iL(0+) = 1A vC(0-) = vC(0+) = 0 Since the circuit is a parallel RLC s2 +

1 1 s+ = s2 + 500s + 40000 = 0 RC LC

»R = 40; C = 0.05e-3;L = 0.5; »si = roots([1 1/(R*C) 1/(L*C)]) si = -400 -100 Define the two roots as: s1 = –100 s2 = –400 Hence, the general form is: iL (t) = K1e −100t + K2e −400t + X f When the capacitor is open and the inductor is a short, Xf = –1 A. Thus, iL (t) = K1e −100t + K2e −400t − 1 From, the initial conditions

iL (0+) = K1 + K2 − 1 = 1

and v (0+) vC (0+) iL '(0) = −100K1 − 400K2 = L = =0 L L »A = [1 1;-100 -400]; »b = [2;0]; »K = A\b K= 2.6667e+00

-6.6667e-01 Hence, iL (t) =

8 −100t 2 −400t e − e −1 A 3 3

SOLUTION TO PROBLEM 9.35. (a) From the problem specs and the continuity property, iL(0-) = iL(0+) = 0.008 A vC(0-) = vC(0+) = 2 At t = ∞, the inductor looks like a short and the capacitor looks like an open; hence iL(∞) = 0 and vC(∞) = 400×0.006 = 2.4 V. The circuit is a series RLC with characteristic polynomial R 1 s2 + s + =0 L LC »R = 400; C = 6.6667e-6; L = 0.2; si = roots([1 R/L 1/(L*C)]) s1 = si(1); s2 = si(2); si = -1.5000e+03 -5.0000e+02 Hence

vC (t) = K1e −1500t + K2e −500t + 2.4 V

Using initial conditions vC(0) = K1 + K2 + 2.4 = 2 iC (0+) iL (0+) 8 × 10 −3 vC '(0) = −1500K1 − 500K 2 = = = = 1200 C C 6.6667 × 10 −6 »A = [1 1;-1500 -500]; »b = [2-2.4;1200] b= -4.0000e-01 1.2000e+03 »K = A\b K= -1.0000e+00 6.0000e-01

Hence

vC (t) = −e −1500t + 0.6e −500t + 2.4 V

(b) vL(t) is going to have the same form as vC(t) above except that vL(∞) = 0 since the inductor is a short at t = ∞. Alternately however, we have diL (t) diC (t) d 2vC (t) v L (t) = L =L = LC = −3e −1500t + 0.2e −500t V 2 dt dt dt SOLUTION TO PROBLEM 9.36. At t = 0, inductor is a short circuit and the capacitor is an open circuit. Since the current source is 0 at t = 0-, and the continuity property, iL(0-) = iL(0+) = –1 A vC(0-) = vC(0+) = 65 V For positive time, we have a series RLC circuit with characteristic equation s2 +

R 1 s+ =0 L LC

»R = 65; C = 0.1e-3; L = 0.1; si = roots([1 R/L 1/(L*C)]) s1 = si(1); s2 = si(2); si = -4.0000e+02 -2.5000e+02 At t = ∞, vC(∞) = 0.6*65 = 39 V. Hence vC (t) = K1e −400t + K2e −250t + 39 V Using initial conditions, vC(0) = K1 + K2 + 39 = 65 i (0+) iL (0+) −1 vC '(0) = −400K1 − 250K 2 = C = = −4 = −10 4 C C 10 »A = [1 1;s1 s2]; »b = [65-39; -1e4];K=A\b K= 2.3333e+01 2.6667e+00

Hence,

vC (t) = 23.333e −400t + 2.6667e −250t + 39 V

SOLUTION 9.37. (a) Rth = 200//50 + R = (40 + R). The characteristic equation is:  2 1 1  s+ s + =0  RthC LC  Critically damped means that both roots are the same, so the discriminant is zero, i.e. 2

 1  4 =0   −  RthC  LC Equivalently, Rth = 40 + R = 0.5

L = 50 Ω C

Thus R = 10 Ω. (b) Short the inductor and open the capacitor. Because the capacitor is in parallel + with the shorted inductor at t = 0-, vC(0 ) = vC(0 ) = 0. The Thevenin equivalent resistance seen by the LC-parallel combination is Rth = 50 Ω from part (a). A simple +

-

calculation indicates that Voc = 0.8*50 = 40 V. Therefore, iL(0 ) = iL(0 ) = 40/50 = 0.8 + A. To find vR(0 ) we use the following equivalent circuit:

Hence, 10 = −8 V. 10 + 40 + To compute the derivative of vR at 0 , consider that v R (0 + ) = −40 ×

d d d −40 − vC (t)  d iC(t) (v R (t)) = (10 × i R (t)) = 10  = −0.2 v (t) = −0.2 ( ) C  dt dt dt 50 dt C

Hence

dv R (t)  iC (0 + ) = −0.2 = −4000iC (0+ )  + dt  t =0 C

But iC (0 + ) =

−40 − i L ( 0+ ) = −0.8 − 0.8 = −1.6 A 50

Therefore,

(c)

dv R (t)  = −4000iC (0 + ) = 6400 V/s  + dt  t =0 Since the circuit is critically damped the roots of the characteristic equation are s1,2 = −

1 = −200 2RthC

According to table 9.2 for t ≥ 0, v R (t) = ( K1 + K2t )e −200t + X F It follows from the circuit and this equation that X F = v R (∞) = −8 V K1 = v R (0 + ) − XF = 0 dv R (t)  = −200K1 + K 2 = K2 = 6400 dt  t =0 + Therefore

(

)

v R (t) = 6400te−200t − 8 u(t) V

A plot of the waveform is given below

SOLUTION TO PROBLEM 9.38. For this problem we first compute the Thevenin equivalent of the circuit to the left of the capacitor for t > 0. Consider

Now observe V − 12 Vs − kv1 Vs − 12 Vs − k (12 − Vs ) Is = I1 + I2 = s + = + 500 500 500 500 It follows that  500  1 + k  Vs =  Is +  12 = Rth Is + Voc  2+ k 2 + k The parallel LC is now driven by this Thevenin equivalent. »L = 0.01; C = 1e-6; »% Critical damping means (1/(Rth*C))2 - 1/(L*C) = 0 »x = sqrt(4/(L*C)); »Rthcrit = 1/(C*x) Rthcrit = 50 »kcrit = (500 - 2*50)/50 kcrit = 8 »% For parallel circuit, larger Rth means less damping »% Hence, smaller Rth means overdamped. Smaller Rth »% means larger k. Therefore k > 8 is the ranger for »% overdamped response. For the critically damped response we have R = Rth = 50 Ω; hence »R = 50; C = 1e-6; L = 0.01; si = roots([1 1/(R*C) 1/(L*C)]) si =

-10000 -10000 in which case vC (t) = (K1 + K 2t)e −50t + X f At t = ∞, vC (∞) = v L (∞) = 0 in which case X f = 0. From the initial conditions, vC (0) = K1 = 0 Voc − i (0+) iC (0+) Rthcrit L vC '(0) = s1K1 + K2 = K 2 = = C C Hence K2 = Therefore

10.8 50 = 2.16 × 105 −6 10

vC (t) = 2.16 × 105 e −50t V

SOLUTION TO PROBLEM 9.39. (a) The series RLC leads to a characteristic equation of the form s2 +

R 1 24 1 5 s+ = s2 + s+ = s2 + 120s + = 0 L LC 0.2 0.2C C

For a critically damped response, 120 2 =

20 . Hence, C = 1.3889 mF. C

(b) »C = 20/120^2 C= 1.3889e-03 »L = 0.2; R = 24; »si = roots([1 R/L 1/(L*C)]) si = -60 -60 Hence iL (t) = (K1 + K2t)e −60t + 0.4 A where iL(∞) = 0.4 because at t = ∞, the capacitor looks like an open and the inductor like a short. Hence all current from the source flows through the inductor.

Using the initial conditions, iL(0) = K1 + 0.4 = 0 ⇒ K1 = – 0.4 v (0+) −vC (0+) − 24(i L (0+) − 0.4) iL '(0) = s1K1 + K2 = 60 × 0.4 + K 2 = L = = 48 L L Hence, K2 = 24 and

iL (t) = (−0.4 + 24t)e −60t + 0.4 A

SOLUTION TO PROBLEM 9.40. This problem differs from 39 in the initial condition calculation. Specifically, iL(0-) = iL(0+) = –0.4 vC(0-) = vC(0+) = 0 Again

iL (t) = (K1 + K2t)e −60t + 0.4 A

and iL(0+) = K1 + 0.4 = – 0.4 ⇒ K1 = – 0.8 v (0+) −vC (0+) − 24 (iL (0+) − 0.4) iL '(0) = 60 × 0.8 + K 2 = L = = 96 L L Hence, K2 = 48 and

iL (t) = (−0.8 + 48t)e −60t + 0.4 A

SOLUTION TO PROBLEM 9.41. (a) At 0-, the capacitor is an open circuit and inductor is a short circuit. So, vC(0-) = vC(0+) = 5 V -3 -3 iL(0-) = iL(0+) = 5×10 – 4×10 = 1 mA Now, at 0+, replace the capacitor by a 5 V voltage source and the inductor by a 1 mA current source. Also, the original independent current source is turned off. Solve the resulting circuit to obtain. -3 -3 iC + 5×10 – 1×10 = 0 ⇒ iC(0+) = –4 mA

vL(0+) = 0 (b) Since the circuit is a parallel RLC, the characteristic polynomial is 1 1 1 1 s+ = s2 + 4 s + = s2 + 2 s + 2 + 2d = 0 RC LC LC 10 C »R = 1e3; C = 0.5e-6; L = 0.184; si = roots([1 1/(R*C) 1/(L*C)]) si = -1.0000e+03 + 3.1416e+03i -1.0000e+03 - 3.1416e+03i s2 +

»wd = imag(si(1)) wd = 3.1416e+03 »sig = -real(si(1)) sig = 1000 -3

Also, at t = ∞, iL(∞) = 5×10 A. Therefore

[

]

iL (t) = e −1000t Acos(π × 103 t) + Bsin(π × 10 3 t) + 5 × 10 −3 A Using initial conditions iL (0) = A + 5 × 10−3 = 10 −3 ⇒ A = −4 × 10 −3 and v (0+) iL '(0) = −1000A + π × 103 B = 4 + π × 103 B = L =0 L Hence, »B = -4e-3/pi B= -1.2732e-03 Finally,

[

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iL (t) = −e −1000t 4 c o s π ( × 10 3 t) + 1.2732sin( π × 103 t) + 5 mA

SOLUTION TO PROBLEM 9.42. The response here coincides with that of problem 41 up to time t = 2s. At this point we need the new initial conditions on the circuit at t = 2+. However, at t = 2, e −1000t = 0 for all practical purposes. Hence, iL (2) = 5 × 10−3 A. Differentiating the expression for iL(t) and evaluating at t = 2 yields zero by inspection.

This follows because Ke −2000 = 0 for K in the range of 1 to 10 . This can also be seen from the circuit because at t = 2 s, the capacitor has charged to 5 V, making v (2+) iL' (2+) = L = 0. L 4

To find steady state current, solve the circuit with the new current source value and with the capacitor and inductor as open and short circuits, respectively: iL (∞) = 5 × 10 −3 + 4 × 10 −3 = 9 × 10−3

[ (

)

)]

(

iL (t) = −e −1000(t −2) A cos π × 103 (t − 2) + B sin π × 10 3(t − 2) + 9 × 10 −3 A Using the new initial conditions iL (2+) = A + 9 × 10 −3 = 5 × 10−3 ⇒ A = −4 × 10−3 and v (0+) iL '(2+) = −1000A + π × 10 3 B = 4 + π × 10 3 B = L =0 L -3

Hence B = –1.2732×10 and for t ≥ 2s,

[ (

)

(

)]

iL (t) = −e −1000(t −2) 4cos π × 10 3(t − 2) + 1.2732sin π × 103 (t − 2) + 9 mA

SOLUTION TO PROBLEM 9.43. At t = 0-, iL(0-) = 10/20 = 0.5 A and vC(0-) = –5 V by the usual considerations. At t = 0+, we have a parallel RLC circuit. Hence s2 +

1 1 1 1 s+ = s2 + 4 s + = s2 + 2 s + 2 + 2d = 0 RC LC LC 10 C

»R = 10; C = 0.05e-3; L = 0.01; si = roots([1 1/(R*C) 1/(L*C)]) si = -1.0000e+03 + 1.0000e+03i -1.0000e+03 - 1.0000e+03i »wd = imag(si(1)) wd = 1.0000e+03 »sig = -real(si(1)) sig = 1000

Also, at t = ∞, vC(∞) = –5 V. Therefore

[

]

vC (t) = e −1000t A cos(103 t) + Bsin(10 3 t) − 5 V Using initial conditions vC (0) = A − 5 = −5 ⇒ A = 0 and

i (0+) −0.5 vC '(0) = −1000A + 1000B = 10 3 B = C = = −10 4 −3 C 0.05 × 10

Hence, B = 10 and vC (t) = −10e −1000t sin(103 t) − 5 V Further

v L (t) = −5 − v C (t) = 10e −1000t sin(103 t) V

SOLUTION TO PROBLEM 9.44. (a) The circuit is a driven series RLC. Hence s2 +

R 1 s+ = s2 + 2 s + 2 + 2d = 0 L LC

»R = 400; C = 0.5e-6; L = 0.2; si = roots([1 R/L 1/(L*C)]) si = -1.0000e+03 + 3.0000e+03i -1.0000e+03 - 3.0000e+03i »wd = imag(si(1)) sig = -real(si(1)) wd = 3000 sig = 1000 Also, at t = ∞, vC(∞) = 400×0.006 = 2.4 V. Therefore vC (t) = e −1000t [A cos(3000t) + B sin(3000t)] + 2.4 V Using initial conditions vC (0) = A + 2.4 = 2 ⇒ A = −0.4 and

i (0+) i L (0+) vC '(0) = −1000A + 3000B = 400 + 3000B = C = = 1.6 × 104 C C Hence, B = 5.2 and vC (t) = e −1000t [−0.4cos(3000t ) + 5.2sin(3000t )] + 2.4 V (b) Consistent with underdamped circuit behavior and because the inductor behaves as a short and the capacitor as an open at t = ∞ (iL(∞) = 0), iL (t) = e −1000t [ Acos(3000t) + Bsin(3000t)] V Using the initial conditions, iL (0+) = A = 0.008 A Further,

v (0+) 400(0.006 − 0.008) − 2 iL '(0) = −1000A + 3000B = −8 + 3000B = L = = −14 L 0.2

Hence, B = −0.002 and iL (t) = e −1000t [0.008cos(3000t) − 0.002sin(3000 t)] V Finally

v L (t) = v R (t) − vC (t) = 400[0.006 − i L (t)] − vC (t) = 2.4 − 400i L (t) − v C (t)

which implies that v L (t) = e −1000t [3.6cos(3000t) − 13.2sin(3000t )] V

SOLUTION TO PROBLEM 9.45. This circuit is the same series RLC as problem 44. Note that at t = ∞, vC(∞) = –2.4 V. Hence vC (t) = e −1000t [A cos(3000t) + B sin(3000t)] − 2.4 V Now, the initial conditions are: vC(0-) = 2.4 = vC(0+) V, iL(0-) = 0 = iL(0+) Thus

vC (0) = A − 2.4 = 2 . 4 ⇒ A = 4.8

Further, i (0+) vC '(0) = −1000A + 3000B = −4800 + 3000B = C =0 C

Hence, B = 1.6 and vC (t) = e −1000t [4.8cos(3000t) + 1.6sin(3000 t)] − 2.4 V

SOLUTION TO PROBLEM 9.46. For all three cases, assuming iL is pointing downward, At t = ∞,

vC (0−) = vC (0+) = 0 and iL (0−) = iL (0+) = 0.1 A vC (∞) = 0 and iL (∞) = −0.2 A

Further iC (0+) = −iL (0+) +

−10 − vC (0+) = −0.1− 0.2 = −0.3 A 50

Lastly, all three cases are for a parallel RLC whose characteristic equation is: s2 +

1 1 s+ =0 RC LC

(a) »R = 50; C = 0.04e-3; L = 0.625; si = roots([1 1/(R*C) 1/(L*C)]) s1 = si(1); s2 = si(2); si = -400 -100 Hence,

vC (t) = K1e −100t + K2e −400t V

From, the initial conditions vC(0+) = 0 = K1 + K2 and

i (0+) −0.3 vC '(0) = s1K1 + s2K 2 = −100K1 − 400K2 = C = = −7500 C 0.04 × 10 −3

»A = [1 1;si(2) si(1)]; »b = [0;-7500]; »K = A\b K= -25 25 Therefore

(

)

vC (t) = −25 e −100t − e −400t V (b) »R = 50; C = 0.04e-3; L = 0.4; »si = roots([1 1/(R*C) 1/(L*C)]) s1 = si(1); s2 = si(2); si = -2.5000e+02 -2.5000e+02 Thus From IC's,

vC (t) = (K1 + K 2t)e −250t V vC (0) = K1 = 0 i (0+) vC '(0) = s1K1 + K2 = K 2 = C = −7500 C

Therefore

vC (t) = −7500te−250t V

(c) »L = 0.2; »si = roots([1 1/(R*C) 1/(L*C)]) si = -2.5000e+02 + 2.5000e+02i -2.5000e+02 - 2.5000e+02i vC (t) = e −250t [A cos(250t) + Bsin(250t)] V From ICs.

vC (0+) = A = 0 i (0+) vC '(0) = −250A + 250B = 250B = C = −7500 C

in which case B = –30. Thus

vC (t) = −30e −250t sin(250t) V

SOLUTION TO PROBLEM 9.47. At t = 0-, he capacitor is open and the inductor is a short. This together with the continuity property implies vC(0-) = vC(0+) = 10 V and iL(0-) = iL(0+) = 1 A by inspection.

Now, for t = 0+, vin = 0, replace capacitor and inductor with a voltage source and a current source, respectively (values are those of the initial conditions). Solve for initial capacitor current and initial inductor voltage to obtain: vL(0+) = -10 V iC(0+) = iL(0+) – iR1 – iR2 = – 1 A Notice that the resulting circuit is an undriven parallel RLC circuit with Req = 10//10 = 5 Ω. »R = 5; C = 0.01; L = 4/3; si = roots([1 1/(R*C) 1/(L*C)]) s1 = si(1); s2 = si(2); si = -15 -5 Hence,

vC (t) = K1e −15t + K 2e −5t V


i (0+) −1 vC '(0) = s1K1 + s2K 2 = −15K1 − 5K 2 = C = = −100 C 0.01

»A = [1 1;-15 -5]; »b= [10; -100]; »K = A\b K= 5.0000e+00 5.0000e+00 vC (t) = 5e −15t + 5e −5t V Verify with dsolve function in matlab: dsolve('D2y+20*Dy+75*y=0,y(0)=10,Dy(0)=100') »dsolve('D2y+20*Dy+75*y=0,y(0)=10,Dy(0)=-100') ans = 5*exp(-5*t)+5*exp(-15*t)

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SOLUTION TO PROBLEM 9.48. For t = 0-, there is no source present nor has there been a non-zero excitation. Hence, vC(0-) = vC(0+) = 0 and iL(0-) = iL(0+) = 0 At t = 0+, replace capacitor and inductor by 0-valued voltage and current sources to obtain: vL(0+) = 10 V, iC(0+) = 1 A For t > 0, we have a driven parallel RLC circuit with vC(∞) = 10 V. Thus »R = 5; C = 0.01; L = 4/3; si = roots([1 1/(R*C) 1/(L*C)]) s1 = si(1); s2 = si(2); si = -15 -5 and

vC (t) = K1e −15t + K 2e −5t + 10 V

From, the initial conditions vC(0+) = 0 = K1 + K2 +10 and

i (0+) 1 vC '(0) = s1K1 + s2K 2 = −15K1 − 5K 2 = C = = 100 C 0.01 »A = [1 1;-15 -5]; »b= [-10; 100]; »K = A\b K= -5.0000e+00 -5.0000e+00 in which case

vC (t) = −5e −15t − 5e −5t + 10 V

To compute iL, note that iL(∞) = 1 A, iL (t) = K1e −15t + K 2e −5t + 1 A From, the initial conditions iL(0+) = 0 = K1 + K2 +1 and v (0+) 10 iL '(0) = s1K1 + s2K2 = −15K1 − 5K2 = L = = 7.5 L 4/3 »A = [1 1;-15 -5]; »b= [-1; 7.5]; »K = A\b K= -2.5000e-01 -7.5000e-01 iL (t) = −0.25e −15t − 0.75e −5t + 1 A

SOLUTION TO PROBLEM 9.49. Input to this circuit is a superposition of the inputs in problems 9.47 and 9.48. So, the output of the circuit here is a superposition of the output of the circuit in problems 9.47 and 9.48: vC (t) = 10e −15t + 10e −5t − 10 V For t > 0, by linearity this is the difference of the zero-input circuit response (i.e., due to the IC's as per problem 47) and the zero-state (zero ICs) as per problem 48.

SOLUTION TO PROBLEM 9.50. (a) vC(0-) = –60 V, iL(0-) = –0.1 A (b) By continuity property, vC(0+) = –60, iL(0+) = –0.1 (c) Replace capacitor by voltage source of value –60 V and inductor by current source of value –0.1 A. vL(0+) + vC(0+) = 60 ⇒ vL(0+) = 120 and iC(0+) + iR1(0+) – 1 – iL(0+) = 0 ⇒ iC(0+) = 1 (d) Req = 120//600 = 100 Ω. »R = 100; C = 1e-3; L = 2; si = roots([1 1/(R*C) 1/(L*C)]) si = -5.0000e+00 + 2.1794e+01i -5.0000e+00 - 2.1794e+01i »sig = -real(si(1)); wd = imag(si(1)); (e) Note that if the excitation of 60 V had remained forever, then iL(∞) would be 0.1 A. Therefore for the interval 0 < t < 1, iL (t) = e −5t [ A cos(21.794t) + B sin(21.794t)] + 0.1 A (f) iL(0+) = A + 0.1 = –0.1 ⇒ A = –0.2 v (0+) iL' (0+) = −5A + 21.794B = 1 + 21.794B = L = 60 L Hence, B = 2.707 and iL (t) = e −5t [−0.2cos(21.794t) + 2.707sin(21.794t)] + 0.1 A (g) For t > 1, the forcing function is zero iL(∞) = 0. Thus, iL (t) = e −5(t −1) [ A cos(21.794(t − 1) ) + Bsin(21.794(t − 1) )] A

From part (f) of the exp(-5t) term, we can guess that iL(1) approximates 0.1 and also that vC(1) approximates its steady state value of 60 V. These may be off by percent or two, but are good enough for our engineering calculations. It follows that vL(1+) approximates –60 V. iL (1+) = A = 0.1 and v (1+) iL' (1+) = −5A + 21.794B = −0.5 + 21.794B = L = −30 L Here B = –1.35. Hence iL (t) = e −5(t −1) [0.1cos(21.794(t − 1) ) − 1.35sin (21.794(t − 1) )] A *SOLUTION 9.51. To find the initial conditions, use the following equivalent circuit at t = – 0 .

By inspection iL (0 + ) = i L ( 0− ) = 1 A and vC (0 + ) = v C( 0− ) = 5 V. To find the characteristic roots, set independent source to zero which means open circuit the independent current source in figure P9.51. This leaves a series RLC with Rth = 10 Ω. Hence R 1 s2 + th s + = s2 + 100s + 2.5 × 10 4 = 0 L LC Using MATLAB, we find »Rth = 10;C = 0.4e-3; L = 0.1; »s12=roots([1 Rth/L 1/(L*C)]) s12 = -50 -50 Since for t > 0, the source is off, we use table 9.1, case 3 to obtain iL (t) = ( K1 + K2t )e −50t A

It follows that 1 = iL (0 + ) = K1 and

di L + 1 (0 ) = −50K1 + K2 = −50 + K2 = v L (0 + ) = 10v L (0 + ) dt L + + To find v L (0 ) we consider the equivalent circuit valid at 0 :

It follows that Hence or K2 = 0. Finally

v L (0 + ) = 5 − 10 × 1 = −5 V −50 + K2 = −50 iL (t) = e−50t u(t) A

SOLUTION TO PROBLEM 9.52. To find the initial conditions, use the following equivalent – circuit at t = 0 .

By inspection iL (0 + ) = i L ( 0− ) = 1 A and vC (0 + ) = v C( 0− ) = 5 V. To find the characteristic roots, set independent source to zero which means open circuit the independent current source in figure P9.51. This leaves a series RLC with Rth = 10 Ω. Hence R 1 s2 + th s + = s2 + 20s + 5 × 10 3 = 0 L LC Using MATLAB, we find »Rth = 10;C = 0.4e-3; L = 0.5; »s12=roots([1 Rth/L 1/(L*C)]) s12 = -1.0000e+01 + 7.0000e+01i -1.0000e+01 - 7.0000e+01i Since for t > 0, the source is off, we use table 9.1, case 2 to obtain iL (t) = e −10t [ Acos(70t) + Bsin(70t)] A It follows that 1 = iL (0 + ) = A and diL + 1 (0 ) = −10A + 70B = −10 + 70B = v L (0+ ) = 2v L (0+ ) dt L + + To find v L (0 ) we consider the equivalent circuit valid at 0 :

It follows that

v L (0 + ) = 5 − 10 × 1 = −5 V Hence −10 + 70B = −10 or B = 0. Finally iL (t) = e −10t cos(70t)u(t) A

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SOLUTION TO PROBLEM 9.53. (a) vC (0−) = vC (0+) = 150 × 10−3 × 80 = 12 V. (b) iL(0-) = iL(0+) = 150 mA (c) For these values at 0+, the independent current source becomes an open circuit. Replace the inductor and capacitor with current and voltage sources to represent the initial conditions. Solve the resulting simple circuit to obtain: vL(0+) = –6 = 12 – 120*0.15 V, iC(0+) = iL(0+) = –150 mA (d) »Rth = 120;C = 2/9 * 1e-3; L = 0.6; »s12=roots([1 Rth/L 1/(L*C)]) s12 = -1.5000e+02 -5.0000e+01 (e)

vC (t) = K1e −150t + K2e −50t V


i (0+) vC '(0) = s1K1 + s2K 2 = −150K1 − 50K 2 = C = −675 C A = [1 1;-150 -50]; b= [12; -675]; K = A\b K= 7.5000e-01 1.1250e+01 in which case

vC (t) = 0.75e −150t + 11.25e −50t V

(f) iL (t) = −iC (t) = −C

[

dvC (t) 2 = × 10 −3 (−150 × 0.75)e −150t + (−50 × 11.25)e −50t dt 9

]

= 0.025e −150t + 0.125e −50t A SOLUTION TO PROBLEM 9.54. (a) At t = 0-, the independent current source is off, the inductor is a short circuit, and the capacitor is an open circuit. By voltage division, vC(0-) = vC(0+) = (80/100)50 = 40 V (b) iL(0-) = iL(0+) = 50/100 = 0.5 A (c) At t = 0+, we have an independent current source. Also, we replace the inductor with a current source and the capacitor with a voltage source to represent the initial conditions. 1.5 – iL(0+) = iC(0+) ⇒ iC(0+) = 1 A Further, vC(0+) + 40 iC(0+) = 80 iL(0+) + vL(0+) ⇒ vL(0+) = 40 V (d) Rth = 120;C = 2/9 * 1e-3; L = 0.6; s12=roots([1 Rth/L 1/(L*C)]) s12 = -1.5000e+02 -5.0000e+01 (e) In steady state, the capacitor is open and the inductor is a short in which case, Xf = 80 * 1.5 = 120 V. vC (t) = K1e −150t + K2e −50t + 120 V

From, the initial conditions vC(0+) = 40 = K1 + K2 + 120 and i (0+) vC '(0+) = −150K1 − 50K2 = C = 4.5 × 103 C A = [1 1;-150 -50]; b= [-80; 4.5e3]; K = A\b K= -5 -75 in which case

vC (t) = −5e −150t − 75e −50t + 120 V

(f) iL (t) = 1.5 − iC (t) = 1.5 − C

dvC (t) = −0.16667e −150t − 0.83333e −50t + 1.5 A dt

SOLUTION TO PROBLEM 9.55. (a) At t = 0-, we replace the inductor by a short circuit and the capacitor by an open circuit; hence iL(0-) = iL(0+) = 1 A and vC(0-) = vC(0+) = (40)(1) – 20 = 20 V (b) At t = 0+, we replace the capacitor by a voltage source of value 20 V and the inductor by a current source of value 1 A. Since the inductor current is 1 A and the independent current source outputs 1 A, no current flows through the branch containing the capacitor. Therefore, iC(0+) = 0 Also, because of the zero current in the branch containing the capacitor, no voltage drop occurs across the resistance in series with the capacitor. Therefore, the voltage across the independent current source is vC(0+). Therefore, vL(0+) = vC(0+) – 40*iL(0+) = – 20 V (c) At steady state (large t), the capacitor becomes an open circuit and the inductor becomes a short circuit. By inspection, vC(∞) = 40 V

(d) Rth = 80;C = 1/15 * 1e-3; L = 0.1; s12=roots([1 Rth/L 1/(L*C)]) s12 = -5.0000e+02 -3.0000e+02 vC (t) = K1e −500t + K2e −300t + 40 V (e) From, the initial conditions vC(0+) = 20 = K1 + K2 + 40 and i (0+) vC '(0+) = −500K1 − 300K 2 = C =0 C A = [1 1;-500 -300]; b= [-20; 0]; K = A\b K= 3.0000e+01 -5.0000e+01 in which case

vC (t) = 30e −500t − 50e −300t + 40 V

SOLUTION TO PROBLEM 9.56. (a) At t = 0-, we replace the inductor by a short circuit and the capacitor by an open circuit; hence iL(0-) = iL(0+) = 1 A and vC(0-) = vC(0+) = (40)(1) – 20 = 20 V (b) At t = 0+, we replace the capacitor by a voltage source of value 20 V and the inductor by a current source of value 1 A. Since the inductor current is 1 A and the independent current source outputs 1 A, no current flows through the branch containing the capacitor. Therefore, iC(0+) = 0 Also, because of the zero current in the branch containing the capacitor, no voltage drop occurs across the resistance in series with the capacitor. Therefore, the voltage across the independent current source is vC(0+). Therefore,

vL(0+) = vC(0+) – 40*iL(0+) = – 20 V (c) At steady state (large t), the capacitor becomes an open circuit and the inductor becomes a short circuit. By inspection, vC(∞) = 40 V (d) Rth = 80;C = 62.5e-6; L = 0.1; s12=roots([1 Rth/L 1/(L*C)]) s12 = -400 -400 vC (t) = (K1 + K 2t )e −400t + 40 V (e) From, the initial conditions vC(0+) = 20 = K1 + 40 ⇒ K1 = – 20 and i (0+) vC '(0+) = −400K1 + K 2 = 8000 + K 2 = C =0 C in which case

vC (t) = − (20 + 8000t)e −400t + 40 V

SOLUTION TO PROBLEM 9.57. Step 1: dvC/dt = iC/C = 2iC diL/dt = vL/L = 0.5vL Step 2: iR = vC/2. From KCL, iL – iR – ic = 0. Therefore, ic = iL – vc/2. Step 3: Similarly, vin – vL – vc = 0, which implies vL = vin – vC. Hence, dvC/dt = 2 iL – vC diL/dt = 0.5 vin – 0.5 vC Step 4. Eliminate terms in vC (see equation 9.47 in text) to obtain: d2iL/dt2 + diL/dt + iL = 0.5 dvin/dt + 0.5 vin

SOLUTION TO PROBLEM 9.58. (a) At t = 0-, vC(0-) = vC(0+) = 0 iL(0-) = iL(0+) = 0 For t between 0 and 1, we have a parallel RLC circuit, with Req being the parallel combination of the two 21.1333 Ω resistors. R = 21.1333/2; C = 15.7729e-3; L = 0.1; si = roots([1 1/(R*C) 1/(L*C)]) si = -3.0000e+00 + 2.5000e+01i -3.0000e+00 - 2.5000e+01i If there were no further switchings, then Xf = 1 A. Hence, iL (t) = e −3t [ A cos(25t) + Bsin(25t)] + 1 A Applying the initial conditions, iL(0+) = A + 1 = 0 ⇒ A = –1 and Hence, for 0 ≤ t < 1,

v (0+) iL' (0+) = −3A + 25B = 3 + 25B = L =0 L iL (t) = −e −3t [cos(25t) + 0.12sin(25 t)] + 1 A

Now, for the next interval, we need initial conditions. These are obtained from the above equation for iL(t) at t = 1. iL(1) = e-3[-cos(25) –3/25 sin(25)] + 1 = 0.9514 and

iL' (1) = 3e −3[cos(25) + 0.12sin(25)] − e −3[−25sin(25) + 0.12 × 25cos(25)] = −0.1671

The circuit is still a parallel RLC circuit, but now there is no source and R = 1.268: R = 1.268; C = 15.7729e-3; L = 0.1; si = roots([1 1/(R*C) 1/(L*C)]) si = -2.5000e+01 + 3.0002e+00i -2.5000e+01 - 3.0002e+00i Hence,

iL (t) = e −25(t −1)[ A cos(3(t − 1)) + B sin(3(t − 1))]

Using initial conditions A = 0.9514 –0.1671 = –25A+ 3B ⇒ B = 7.8726 Thus, for 1 ≤ t < 2 iL (t) = e −25(t −1) [0.9514cos(3(t − 1)) + 7.8726sin(3( t − 1)) ] A (b) In period between 1 and 2 seconds, the response has a time constant of 1/25. So, when t = 2, 25 time constants would have passed from the time the switch is turned (t = 1). This means that the L and C currents and voltages would have settled almost identically to their values at 0–. A similar argument can be made for the other cycle. Thus the overall response effectively becomes a periodic response equal to the response over 0 ≤ t < 2 that reflects the periodicity of the switching. (c) 1.8 1.6

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SOLUTION TO PROBLEM 9.59. (a) dvc/dt = ic/C = 0.707ic diL/dt = vL/L = 0.707vL To find expressions for iC and vL we use the following figure.

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From this resistive circuit, iC = – vC/1 + iL and vL = vin – 1×iL – vC v˙C = −0.707v C + 0.707iL i˙L = −0.707vC − 0.707iL + 0.707v in Using equation 9.47, d 2vC dv + 1.414 C + v C = 0.5vin 2 dt dt »si = roots([1 sqrt(2) 1]) si = -7.0711e-01 + 7.0711e-01i -7.0711e-01 - 7.0711e-01i (b) At steady state (large t), vC = 0.5 vC(t) = e-0.707t[A cos(0.707t) + B sin(0.707t)] + 0.5 V At t = 0–, vC(0-) = vC(0+) = A + 0.5 = 0 ⇒ A = –0.5 i (0) iL (0) − vC (0) vC' (0) = C = = 0 = −0.707A + 0.707B C C ⇒ B = –0.5 Thus vc(t) = e-0.707t[–0.5 cos(0.707t) – 0.5 sin(0.707t)] + 0.5 V

SOLUTION TO PROBLEM 9.60. (a) dvC = 3iC dt diL v L = dt 3

To eliminate iC and vL, consider

Hence iC = iin – vC/2 – iL vL = vC – 12iL v˙C = −1.5v C − 3iL + 3iin 1 i˙L = v C − 4iL 3 Using equation 9.47, d 2vC dv C ' + 5.5 + 7v C = 3iin + 12iin 2 dt dt »si = roots([1 5.5 7]) si = -3.5000e+00 -2.0000e+00 Note: iin’(t) = 0 for t>0. At t = 0-, current source is off, inductor is a short circuit, and capacitor is an open circuit. vc(0-) = vc(0+) = 0 iL(0-) = iL(0+) = 0 At t = 0+, the current source is on. Replace the inductor and capacitor by current and voltage sources to represent the initial conditions. Hence, iC(0+) = 1 A. At t = ∞, vC (∞) = 1× (12 / / 2) = 1.7143. Thus, vC (t) = K1e −3.5t + K2e −2t + 1.7143 V To find the constants,

vC (0) = K1 + K 2 + 1.7143 = 0

and i (0) vC' (0) = C = 3 = −3.5K1 − 2K 2 C

»A = [1 1;-3.5 -2]; »b = [-1.7143;3]; »K = A\b K= 2.8573e-01 -2.0000e+00 Therefore

vC (t) = 0.28573e −3.5t − 2e −2t + 1.7143

(b) From equation 9.47b, d 2iL di + 5.5 L + 7vC = iin 2 dt dt

Since iL (∞) = 0.14286 A,

iL (t) = K1e −3.5t + K2e −2t + 0.14286 A

Using the initial conditions, iL (0) = K1 + K 2 + 0.14286 = 0 and

v (0) vC (0) − 12iL (0) iL' (0) = L = = 0 = −3.5K1 − 2K2 L L

»A = [1 1;-3.5 -2]; »b = [-1/7; 0]; »K = A\b K= 1.9048e-01 -3.3333e-01 Therefore,

iL (t) = 0.19048e −3.5t − 0.3333e −2t + 0.14286 A

SOLUTION TO PROBLEM 9.61. (a) dvC1/dt = 0.5iC1 dvC2/dt = 0.5iC2 Writing a node equation, iC1 + vC1/0.5 + (vC1 – vC2)/0.5 = 0 ⇒ iC1 = –4vC1 + 2vC2 By symmetry, iC2 = –4vC2 + 2vC1 Hence

v˙C1 = −2v C1 + vC 2 v˙C 2 = v C1 − 2v C2 From equation 9.47a, d 2vC dv C + 4 + 3vC = 0 dt dt 2 (b) »si = roots([1 4 3]) si = -3 -1 Hence

vC (t) = K1e −3t + K 2e −t

(c) vC1(0) = 2 ⇒ K1 + K2 = 2 and

Hence

i (0) 2(v C 2 (0) − vC1(0)) − 2vC1 (0) 8 − 8 ' vC1 (0) = C1 = = = 0 = −3K1 − K2 C C C vC (t) = −e −3t + 3e −t V

*SOLUTION P9.62. (a) For this problem we need to define a voltage at the output of the first op amp as shown below.

Also, let us relate the input and output voltages for an arbitrary leaky integrator as shown below.

We now write a node equation at the inverting terminal of the op amp. Here va vb dv + +C b = 0 Ra Rb dt Equivalently dv b v v =− b − a dt RbC Ra C

(1)

Now we apply the formula of (*) to the second stage of our given op amp circuit to obtain: dvout v v (2) = − out − 1 dt RC2 RC2 where R = 1 MΩ and C2 is to be determined. Applying the formula of (*) to the first stage we obtain: dv1 v v (3) =− 1 − s dt RC1 R1C1 The equations (2) and (3) form a coupled set of state equations which we can write as 1  − 1 − 0 ˙  v  out  RC2 RC2   vout   −1  v  v˙  =  1   v1  +   s  1   0 − R C  1 1  RC1  Using equation 9.47a of text we can write down the characteristic equation as  1  1  1 1  1  s2 +  + = s + s + s +  =0  RC2 RC1  RC1 RC2  RC2   RC1  We require that the natural frequencies be –4 and –12 in which case

1 1 = 4, = 12 RC2 RC1 From MATLAB »R = 1e6; »C2 = 1/(R*4) C2 = 2.5000e-07 »C1 = 1/(R*12) C1 = 8.3333e-08 (b) For this part, the overall dc gain must be 10. The dc gain of the second stage is –1. 106 The dc gain of the first stage must be −10 = − ; hence R1 = 100 kΩ. R1 (c) Since the roots are distinct and real, for t > 0 vout (t) = K1e −4t + K 2e −12t + X f = K1e −4t + K2e −12t + 10 V where X f = 10 V by part (b). The problem states that the capacitor voltages are initially zero. Hence 0 = vC2 ( 0 ) = vout (0) = K1 + K2 + 10 Equivalently

K1 + K2 = −10

(1)

Also, dvout 1 (0) = i ( 0 ) = −4K1 − 12K2 = 0 dt C2 C2

(2)

because iC2 ( 0 )= 0 . This is so because vC1( 0 )= 0 = v1( 0 ) means no current flows through the 1 MΩ input resistor to stage 2. This fact and the fact that vC2 (0) = 0, means that no current flows through C2. From equation 2, K1 = –3 K2. Substituting into equation (1) yields K2 = 5 and hence K1 = –15. Finally,

vout (t) = −15e −4t + 5e −12t + 10 V

*SOLUTION 9.63. This problem requires the characteristic equation in terms of A. For this we may set vin = 0 and the circuit becomes the one given below. Note the new label v1.

The first step is to write a node equation at v1: v1 d 1  v + 10 −4 ( v1 − vout ) + v1 − out  = 0  100 dt 100 A Equivalently dv1 dvout 100 − = v − 200v1 dt dt A out

(*)

Now we write a node equation at vg = vout/A. Here, 10 −4 dvout 1  vout + − v1 = 0  A dt 100 A Equivalently, dvout = −100v out + 100 Av1 (**) dt Let us put (*) and (**) in matrix form to obtain: 1 0 

−1  v˙1   −200 100 / A   v1  = 1   v˙out  100A −100   vout 

We can now solve this to obtain the state equations  v˙1  1 1  −200 100 / A  v1   −200 + 100A −100 + 100 / v˙  = 0 1 100A −100   v  =  100 A −100  out      out  

A  v1   v    out 

Compare these equations with equation 9.37 and use the formula of 9.47b to obtain the following second order differential equation in vout: d 2 vout dvout 4 2 + (300 − 100A ) dt + 10 vout = 0 dt The discriminant of this characteristic equation is plotted below for 0 < A < 3. For values of A > 3, the circuit is unstable. A negative value of the discriminant indicates underdamped (1 < A < 3) and a positive value overdamped (0 < A < 1). For A = 1, we have critical damping.

Solution 9.64 We can write two state equations as follows: (i) From the definition of a capacitor, dvC1 = 10 7 iC1 dt dvC 2 = 109 iC 2 dt (ii) From KVL and Ohm's law dvC1 = 10 6 (vi − vC1 − v C 2 ) = −10 6 vC1 − 10 6 vC 2 + 106 v i dt dvC 2 = 108 (vi − vC1 − v C 2 − 0.01vC 2 ) = −108 v C1 − 1.01× 108 v C2 + 108 vi dt Casting these two equations into a second order differential equation, as described in the text: d 2vC 2 dv dv + 1.02 × 108 C 2 + 1012 vC 2 = 108 i 2 dt dt dt The characteristic equation for this differential equation has real roots: »si = roots([1 1.02e8 1e12]) si = -1.0199e+08 -9.8049e+03

Since the capacitors become open circuits, vC2(∞) = 0 and vC1(∞) = vi. 8 3 vC 2 ( t) = K1e −1.02×10 t + K2e −9.8×10 t V

Applying IC's: vC 2 (0) = K1 + K2 = 0 Also, i (0+) 0.1 vC' 2 (0) = C 2 = −9 = −1.02 × 108 K1 − 9.8 × 10 3K 2 C2 10 Thus

vC' 2 (0) = 1 = −1.02K1 − 9.8 × 10−5K 2

»b = [0; 1]; »A = [1 1; -1.02 -9.8e-5]; »K = A\b K= -9.8049e-01 9.8049e-01 8 3 vC 2 ( t) = −0.9805e −1.02×10 t + 0.9805e −9.805×10 t V

»t = 0:1/(abs(s1)*100):100/abs(s1);vc = -0.9805*exp(s1*t) + 0.9805*exp(s2*t); »plot(t,vc) »grid »xlabel('Time in s') »ylabel('vout in V') 1

vout in V

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Solution 9.65. First, derive the differential equation by writing state equations: dvC = 3iC dt diL v L = dt 3 Now, assume that the capacitor is a voltage source and the inductor is a current source, and write by KCL v v iC = − C + C − iL 2 RN And by KVL: v L = vC − 12i L Substitute into the differential equations: dvC  3 3  = − + v − 3iL dt  2 RN  C diL 1 = v − 4i L dt 3 C Using equation 9.47 we obtain  dvC  3  d 2vC  3 3  3  − − + − 4 + − + ( −4 ) + 1    vC = 0     dt 2  2 RN   dt  2 RN   or equivalently d 2vC  3  dvC  3  +  5.5 − + 7 − v =0  2  RN  dt  RN  C dt For constant amplitude oscillations, the middle term should be zero, which means that RN = 3/5.5 = 0.54545 Ω. Thus the negative resistance is – RN = –0.54545 Ω. SOLUTION 9.66. The problem data is i1(t) = Im sin(ωt + θ) A R1 = 500 + 100(Im – 0.01), R2 = 500

Suppose it starts out with exponentially growing amplitude. R1 will increase with increasing amplitude. This changes the location of the roots until equilibrium is reached where the roots and the amplitude are stable. This is achieved when the roots of the characteristic equation describing the output voltage are purely imaginary, i.e., (R1 – R2)/(R1R2C) = 0 ⇒ R1 = R2 = 500 = 500 + 100(Im – 0.01) ⇒ Im = 0.01 (a)

ω0 = 1/[(500)(1µ)] = 2 k-rad/s

(b) Amplitude of i1 is 0.01. Thus i1 = 0.01 sin(ω0t) A. Let v1 = Vm cos(ω0t + φ) V. Then dv1/dt = –ω0Vmsin(ω0t + φ) = (0.01/C) sin(ω0t) ⇒ Vm = 0.01/(ω0C) = 5 V and φ = π rad Now, v2 + v1 + i1R1 – 3v2 = 0 ⇒ v2 = v1/2 + i1R1/2. Finally, vout = 3v 2 = 1.5v1 + 1.5R1i1 = 7.5cos( 0t + π) + 7.5sin( 0 t) V Hence the amplitude of vout(t) is: 7.5 2 V.

SOLUTION 9.67. (a) n

=

1 = 10 4 rad/s RC

(b) From equation 9.47a, d 2 v1 1 2 + 2 2 v1 = 0 dt R C Hence,  1   1  v1( t) = A cos t + Bsin t V RC RC  Using the value of v1(0) we have, v1(0) = A = 5 To compute the second initial condition,

i (0+) i (0+) v1'(0 + ) = C1 = R1 −6 = 104 B C 0.1× 10 But, iR1 (0+) =

3v 2 (0+) − v 2 (0+) − v1(0+) 2v2 (0+) − v1(0+) = = −0.005 Amps R1 103

Hence v1'(0

−5 × 10 −3 +) = = 104 B −6 0.1× 10

Hence B = –5. v1( t) = 5cos (10,000t ) − 5sin (10,000 t) V

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*SOLUTION 9.68. For this problem, R2 = 10 kΩ should be Rf = 10 kΩ and R1 = 1 kΩ should be R2 = 1 kΩ. (a) For sustained sinusoidal oscillation, R1 = R2 = 1 kΩ. From equation 9.59, ω0 =

1 = 10 4 rad/s or 1.5915 kHz R1C

(b) From figure P9.68, to obtain an R1 = 1 kΩ, IR1,peak = 0.2 mA. Therefore, dv iR1(t) = 0.2sin(ω 0t + θ) mA for appropriate θ. Since C 1 = iC1 = iR1 = 0.2sin(ω 0t + θ) dt mA, we know that v1 has the following form: v1(t) = V1m cos(ω 0t + φ) In which case dv d C 1 = C (V1m cos(ω 0 t + φ)) = CV1mω 0 sin(ω 0 t + φ+ π) = 0.2sin(ω 0t + θ) mA dt dt Therefore CV1m ω 0 = 0.2 × 10 −3 . It follows that V1m = 0.2 volts. Here V1m is the peak value of the sinusoid. However, the op amp peak output voltage with respect to ground, as shown in problem 66, is 1.5 2V1m . Also, for such a small amplitude, we expect the output waveform to be quite close to sinusoidal. By choosing a different lamp (R1) with a different characteristic, we can obtain larger peak output voltages.

SOLUTION 9.69. (a) Note that the capacitor is like an open circuit and the inductor is like a short circuit at t=0-. Thus, we can obtain the capacitor voltage by voltage division: vC (0 − ) = 10 ×

4 = 8 = vC (0 + ) 5

Similarly, the inductor current is obtained by applying Ohm’s Law: iL (0 − ) =

10 = 2 = iL (0 + ) 5

(b) Here, we note that the new initial conditions are just 2.5 times the values that we just obtained in part (a). This can be achieved by simply changing the input voltage source, from 10 to 25 V.

PROBLEM SOLUTIONS CHAPTER 10 SOLUTION 10.1 Using KCL, we can write C

dvC v C + = iin ( t) dt R

Dividing by C: dvC 1 i (t ) + v C = in dt RC C We know that iin (t ) = 20sin (400t ) mA, which can be represented by a complex exponential,

[

]

iin (t ) = Re 20e j 400t e − jπ/2 mA. For convenience we will simply let iin (t ) = 20e j 400te − jπ/2 mA, knowing that we must take the real part to complete our solution. The output voltage will also be reparesented as a complex exponential: v (t ) = V e j ( 400t + ) = V e j 400t e j C

m

m

Substituting this expression into the differential equation and canceling e j400t : V 20 × 10−3e − jπ/2 j400Vme j + m e j = RC C Thus −3 − j4000  1  20 × 10 Vme j  + j 400 = ⇒ Vm e j = = 3.714∠− 111.8o RC C 1000 + j400

where the values for R = 100 Ω and C = 5 mF were substituted in. Thus, Vm = 3.714  400  = 0 − tan−1  = −111.8° 1000 Taking into account a 90o phase shift we obtain vC (t ) = 3.714cos( 400t − 111.8°) = 3.714sin ( 400t − 21.8°) V and iout ( t) = 18.57sin (400t − 21.8°) mA

SOLUTION 10.2 From KCL and component definitions: 1

v 0.1 diL diL iin (t ) − L − iL = 0 ⇒ + iL = iin (t ) ⇒ + 250iL = 250iin (t) 25 25 dt dt We represent the input signal by the complex exponential: iin (t ) = 0.2e j 250t A and the unknown current can be represented as iL (t) = IL e j ( 250t + ) . Substituting this into the differential equation and canceling e j250t : j250IL e j + 250IL e j = 50 Thus IL e j

( j250 + 250) = 50

50 ⇒ IL e j = = 0.14142∠ − 45o 250 + j250

and IL = 0.141,

= −45° ⇒ iL (t) = 0.141cos(250t − 45°) A

SOLUTION 10.3. Construct differential equation by KVL and device definitions: diL diL − 200iL = 0 ⇒ + 400iL = 2vin ( t) dt dt

vin (t) − 0.5

We represent vin (t) as the complex exponential function, vin (t) = 20e j 400t V. The current in the inductor has the form: iL = IL e j( 400t + ) . Substituting into the differential equation and canceling e j400t : j400IL e j + 400IL e j = 40 Thus IL e j

(j

40 + 400) = 40 ⇒ IL e j = = 0.070711∠− 45o 400 + j 400

and IL = 0.0707,

= −45°, ⇒ i L (t) = 70.7cos (400t − 45°) mA

Hence, vout (t ) = 14.14cos (400t − 45°) V

SOLUTION 10.4. Construct differential equation using KVL and device definitions:

2

vin (t) − v C − C

dv C dv R = 0 ⇒ RC C + v C = v in (t ) dt dt

The output voltage is defined as: vout (t ) = vin ( t) − v C (t ) This means that finding vC is enough to be able to obtain the output voltage. The input voltage is represented by the complex exponential: vin (t) = 20e j 250t e − jπ/2 V and vC (t ) = Vm e j ( 250t + ) . Substituting into the differential equation, dividing by e j250t , and rearranging: j250RCVC e j + VC e j = − j20 ⇒ VC e j ( j250RC + 1) = − j20 − j20 ⇒ VC e j = = 14.142∠− 135o 1+ j Now Vout e j = − j20 − VC e j

⇒ Voute j = 10 − j10 = 14.142∠ − 45o

Thus, in the time-domain, vout (t ) = 14.142cos(250t − 45°) V

SOLUTION 10.5. The circuit is identical to that of problem 10.1. Thus,

RC

dv s dv + v s = Ris (t ) ⇒ Ris (t ) − RC s = v s dt dt

Moreover, the complex exponential solution is given by v s (t) = Vm e j = 223.6e − j 63.43 V o

Hence R − jRC223.6e − j 63.43 = 223.6e − j 63.43 = 100 − j200 o

o

i.e., R − jRC(100− j200) = R − 200RC − j100RC = 100 − j200

3

Thus RC = 2 s, R = 100 + 200RC = 500 Ω, and C = 4 mF.

SOLUTION 10.6. First use KVL di 1 vin − v L − vc − vout = 0 ⇒ vin − L L − dt C

t

∫ i Ldt − RiL = 0

−∞

Differentiating this equation, rearranging, and dividing by L, d 2iL R diL 1 1 dvin + + i = L L dt LC L dt dt 2 = 104 as vin (t) = 100e j t e − jπ/2 V and iL (t) = IL e j t e j A.

We represent the input signal with

Substituting these two expressions into the differential equation and dividing out e j t : R 1 − j100× j − 2IL e j + j IL e j + IL e j = L LC L ⇒ IL e j =

108 −75 × 106

+

j108

R   1 ⇒ IL e j  − 2+ j = 108 LC L 

= −0.48 − j0.64 = 0.8∠− 126.87o

Solving for the magnitude and angle (by hand or using MATLAB): iL ( t) = 0.8cos (10,000t − 126.87°) = 0.8sin (10,000t − 36.87°) A and vout (t ) = 80sin (10,000t − 36.87°) V

SOLUTION 10.7. Using standard reference directions, from KCL and component definitions, t

v 1 dv iin = iR + iL + iC = out + iL (0) + ∫ vout ( ) d + C out R L dt 0

Taking a second derivative and dividing by C yields d 2vout 1 dvout 1 1 diin + + v = out RC dt LC C dt dt 2 4

We now let ω = 2500 rad/s and represent iin (t) by the real part of the complex exponential 0.02e j t . Further we represent vout (t) as the real part of the complex exponential Vme j( t + ) = Vm e j te j . Substituting these expressions into the differential equation and taking the indicated derivatives yields V 1 0.02 Vm ( j ) 2 e j te j + m j e j te j + Vm e j te j = j ej t RC LC C Observe that e j t cancels out on both sides of this equation producing j  j0.02  1 Vme j  − 2+ = LC RC  C Hence Vme j =

j0.02 = 1.28 + j0.96 1 j   C − 2+  LC RC 

This was obtained using MATLAB as follows: »w = 2500; »L = 40e-3; »C = 1e-6; »R = 100; »Vout = j*0.02*w/(1/L - C*w^2 +j*w/R) Vout = 1.2800e+00 + 9.6000e-01i »magVout = abs(Vout) magVout = 1.6000e+00 »angVout = angle(Vout)*180/pi angVout = 3.6870e+01 Therefore vout (t) = 1.6cos(2500t + 36.87o) V

SOLUTION 10.8. (a) From KCL: 3∠15° − 5∠45°− I = 0 ⇒ I = 3∠15°− 5∠45°. In MATLAB, »Ibar = 3*exp(j*15*pi/180) - 5*exp(j*45*pi/180) 5

Ibar =-6.3776e-01 - 2.7591e+00i »abs(Ibar) ans =2.8318e+00 »angle(Ibar)*180/pi ans =-1.0302e+02 Therefore, i( t) = Re {Ibar} = 2.83cos( t − 103°) A (b) From KCL, »Ibar = (1 + 2*j) - (-2 + j*6) Ibar = 3.0000e+00 - 4.0000e+00i »abs(Ibar) ans = 5 »angle(Ibar)*180/pi ans =-5.3130e+01 Therefore i( t) = 5cos (50πt − 53°) A.

SOLUTION 10.9. We define a Gaussian surface encompassing the three bottom nodes. Thus, KCL dictates that the sum of 4 currents be zero: »Ibar = -(-2-j*8) + (3 +j*12) + 10 Ibar = 1.5000e+01 + 2.0000e+01i »abs(Ibar) ans = 25 »angle(Ibar)*180/pi ans = 5.3130e+01 Therefore i( t) = 25cos (1000t + 53.13°) A.

SOLUTION 10.10. First represent the time-domain functions as phasors: 6

V1 = 2∠0°, V2 = 2 2∠− 45° Then, by KVL »V1 = 2; V2 = 2*sqrt(2)*exp(-j*pi/4); »VL = V1 - V2 VL = 0 + 2.0000e+00i Therefore, VL = 2∠90° V and v L (t ) = 2cos( t + 90°) = −2sin( t) V

SOLUTION 10.11. Apply KVL by simply following the loop defined by the independent voltage sources: Vx = 4 j − 2 j − 1− 1+ (1− j) − (1+ j) = −2

V SOLUTION 10.12. First note that Iout = R . VR = − j10 + (5 − j5) − 10 ⇒ VR = −5 − j15 V. R −5 − j15 Thus, Iout = = −1− j 3 = 3.16∠ − 108.4° ⇒ iout (t) = 3.16cos (500 t − 108.4°) A. 5

SOLUTION 10.13. Using a Gaussian surface, Iy = −(2 + j3) − (1+ j2) + (1− j5) = −2 − j10 = 10.2∠− 101.3o A. Therefore, i( t) = 10.2cos (2000 t − 101.3°) A. Now applying KVL, Vx = (2 + j2) + (2 + j 3)− (1− 4 j) = 3 + 9 j »Vx=2+2*j+2+j*3-1+4*j Vx = 3.0000e+00 + 9.0000e+00i »abs(Vx) ans = 9.4868e+00 »angle(Vx)*180/pi 7

ans = 7.1565e+01 Thus, v x ( t) = 9.487cos (2000 t + 71.6°) V.

SOLUTION 10.14. (a) At

= 1000 rad/s, YC ( j ) = j C = j4.7 × 10 −3 ⇒ C = 1.496 F

= 50 , YC ( j ) = j C = j2.3499 × 10−4 S ⇒ ZC (j ) = − j4255 Ω . (b) At = 1000 rad/s, ZL = j L = j18.85 ⇒ L = 6 mH. Since impedance is proportional to So, at

frequency, multiplying the frequency by 20 means the impedance is multiplied by 20. Thus, at = 20 × (103 π) rad/s, ZL = j18.85 × 20 = j 377 ⇒ YL = − j2.652 mS

SOLUTION 10.15. The input voltagephasor as 2∠0°. By inspection: I1 = j C VC = j10 × 0.1× 2 = j2 From the definition of a dependent V-source and an inductor:

I2 =

5I1 j10 = =5 j L j2

Finally, Vout = 5 × 2(I1 + I2 ) = 10(5 + j2) = 50 + j20 = 53.85∠21.8° vout (t ) = 53.85cos(10t + 21.5°) V

SOLUTION 10.16. Using KCL and the definition of a resistor: V10Ω = 10(6∠0° − 3∠90°) = 10(6− 3j) = 60 − j 30 V Thus, Z( j 0 ) =

60 − j 30 = −10 − j20 Ω j3 8

And, the combination of this Z( j 0 ) with the 10 Ω resistance, at this frequency, is 10 – j5. »Z=-10-j*20; R = 10; »ZZ = R*Z/(R+Z) ZZ =1.0000e+01 - 5.0000e+00i »C = 1/(-5*j*j*2000*pi) C =3.1831e-05 This is equivalent to a 10 Ω resistor in series with a 31.83 µF capacitor.

SOLUTION 10.17. »w = 2*pi*60; »VL = 3 +12*exp(-j*30*pi/180) + 6 -12*exp(j*30*pi/180) VL =9.0000e+00 - 1.2000e+01i »ZL = j*w/60 ZL =0 + 6.2832e+00i »IL = VL/ZL IL =-1.9099e+00 - 1.4324e+00i »abs(IL) ans =2.3873e+00 »angle(IL)*180/pi ans =-1.4313e+02 Therefore, iL (t) = 2.387cos(120πt − 143.1o ) A.

Solution 10.18 (a) First represent the inputs with their phasors: Is1 = 10∠30° = 8.66 + j5 and Vs2 = 100∠0°. 1 The admittance of the RC combination is: Y1 = + j C = 0.1+ j0.1. Using superposition and noting that R a 0-volt V-source is a short circuit and a 0-amp current source is an open circuit, Ix = Y1Vs2 − Is1 = 1.3397 + j5 = 5.1764∠75o A Therefore, i x (t) = 5.176cos (100t + 75°) (b) Let i x1 be the contribution to i x generated by the current source and i x2 the contribution generated by the voltage source. Then i x1 (t) = −10cos(50t + 30o ) A, and i x2 (t) = −14.142cos(100t + 45o ) A since 9

»Vs2 = 100; »w = 100; »R = 10; C = 1e-3; »Y1 = 1/R + j*w*C Y1 =1.0000e-01 + 1.0000e-01i »Ix2 = Y1*Vs2 Ix2 =1.0000e+01 + 1.0000e+01i »abs(Ix2) ans =1.4142e+01 »angle(Ix2)*180/pi ans =45 Therefore i x (t) = −10cos(50t + 30o ) − 14.142cos(100t + 45o ) A.

Solution 10.19 First of all, write out the given values: Vs2 = 4∠0° . From KVL Vs1 = 3I1 + j LI1 +

= 200 rad/s, I1 = 0.5∠90° = 0.5 j A, and

I1 + Vs2 which leads to: j C

»w = 200; I1 = 0.5j; Vs2 = 4; »Vs1 = 3*I1 + j*w*0.04*I1 + I1/(j*w*1e-3) +Vs2 Vs1 =2.5000e+00 + 1.5000e+00i »abs(Vs1) ans = 2.9155e+00 »angle(Vs1)*180/pi ans = 3.0964e+01 Therefore v s1 (t) = 2.9155cos(200t + 30.964 o ) V.

Solution 10.20 The phasor for the input can be written as

= 1000 rad/s and Vin = 10∠60° = 5 + j8.66

V. The currents can be obtained easily by applying Ohm’s law for phasors:

10

IR =

10∠60 10∠60 = 0.02∠60° A, IL = = 0.04∠− 30° A, and 500 j1000 × 0.25

(

)

IC = 10∠60 × j1000 × 2 × 10−6 = 0.02∠150° A. Thus Iin = 0.02∠60° + 0.04∠ − 30° + 0.02∠150° = 0.0283∠15° A and iin (t ) = 0.0283cos(1000t + 15°) A using the following MATLAB code: »Vin = 10*exp(j*60*pi/180) Vin =5.0000e+00 + 8.6603e+00i »R = 500; L = 0.25; C = 2e-6; »w = 1e3; »IR = Vin/500 IR =1.0000e-02 + 1.7321e-02i »IL = Vin/(j*w*L) IL =3.4641e-02 - 2.0000e-02i »IC = j*w*C*Vin IC =-1.7321e-02 + 1.0000e-02i »Iin = IR + IL + IC Iin =2.7321e-02 + 7.3205e-03i »abs(Iin) ans =2.8284e-02 »angle(Iin)*180/pi ans =1.5000e+01

Solution 10.21 In MATLAB »Iin = -100*j*1e-3; R = 100; »L = 0.04; C = 2e-6; w = 2500; »VR = R*Iin VR =0 - 1.0000e+01i »VL = j*w*L*Iin VL =10 »VC = Iin/(j*w*C) VC =-20 »Vin = VR + VL + VC 11

Vin =-1.0000e+01 - 1.0000e+01i »abs(Vin) ans =1.4142e+01 »angle(Vin)*180/pi ans =-135 Therefore, vin (t) = 14.14cos(2500t − 135°) V. Solution 10.22 (a) Here, i1 (t) = 0.6cos (200t) A and vout (t ) = 20sin (200t) = 20cos (200t − 90°) V. For = 200 rad/s, the phasors are by inspection: I1 = 0.6∠0° A, Vout = 20∠ − 90° V . (b) Write down the resistor, inductor, and capacitor current phasors, given Vout : 20∠− 90 = 0.6∠ − 90° = − j0.6 1/0.03 20∠− 90 IL = = 1∠ − 180° = −1 j200 × 0.1 IR =

(

)

IC = 20∠ − 90 × j200 × 0.4 × 10 −3 = 1.6∠0° Now, by KCL I2 = IR + IL + IC − I1 = − j0.6 where we have substituted the above values of branch currents. The time-domain function is: i2 ( t) = 0.6cos (200t − 90°) = 0.6sin (200t ) A

Solution 10.23 (a)

= 400 rad/s and Vin = 20∠0° V. We can easily use the voltage divider formula for

phasors and substitute values to obtain: Vout =

200 20 Vin = = 14.14∠ − 45° 200 + j L 1+ j

in which casevout (t ) = 14.14cos (400t − 45°) V (b)

= 250 rad/s and Vin = 20∠− 90° = − j20 V. Again, we can easily use voltage division: Vout =

400

1 Vin = 400 + j C

20∠ − 90° = 14.14∠− 45° V 1− j

Thus, in the time-domain, vout (t ) = 14.142cos(250t − 45°) V 12

= 10,000 rad/s and Vin = 100∠− 90° = − j100 V. Apply the voltage divider

Solution 10.24 (a) formula: Vout =

100 100 Vin = V = 80∠ − 126.87° 1 100 + j100 − j0.25 in 100 + j L + j C

The steady-state response is thus, vout (t ) = 80sin (10,000t − 36.87°) V The phasor method provides for a much easier way of obtaining the steady-state response. (b) Here,

= 2500 rad/s and Iin = 0.02∠0° A. Now, apply current division: IR =

0.01 0.01 I = I = 0.016∠36.86° in 0.01 + j0.0025 − j0.01 in 0.01 + j C + 1 j L

By Ohm’s law: Vout = 100IR = 1.6∠36.87° V Therefore vout (t) = 1.6cos(2500t + 36.87o) V

Solution 10.25 (a) It is easier to find the admittance first: Yin ( j100) =

1

+ j C = − j0.2 + j12.5 = j12.3 j L ⇒ Zin ( j100) = − j0.0813 (b) Yin ( j100) =

1 1 = + j100C . Hence, in MATLAB, 25 j j100 × 0.05

»C = (1/(25*j) - 1/(j*w*L) )/(j*100) C = 1.6000e-03

Solution 10.26 (a) Zin ( j ) =

1 j C

+

R× j L − j j RL = + = 9.975∠− 0.0072° R+ j L C R+ j L

»R = 10; L = 0.1; C = 1e-3; w = 2e3; »Zin = -j/(w*C) + j*w*R*L/(R + j*w*L) Zin = 9.9751e+00 - 1.2469e-03i 13

»abs(Zin) ans = 9.9751e+00 »angle(Zin)*180/pi ans = -7.1620e-03 (b) As the frequency increases, the capacitor becomes a short circuit and the inductor becomes an open circuit. Thus, the impedance approaches R. Analytically, 1 R lim →∞ Zin ( j ) = lim →∞ + lim →∞ R =R j C +1 j L  − j j RL(R − j L) LR 2 1  R 2 L2 (c) Zin ( j ) = + = j 2 − + . It follows that w must satisfy C C  R 2 + 2 L2 R 2 + 2 L2  R + 2L2

2

LR2C = R 2 + 2L2 or equivalently

2

=

R2 1 = . For the given component LR2C − L2 LC1− L     CR 2 

 L  values 1− = 0; hence there is no finite value of frequency for which the impedance is real.  CR 2 

SOLUTION 10.27. We note that the input admittance is given by: Yin ( j ) =

1 100 − j 0.1 −6 + j × 10−6 = 4 2 + j × 10 100 + j 0.1 10 + 0.01 100 0.1 −6 − j = 4 + j × 10 10 + 0.01 2 10 4 + 0.01 2

Thus, ω must satisfy 10 5 = 10 4 + 0.01 2 or equivalently,

= 107 − 10 6 = 3000 rad/s.

SOLUTION 10.28. (a) As usual we will use MATLAB. »R1 = 20; R2 = 10; »L = 0.04; C = 0.6e-3; »w = 250; 14

»Yin = 1/R1 + 1/(R2 + j*w*L) + j*w*C Yin = 1.0000e-01 + 1.0000e-01i Hence Yin ( j250) = 0.1+ j0.1 S (b) For this part we observe that Yin ( j250) =

1 1 + + jωC = 0.1 − 0.05 j + j250C R1 R2 + jωL

For this to be real, the imaginary part must be zero, i.e., 0.05 1 C= = = 0.2 mF 250 5000

Solution 10.29 (a) We can derive an expression for the input impedance by noting that it is the series combination of the resistance and the inductor/capacitor pair connected in parallel. Thus, 1

×j L

L C

j C =R− j 1 1 +j L L− j C C

Zin ( j ) = R +

Equating the real and imaginary parts of the given impedance, R = 4 Ω and −L LC − 1/

=

−L = 2 ⇒ L = 0.1 H 2L − 0.25

(b) At zero inductance, the above reactance is zero. Also, at L = 0.125, the denominator of the above reactance is zero, which means that the reactance is infinite.

Solution 10.30 (a) First derive an expression for the input impedance as a function of frequency: Zin ( j

)=5+ j

L+

1 1   = 5 + j L − j C C

We want the imaginary part to be equal to zero. Thus, L−

1 =0 ⇒ C

2

=

1 ⇒ LC

= 2500 rad/s

15

where we have substituted the values of L and C. The magnitude of the impedance is minimum when the imaginary part is zero, and Zin (j2500) = 5. (b) Derive an expression for the admittance: Yin ( ) =

1 1 1 1  + j C+ = + j C − 5 j L 5 L

Again, the imaginary part is equal to zero when

=

1 or LC

= 2500, at which point the admittance is

0.2.

Solution 10.31 The input admittance is Yin ( ) = admittance at

1 + j C = 0.008+0.004. Equating this to the given R

= 500yields R = 125 Ω, ωC = 0.004 or C = 8 µF.

Solution 10.32 Derive an expression for the admittance at Yin ( j ) =

1 j L

+

= 1000:

0.25 × j C j0.125 = − j0.5 + = 0.2 − j0.4 S 0.25 + j C 0.25 + j0.5

Note that this is equivalent to a 0.2 S conductance (i.e. 5 Ω resistance) in parallel with a 2.5 mH inductance (at the given frequency!). Now, the impedance is:

Zin =

1 = 1+ j2 Ω 0.2 − j0.4

This is equivalent to a resistance of 1 Ω in series with a 2 mH inductance (at the given frequency).

Solution 10.33 The current is zero when the input impedance of the parallel combination of inductor and capacitor is infinite. The latter is given by: 1 L ×j L j C C ZLC ( j ) = 1 = −j 1 +j L L− j C C 1 1 The magnitude of this is infinite when r L − = 0 ⇒ r2 = ⇒ C LC Vs Is = =0 R + ZLC ( j r )

r

= 15,811 rad/s. Observe that

16

Hence is (t) = 0 at r = 15,811 rad/s. At this frequency, the voltage across the LC tank is equal to the input voltage (since there is no drop across the resistor).

Solution 10.34 (a) By inspection: 2 × j 0.4 j = 2 + j 0.4 2 + 1 1 Yin = + = 0.5 − R j L Zin =

0.8 j 0.4 2.5 j

(b) Use the current divider formula, and substitute the given frequency, to obtain: IL =

R 2 2 2= = 1∠ − 45° R+ j L 2 + j2

and iL (t) = cos(5t − 45o ) A.

Solution 10.35 (a) Using voltage division, Vout =

R j RC Vin = V which j 1+ j RC in R− C

⇒ ∠Vout = 90o − in − tan−1 ( RC ) and Vout =

(b) For this part, we need to make sure that tan−1( RC) = 45° ⇒

RC Vin 1+ 2 R 2C 2

RC = 1 ⇒

= 1/ RC .

(c) At this frequency, 1 RC Vout 1 RC = = Vin 1 2 1+ 2 2 R 2C 2 R C

1 V j C 1 1 Solution 10.36 Here, = 1/ RC and C = 1 = = . Therefore Vin 1 + j RC 1 + j +R j C 1 VC = V = 0.707Vm∠ − 45° 1 + j in The time-domain function is  1  vC (t ) = 0.707Vm cos t − 45° V RC 17

Solution 10.37 (a) The magnitude of the capacitor voltage is 10/14.14 = 0.707 times the magnitude of the input signal. We just showed in the above problem that VC 1 VC 1 = ⇒ = Vin 1+ j10RC Vin 1 + 100R 2C 2 And we also showed that the ratio is 0.707 when the frequency is 1/RC. So, C = 1/(10R) = 0.01 F. (b) Again, from the results of the previous problem, the angle is –45 degrees.

Solution 10.38

= 1000 rad/s and Iin = 2∠45° A. The equivalent admittance is Yeq ( j

)=

1 1   + j C − = 0.25 + j (0.25 − 0.25) = 0.25 S R L

In MATLAB »Yeq = 0.25; »Iin = 2*exp(j*pi/4) Iin = 1.4142e+00 + 1.4142e+00i »Vout = Iin/Yeq Vout = 5.6569e+00 + 5.6569e+00i »abs(Vout) ans = 8 »angle(Vout)*180/pi ans = 4.5000e+01 »% Using Current Division »IL = (1/(j*1000*4e-3)/Yeq)*Iin IL = 1.4142e+00 - 1.4142e+00i »abs(IL) ans = 2 »angle(IL)*180/pi ans = -45 18

Therefore, vout (t) = 8cos(1000t + 45o) V and iL (t) = 2cos(1000t − 45o ) A. (b) If

= 618 rad/s, then Yeq ( ) =

1 − j0.405 + j0.1545 = 0.25 − j0.25 = 0.3536∠ − 45° 4

and »Vout = Iin/Yeq Vout = -3.4779e-04 + 5.6565e+00i »abs(Vout) ans = 5.6565e+00 »angle(Vout)*180/pi ans = 9.0004e+01 »IL = Iin*(1/(j*618*4e-3))/Yeq IL = 2.2882e+00 + 1.4069e-04i »abs(IL) ans = 2.2882e+00 »angle(IL)*180/pi ans = 3.5228e-03 Therefore, vout (t) = 5.657cos(1000t + 90o ) = −5.657sin(1000t) V and iL (t) = 2.288cos(1000t) A.

Solution 10.39 Write the input phasor:

= 1000 rad/s and Iin = 0.01 2∠60°A.

»w =1000; »Iin = 0.01*sqrt(2)*exp(j*60*pi/180) Iin = 7.0711e-03 + 1.2247e-02i »Yeq = 1/500 +1/(j*w*0.25) +j*w*2e-6 Yeq = 2.0000e-03 - 2.0000e-03i »Vin = Iin/Yeq 19

Vin = -1.2941e+00 + 4.8296e+00i »abs(Vin) ans = 5 »angle(Vin)*180/pi ans = 1.0500e+02 »IR = Iin*(1/500)/Yeq IR = -2.5882e-03 + 9.6593e-03i »abs(IR) ans = 1.0000e-02 »angle(IR)*180/pi ans = 1.0500e+02 »IL = Iin*(1/(j*w*0.25))/Yeq IL = 1.9319e-02 + 5.1764e-03i »abs(IL) ans = 2.0000e-02 »angle(IL)*180/pi ans = 1.5000e+01 »IC = Iin*j*w*2e-6/Yeq IC = -9.6593e-03 - 2.5882e-03i »abs(IC) ans = 1.0000e-02 »angle(IC)*180/pi ans = -1.6500e+02

20

Therefore, vin (t) = 5cos (1000t − 105°) V, iC ( t) = 0.01cos(1000t − 165°) A, iL ( t) = 0.02cos (1000t + 15°) A, and iR (t) = 0.01cos(1000t + 105°) A.

SOLUTION 10.40.

V1 =

= 4 . Using voltage division R R+

1 j C

+j L

Vin =

2 (−8 j ) = 5.657∠− 45o 2−4 j +2j

Converting back to time: vout (t ) = 5.657cos( 4t − 45°) V

SOLUTION 10.41.

= 25 rad/s, Vs = 10∠0° V. By voltage division, −j

VC =

0.02 × 25 10∠0° = −2 j = 2∠ − 90° 10 + j0.08 × 25 − j 0.02 × 25

Thus, vC (t ) = 2cos(25t − 90°) V.

2 Vout 1 1   SOLUTION 10.42. = = 0.2 ⇒ 25 = 3 + j  8 − . Vin 4C  3 + j8 + 1 j 4C

1 2 1  Thus  8 − = 25 − 9 ⇒ C = = 0.0625 F.  4C 16

SOLUTION 10.43. Here,

= 3.33 × 103 rad/s and Vin = 50∠0° V. Using phasors, VR =

400 400 −

j

× 50 = 40∠36.897° V

3.33× 10 3 × 10 −6

and Vout =

100VR j C 100VR = = 3985∠32.1° V 1 j100 C + 1 100 + j C 21

Hence, vout (t) = 3985cos(3.33× 10 3 t + 32.1°) V.

SOLUTION 10.44. Here, Z1 =

= 104 rad/s, Iin = 0.01∠0° A.

100 × j 0.1 = 1+ j1000 ⇒ VL = ZL Iin = 0.01× ZL = 0.01 + j10 100 + j 0.1

Now, in MATLAB »w=1e4; R = 100;L = 0.1; C = 0.1e-6; »Z1 = R*j*w*L/(R+j*w*L) Z1 = 9.9010e+01 + 9.9010e+00i »Iin = 0.01; »VL = Z1*Iin VL =9.9010e-01 + 9.9010e-02i »Z2 = 1/(1/R + j*w*C) Z2 = 9.9010e+01 - 9.9010e+00i »VC = Z2*VL VC = 9.9010e+01 + 1.7764e-15i »abs(VC) ans = 9.9010e+01 »angle(VC)*180/pi ans = 1.0280e-15 Thus, vC (t ) = 99cos(10000t) V.

SOLUTION 10.45. Here = 40 rad/s and Vin = 120∠0°. This problem is best done in MATLAB using parallel impedance computation, voltage division, and Ohm's law for phasors: »R1 = 500; R2 = 80; »C = 0.1e-3; L = 2; »w = 40; Vin = 120; »Z1 = R1/(j*w*C)/(R1 + 1/(j*w*C)) Z1 = 1.0000e+02 - 2.0000e+02i »Z2 = R2*j*w*L/(R2 + j*w*L) Z2 = 4.0000e+01 + 4.0000e+01i » Use voltage division »VC = Z1*Vin/(Z1+Z2) VC = 1.2212e+02 - 3.1858e+01i »abs(VC) ans = 1.2621e+02 »angle(VC)*180/pi ans = -1.4621e+01 22

» Use voltage division and Ohm's law for inductors »VL = Z2*Vin/(Z1+Z2) VL =-2.1239e+00 + 3.1858e+01i »IL = VL/(j*w*L) IL =3.9823e-01 + 2.6549e-02i »abs(IL) ans = 3.9911e-01 »angle(IL)*180/pi ans = 3.8141e+00 Therefore, vC (t) = 126.21cos(40t − 14.621o ) V and iL (t) = 0.399cos(40t + 3.814) A.

SOLUTION 10.46. Here MATLAB.

= 40 rad/s and Iin = 0.120∠0° A. This problem is best solved using

»w = 40; Iin = 0.12;R = 5; »C = 0.004; L = 0.1; »Y1 = 1/R + j*w*C Y1 = 2.0000e-01 + 1.6000e-01i »Y2 = 1/R + 1/(j*w*L) Y2 = 2.0000e-01 - 2.5000e-01i »% USING CURRENT DIVISION »IL = Iin*Y2/(Y1 + Y2) IL = 7.3171e-02 - 5.8537e-02i »abs(IL) ans = 9.3704e-02 »angle(IL)*180/pi ans = -3.8660e+01 »% AGAIN USING CURRENT DIVISION »IC = Iin*Y1/(Y1 + Y2) IC = 4.6829e-02 + 5.8537e-02i »% USING OHM'S LAW FOR CAPACITORS »VC = IC/(j*w*C) VC = 3.6585e-01 - 2.9268e-01i »abs(VC) ans = 4.6852e-01 »angle(VC)*180/pi ans = -3.8660e+01

SOLUTION 10.47 Here,

= 25 rad/s and IS = 2∠0° A. Now perform a source transformation. The

combination of current source in parallel with resistor is changed into a voltage source in series with the same resistor. The voltage source value is: VS = 1× IS = 2∠0° V. Apply Ohm’s law to obtain: I1 =

2∠0° = 0.5 + j0.5 = 0.707∠45° 2 + 1 j25 × 0.02 23

Hence, i1 (t) = 0.707cos(25t + 45o ) A.

SOLUTION 10.48 Apply a source transformation to obtain:

The impedance of the inductor branch is ZL ( j 4) = 2 + j4 × 0.5 = 2 + j2 Ω and YL ( j 4) =

1 = 0.25 − j0.25 S. Now, the total admittance seen by the source: ZL ( j4) Yeq ( j 4) =

Thus, by inspection, VC =

1 + j0.25 + YL ( j4) = 0.75 S 2

6∠0° = 8∠0° V and IC = 0.25∠90°× 8∠0° = 2∠90° A. Therefore 0.75

iC (t) = 2cos(4t + 90o ) = −2sin(4 t) A.

SOLUTION 10.49. Apply a source transformation to obtain:

Then, by inspection, Yeq ( j500) =

1 1 1 1 + + − = 1.5 S. Thus, 2 1 j2 j2

24

VC =

2.5∠0° = 1.667∠0° V = VL ⇒ IL = 1.667/ j2 = 0.833∠− 90° A 1.5

In the time-domain: vC (t ) = 1.667cos(500t) V and iL ( t) = 0.833cos(500t − 90°) = 0.833sin(500t ) A.

25

PROBLEM SOLUTIONS CHAPTER 10 SOLUTION 10.50. The input voltage phasor is

= 2000 rad/s and VS = 20∠0° V. Now, do a source

transformation on the phasor circuit:

where V 20∠0° IS = S = = 1∠− 90° = − j A ZL j2000 × 10 × 10 −3 Now, Yeq =

1 1 + + j2000 × 50 × 10 −6 = 0.0707∠45° S 20 j20

and I 1∠− 90° VC = S = = 14.14∠ − 135° V Yeq 0.0707∠45°

SOLUTION 10.51. Use superposition. First, find response to current source using circuit below:

3× (− j 3) Vx_1 = Is1ZRC = Is1 = 2∠0°× 2.121∠− 45° = 4.242∠− 45° 3 − j3 Now, find the response due to the voltage source using the following circuit: 1

The voltage across the inductor is the same as the input source, and this voltage divides between the series combination of capacitor and resistor: Vx_2 =

j3 j3 Vs2 = 3∠90o = 2.121∠ − 135o V 3 − j3 3 − j3

Combining the two contributions implies that: Vx = 4.242∠ − 45o + 2.121∠− 135o = 4.74∠ − 71.6o V

SOLUTION 10.52. (a) As stated, VL = aVs 1 + bIs 2 . To find a, set Is 2 = 0 and use voltage division: VL = −

j30 V = −0.5Vs 1 = −aVs 1 j30 + j30 s 1

To find b, set Vs 1 = 0 and use parallel impedance and Ohm's law: VL = ( j30/ / j30)Is 2 = j15Is 2 = bIs 2 Hence VL = –0.5Vs 1 +j15Is 2 (b) For this part, Vs 1 = 10 and Is 2 = 0.5∠ − 90o . Hence from the formula, VL = − 0.5Vs1 + j15Is 2 = −0.5 × 10 + j15 × 0.5(− j) = − 5 + 7.5 = 2.5 V 2

Therefore vL(t) = 2.5 cos(100πt) V.

SOLUTION 10.53. In this problem, we can make use of the linearity property for phasors. Specifically, from the given information, we can write V1 20∠45° V 5∠90° = = 2∠45° = a and 1 = = 0.5∠45° = b Iin 10∠0° Vin 10∠45° Hence, V1 = aIin + bVin Substituting the new values of input current and voltage, we obtain: V1 = 10∠0° + 10∠45° = 18.48∠22.5° V

SOLUTION 10.54. (a) For Vout to be zero, we want RC

j L L = ⇒ R R = C L j RL + j L C RC − C (b)

Substituting RC C = 2 s and RL = 3 Ω into the above expression gives: L = RCCRL = 6 H.

(c) The bridge circuit can be represented by an impedance Zbridge(jω). The voltage that appears across the bridge, say Vbridge, is obtained by voltage division. Hence, by the voltage substitution theorem, the problem may be solved as in part (a) with this new source voltage Vbridge appearing across Zbridge(jω).

SOLUTION 10.55 The input phasor is: = 1000 rad/s and Iin = 2∠45° A, assuming peak value. First compute Yth ( j103 ) = 0.25 −

j + j1000 × 0.25 × 10 −3 = 0.25S ⇒ Zth ( j10 3) = 4 Ω −3 1000 × 4 × 10

Then, Voc = Iin × 4 = 8∠45° V

3

in which case voc (t) = 8cos(10 3 t + 45o ) V. The final equivalent is a voltage source (having value Voc ) in series with a resistance of 4 Ω.

SOLUTION 10.56. (a) First note that the frequency is given in Hz, so, = 1281.77rad/s and Iin = 10∠0° A. Then, in MATLAB, »R = 0.25; L = 1.17e-3; C = 520e-6; »w = 2*pi*204; »Yin = j*w*C +1/(R + j*w*L) Yin = 1.0815e-01 + 1.7737e-02i »Zin = 1/Yin Zin = 9.0039e+00 - 1.4766e+00i »abs(Zin) ans = 9.1242e+00 »angle(Zin)*180/pi ans = -9.3134e+00 Therefore Zth = 9.1242∠− 9.313° Ω. Finally, Voc = ZthIin = 91.2∠− 9.313° V. (b) Now, the circuit looks like the following:

Simple voltage division can yield: VL =

ZL V = 46.219∠0o ⇒ v L ( t) = 46.2cos(1281.77t) V ZL + ZTH oc

4

SOLUTION 10.57. For this problem we short the V-source and compute Zth and use voltage division to find Vo c . Specifically Zth =

1

= 103 − j10 Ω 1 jωC + 0.1+ jωL

and Voc =

1 jωC

× 2 = −200 j V 1 + 0.1 + jωL jωC

where »w=1000; »L = 0.01; »C = 0.1e-3; »Zth = 1/(j*w*C + 1/(0.1 + j*w*L)) Zth = 1.0000e+03 - 1.0000e+01i »Voc = 2*(1/(j*w*C))/(0.1 + j*w*L + 1/(j*w*C)) Voc = 0 - 2.0000e+02i To compute the load voltage, define Zload = 103 + j10 −2 ω = 103 + j10 Ω Again using voltage division and MATLAB we have »Zload = 1e3 + j*10; »Vload = Voc*Zload/(Zload + Zth) Vload = 1.0000e+00 - 1.0000e+02i »magVload = abs(Vload) magVload = 1.0000e+02 (volts)

5

SOLUTION 10.58. Here,

= 3.33 × 103 rad/s and Vin = 50∠0° V. Here we note that we already found

Voc in Problem 10.43. Thus, Voc = 3985∠32.1° V. In order to find Zth, we introduce a fictitious 1 A current source at the A and B terminals:

Now, we note that the VR voltage phasor is zero. Thus, the dependent source has zero volts across it. This way, the temporary current source sees the parallel combination of a resistor and a capacitor:

VAB =

100 ×

−j 3.33×10 × 0.25×10 −6 3

100 − j1200

= 99.65∠ − 4.77° V

Thus, Zth = 99.65∠− 4.77° Ω. The Thevenin equivalent consists of Voc in series with Zth = 99.65∠− 4.77° = 99.3 − j8.2865 Ω. Thus the Norton equivalent is the parallel combination of Zth and the current source with value

Isc =

Voc 3985∠32.1° = = 39.99∠36.91o A. Zth 99.65∠ − 4.77°

SOLUTION 10.59. = 10,000 rad/s and Vin = 10∠0° V. Again, we have already found Voc in Problem 10.44: Voc = 99∠0° V. Now, to compute the impedance, we introduce the temporary current source of 1 A:

6

Again, the inductor has no voltage across it. So, the dependent source generates no current. Hence, the independent source sees the parallel combination of a resistor and a capacitor:

VAB =

100 × (− j1000) = 99.5∠− 5.71° V 100 − j1000

Thus Zth ( j104 ) = 99.5∠ − 5.71° Ω. So, the Thevenin equivalent is the series combination of the Voc source and the above Zth ( j104 ) = 99.5∠ − 5.71° Ω.

SOLUTION 10.60. Inject a current source at terminals A and B. Then, write a KCL equation at node A: IR = IS − I1 ⇒ VR = IS − I1 ⇒ VC = VAB = VR + I1 = IS − I1 + I1 = IS Since the voltage across the current source is equal to its current, the equivalent impedance across this current source is 1 Ω.

SOLUTION 10.61. Inject a current source as usual. Then, write Ohm’s law for phasors for the equivalent series RLC circuit. Note that the controlling current for the dependent source is the input current:

VAB = j0.01× 200 × IS −

j I − 2IS = (2 + j)IS 200 × 0.005 S

V Therefore, Zth = AB = 2 + j Ω. Is

SOLUTION 10.62. Inject a current source Is :

7

Apply KCL:

j2I − 6I 6 = Is − I ⇒ I = I . Now, VAB is the voltage across the inductor: 6 j2 s VAB =

Solution 10.63.

6 × j2 I = 6Is ⇒ Zth = 6 Ω. j2 s

= 2000 rad/s and VS = 25∠0° V. Now, inject a current source, and express VAB as a

function of this current source and Voc :

Now, write two nodal equations at A and the top of the dependent current source: V1 − VS 10 3

+ gm VAB + j C( V1 − VAB ) = 0

V j C(VAB − V1 ) + AB = IS j L Substituting values, these can be cast into the following matrix equation:

8

j 3 − j   V1  10-3 VS  =   −j − j   VAB   IS  

10

−3 1 +

This can be solved in MATLAB to obtain:  V1  −2.5 − j2.5 −5 + j10 10-3 VS  = 100 ×  V   2.5 + j2.5  5  AB     IS  Thus VAB = (0.25 + j0.25)Vs + 500Is = 500Is + 6.25 + j6.25 Therefore Zth = 500 Ω and Voc = 6.25 + j6.25 = 8.839∠45o V. For the Norton equivalent we need Isc =

Voc 8.839 = ∠45o = 0.01768∠45o A. Zth 500

SOLUTION 10.64. Inject an upward current source, IS2 , at VAB. Then, write the following two nodal equations at Vx and VA: let R1 = R2 = R, then Vx + j C1Vx + j C2 (Vx − VAB ) = IS R V j C2( VAB − Vx ) + gm Vx + AB = IS2 R which after grouping terms becomes 1  + j C1 + j C2 Vx − j C2VAB = IS R 

(− j

1  C2 + gm )Vx + j C2 + V =I  R  AB S2

In MATLAB »R = 100e3; C1 = 1e-9; C2 = 1e-10; »gm = 0.1e-3; w = 1e3; »Nodal = [1/R+j*w*C1+j*w*C2 -j*w*C2;-j*w*C2+gm j*w*C2+1/R] Nodal = 1.0000e-05 + 1.1000e-06i 0 - 1.0000e-07i 9

1.0000e-04 - 1.0000e-07i 1.0000e-05 + 1.0000e-07i »Nodalinv = inv(Nodal) Nodalinv = 9.5680e+04 - 2.0070e+04i 2.1024e+02 + 9.5470e+02i -9.5449e+05 + 2.1120e+05i 9.7783e+04 - 1.0523e+04i Thus 0.21+ j0.955   40 × 10-6   Vx  3  95.68 − j20 = 10 ×  V  −954.5 + j211.2 97.78 − j10.52   AB     IS2  Therefore VAB = −38.18 + j8.448 + (97.8 − j10.5) × 103 IS2 from which we identify Voc = −38.18 + j8.448 = 39.1∠167.5o V and Zth = (97.8 − j10.5) × 103 = 98.35 × 103 ∠− 6.14 o Ω.

SOLUTION 10.65. We solve this problem by the method illustrated in example 6.3 where a fictitious source is applied and the repsonse is calculated. One can either apply a voltage source or a current source (see figure 6.10 a and b). Generally speaking, neither choice can be claimed as superior to the other. But for a specific circuit, one choice can lead to a much simpler solution than the other. To illustrate this point, we solve the problem with both choices below. Note that although the first method is much simpler than the second, it lacks the generality. If one more resistor were inserted into the circuit, the simplicity of solution of solution may disappear totally, whereas the second method will proceed with very few changes. Method 1. For this solution we apply an arbitrary voltage source, labeled in phasor form as Vs as indicated in the circuit below.

For this circuit we will compute an equation of the form ofequation 6.11:

10

IA =

1 V − Isc Zth s

(1)

By inspection of the circuit, V IA = IC + I L − 1 = jωCVs + L − 1 = j2Vs − jVL − 1 jωL But

VL = 0.25IC + Vs = 0.25 jωCVs + Vs = (0.5 j + 1)Vs

Substituting (3) into (2) produces

IA = (0.5 + j )Vs − 1

(2)

(3)

(4)

By comparing (4) with (1), we obtain the answers for the Norton equivlent circuit: Is c = 1 A and Zth =1/(0.5 + j) = 0.4 – j0.8 Ω. Method 2: For this method we apply an arbitrary current source, labeled in phasor form as Is as indicated in the circuit below.

Notice that we have added a current label Ix as we plan to use modified nodal analysis method to obtain the desired answer. For this we will compute an equation of the form VAB = Zth Is + Voc Because of the addition of Ix , we can write the modified nodal equations more or less by inspection: 1/ jωL  0   1

0 1   VL   − j 0 1   VL   1        jωC −1  VAB  =  0 2j −1  VAB  =  Is  −(1+ j0.25ωC) 0   I x   1 −1 − 0.5 j 0   I x   0 

where from MATLAB,

11

»w=2000; »L = 0.5e-3; »C = 1e-3; »Y11 = 1/(j*w*L) Y11 = 0 - 1.0000e+00i »Y22 = j*w*C Y22 = 0 + 2.0000e+00i »Y32 = -(1+j*0.25*w*C) Y32 = -1.0000e+00 - 5.0000e-01i Solving the equations in MATLAB produces, »A = [-j 0 1;0 2*j -1; 1 -1-0.5*j 0]; »Ainv = inv(A) Ainv = 8.0000e-01 - 6.0000e-01i 8.0000e-01 - 6.0000e-01i 1.6000e+00 + 8.0000e-01i 4.0000e-01 - 8.0000e-01i 4.0000e-01 - 8.0000e-01i 8.0000e-01 + 4.0000e-01i 1.6000e+00 + 8.0000e-01i 6.0000e-01 + 8.0000e-01i -8.0000e-01 + 1.6000e+00i Multiplying the second row of Ainv times the right-most vector of our equations produces VAB = (0.4 − j 0.8)Is + (0.4 − 0.8j ) This implies that Zth = 0.4 – j0.8 Ω and Vo c = 0.4–j0.8 V. For the Norton equivalent we need V Isc = oc = 1 A Zth

SOLUTION 10.66. (a)

= 1000 rad/s and VS = 20∠45° V. By voltage division: Vout =

1 20∠45° = 14.14∠0° V ⇒ IL = 14.14∠0° A 1+ j

(b) At dc, Vout = Vin. We want the frequency at which Vout = 0.1Vin . Thus, we want:

12

Vout 1 1 1 = = 0.1 ⇒ = ⇒ 2 −6 Vin 1+ j × 0.001 100 1+ × 10

= 9950 rad/s

SOLUTION 10.67. Z = 25 − j20 Ω. By KCL: VZ − 14∠0° VZ V + 8∠90° + + Z =0 j15 25 − j20 − j20 This equation simply needs to be manipulated in order to obtain: VZ = 41.35∠ − 73.46°

SOLUTION 10.68. In this problem, we note that the impedance, jX, is in series with the parallel RLC circuit to the right. Thus, all we need to do is to find an expression for the equivalent impedance of the parallel RLC circuit: 1 j − + j0.025 ⇒ ZRLC = 19.2 + j14.4 30 20 Now, the total impedance seen by the source is 19.2 + j14.4 +jX. Therefore, for this to be real, the YRLC =

unknown reactance has to be –14.4 Ω. Also, the input current now is I = 96/19.2 = 5 A. Hence i(t) = 5cos(10t) A.

1 SOLUTION 10.69. First, compute the current through the series RC section: IC = A. 1− j Now, by KCL, we can write 0 = IC − Ix +

SOLUTION 10.70. (a) Since Yin =

1− 2Ix 1 1− 2Ix = − Ix + j 1− j j

⇒ Ix = 0.3 + j0.1= 0.316∠18.44 o A

= 20 rad/s and Yin = 0.05 + j0.0866 S.

1 j − + j C , equating the real parts of the above two expressions implies that R = R L

1/0.05 = 20 Ω. (b) Similarly, equating the imaginary parts and substituting, we obtain: 13

1   j C − = j0.0866 ⇒ L = 3.73 H L (c)

(

)

I 20∠30° VC = in = = 200∠− 30° V ⇒ vC ( t) = 200cos 20t − 30o V Yin 0.05 + j0.0866 (d) IL =

(

)

200∠30° = 2.68∠ − 120°A ⇒ i L ( t) = 2.68cos 20t − 120o A j20 × 3.73

1 (e) Voc is just the voltage VC, which was obtained in (c), and Zth = = 10∠− 60° Ω. Y in

(f) Zth = 5 – j8.66 Ω. This is equivalent to a series combination of a 5 Ω resistance and a 5.77 mF capacitance at the given frequency ω = 20 rad/s.

SOLUTION 10.71. (a) Vs = 2 + j0V,

= 1000 rad/s. By KCL

VA VA V − Vs + + A = 0 ⇒ VA = 0.5 − j0.5 = 0.707∠ − 45° V − j1.33 2 + j2 4 The time-domain expression is:

(

)

vC (t ) = v A (t) = 0.707cos 1000t − 45o V (b) IL =

(

)

0.707∠ − 45° = 0.25∠− 90° A implies that iL ( t) = 0.25cos 1000t − 90o = 0.25sin(1000t) A. 2 + j2

(c) We already determined Voc in part (a). Now, turn off the source to compute the equivalent impedance: Yth = 0.75 j +

1 + 0.25 = 0.5 + j0.5 ⇒ Zth = 1− j Ω. This is the series connection of a 1 Ω resistor 2 + j2

with a capacitor of value C =

1

= 10 −3 F. This completes the definition of the Thevenin equivalent.

SOLUTION 10.72. In MATLAB, »R = 5; L = 1e-3; C = 20e-6;Vs1 = 5; Is2 = 0.5*j; »G = 1/R; w = 10e3; »YL = 1/(j*w*L) YL = 0 - 1.0000e-01i 14

»YC = j*w*C YC = 0 + 2.0000e-01i Hence G(VC − Vs1 ) + YC VC + YL (VC − Vs1 − 2VR ) = Is2 Substituting for VR, G(VC − Vs1 ) + YC VC + YL (VC − Vs1 − 2(Vs1 − VC )) = Is2 Therefore

(G + YC + 3YL )VC = (G + 3YL )Vs1 + Is2 Again, using MATLAB, »a = G + YC + 3*YL a= 2.0000e-01 - 1.0000e-01i »b = (G+3*YL)*Vs1 + Is2 b= 1.0000e+00 - 1.0000e+00i »VC = b/a VC = 6.0000e+00 - 2.0000e+00i »abs(VC) ans = 6.3246e+00 »angle(VC)*180/pi ans = -1.8435e+01 Hence, vC (t) = 6.325cos(104 t − 18.44o ) V.

SOLUTION 10.73. Denote by vC1 , the node voltage of the 2.5 mF capacitor . Note that at ω = 800 rad/s, o

Vs = 20∠0 V. From this we write a set of nodal equations by inspection after observing the following from MATLAB:

15

»w = 800; »C1 = 2.5e-3; »L = 1.25e-3; »Y1 = 0.5+j*w*C1-j/(L*w) Y1 = 5.0000e-01 + 1.0000e+00i »Yoff=j/(L*w) Yoff = 0 + 1.0000e+00i »Y2 = 0.25+j*w*3.75e-3 - j*w*1.25e-3 Y2 = 2.5000e-01 + 2.0000e+00i This information leads to the following matrix nodal equation: j 0.5 + j   VC1   0.5Vs  =  j 0.25 + j2   Vout   0.25Vs   To solve these equations we again use MATLAB: »A = [Y1, Yoff;Yoff, Y2] A= 5.0000e-01 + 1.0000e+00i 0 + 1.0000e+00i 0 + 1.0000e+00i 2.5000e-01 + 2.0000e+00i »b = [0.5*20;0.25*20] b= 10 5 »Vnodes = inv(A)*b Vnodes = 7.1141e+00 - 6.9799e+00i -3.6242e+00 + 5.3691e-01i »magVnodes = abs(Vnodes) magVnodes = 9.9664e+00 16

3.6637e+00 »angVnodes = angle(Vnodes)*180/pi angVnodes = -4.4454e+01 1.7157e+02 Therefore vout (t) = 3.664cos(800t + 171.57o ) V

SOLUTION 10.74. For this problem we use loop analysis with loops indicated in the figure below.

Since there are no controlled sources, we can write down the loop equations by inspection:  180 + 103.92  10 + 9 j 6 + 5j  I A   120 − 120∠ − 120o =  6 + 5 j 12 + 9 j   I  =      C  120∠120 o − 120∠− 120o   j207.85  The solution of this equation is done in MATLAB as follows: »b1=120-120*exp(-j*2*pi/3) b1 = 1.8000e+02 + 1.0392e+02i »b2=120*exp(j*2*pi/3)-120*exp(-j*2*pi/3) b2 = 0 + 2.0785e+02i 17

»A = [10+j*9, 6+j*5;6+j*5,12+j*9] A= 1.0000e+01 + 9.0000e+00i 6.0000e+00 + 5.0000e+00i 6.0000e+00 + 5.0000e+00i 1.2000e+01 + 9.0000e+00i »I=inv(A)*[b1;b2] I= 1.4472e+01 - 1.3469e+01i 4.2926e-01 + 1.7703e+01i »% Please note that using the commands I=inv(A)*[b1,b2]' »% will lead to the wrong answer because a conjugate is »% inserted along with the transpose. »magI = abs(I) magI = 1.9770e+01 1.7708e+01 »angleI = angle(I)*180/pi angleI = -4.2944e+01 8.8611e+01 »IB = -I(1)-I(2) IB = 1.8288e+00 + 1.7234e+01i »magIB = abs(IB) magIB = 1.7331e+01 »angleIB = angle(IB)*180/pi angleIB = 8.3943e+01 Changing the sign on each source amounts to multiplying its value by "–1". This means that all o

o

magnitudes remain the same, but there is a 180 phase shift for each current, i.e., add 180 to each current angle. 18

SOLUTION P10.75. For this problem we have both a transient component to the response and a steady state component. The steady state component is computed in the usual way because the circuit is stable, i.e., the time constant is positive. Once the steady state part is computed, we use initial conditions to obtain the coefficient B in the response. Part 1: Compute steady state response. For this we use MATLAB, »R = 0.5; L = 0.866; »Vs = 10; »w = 1; »Zin = R + j*w*L Zin = 5.0000e-01 + 8.6600e-01i »IL = Vs/Zin IL = 5.0002e+00 - 8.6604e+00i »magIL = abs(IL) magIL = 1.0000e+01 »angIL = angle(IL)*180/pi angIL = -5.9999e+01 Hence iL (t) = 10cos(t − 60o ) + Be−0.577t A Part 2: From the initial conditions we have

[

iL (0) = 1 = 10cos(t − 60 o ) + Be−0.577t

]t =0 = 10cos(−60o ) + B = 5 + B

Hence B = –4. It follows that iL (t) = 10cos(t − 60o ) − 4e −0.577t A

SOLUTION 10.76. KCL dictates that: IL + IR + IC = IS ⇒ IL = IS – IR – IC We can perform this sum graphically as follows:

19

j2

-IC IL=-j -IR

SOLUTION 10.77. (a) Note that VS = VR + VL, but that VL leads VR by 90 degrees. Similarly, IS = IR + IC, but IC leads VS by 90 degrees. Also note that the inductor current is also IR, and the capacitor voltage is V S.

IS

IC VS VL IR

VR

(b) Using graph paper to construct the phasor diagram to scale, we find the difference between the phase angles of Is and Vs is zero.

Solution 10.78. First note that VC, the capacitor voltage, will lag IS by 90 degrees. Now, VC plus the unknown element voltage should result in a vector that runs diagonally between the two vectors. From the following illustration, it follows that the unknown voltage should have the same phase as the input current:

IS VS VC

VC

Vu 20

This means that the unknown element is a resistor. The 45o phase difference implies that VC = Vu or Is 1 1 = Is R . Therefore R = = 3 = 1000 Ω. C C 10 × 10 −6

SOLUTION 10.79. The student can construct the phasor diagram using graph paper. The diagram is going to look like that in the problem statement, except that the proper lengths and angles will be used.

SOLUTION 10.80. As the frequency approaches infinity, the capacitor shorts and the inductor opens. So, the output voltage is zero. As the frequency approaches zero, the capacitor opens, but the inductor shorts, so the output is also zero. A plot of the complete response is shown below. (Note that the magnitude response at 10rad/s is infinite): »L = 0.04; C = 0.25; »w = 0: 30/300:30; »% Vout = Zin * Iin »Zin = j*w*L ./(j*w*L*j.*w*C + 1); »plot(w, abs(Zin)) »grid »ylabel('Magnitude Zin') »xlabel('Frequency in rad/s') »plot(w,angle(Zin)*180/pi) »grid »xlabel('Phase in degrees') »ylabel('Phase in degrees') »xlabel('Frequency in rad/s')

21

100

25

80 60

Phase in degrees

Magnitude Zin

20

15

10 TextEnd

40 20 0 -20 TextEnd

-40 -60

5

-80 0

0

5

10 15 20 Frequency in rad/s

25

30

-100

0

5


25

30

SOLUTION 10.81. At infinite frequency, the resistor current is zero (because the inductor opens). So, the output voltage is zero. At DC, the inductor is short, and the output voltage is equal to the input voltage. The plot of the frequency response is shown below (a logarithmic x-axis is used):

SOLUTION 10.82. The circuit inside the black box is

22

At DC, the capacitor is an open circuit. Thus, the voltage across the resistor is 1mA×R. But we know that this voltage is 1 from the graph. This means that R = 1 kΩ. Now, in general for the above diagram: V R = I 1+ j RC The magnitude of this function is R/sqrt(2) when

= 1/ RC . Substituting the frequency from the graph

(1000 rad/s), we get C = 1 µF.

Solution 10.83. The admittance of the parallel RLC circuit is: 1 1 + +j C R j L 1 V Zin = = s Yin Is

Yin =

The function we want to find the frequency response for is nothing but the input admittance of the circuit. Using MATLAB, the following plot can be obtained: »R = 100; L = 0.1; C = 1e-3; »w = 0:0.5:300; »w = 0.01:0.5:300; »Yin = 1/R + 1. ./(j*w*L) + j*w*C; »Zin = 1 ./Yin; »plot(w,abs(Zin)) »grid »xlabel('Frequency in rad/s') »ylabel('Magnitude Zin')

23

100 90 80 Magnitude Zin

70 60 50 40 TextEnd

30 20 10 0

0

50


250

300

250

300

»plot(w,angle(Zin)*180/pi) »grid »xlabel('Frequency in rad/s') »ylabel('Phase Zin in degrees')

100 80

Phase Zin in degrees

60 40 20 0 -20 -40

TextEnd

-60 -80 -100

0

50


24

SOLUTION 10.84. The circuit inside the box is a series RLC circuit. It cannot be a parallel RLC, because as per problem 83, the admittance of a parallel RLC does not approach zero as w approaches infinity. Thus, I 1 = Yin = 1 V R+ +j L j C The resonance frequency is 50 rad/s and is determined by 1/sqrt(LC). Given L = 0.4 H, C = 1 mF. To obtain R, we make use of the fact that, from the given graph at ω = 57 rad/s, the current magnitude is approximately 0.2 times the peak magnitude. Therefore 1 R2

1 2  + 57L −  57C 

0.2

≅ R2

1 2  + 50L −  50C 

=

0.2 R

Hence 2

 1  R2 R +  57 × 0.4 − = 25R 2  = −3  0.04 57 × 10 2

R2 + 57x0.4 -

From which R =

1 R2 = 25R 2 2 ≅ 57x0.001 0.04

5.2561 = 1.0728 Ω. 24

SOLUTION 10.85. Create three mesh currents in the three planar loops. All currents are clockwise: I1 in the voltage source loop, I2 in the top bridge loop, and I3 in the bottom one. The three mesh equations are: V − I1R1 − R2 (I1 − I2 ) − R3 (I1 − I3 ) = 0 R2(I2 − I1) +

1 I + Rmeter (I2 − I3 ) = 0 j C1 2 1 R3(I3 − I1 ) + Rmeter (I3 − I2 ) + I =0 j C2 3 The plots of the magnitude and phase of VB − VC = Rmeter (I3 − I2 ) are shown in the text. 25

SOLUTION 10.86. We can see all ranges by plotting on a logarithmic scale:

Note that the output will decay when we start to reach the bandwidth of the op-amp. In other words, the inverting amplifier says that the output is –1 times the input (provided the op amp works properly). Once the op amp’s gain starts dropping, the output voltage also decays with it.

SOLUTION 10.87. Correction: Change the 0.01 µF capacitor to 1 µF. (a) For this part consider the diagram below,

o

From the problem statement, ω = 320π rad/s, and Iin = 0.01∠0 A. Observe that the 50 kΩ resistor input to the inverting op amp terminal is in parallel with the 100 Ω resistor because of the virtual ground at the op amp terminals. However, for all practical purposes, this has no effect on the 100 Ω resistor, hence from Ohm's law 26

VL =

100 × j 0.1ω I = (50.265 + j50) × 0.01 = 0.50265 + j0.5 100 + j0.1ω in

From the inverting op amp configuration, Vˆ =-10VL = −5.0265 − j5 From voltage division, VC =

10−3 o ˆ −3 −9 V = (0.49735 − j0.5) × (−5.0265 − j5) = 5∠179.7 V 10 + jω10

Therefore vC (t) = 5cos(320πt + 179.7o )V

2

Parts (b) and (c). For the SPICE simulation we have the following circuit in B -SPICE:

which leads to the response below

27

Prb10-87-Small Signal AC-2 +1.000

(V)

+10.000

+100.000

Frequency +1.000k

(Hz)

+10.000k

+500.000m

+400.000m

+300.000m

+200.000m

+100.000m

+0.000e+000

MAG(V(IVM))

The magnitude at 160 Hz is 0.499 for a 1 mA current input. Thus a 10 mA input current should lead to 4.99 V by linearity which approximates the 5 V computed analytically in part (a). Hence with a 15 V saturation limit, the input magnitude may increase by a factor of 3 to 30 mA.

SOLUTION 10.88. (a)

= 400 rad/s and Vin = 10−3∠ − 90° V. By the virtual short property: Iin =

Vin = j CVin 1/ j C

All this current flows through the 1 MΩ resistor: Vin = − j C106 Vin = 0.4∠ − 180°. Thus, vout(t) = – 0.4cos(400t) V (b)

= 200 rad/s, Vin = 10−3∠0° V, and Vout = 10−3∠− 90° V. Again, Vout = − j10 −3 = − j C × 4 × 105 Vin = − j80 × 10 3C ⇒ C = 12.5 nF

28

SOLUTION 10.89. (a)

= 800rad/s and Vin = 1∠− 90°V ⇒ Iin =

Vin . No current flows into Op106

Amp terminals: Vout = −Iin

1 j × (− j) = = 1.25 × 10−3 V j C 800

Thus, vout(t) = 1.25cos(800t) mV. (b) Again, Iin =

Vin 1 j × (− j) A, and Vout = −Iin ⇒ 10 = ⇒ C = 2.5 nF . 3 j C 200 × 10 C2 × 105

Solution 10.90 (a)

= 2 700rad /s and Vin = 1∠0° V ⇒ Iin = −j

1 A. Further, 150 × 103 −j

150 × 103 C 150 × 103 C −1 o Vout = − j I in = 150 × 103 × j = 0.0015∠90 V 3 3 150 × 10 − C 150 × 10 − C Thus, vout(t) = 1.5cos(2π700t + 90 o) mV. (b)

(c) The output lags the DC response by 45 degrees (note that at DC, the amplifier is inverting, or has a phase of –180 degrees). Now, the frequency response is really determined by the RC circuit in the feedback path of the op-amp. The first resistance at the input simply converts the input into a current that

29

drives this RC circuit. It can be shown that a 45 degree phase shift occurs in an RC circuit when the frequency is 1/RfC (directly from the results of an analysis on an RC circuit). So, RfC = 1/2000π ⇒C = 0.016 nF (We know Rf = 10 MΩ.) 7

Also, at this frequency, the response is 0.707×DC response. The DC response is –10 /R. So, the DC gain is 14.14. Thus, R = 7.07 kΩ.

Solution 10.91 (a) Use the virtual short property: Iin = 4

feedback path: Vout = −10 Iin =

10 4 Vin 3

10 −

j C

⇒

Vin . All of current flows through the + j 1C

10 3

Vout = 7.07∠ − 135o . Vin

(b)

(c) The spice result looks the same at low frequencies. However, at high frequencies, the response falls back to zero as the op-amp non-ideal frequency response starts to affect the behavior of the circuit.

V SOLUTION 10.92. (a) The negative terminal of the op-amp is at Vs. This implies IR = S . By KVL, R

30

− j 3R

V 1 3 C VS + ZRCIR = Vo ⇒ o = 1 + = 1 + VS R 3R − j j3 RC + 1 C The MATLAB plot for the given values is:

(c) The spice result looks pretty much the same, especially since the cut-off frequency of this circuit is much lower than the frequency at which the op-amp ceases to operate as an ideal op-amp.

SOLUTION 10.93. To compute the gain as a function of ω we observe that by the properties of an ideal op amp, Vout 2 × 10−5 + j0.5 × 10−6 ω 20 + j0.5ω Gain = =− = Vin 100 + j 0.1ω 10−4 + j0.1 × 10 −6 ω In MATLAB »G1 = 1/50e3; »G2 = 1/10e3; »C1 = 0.5e-6; »C2 = 0.1e-6; »w = logspace(-1,5,1500); »Y1 = G1 +j*w*C1; »Y2 = G2 + j*w*C2; »H = Y1 ./ Y2; »semilogx(w,abs(H)) »grid

31

SOLUTION 10.94. (a) The equations are: V1 − Vin V1 − V2 V1 − V2 + + =0 1000 1000 1/ jwC f V2 − V1 V2 + =0 1000 1/ j C2 V2 = Vout Substituting values and solving for V1 and V2 , V1 = 0.8535∠ − 82o Vin V2 = 0.723∠− 114.7o Vin The second one is the relation that we are looking for. (b) At 100 Hz, V1 = 1.00∠ − 3.68o Vin V2 = 1.00∠ − 7.3o Vin At 3000 Hz, V1 = 0.15∠ − 102o Vin V2 = 0.0728∠ − 164o Vin 32

(c) The response is that of a low-pass filter, as predicted from the results of part (b) above.

SOLUTION 10.95. (a) The two nodal equations: Vx − Vin + j C1Vx + j C2 ( Vx − Vout ) = 0 200 Vout j C2 (0 − Vx ) − =0 28 × 103 In MATLAB, »w = 2*pi*1.34e3; C = 0.05e-6; R1 = 200; R2 = 28e3; »A = [1/R1+j*w*C+j*w*C -j*w*C;-j*w*C -1/R2] A= 5.0000e-03 + 8.4195e-04i 0 - 4.2097e-04i 0 - 4.2097e-04i -3.5714e-05 »b = [1/R1; 0]; »V = A\b V= 2.6664e-01 - 5.9266e+00i -6.9859e+01 - 3.1429e+00i »Vout = V(2) Vout = -6.9859e+01 - 3.1429e+00i »abs(Vout) ans = 6.9929e+01 »angle(Vout)*180/pi ans = -1.7742e+02

Hence,

Vout = 69.93∠− 177.4 o . Vin

When the capacitors are shorts, the output is shorted to the virtual ground input, at 0 V. Similarly, when they are opens, the virtual ground makes sure that vout is zero, since there is no drop across the feedback resistor. 33

(b) The band-pass response can be computed using any SPICE program.

SOLUTION 10.96. First, note the input-output relationship: Zf Vout =− Vin R1 where Zf is the impedance of the parallel RLC circuit. We have studied this circuit extensively earlier, and we have shown that at “resonance”, the impedance of this circuit is going to be real and equal to the value of resistance, in this case R2 . That’s exactly the requirement of this problem, since we want

Zf Vout =− Vin R1

to simply be equal to R2 /R1 . It remains to note that this resonance occurs at a frequency ω = 1/sqrt(LC).

SOLUTION 10.97. (a) First, analyze the feedback amplifier circuit. The output of this op-amp circuit is:

Vop2 = −

1/ j C Vout 2 × 10 6

Also, by voltage division, the voltage at the resistive voltage divider (+ terminal of first op-amp): 2 × 103 1 −0.01 VRR = ×− Vout = Vout 5 6 2 × 10 j C2 2 × 10 j C2 2 × 106 Now, the first op-amp circuit is an inverting amplifier, but it’s + terminal is at VRR now. Thus, Vout = −10(Vin − VRR ) + VRR Substituting the above VRR means that: Vout =

−10 1.1 Vin 1+ j

Since the input voltage has unity magnitude and zero phase, the above expression gives the required magnitude and phase of the output voltage. (b) 34

10 9 8

Magnitude

7 6 5 TextEnd

4 3 2 1 0

0

0.2

0.4

0.6

0.8 1 1.2 Frequency Hz

1.4

1.6

1.8

2

(c) As can be seen, the response to zero frequency (i.e. DC) is zero. Also, the circuit goes back very quickly (less than 2 Hz) to provide the required operation, which is to achieve a gain of 10.

35

Complex Power Prbs 2/5/00

CHAPTER 11.

P11-1

@ DeCarlo & P. M. Lin

PROBLEM SOLUTIONS 1

1  e 2t  1 SOLUTION 11.1. Using equation 11.3, Pav = ∫ (e t − 1)2 Rdt =  + t − 2e t  = 0.758 W. 2  2  0

0

SOLUTION 11.2. V2 (a) From 11.6, Pav = m = 50 mW for a sinusoidal input. 2R (b) From 11.3, R Pav =  ∫ (10cos(10t)) 2 dt + 2  0 /20

3 /20

∫

 2  (10cos(10t)) dt = ∫  3 /20  2 /10

2

(−10cos(10t)) dt +

/20

2 Im R = 50 mW 2

just as the previous case since the square of the absolute cos(10t) is the same as the square of cos(10t). (c) 10R Pav = 2

2 /10

∫

0

[

10 −3 0.01cos (10t) dt = R 2 2

]

2

2 /10

∫ [0.5cos(20t) + 0.125cos(40t) + 0.375]dt 0

2 /10 1  0.5 0.125  = sin(20t) + sin(40t) + 0.375t  = 37.5 mW 2  20 40 0

(d) t=0:1/1000:1; R=1e3; pta= (0.01*cos(10.*t)).^2.*R; ptb= (0.01*abs(cos(10.*t))).^2.*R; ptc= (0.01*cos(10.*t)).^4.*R; subplot(3,1,1); plot(t,pta); grid ylabel('W'); subplot(3,1,2); plot(t,ptb); grid ylabel('W'); subplot(3,1,3); plot(t,ptc); grid ylabel('W'); xlabel('time in seconds');


P11-2


W

0.1

0.05 TextEnd

0

0

0.2

0.4

0.6

0.8

1

0.2

0.4

0.6

0.8

1

0.2

0.4 0.6 time in seconds

0.8

1

W

0.1

0.05 TextEnd

0

W

1

0 x 10-5

0.5 TextEnd

0

0

1 2 v 2 (t) 1 SOLUTION 11.3. (a) For figure a, the period is 2, and Pav = ∫ dt = (400 + 100) = 25 W. 2 0 R 2R 1 1 (20t )2 400  t 3  In figure b, the period is 1, and Pav = ∫ dt =   = 13.3 W. 0 R R  3  0 (b)


P11-3


40 Instantaneous Power in W

35 30 25 20 TextEnd

15 10

0

0.5

1

1.5

2 Time in s

2.5

3

3.5

4

2.5

3

3.5

4

40 Instantaneous Power in W

35 30 25 20 15 10 TextEnd

5 0

0

0.5

1

1.5

2 Time in s

SOLUTION 11.4. (a) For (a), looking the definition for the effective voltage, one sees graphically that the integral over one period, 2, of the squared waveform, is 500. Dividing by the period, and taking the 1

square root, Veff = 15.81 V. For fig. b, Veff =

∫ (20t )2 dt = 0

2

 Veff  2 (b) Pav = Ieff R= 8 = 20 W.  10  2

 Veff  (c) Pav =  8 = 10.67 W  10 

400

1 = 11.55 V 3


P11-4


SOLUTION 11.5. (a) This can be done graphically quite easily. The period of fig a, is 9s. The total 75 area of one period of the squared waveform is 75. This yields Ieff = = 2.89 A. In fig. b, the area 9 25 over one period is 25 which yields Ieff = = 2.89 A. 3 60 2  (b) Using current division, Pav =  Ieff 30 = 111.36 W. 90 (c) The same result as (b) is obtained since the effective current is the same.

(

)

2 1 1 t e − 1 dt = 0.615 A. ∫ 2 0

SOLUTION 11.6. (a) Ieff = 2 (b) Pav = Ieff R = 0.758 W.

SOLUTION 11.7. (a) 2 /20 20 2 /20 20  2  2 2 Veff = (10 + 2cos(20t)) dt = 100t + 2t + sin(40t) + 2sin(20t) = 102.01 2 ∫0 2  40 0 Hence Veff = 10.1 V. (b) 2 Veff =

Hence Veff = 7.91 V.

1

1 ∫0 (10cos(2t) + 5cos(4t))2 dt = [62.5t]0 = 62.568

(c) Without going into detailed calculation, note the following fact about v 32(t) . Only the product terms that have the same frequency will produce a non-zero result when integrated. Thus the integral reduces to the following: 1 2 Veff = ∫ 100cos 2 (2t) + 25cos 2(4 t) + 25cos 2 (4t − π /4) + 50cos(4 t)cos(4t − π /4) dt 0

(

)

Hence Veff =

[50 + 12.5 + 12.5 + 25cos(−π / 4)] = 9.627 V

SOLUTION 11.8. The voltage is V = 50∠0 V and the impedance Zeq = 100 − 100 j = 141.42∠ − 45o 2

Ω. Thus I =

V  0.3536 2 = 353.6∠45o mA. Hence, Pav = R Ieff = 100 ×   = 6.2516 W.  Zeq 2 


P11-5

SOLUTION 11.9. (a) The equivalent load seen by the source, Zeq =


1 = 2∠ − 36.87o . Thus j C + 1/ R

VL = IZeq = 10∠− 36.87o V and v L (t) = 10cos(30t − 36.87o) V. V I 10 × 5 (b) Pav = m m cos( v − i ) = cos(−36.87o ) = 20 W, and from 11.4 2 2 pL (t) = 25cos(−36.87o ) + 25cos(60t − 36.87o ) W SOLUTION 11.10. (a) First find the impedance Zeq = 3 + j4 = 5∠53.13o Ω, then V Is = s = 10∠ − 143.13o A. The magnitude is 10 A rms or 14.14 A peak-value. Zeq (b) Pav = 10⋅ 50cos(53.13o ) = 300 W. (c) 3Is

2

= 300 W

SOLUTION 11.11. (a) By KVL, Vs = 10IL + j100IL + 9 ⋅10IL and Vs 100∠0 1 IL = = = ∠− 45o A. 100 + j100 141.42∠45o 2 (b) PavVs = Vs IL cos(45o ) = 50 W and Pav9V1 = 9⋅10IL IL cos(180o ) = −45 W. SOLUTION 11.12. (a) By KCL the current through the resistor is 5I. So by KVL, Vs = j1000I − j500I + 200⋅ 5I and hence I=

Vs = 107.33∠ − 26.57o A 1000 + j500

The power absorbed by the resistor is, Pav = 200⋅ 5I 5I = 57.6 W. (b) PavVs = Vs I cos(26.57o ) = 11.52 W and Pav4I = 4I 200 ⋅5I cos(0) = 46.08 W. SOLUTION 11.13. First find the input impedance, Zeq = 2 − j8 + 6 + j2 = 8 − j6 Ω. Then calculate the 100 current IL = = 8 + j6 A. The complex power is then computed: 8 − j6 »ZL = 6 + j*2; »IL = 8 + j*6; »VL = ZL*IL VL = 3.6000e+01 + 5.2000e+01i »SL = VL * conj(IL) SL = 6.0000e+02 + 2.0000e+02i


P11-6


»abs(SL) ans = 6.3246e+02 Thus the apparent power is 790 VA, the average power 600 W, the reactive power 200 var, and the apparent power is 632.46 VA. SOLUTION 11.14. (a) Using MATLAB »Vseff = 100*exp(j*pi/6) Vseff = 8.6603e+01 + 5.0000e+01i »ZL = 350 +j*1*300; »Zin = 50 + ZL Zin = 4.0000e+02 + 3.0000e+02i »ILeff = Vseff/Zin ILeff = 1.9856e-01 - 2.3923e-02i »ILpk = sqrt(2)*abs(ILeff) ILpk = 2.8284e-01 »ILang = angle(ILeff)*180/pi ILang = -6.8699e+00 »SL = ZL*ILeff*conj(ILeff) SL = 1.4000e+01 + 1.2000e+01i Therefore, iL (t) = 2 (0.2) cos(300t − 6.87o ) = 0.2828cos(300t − 6.87o) A. SL = 14 + j12 VA, and the average power is 14 W.

SOLUTION 11.15. First use voltage division and observe that VL = complex power is SL = VL

1 V = 60∠60° V. Now the 2 s

VL* = 360∠53.13° = 216 + j288 VA. Thus the average power is 216 (6 + j8)*

W. *S OLUTION 11.16. (a) To find V2 we write a node equation. First we note that Yi = 1/Zi is the corresponding admittance. Hence Y1(V2 − Va ) + Y2 V2 + Y3( V2 − Vb ) = 0 Hence

Y V + Y3Vb V2 = 1 a = 100 + j50 = 111.8∠26.57o V Y1 + Y2 + Y3

obtained using MATLAB as follows »Z1 = 0.1+j*0.1; Z2 = 0.4+j*2.2; »Z3 = 0.2 + j*0.2; Va = 104 + j*50; Vb = 106 + j*48; »Y1 = 1/Z1


P11-7


Y1 = 5.0000e+00 - 5.0000e+00i »Y2 = 1/Z2 Y2 = 8.0000e-02 - 4.4000e-01i »Y3 = 1/Z3 Y3 = 2.5000e+00 - 2.5000e+00i »V2 = (Y1*Va + Y3*Vb)/(Y1 + Y2 + Y3) V2 = 1.0000e+02 + 5.0000e+01i »magV2 = abs(V2) magV2 = 1.1180e+02 »angleV2 = angle(V2)*180/pi angleV2 = 2.6565e+01 (b) Again working strictly in MATLAB we have the following complex powers of the loads and the two sources: »Sz1 = (V2 - Va)*conj((V2-Va)*Y1) Sz1 = 8.0000e+01 + 8.0000e+01i »Sz2 = V2*conj(V2*Y2) Sz2 = 1.0000e+03 + 5.5000e+03i »Sz3 = (V2 - Vb)*conj((V2-Vb)*Y3) Sz3 = 1.0000e+02 + 1.0000e+02i »Sva = Va*conj((Va - V2)*Y1) Sva = 1.0800e+03 + 3.0800e+03i »Svb = Vb*conj((Vb - V2)*Y3) Svb = 1.0000e+02 + 2.6000e+03i (c) To verify conservation of power observe that: »TotSrsPwr = Sva + Svb TotSrsPwr = 1.1800e+03 + 5.6800e+03i »TotLdPwer = Sz1 + Sz2 + Sz3 TotLdPwer = 1.1800e+03 + 5.6800e+03i which provides the desired verification. SOLUTION 11.17. Use MATLAB and refer to the following figure:


P11-8

Z1

Z4 Z2

Z3

%(a) %Bundle the impedances as per the following figure and %obtain the following. Z1=2+2*j; Y1=1/Z1; Y2=2+0.5*j; Z2=1/Y2; ZL3=4*j; YL3=1/ZL3; Y3=4+0.25*j; Z3=1/Y3; Z4=4+4*j; Y4=1/Z4; V1=10+2*j; V2=12+2*j; %Write out KCL for node 1 and 2 %(Va-V1)*Y1=V1*Y2+(V1-V2)*YL3 %(Vb-V2)*Y4=V2*Y3+(V2-V1)*YL3 Va= (V1*Y2+(V1-V2)*YL3)/Y1+V1 Vb= (V2*Y3+(V2-V1)*YL3)/Y4+V2 Va = 29.0000 +59.0000i Vb = 1.6000e+02+ 2.3400e+02i (b) Sr3=( (Va-V1)*Y1*2)*conj((Va-V1)*Y1) Sl1= ((Va-V1)*Y1*(2*j))*conj((Va-V1)*Y1) Sc1= V1*conj(V1*0.5*j) Sr1= V1*conj(V1/0.5) Sl3= (V1-V2)*conj((V1-V2)*YL3) Sc2= V2*conj(V2*0.25*j) Sr2= V2*conj(V2/0.25) Sl2= (V2-Vb)*Y4*4*j*conj((V2-Vb)*Y4) Sr4= (V2-Vb)*Y4*4*conj((V2-Vb)*Y4) SVA = Va * conj((Va-V1)*Y1) SVB = Vb * conj((Vb-V2)*Y4) Sr3 = 9.0250e+02 Sl1 = 0 + 9.0250e+02i Sc1 = 0 - 5.2000e+01i



P11-9


Sr1 = 208 Sl3 =

0 + 1.0000e+00i

Sc2 = 0 - 3.7000e+01i Sr2 = 592 Sl2 =

0 + 9.4660e+03i

Sr4 = 9466 SVA = 1.1115e+03 + 8.4550e+02i SVB = 1.0057e+04 + 9.4350e+03i (c) Take the real part of each of the complex power found in (b). The only components with non-zero average power will be the resistors which have 208 W, 592 W, 902.5 W, and 9466 W average power respectively. %(d) Total_passive=Sr1+Sr2+Sr3+Sr4+Sl1+Sl2+Sl3+Sc1+Sc2 Total_active= Va*conj((Va-V1)*Y1)+Vb*conj((Vb-V2)*Y4) Total_passive = 1.1168e+04+ 1.0280e+04i Total_active = 1.1168e+04+ 1.0280e+04i which verifies the conservation of power. SOLUTION 11.18. (a) From conservation of energy, the complex power is the sum of the complex power absorbed by every circuit elements. Thus Ss = 1240 + j145 VA, and the apparent power is 1248.4 VA. The average power is 1240 W. S (b) From S s = Vs I*s , Is = s = 5.428 A. Vs SOLUTION 11.19. (a) The complex power delivered by the source is the sum of the complex power consumed by the circuit elements. Thus Ss = 44 + j28 kVA. (b) Is =

Ss = 22.675 A Vs

(c) The total power delivered to the three groups of impedance following V1 is S1 = 41.5 + j22. From the current obtained in (b) V1 =

S1 + S 2 + S4 = 2071.5 V. Is

(d) From V1 , and the total power delivered to Z4 and Z2 , I2 = V2 =

S2 = 1805.5 V. I2

S 4 + S2 = 12.385 A. Finally V1


P11-10


*  44 + j28 SOLUTION 11.20. (a) Is =  = 22.675∠− (32.47 − 0) = 22.675∠− 32.47° A. 2.3  S + S2 + S4  (b) V1 =  1  = 2071.5∠ − 4.54° V I*s   *

S + S2  (c) Similarly as before I2 =  1  = 12.385∠ − 37.61° A, and V2 = Z2I2 = 1805.5∠− 11.05° V.  V1 

SOLUTION 11.21. From equation 11.30, we find Q = P

1 − 1 = 455.61 var; thus pf 2

S = 1000 + j455.61 = 1099∠24.5° VA. SOLUTION 11.22. (a) S1 = 1000 W, and S2 = 800 + j600 VA. Thus the total power delivered by the S* source is Stot = 1800 + j600 = 1897.37∠18.43° VA, and Is = tot = 15.81∠− 18.43° = 15 − j5 A rms. Vs* S (b) V1 = *1 = 63.25∠ − 18.43° V. Is S (c) V2 = *2 = 63.25∠18.43° V. Is P 3000 SOLUTION 11.23. Using 11.25, find SL = ave = = 4000 VA, and the load current pf 0.75 IL =

SL VL

=

4000 = 33.33 A. The power absorbed by the transmission line is then from 120

Pline = R IL = 0.2 × ( 33.33) 2 = 222.2 W. 2

SOLUTION 11.24. The capacitor must absorb a reactive power of Qnew − Qold = −17.9 kvar. Thus −QC 2 jQC = − j17.9 = − j CV s , and C = 2 = 0.897 mF. Vs 1 − 1 = 31.43 VA. Thus the reactive (0.94)2 −QC power absorbed by the capacitor is –18.57 var. Hence C = 2 = 3.714 µF. Vs SOLUTION 11.25. From equation 11.30, Qnew = 86.6


P11-11


SOLUTION 11.26. Device 1 has a complex power of S1 = P 1 + jQ1 = 360 + j 0 VA. Recall equation P 1 11.29, pf = ave , and equation 11.30 Qnew = Pave −1 S (pf ) 2 where with a lagging pf, Q > 0, and with a leading pf for Q < 0. Using equation 11.30, we have for device 2: Q2 = 1440 S2 = 1440 + j 1080 VA.

1 − 1 = 1080 var (0.8)2

S1,2 = S1 + S2 = 1800 + j1080 VA

As an aside we compute the magnitude of the current without the capacitor attached. Is =

S1,2 120

= 17.493 A

The capacitor is used to achieve a lower source current with the same average power. The first step is to find the desired QC. Here Hence

Stot = S1 + S2 + SC = 1800 + j1080 + jQC Stot 1800 + j1080 + jQC = = 15 120 120

In MATLAB we have: »QC = sqrt((15*120)^2 - 1800^2) - 1080 QC = -1080 From the formula on page 451 of the text, QC = − C Vsource

2

Hence C = 0.2 mF. Finally pf =

Pave 1800 = =1 S1 + S2 + SC 1800

1 − 1 = 8.1839 kvar. Therefore, the (0.65)2 power absorbed without the capacitor bank is Sold = 7 + j8.1839 kVA. When the bank is added we 1 want Qnew = 7 − 1 = 5.25 kvar, and henceSnew = 7 + j5.25 kVA. Thus the reactance that must (0.8)2 −QC be absorbed by the bank is –2.934 kvar, and Ceq = 2 = 0.13511 mF. Vs (b) As was just determined 5.25 kVA. SOLUTION 11.27. (a) From equation 11.30, Qold = 7


(c) Sold =

P11-12


7 7 = 10.77 kVA. Snew = = 8.75 kVA. The kVA saving is: 10.77 – 8.75 = 2.02 0.65 0.8

kVA. »Savings = 20*2.02*12 Savings = 4.8480e+02 i.e., $484.80. SOLUTION 11.28. (a) The apparent power is simply 94kW/0.78=120.51 kVA. (b) Sm = 120.51∠38.74o = 94 + j 75.41 kVA. (c) 75.41 kvar. S* (d) Ieff = m * = 523.96∠− 38.74 A. Veff (e) By KVL, Vs = RlineIeff + Veff = 516.08 − j229.52 V. (f) Ss = VsI*eff = 295.94∠14.76° kVA. 120.51cos(38.74) (g) The efficiency is = ⋅100 = 32.96% . Note that the line resistance of 0.7 Ω is much 295.94cos(14.76) to large for practical usage. This value is chosen for pedagogical reasons. (h) With a power factor of 0.94, Snew m = 94 + j 34.12 kVA. The average power of the motor must be kept the same. The reactance that must be provided by the capacitor is, Qnew − Qold = QC = −41.29 kvar. The 2

Veff j42.29k 1 Veff proper capacitor current will be IC = , and Z = = = . Solving for C * j C IC Veff j 42.29 × 103 C=

42.29 × 10 3 Veff

2

= 2.12 mF. *

(i)

Inew eff

 Snew  =  m  = 434.79∠ − 19.95° A  Veff 

o (j) By KVL, Vs = RlineInew eff + Veff = 516.1− j103.84 = 526.43∠− 11.38 V.

(

)

new new * (k) Snew = V Ieff = 228.89∠8.57° = 226.33 + j 34.12 kVA, and the efficiency is 41.5%. s s

SOLUTION 11.29. (a) The Thevenin equivalent seen at the output is, Zth = 5Ω− j /(0.1 ). For VOC * maximum power transfer, ZL = Zth = 5 + j . Note that VOC = 50 V rms, and that Ieff = =5 A Zth + ZL rms. Thus SL = ZL Ieff I*eff = 125 + j25 VA, and the average power is 125 W. SOLUTION 11.30. First find the following Thevenin equivalent, 10 2 Zth = 2 + j 4 Ω, Voc = V. 3 By the maximum transfer property, RL = 2Ω , and C = 1/(4 ) = 0.25 mF.


P11-13


SOLUTION 11.31. (a) The thevenin equivalent left of the load is by KCL, Itest + Is = VR ( j0.001+ 0.001) and Vtest = 3VR + j500Itest V. Substituting for VR , Vtest = Itest[1500 − j1000] + Is[1500 − j1500] V, and Zth = 1500 − j1000 Ω with Voc = Is × 2121.3∠ − 45° V. The value of the load for maximum power transfer is then, ZL = 1500 + j1000 Ω. Voc (b) The complex power absorbed by the load is, SL = ZL Zth + ZL power 750 W.

2

= 750 + j500 VA, and the average

SOLUTION 11.32. Consider V1

V2

(a) Using nodal analysis get the following equations: V1 − V2 = 4(V1 /8) Is + Itest = V1(1/8 + j0.25) + V2 (1/16 + j0.5) Vtest = V2 + 8Itest Using MATLAB we get the following expression, Vtest = (8.2847- j0.9110)Itest + (0.2847- j0.9110)Is From this expression Voc = 28.47 − j91.1 V, and Zth = 8.2847- j0.9110 Ω. The phasor equivalent of this circuit is a Voc source in series with a 8.28 Ω resistor and a 7.32 mF capacitor. (b) ZL = 8.2847+ j0.9110 , which is a 8.28 Ohm resistor in series with a 6.07mH inductor. Same configuration as before. Voc (c) SL = ZL Zth + ZL

2

= 275.2 + j30.26 VA and the average power is 275.2 W.

SOLUTION 11.33. (a) The Norton equivalent may be found by inspection as Zth = 10 + j20 Ω and Isc = 10 A. Thus for maximum power transfer, ZL = 20 − j20 Ω. This is a 10 Ω resistor and a 0.005 F 2

I Z capacitor in series. The maximum power is SL = ZL sc th = 1250 − j2500 VA. The maximum Zth + ZL average power is 1250 W. (b) If R is set to 20 Ω, the closest that can be achieved to maximum power transfer is ZL = 20 − j20 , or C equal to 0.005 F. With ZL as above, by current division


Iload = 10 ×

P11-14


10 + j20 = 3.333 − j6.6667 = 7.45∠63.434o A (10 + j20) + (20 − j20) 2

The maximum average power then 20 × Iload = 1111 W. 2

I Z (c) Using 11.38, RL = 31.62 Ω. Pav = RL sc th Zth + ZL

= 600.63 W.

SOLUTION 11.34. (a) From Thevenin Zth = 19.2 − j14.4 Ω, and from 11.38, set RL = 24 Ω. 2 Vs(− j 40) Voc Voc = = 80∠− 36.87°. The maximum power is Pav = R = 74.07 W. 30 − j40 Zth + RL L RL (b) The voltage is V = Voc , from this relationship, one sees that as the load resistance increase Zth + RL to infinity the output voltage goes to Voc , which is the maximum output voltage. SOLUTION 11.35. Correction: the inductor symbol in the load should be a resistor. Since the source resistance is variable, example 6.21 serves as a reference suggesting that R = 0 is the answer. To see this consider that 2 × 100 2 × 100 2 P = 2Iload = 2 = (R + 2) 2 + (2 − 2) 2 1   (R + 2)2 + L −  C Hence, decreasing R produces increasing power and the maximum power is transferred when R = 0 with Pmax = 50 W assuming that the source voltage is given in rms V. SOLUTION 11.36. As per problem 35, 2

P = 10Iload =

(

10 × 50 (R + 10)2 +

2

)2

1   L−  C

2 =

12.5 × 103 (R + 10) 2 +

1 2  L−  C

Here, again R = 0 with C chosen to eliminate the reactive term maximizes power transfer. Hence C= with Pmax = 125 W.

1 2L

= 0.01 F

SOLUTION 11.37. (a) By the maximum power transfer theorem, P1 is maximized when ZL is chosen as the conjugate of Zsource, i.e. ZL = 10 + j1000 Ω (b) To find the appropriate values of L and C observe that


P11-15


1 10 −4 − j C 7 ZL ( j ) = j L + −4 = j10 L + −8 10 + j C 10 + 1014 C 2 10 −4  C  = −8 + j  107 L − −8  = 10 + j1000 14 2 14 2  10 + 10 C 10 + 10 C  Equating real parts leads to: »%10^-4 = 10^-7 + 10^15*C^2 » »C = sqrt(1e-4 - 1e-7)/sqrt(1e15) C = 3.1607e-10 Thus C = 0.31607 nF. Equating imaginary parts using the above value of C leads to: »w = 1e7; »L = (1e3 + w*C/(1e-8 + 1e14*C^2))/1e7 L = 1.3161e-04 Thus L = 0.1316 mH. (c) In part (b), L and C are chosen to maximize P1 , the power delivered to ZL. Since L and C consume no average power, this maximum power is transferred to the 10 kΩ fixed resistor with the computed values of L and C. Thus ZL is the same as in part (b) or ZL = 10 + j1000 Ω. Since we know ZL, Pmax =

(0.1) 2 = 0.25 mW 4 × 10

This is the average power consumed by the 10 kΩ resistor. Therefore V2 = 0.25 × 10−3 × 10 4 = 2.5 2

It follows that V2 = 1.5811 V. Power to the 10 kΩ fixed resistor is P10kΩ =

V2

2

104

Thus if P10kΩ is maximized, then V2 is maximized.

SOLUTION 11.38. (a) From equation 11.4, if we substitute

2 ⇒ T

, then the resulting instantaneous

V I V I 4 power will be p(t) = m m cos( v − i ) + m m cos( t + v + i ) W where it is clearly seen that the 2 2 T fundamental period will now be halved. Note that by the same argument the fundamental frequency of the instantaneous power is double that of the voltage and current.


P11-16


(b) As a sinusoid, the fundamental period is 2π/10, any integer multiple of this period will also be periodic. (c) This is the same as (b) with an offset of 1 V added. 1 to +T 2 SOLUTION 11.39. First, Feff = f (t)dt . Without going into detailed calculation, f 2 (t) will ∫ T t0 give a summation of two types of products, a product of each element with themselves, and products of each element with the other element. In the later case, we know that two cosines multiplied with one another and integrated over one period will yield zero if their angular frequency are different. As for the former case the integral will yields the result we are looking for. For example look at the first two terms, 1 to +T 2 f (t)dt = T ∫t 0

Feff =

(

)

1 t0 +T 2 Fo + 2F12 cos2 ( 1t + 1) + ... dt T ∫t 0

(

)

=

1 t0 +T 2 Fo + F12 + F12 cos(2 1t + 2 1 ) + ... dt T ∫t 0

=

T 1 2 Fo t + F12 t + F12 sin(2 1 t + 2 1 ) + ... = Fo2 + F12 ... 0 T

[

]

SOLUTION 11.40. (a) We are given that vC (t) = Vm sin( t) V. Hence iC (t) = C

dvC = CVm cos( t) A dt

It follows that p(t) = v C (t)iC (t) = CVm2 sin( t)cos( t) = 0.5 CVm2 sin(2 t) Watts Clearly, p(t) has a peak value of 0.5 CVm2 and the integral of the sign over one period is zero implying that the average value of p(t) is zero. (b) Here WC (t) = 0.5CvC2 (t) = 0.5CVm2 sin2 ( t) = 0.25CVm2 (1− cos(2 t)) J Here the peak value occurs when cos(2ωt) = –1 in which case the peak value is 0.5CVm2 . Further, the average value of cos(2ωt) over one period, T = π/ω, is zero whereas the average of a constant over the same period is simply the constant. Hence, WC,ave = 0.25CVm2 J. (c) From example 11.6, 2 QC = −IC ,eff VC ,eff = − CVC,eff = −0.5 CVm2 = −2

Q Therefore, WC,ave = − C . 2

(0.25CVm2) = −2

WC,ave .


P11-17


SOLUTION 11.41. (a) We are given that iL (t) = I m sin( t) A. Hence v L (t) = L

diL = LIm cos( t) V dt

It follows that 2 2 p(t) = v L (t)iL (t) = LIm sin( t)cos( t) = 0.5 LIm sin(2 t) watts 2 Clearly, p(t) has a peak value of 0.5 LIm and the integral of the sign over one period is zero implying that the average value of p(t) is zero.

(b) Here 2 2 WL (t) = 0.5LiL2 (t) = 0.5LIm sin2 ( t) = 0.25LIm (1− cos(2 t)) J 2 Here the peak value occurs when cos(2ωt) = –1 in which case the peak value is 0.5LI m . Further, the average value of cos(2ωt) over one period, T = π/ω, is zero whereas the average of a constant over the 2 same period is simply the constant. Hence, WL,ave = 0.25LIm J.

(c) As an extension to example 11.5, 2 2 QL = IL,eff VL,eff = LI L,eff = 0.5 LI m =2

(0.25CI m2 ) = 2

WL,ave .

Q Therefore, WL,ave = L . 2 SOLUTION 11.42. (a) The complex power absorbed by the load is, SL = VI* = ZII* = Z I 2 . Now note that the average power is the real part of the complex power. Also note that a complex number multiplied by its complex conjugate will yield a real value. Therefore the real part of Z I 2 is just the real part of Z, R, multiplied by I 2 , Pav = R I 2 . With the same reasoning, the reactance is the imaginary part of Z, X, multiplied by I 2 , Q = X I 2 . (b) The complex power absorbed by the load is SL = VI* = V(YV)* = Y V 2 . The same reasoning as in (a) holds thus the real part of the admittance times V 2 , yields Pav = G V 2 . Using the imaginary part, Q = B V 2 . SOLUTION 11.43. (a) The equivalent resistance seen by the source is Req = 6 – j9 Ω. So the current delivered by the source is: Is =

Vs − j110 = = 10.17∠− 33.69° A 6 − j9 6 − j9

or iL (t) = 10.17 2 cos(120 t − 33.69o ) A. Similarly VC = − j15 × Is =

−15 × 110 = 152.54∠− 123.69° V 6 − j9


P11-18

or vC (t) = 152.54 2 cos(120 t − 123.69o ) V. Further, L =


6

= 15.9 mH and C =

1 = 176.8 µF. 15

The instantaneous energy stored in the inductor is WL (t) = 0.5LiL2 (t) = 1.646cos 2 (120 t − 33.69o ) J and the instantaneous energy stored in the capacitor is WC (t) = 0.5CvC2 (t) = 4.115cos 2 (120 t − 123.69o ) J The source voltage is zero when t = 0. Therefore

and

WL (0) = 1.646cos 2 (−33.69o ) = 1.1395 J WC (0) = 4.115cos 2 (−123.69o ) = 1.266 J

(b) and (c) Observe that iL(t) and vC(t) are 90o out of phase. When one is zero, the other has a peak value. Therefore whenWC = 0 implies vC(t0 ) = 0 for appropriate t0 ; hence WL (t 0 ) = 1.646 J. Similarly, when WL = 0, say at t0 , then WC (t 0 ) = 4.115 J. SOLUTION 11.44. In order to solve this problem, we want to express the power in terms or R's and L's in both circuits. First, looking at the circuit with just the coil and the 110 V source: I = V / Zcoil , Zcoil = R + j L , and 110 2 2 Pcoil = I R = 2 R = 300 watts (*) R + 2L2 Next, looking at the circuit when a resistance is added in series with the coil, I = V /(8 + Zcoil ), Zcoil = R + j L , 220 2 Pcoil = I 2 R = R = 300 watts (**) (8 + R)2 + 2L2 To find R, solve equation (*) for R 2 + 2 L2 and substitute into equation (**) to obtain »R = 300*64/(220^2 - 300*16 -110^2) R = 6.0952e-01 Substituting R into equation (*) yields L = 13.05 mH. SOLUTION 11.45. The average power consumed by the 2.7 Ω resistor is 250 watts. This allows us to compute the magnitude of | Icoil |. We know that | Vcoil | is 150 Vrms. Thus we can compute the magnitude of the coil impedance and hence L as follows: »magIcoil = sqrt(250/2.7) magIcoil = 9.6225e+00 »magZcoil = 150/magIcoil magZcoil = 1.5588e+01


P11-19

»% magZcoil^2 = 2.7^2 + (w*L)^2 » »w = 2*pi*60; »L = sqrt( magZcoil^2 - 2.7^2)/w L= 4.0725e-02 »% Magnitude of impedance seen by 220 V source is: » »magZin = 220/magIcoil magZin = 2.2863e+01 »% magZin^2 = (R + 2.7)^2 + (w*L)^2 » »Realpart = sqrt(magZin^2 - (w*L)^2) Realpart = 1.6941e+01 »R = Realpart - 2.7 R = 1.4241e+01


9/26/01

P12-1


CHAPTER 12. PROBLEM SOLUTIONS SOLUTION P12.1. By conservation of energy, the instantaneous power consumed by each load when summed together is equivalent to the total power consumed by the three phase load. Thus writing out ptot (t) = pAB (t) + pBC (t) + pCA (t) = v AB (t)i AB (t) + vBC (t)iBC (t) + vCA (t)iCA (t) V 2 V 2 = VL 2 cos( t + v )⋅ L cos( t + i ) + VL 2 cos( t + v − 120o ) ⋅ L cos( t + i − 120o ) Z Z V 2 + VL 2 cos( t + v + 120o ) ⋅ L cos( t + i + 120o ) Z

(

V2 V2 = L (cos( v − i ) + cos(2 t + v + i )) + L cos( v − i ) + cos(2 t + v + i + 120o ) Z Z

(

VL2 + cos( v − i ) + cos(2 t + v + i − 120o ) Z

)

)

3VL2 3VL2 = cos( v − i ) = × pf Z Z

SOLUTION P12.2. To justify the point of this problem we equate the following two equations: (i) For the 3 phase system: 3 PL2 ' PL2 ' = 3× IL R = 3 × × 2 R = 2 R 9 VL VL (ii) For the single phase system: PL2 2 Ploss = 2 × IL R = 2 2R VL It follows that R' = 2R. Since both systems have the same distance of transmission and the resistance of a wire is inversely proportional to the cross sectional area, the condition R' = 2R implies that the cross section A' of each wire in the three-phase system need only be half of the area A of the wire in the single phase system. But there are two wires in the single phase system and three wires in the three-phase system. Therefore the ratio of the materials used is: ' Ploss

2 '

(

)

material in 3 − phase system 3A ' 3 1 = = × = 75% material in 1 − phase system 2A 2 2

SOLUTION P12.3. For row 1 of table 12.1, the impedance in (a) seen between each pair of terminal is Z Zik = Z∆ || 2Z∆ = 2 ∆ . In (b) the impedance seen between any two terminals is 3 Z∆ Z∆ Z∆ Z jk = + =2 . 3 3 3

9/26/01

P12-2


In row 2, the impedance between any two terminals is Z jk = ZY + ZY = 2ZY for (c), and Zik = 3ZY || 6ZY = 2ZY for (d).

SOLUTION P12.4. Consider the ∆-Y relationship of the figures below (row 3 of table 12.1):

Let us consider the terminal pair (1,2): Z (i) For the Y-connected case, Zth = 2 × ∆ and 3 Voc =

Vp 3

∠

(

)

− 30o −

Vp 3

(

)

∠ − 150o =

Vp

(

)

∠ 1∠− 30o − 1∠ − 150o = Vp ∠ 3

(ii) For the D-connected case, 2(Z∆ ) 2 2 = Z∆ 3Z∆ 3 Now note that there is no load connected to the ∆-configuration. Applying KVL to the indicated loop implies that: 0 = 3Z∆ Iloop + V p∠ + V p∠ − 120o + V p∠ + 120o = 3Z∆Iloop Zth = Z∆ / /(2Z∆ ) =

Hence Iloop = 0 . Finally, Voc = V12 = Z∆Iloop + V p∠ = V p∠ . Therefore, looking into terminals 1-2, both the ∆-configuration and the Y-configuration have the same Thevenin equivalent. For terminal pairs (1-3) and (2-3), the proof is virtually the same. Hence this establishes the equivalence in row 3 of table 12.1. To establish the equivalence in row 4 of table 12.1, we do all the same computations with the slightly different labeling to obtain the same result, i.e., the circuits are equivalent.

V V  SOLUTION 12.5. For (a) first note the following relationship VN =  1 + 2  ( Z1 || Z2 || Z3) , which is  Z1 Z2  obtained by KCL at the center node, N, with node 3 as the reference node.. Write out KCL at terminal 1,     V1  V1 V2  Z3 + Z2 Z3 I1 = − + ( Z || Z || Z ) = V1 Z Z + Z Z + Z Z  − V2   . Do the Z1  Z1Z1 Z2Z1  1 2 3 2 3 1 3 1 2  Z1Z2 + Z2 Z3 + Z3 Z1 

9/26/01

P12-3


 1 1  V2 same for terminal 1 in (b), I1 = V1 + − . Now substitute the formulas in the problem  Z31 Z12  Z12     Z2 + Z3 Z3 statement into the later equation, I1 = V1  − V2   , which is  Z1Z2 + Z2Z3 + Z3Z1   Z1Z2 + Z2 Z3 + Z3 Z1  the same as the equation for (a). Using the same method, the same result is obtained for node 2. So the fact that the substitution of the equivalence in (b) yields the same equation as in (a) proves that the equivalences are accurate.

Vp 200∠0° = = 17.89∠− 26.57° A. By the same relationship Z 10 + j5 IB = 17.89∠ − 146.57° A, and IC = 17.89∠93.43° A. The neutral line current is the sum of the other three and is zero. The power of each phase is the same, as they have the same load, and current SOLUTION 12.6. By Ohm’s law IA =

2

magnitude. Using the current at terminal A, the total power is, P = 3R IA = 9602 W.

SOLUTION 12.7. For this balanced Y-Y connection,we follow the method of example 12.6. Using MATLAB: »Z = 20 + j*10' Z = 2.0000e+01 + 1.0000e+01i »VAN = 200; »IA = VAN/Z IA = 8.0000e+00 - 4.0000e+00i »IAmag = abs(IA) IAmag = 8.9443e+00 »IAangle = angle(IA)*180/pi IAangle = -2.6565e+01 »% Use phase inference to obtain: »IBmag = IAmag IBmag = 8.9443e+00 »IBangle = IAangle - 120 IBangle = -1.4657e+02 »ICmag = IAmag ICmag = 8.9443e+00 »ICangle = IAangle + 120 ICangle = 9.3435e+01 »Ptotave = 3*real(Z)*IAmag^2 Ptotave = 4.8000e+03

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P12-4


SOLUTION 12.8. Using the same approach as in the example 12.6 we compute VAN = VsA ×

Thus the drop is:

Z 4 + j3 = 120 × = 116.94 V Z + Zg 4 + j3 + 0.1 + j0.2

120 − 116.94 × 100 = 2.55 % . 120

SOLUTION 12.9. A 5.5 % drop corresponds to VAN = 113.4 V. This means V Z 0.945 × 120 = AN = = 0.945 120 120 Z + Z1 + Zg From this equation, Z + Z1 + Zg =

Z = 5.291. 0.945

So (4.05 + 0.1)2 + (3.15 + 0.2)2 = 27.99 . Solving the quadratic equation results in

SOLUTION 12.10. Solving the single phase equivalent circuit for phase A, IA =

V p − V p∠120° = 58.87∠− 56.57° 10 + j5

Using phase inference, IB = IA × ∠− 120° = 58.87∠ − 176.57° IC = IA × ∠120° = 58.87∠64.43° Alternately, using much more work, IB =

V p∠ − 120°− V p

= 58.87∠ − 176.57° 10 + j5 V p∠120° − V p∠ − 120° IC = = 58.87∠64.43° 10 + j5 The total power, using the previously derived relationship, is

where

P = 3VL IL pf = 3 × 380 × 58.87 × pf = 34.655 kW   5  pf = cos tan−1   = 0.8944   10 

= 0.7893.

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P12-5


SOLUTION 12.11. (a) By KVL and Ohm's law, IAB =

Vp = 6.57∠− 26.57° 30 + j15

By phase inference, IBC = IAB × ∠ − 120° = 6.57∠ − 146.57° ICA = IAB × ∠120° = 6.57∠93.43° Alternately, IBC = ICA =

V p∠ − 120° 30 + j15 V p∠120° 30 + j15

= 6.57∠ − 146.57°

= 6.57∠93.43°

To compute the line currents we have in Amps: IA = IAB − ICA = 11.36∠ − 56.57° IB = IBC − IAB = 11.36∠− 176.57° IC = ICA − IBC = 11.36∠63.43° (b) The average power is P = 3 × 30 IAB

2

= 3.88 kW, and the total reactive power is

2

Q = 3 × 15 IAB = 1.94 kvar.

25 ⋅ 746 = 0.85 , so the total power being Pdeliverd 21941 = 63.31 A. 3 ⋅VL ⋅ pf

SOLUTION 12.12. From the efficiency we know that delivered to the load is 21941 W. Using 12.4, IL =

SOLUTION 12.13. (a) Using the power efficiency relationship, Ptot = (b) Using 12.4, IL =

300 ⋅ 746 = 239.36 kW. 0.935

239.36 × 10 3 = 68.28 A. 3 ⋅VL ⋅ pf

SOLUTION 12.14. The magnitude of the power delivered to each delta connected load is 79.787 kW. We can now perform the analysis on a single phase. Using equation 11.30, we obtain

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P12-6


2  1  Qold = 79787  − 1 = 43064 vars 0.88 

and 2  1  Qnew = 79787  − 1 = 26225 vars 0.95 

Hence Sold = 79787 + j 43064 Snew = 79787 + j26225 So the reactive power to be supplied by the capacitor is QC = 16839 = 43064 – 26225. The capacitance QC C= 2 = 8.44 uF. Vl

SOLUTION 12.15. Replacing the delta configuration of the source by its Y-equivalent, the following voltages (in V) are obtained, VAN =

Vp 3

∠ − 30°, VBN =

Vp 3

∠− 150°, VCN =

Vp 3

∠90°

From the single phase equivalent, IA = From phase inference

VAN = 19.63∠− 56.57° Z

IB = IA × ∠− 120° = 19.63∠− 176.57° IC = IA × ∠120° = 19.63∠63.43°

The power factor of the load is   5  pf = cos tan−1   = 0.8944   10  Hence, the total power in the balanced system is P = 3VL IL ⋅ pf = 3 × 380 × 19.63 × 0.8944 = 11.556 kW.

SOLUTION 12.16. First replace the delta connected load by it Y-equivalent. Then analyzing a single phase and using phase inference, we obtain in amps,

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P12-7

IA =


Vp

= 34.08∠− 26.56° (Z / 3) IB = IA × ∠− 120° = 34.08∠− 146.56° IC = IA × ∠120° = 34.08∠93.43° The power factor of the load is   5  pf = cos tan−1   = 0.8944   10  The total power is P = 3VL IL ⋅ pf = 3⋅127⋅ 34.08 ⋅ pf = 11.61 kW. The voltage across each load is 3 ⋅V p = VL = 219.97 V.

SOLUTION 12.17. (a) Referring to figure 12.2, the following relationships may be pointed out: VAB = VAN + VNB = VAN − VBN VBC = VBN + VNC = VBN − VCN VCA = VCN + VNA = VCN − VAN (b) Perform a delta to Y-transformation and use Ohm’s law and phase inference to obtain in amps: 220∠− 30° = 11.36∠ − 56.57° 3⋅Z IB = 11.36∠ − 176.57° IC = 11.36∠63.43° IA =

(c) First,   5  pf = cos tan−1   = 0.8944   10  The average power, Pav = 3VL IL pf = 3 × 220 × I A × 0.8944 = 3.87 kW. 2

The reactive power may be found as follows, Q = 3 × 5IL = 1.936 kvar.

SOLUTION 12.18. Part (a) Apply ∆−Y transformation to the source. Apply the transformation of row 3, Table 12.1 to the source. The result is the circuit shown in figure 12.19 of example 12.8. Therefore we may use many of the calculated values in example 12.9. In particular, the line currents are IA = 19.49/-68.9o A, IB = 19.49/-188.9o A, and

IC = 19.49/51.09o A

Part (b). Compute line-to-neutral voltage. From figure 12.21 and by Ohm's law

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P12-8


VAN = IA ZY = 19.49/-68.9o × 5∠36.87o = 97.45∠−32.03o V By inference on phase angles VBN = 97.45∠−152.03o V and VCN = 97.45∠87.97o V Part (c). Compute line voltages VAB , VBC , and VCA . From the voltage phasor diagram of figure 12.2. VAB = 3 VAN e j 30 = 168.8e − j 2.03

V

VBC = VABe − j120 = 168.8e − j122.03

V

VCA = VABe j120 = 168.8e − j117.97

V

o

o

By inference on phase angles, o

o

o

o

Part (d). Compute the total power. From equation 12.4 Ptotal = 3 × VL × IL × pf = 3 × 168.8 × 19.49 × 0.8 = 4558 W

SOLUTION 12.19. Perform a delta to Y transformation on the source. This yields Vp 380 V1N = ∠− 30o = ∠ − 30o V. Perform a delta to Y-transformation on the load. This yields a 3 3 Z new impedance Znew = = 4 + j 3 Ω. Now analyzing a single phase and finding the others by 3 inference yields, in amps, V1N IA = = 42.18∠− 67.97° Z1 + Znew IB = 42.18∠− 187.97° IC = 42.18∠52.03° 2

The average power delivered to the load is Pav = 3 × IA × 4 = 21.35 kW.

SOLUTION 12.20. Performing a delta-to-Y-transformation on the source and the load, the new source magnitudes are multiplied by 1 3 , and the loads are divided by 3. Observe that Zg = 0.15 + j0.45 Ω, Z1 = 0.1 + j0.2 Ω, and Z = 12 + j9 Ω. Using Ohm’s law on phase one, and then inference on the phase angles for the other currents, yields in amps, IA =

V p∠− 30° = 41.14∠− 68.91°  Zg Z 3 + Z1 +   3 3

IB = 41.14∠ − 188.91° IC = 41.14∠51.09°

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P12-9


 12 2 The total power may be calculated as follows, Pav = 3  I L = 12 × (41.14)2 = 20.310 kW. 3

SOLUTION 12.21. Perform delta-to-Y-transformation on the delta load. Because of the property of the neutral to be ground in balance circuit, the two loads in the same phase combine in parallel. Thus the new impedance seen in one branch is Z (5 + j5) × (5 − j 3) ZYnew = ZY || ∆ = = 4.044 + j0.1923 Ω 3 (5 + j5) + (5 − j3) Now looking at a single phase and inferring for the others, IA =

V p∠ − 30°

230∠ − 30° = 32.84∠ − 32.73° A 3 ⋅ ZYnew 3( 4.044 + j0.1923) IB = 32.84∠ − 152.73° A IC = 32.84∠87.27° A =

2

The complex power is S = 3⋅ ZYnew IL = 13.07 + j0.615 kVA.

SOLUTION 12.22. For this problem we do not need to use any 3-phase knowledge if we are clever. We will provide a clever solution here. VDN = VDB – VNB . Using B as the reference we write a node equation at N to obtain: V V V V V − VAB VNB − VCB 0 = NB + NA + NC = NB + NB + Z1 Z1 Z1 Z1 Z1 Z1 Therefore 3VNB = VAB + VCB = 115 − 115∠− 120o = 172.5 + j99.59 From voltage division VDB =

Z3 2 + j4 VAB = 115 = 57.5 + j57.5 V Z3 + Z2 2 + j4 + 4 − j2

Now using MATLAB we have »115-115*exp(-j*2*pi/3) ans = 1.7250e+02 + 9.9593e+01i »VNB =ans/3 VNB = 5.7500e+01 + 3.3198e+01i »VDB = 115*(2+j*4)/(6+j*2) VDB = 5.7500e+01 + 5.7500e+01i »VDN = VDB - VNB VDN = 7.1054e-15 + 2.4302e+01i »magVDN = abs(VDN) magVDN = 2.4302e+01 »angVDN = angle(VDN)*180/pi angVDN = 9.0000e+01

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P12-10


SOLUTION P12.23. First we write the following two loop equations,

(

)

(

VsA − VsA∠− 120 = IA Zg + Z1 + ZA + ( IC + IA) Z + Z1 + Zg

(

VsA∠120 − VsA∠ − 120 = (IA + 2IC ) Zg + Z1 + Z and solve in MATLAB: (a) %Problem 12.23a Zg=0.05+0.15i; Z=4+3i; Z1=0.1+0.2i; ZA=5+2i; VA= 120; VB= 120*exp(-120*i*pi/180); VC= 120*exp(120*i*pi/180); %Write mesh equations in matrix form and solve: % i.e. [IA;IC] = A - 1*C A=[ 2*(Z1+Zg)+ZA+Z Z+Z1+Zg; Zg+Z1+Z 2*(Zg+Z1+Z)]; Ainv= A^(-1); C=[VA-VB;VC-VB]; B=Ainv*C; IA= B(1); IC= B(2); IB= -IA-IC; IA abs(IA) 180*angle(IA)/pi IB abs(IB) 180*angle(IB)/pi IC abs(IC) 180*angle(IC)/pi %And By Ohm's law, VAN= IA*ZA abs(VAN) 180*angle(VAN)/pi VBN= IB*Z abs(VBN) 180*angle(VBN)/pi VCN= IC*Z abs(VCN) 180*angle(VCN)/pi IA = 21.76∠–29.12° A IB = 23.88∠–155.60° A IC = 20.64∠ 82.39° A

)

)

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P12-11


VAN = 117.20∠-7.32° V VBN = 119.4∠–118.73° V VCN = 103.2∠119.26° V (b) % From phasor diagram VAB = VAN - VBN abs(VAB) 180*angle(VAB)/pi VBC = VBN - VCN abs(VBC) 180*angle(VBC)/pi VCA = VCN-VAN abs(VCA) 180*angle(VCA)/pi VAB = 195.5 ∠27.34° VBC = 194.9 ∠–92.04° VCA = 196.99 ∠147.8° (c) Calculating the average power delivered to each load using the general formula Ptot = ∑ R I

2

= 5 × (21.76) 2 + 4 × (23.88)2 + 4 × (20.64) 2 = 6.353 kW

SOLUTION P12.24. Doing this problem in MATLAB, %Problem 12.24 ZDg=0.15+0.45i; ZD=12+9i; Z1=0.1+0.2i; ZDA=15+6i; %Apply the delta to Y transformation to the load as follows, ZA=(ZD*ZDA)/(2*ZD+ZDA); ZB=(ZDA*ZD)/(2*ZD+ZDA); ZC=(ZD*ZD)/(2*ZD+ZDA); %Apply delta to Y trans. to the source, V1= 180/sqrt(3)*exp(-30*i*pi/180); V2= 180/sqrt(3)*exp(-150*i*pi/180); V3= 180/sqrt(3)*exp(90*i*pi/180); Zg= ZDg/3; %Write nodal equations in matrix form and solve %i.e. [IA;IC]=A-1*C A=[ 2*(Z1+Zg)+ZA+ZB ZB+Z1+Zg; Zg+Z1+ZB 2*(Zg+Z1)+ZC+ZB]; Ainv= A^(-1); C=[V1-V2;V3-V2]; B=Ainv*C; IA= B(1); IC= B(2); IB= -IA-IC; IA abs(IA)

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P12-12


180*angle(IA)/pi IB abs(IB) 180*angle(IB)/pi IC abs(IC) 180*angle(IC)/pi IA = 17.4 ∠–63.1° A IB = 20.1 ∠178.9° A IC = 19.5 ∠51° A (b) % By Ohm's law, VAN= IA*ZA; VBN= IB*ZB; VCN= IC*ZC; %From phasor diagram VAB=VAN - VBN abs(VAB) 180*angle(VAB)/pi VBC = VBN-VCN abs(VBC) 180*angle(VBC)/pi VCA =VCN - VAN abs(VCA) 180*angle(VCA)/pi VAB = 170.40 ∠–2.5° V VBC = 168.58 ∠–122.4° V VCA = 169.75 ∠118.1° V (c) % by Ohm's law IAB=VAB/ZDA abs(IAB) 180*angle(IAB)/pi IBC=VBC/ZD abs(IBC) 180*angle(IBC)/pi ICA=VCA/ZD abs(ICA) 180*angle(ICA)/pi o

IAB = 10.55 ∠–24.3 A IBC = 11.24 ∠–159.3° A ICA = 11.32 ∠81.2° A (d) Ptot = ∑ R I

2

= 15 × (10.55) 2 + 12 × (11.24) 2 + 12 × (11.32) 2 = 4.721 kW

9/26/01

P12-13


SOLUTION P12.25. The average power that the load draws is P = 440 ⋅ 40⋅ pf = 8.8 kW. From this we obtain the old and new complex powers: Sold = 8.8 + j15.242 kVA and Snew = 8.8 + j15.242 + jQC kVA 2

Here the reactance supplied by the capacitor is QC = − CV l = −7299 var. So the new power factor is   Q  pf = cos tan−1  new   = 0.742 . From S = Vl Il = 11.855 kVA, we solve for  8800  Il = IA = IB = 26.94 A.

SOLUTION P12.26. Using Ohm’s law and the power factor of the load, IC = −440 ⋅ C∠90° = 16.59∠ − 90° A, IA = 40∠60° A, and IB = −IC − IA = 26.94∠ − 137.9° A.

SOLUTION P12.27. By simply applying Ohm’s law to each load and then KCL at node N, we get the following: V V V IA = A = 4.4 A, IB = B = 4.4∠ − 120° A, IC = C = 0.88∠120° A, and 50 50 250 IN = IA + IB + IC = 3.52∠− 60° A. Finally, PA = 50 × (4.4) 2 = 968 W, PB = 50 × (4.4 )2 = 968 W, PC = 250 × (0.88) 2 = 193.6 W.

SOLUTION P12.28. Using MATLAB, VA = 220 VB = 220*exp(-120*pi*i/180); VC = 220*exp(120*pi*i/180); %Use KVL to solve for the different currents IA = (VA - VB)/50 + (VA - VC)/50 abs(IA) 180*angle(IA)/pi IB = (VB - VC)/100 + (VB - VA)/50 abs(IB) 180*angle(IB)/pi IC = (VC - VB)/100 + (VC - VA)/50 abs(IC) 180*angle(IC)/pi IA = 13.2 A IB = 10.1 ∠–130.9° A

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P12-14


IC = 10.1 ∠130.9° A 2

V The total power delivered to each resistor may be calculated using the following relationship P = l , R 2

where Vl = 3V p PCA = 2904 W.

2

using Y to Delta source transformation. Thus PAB = 2904 W, PBC = 1452 W, and

SOLUTION P12.29. Consider the circuit of figure 12.29 with two additional node labeling:

Choose N as the reference node and apply KCL to node M. This yields the following node equation: o o 0.02(VM − 220) + 0.02 VM − 220e − j120  + 0.004  VM − 220e j120  = 0

Equivalently

2.2VM = 88 − j152.42 or VM = 40 − j69.282 V

The remaining calculations proceed in MATLAB as follows: »% Compute node voltage VM »X = +220 +220*exp(-j*2*pi/3) + 0.2*220*exp(j*2*pi/3) X = 8.8000e+01 - 1.5242e+02i »VM = X/2.2 VM = 4.0000e+01 - 6.9282e+01i »% Compute line currents »IA = 0.02*(220 - VM) IA = 3.6000e+00 + 1.3856e+00i »magIA = abs(IA) magIA = 3.8575e+00 »angIA = angle(IA)*180/pi angIA = 2.1052e+01 »IB = 0.02*(220*exp(-j*2*pi/3) - VM) IB = -3.0000e+00 - 2.4249e+00i »magIB = abs(IB) magIB = 3.8575e+00 »angIB = angle(IB)*180/pi angIB = -1.4105e+02 »IC = 0.004*(220*exp(j*2*pi/3) - VM) IC = -6.0000e-01 + 1.0392e+00i »magIC = abs(IC)

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P12-15


magIC = 1.2000e+00 »angIC = angle(IC)*180/pi angIC = 1.2000e+02 »% Compute total power »Ptot = 50*magIA^2 + 50*magIB^2 + 250*magIC^2 Ptot = 1.8480e+03

SOLUTION P12.30. (a) The equivalent resistance of a 40 ft size #14 wire is 2.575 Req1 = 40⋅ = 0.103Ω . The voltage at the light and appliance is then 1000 Vapp = 115 − 2⋅ Req110 = 112.94 V, which is equivalent to a 1.8% drop, thus the wire gauze is appropriate. 1.619 = 0.08095 Ω. 1000 Vapp = 115 − 2⋅ Req214 = 112.7334 V, which corresponds to 1.97 %. So the wire gauze is appropriate.

(b) Repeating the previous calculations, Req2 = 50⋅

(c) Under normal operating condition the current in the live wire should equal the current in the neutral wire; thus no current should be in the ground wire. (d) Note the following relationship, VAB = VAN + VNB = VAN − VBN . This means that VAN = 115 V and VBN = 115∠180° V.

SOLUTION P12.31. Assuming a very large resistance for the person who touches the prong. For prong A, the voltage is approximately V = 115 V. For prong N, the voltage is approximately V = 0. For the ground prong G, no current flows through the wire, so the voltage should be ground or zero.

SOLUTION P12.32. For an approximate analysis, we use the circuit models below given the following assumptions: (i) The resistance from body to ground is very large (possibly due to rubber shoes); (ii) The resistances of all connecting wires are negligible. (iii) The motor winding is represented by R1 in series with R2 . Here we can further estimate that. R1 + R2 = 115/3 = 38.3 Ω, and R1 = R2 = 19.15 Ω. (iv) The trigger switch of the drill has been depressed. (a) The point P is connected to the metal case as shown below. It is obvious that Vcase = Vp = 0.

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P12-16


(b) The point M is connected to the metal case as shown below. The current through the hot wire is 115/19.15 = 6 A. If the fuse capacity is smaller than 6 A, then it will blow and Vcase = VM = 0. On the other hand , if the fuse capacity is greater then 6 A, then R2 is nearly shorted and Vcase = VM = 0.

(c) The point Q is connected to the metal case as shown below.

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P12-17


Both R1 and R2 are nearly shorted. The line current is so large that it will blow the fuse. Hence Vcase = VQ = 0. CONCLUSION: In all three cases, the person touching the metal case of the defective appliance will experience zero or a very low voltage. The circuit is safe.

SOLUTION P12.33. For an approximate analysis, we use the circuit models below given the following assumptions: (i) The resistance from body to ground is very large (possibly due to rubber shoes); (ii) The resistances of all connecting wires are negligible. (iii) The motor winding is represented by R1 in series with R2 . Here we can further estimate that. R1 + R2 = 115/3 = 38.3 Ω, and R1 = R2 = 19.15 Ω. (iv) The trigger switch of the drill has been depressed. (a) The point P is connected to the metal case as shown below. It is obvious Vcase = Vp = 0.

(b) The point M is connected to the metal case as shown below. Simple voltage divider action leads to

9/26/01

P12-18


Vcase = VM = 0.5×115 = 57.5 V

(c) The point Q is connected to the metal case as shown below.

It is obvious that Vcase = VQ = 115 V. CONCLUSION: In all three cases, the fuse will not blow. For cases 2 and 3, the voltage appearing on the metal case may cause serious injury to the person.

Prbs Ch 13

March 18, 2002

P13-1

©R. A. DeCarlo, P. M. Lin

PROBLEM SOLUTIONS CHAPTER 13 S OLUTION 1 3 . 1 . Given i ′′(t) + 16i′ (t) + 4Bi(t) = v ′ (t) + 8v(t) (a) with v(t) = v ′(t) = 0 and i1 (0) + i2 (0) = 0 (i) the characteristic equation is s2 + 16s + 48 = 0 (ii) the characteristic equation has factors (s + 4) (s + 12) = 0 and hence s1,s2 = −4,−12 (iii) Equivalent circuit at t = 0

(iv) Here by KCL i1 (0) = i 2 (0) = 6A At point (1) i1 (0) + i2Ω(0) = i2 (0) 6A + i2Ω(0) = 6A i2Ω(0) = 0 and v L1 = v R2 (0) = 2i 2Ω (0) = 0V Then by KVL at t = 0 v L1 (0) − v 6Ω (0) + v L 2 (0) = 0 ⇒ −v L 2 = −36V di2 = −36V dt i ′(0) = i2′ (0) = −36A / s L2

(v) from part ii i(t) = Ae −4t + Be−12t and then i ′(t) = −4 Ae−4t − 12Be −12t Then using the initial conditions A+ B=6 −4A − 12B = −36 1

Prbs Ch 13

March 18, 2002

P13-2


solving yields B = 1.5 and A = 4.5. Then i(t) = 4.5e −4t + 1.5e −12t A (b)

If v(t) = 12V and i(0) = i ′(0) = 0 , then v ′(t) = 0 and i ′′(t) + 16i′ (t) + 48i(t) = 8v(t)

with i(t) = Ae −4t + Be−12t i′(t) = 4Ae −45 − 12Be−12t i ′′(t) = 16Ae −4t + 144Be−12t Then at t = ∞ when v(∞) = 12V i(∞) = C, i′ (∞) = 0 and i′ ′(∞) = 0 Thus 48C = 96

⇒

C=2

Then i(t) = Ae −4t + Be−12t + 2 and i(0) = 0 = A + B + 2 i ′(0) = 0 = −4 A − 12B Multiply the first of these by 4 yields 4 A + 4B = −8 −4A − 12B = 0 Solving yields B = 1 and A = -3. Thus for t > 0 i(t) = −3e −4t + e −12t + 2 A (c) If v(t) is as in fig. 13.1(b) then 2t 0 < t ≤ 2 v(t) =  t≥2 0 The value of C in part (b) will change and at t = 2s, a new set of initial conditions will be required (obtainable from the solution at t = 2s) and these would be used in the decay portion described by i ′′(t) + 16i′ (t) + 48i(t) = 0 t ≥ 2s

2

Prbs Ch 13

March 18, 2002

P13-3


S OLUTION 1 3 . 2 . (a) Use the figure with the currents i1 through i5 designated in the circuit below. Then work from v 0 to vin using repeated applications of KVL, KCL and the elemental equations: i1 = 2vo i2 = 2vo′ ′

i3 = i1 + i2 = 2(v ′o + v o)

with v 2 = vo

v1 = 0.5i3 + v 2 = 0.5[2(vo′ + vo ) + vo ] = vo′ + 2v o

then

(

)

d 1 v + 2vo = 2(vo′′ + 2vo′ ) dt o i5 = i3 + i4 = 2(v o′ + v o) + 2(vo′′ + 2vo′ ) = 2vo′′ + 6vo′ + 2vo

i4 = 2v1′ = 2

Finally vin = 0.5i5 + v1 = 0.5(2vo′′ + 6vo′ + 2vo ) + v ′o + 2v o = vo′′ + 4v ′o + 3v o Hence v˙˙out (t) + 4v˙out (t) + 3v out (t) = vin (t) (b) Note from part (a) that vout (t) = v 2 (t) and v1(t) = v˙out (t) + v out (t) Hence vout (o) = v 2 (o) = 1V v˙out (o) = v1(o) − vout (o) = 7 − 1 = 6V (c) The characteristic equation s2 + 4s + 3 = 0 has factors

(s + 1)( s + 3) = 0 and roots s1,v 2 = −1,−3 3

Prbs Ch 13

March 18, 2002

P13-4


Thus, because of the input, vin (t) = 6V vout (t) = Ae−t + Be −3t + C v˙out (t) = − Ae−t − 3Be−3t v˙˙out (t) = Ae−t + 9Be −3t and at t = ∞ v˙˙out (∞) + 4vout (∞) + 3v out (∞) = 6V 0

+

0 + 3C C = 2V

= 6V

and vout (t) = Ae−t + Be −3t + 2V

S OLUTION 1 3 . 3 . (a) From the given differential equation, the characteristic equation is s3 + 14s2 + 52s + 24 = (s + 6)(s2 + 8s + 4) = 0 Therefore the roots a = −6, b = −4 + 2 3 = −0.5359,c = −4 − 2 3 = −7.4641. (b)

(i) v(0) = v C3 (0) = 6 V, as given. (ii) To compute v'(0) we write a nodal equation at node 3. In particular

(v − vC1 ) + (v − v C2 ) + 0.5v'+v = 0 which implies that v'(t) = 2vC1 (t) + 2vC 2 (t) − 6v(t)

(*)

Hence v'(0) = 2vC1 (0) + 2vC 2 (0) − 6v(0) = 24 + 18 − 36 = 6 V/s. (iii) To compute v''(0) we first differentiate equation (*). This yields v''(t) = 2v'C1 (t) + 2v'C 2 (t) − 6(2vC1 (t) + 2vC 2 (t) − 6v(t))

(**)

To express v'C1 (t) and v'C 2 (t) in terms of the node voltages we write node equations at nodes 1and 2 respectively. At node 1 ' (vC1 (t) − vC 2 (t)) + (vC1 (t) − v(t)) + 0.5vC1 (t) = i(t)

and at node 2 (vC 2 (t) − vC1 (t)) + (vC 2 (t) − v(t)) + 0.5v 'C 2 (t) = 0 Hence v'C1 (t) = −4v C1(t) + 2v C 2 (t)) + 2v(t) + 2i(t) 4

Prbs Ch 13

March 18, 2002

P13-5


and vC' 2 (t) = 2vC1 (t)) − 4vC 2 (t) + 2v(t) Substituting these two equations into (**) yields the desired result when t is set to 0. However, this quantity has no direct physical meaning. v''(t) = 2(−4vC1(t) + 2vC 2 (t)) + 2v(t) + 2i(t)) + 2(2vC1 (t)) − 4vC 2 (t) + 2v(t)) − 6(2vC1 (t) + 2vC 2 (t) − 6v(t))

2

Hence, v''(0) = 2(–48 + 18 + 12 + 2i(0)) + 2(24 – 36 + 12) – 6(24 + 18 – 36) = –72 + 4i(0) V/s . Finally, using the characteristic roots found in part (a) and assuming a constant input, the form of the solution is v(t) = Ae−6t + Be−0.5359t + Ce −7.4641t + D Following the methods of example 13.2, A = 0, B = (7.3301 – 1.0774×i(0)), C = (–1.3301 + 0.0774×i(0)), and D = i(0). (c)

(i) Not proportional to a single voltage but it is proportional to iC3 (0).

(ii) Much more complex. (iii) No. (iv) No. This is why we use the Laplace transform approach.

S OLUTION 1 3 . 4 . (a) f (q + T0 )]q=t = f (t + T0 ) (b) e −5q cos(0.5πq + 0.25π )

]q=2t = e−10t cos(πt + 0.25π)

S OLUTION 1 3 . 5 . (a) Let T = 1.

5

Prbs Ch 13

March 18, 2002

P13-6


2

1.5

1

0.5

0 -1

-0.5

0

0.5

1

1.5

2

2.5

3

(b) Again let T = 1. 0

-0.5

-1

-1.5

-2 -1

-0.5

0

0.5

1

1.5

2

2.5

3

-0.5

0

0.5

1

1.5

2

2.5

3

(c) 0.25 0.2 0.15 0.1 0.05 0 -1

(d)

6

Prbs Ch 13

March 18, 2002

P13-7


1 0.8 0.6 0.4 0.2 0 -1

-0.5

0

0.5

1

1.5

2

2.5

3

(e) 7 6 5 4 3 2 1 0 -1

r

(f)

∑i

-0.5

0

0.5

1

1.5

2

-0.5

0

0.5

1

1.5

2

2.5

3

(t − i)

i =0

(g) 5 4 3

2 1

0 -1

2.5

3

(h) Pulses of height 1 and width T. 7

Prbs Ch 13

March 18, 2002

P13-8


S OLUTION 1 3 . 6 . (a) L[ f1(t)] = F1(s) = ∫

∞

f (t)e 0− 1

−st

dt ==

T 2 −st e dt = T1

∫

T2

e −st   −s 

=

T1

(

1 −sT 1 e − e −sT 2 s

)

(b) f 2(t) = f1(t) . Hence, the answer is the same as in (a). (c) L[ f 3(t)] = F3(s) = ∫ (d) F4 (s) = ∫ (e) F5(s) = ∫

∞ 0−

∞ 0−

∞ 0−

f 3 (t)e −st dt = −2 ∫

f 4 (t)e −st dt = −2∫ f 5(t)e −st dt = ∫

∞

0−

∞ 0−

[

∞ 0−

(t)cos(4 πt − 0.25π)e −st dt = −2

2 =− 2 2

(t − T )cos(4πt − 0.25π)e −st dt = −2cos(4 πT − 0.25π)e −sT

(t) − (t − T )]e −st dt = 1− e −sT

S OLUTION 1 3 . 7 . 0−

f (t)e

−st

dt = ∫

∞

0−

5e

−4t −st

e

e −(s+ 4)t  dt = −5  s + 4 

∞

=

5 s+ 4

1.4

1.6

0−

5 4.5 4 3.5 3

f(t)

(a) F(s) = ∫

∞

2.5 TextEnd 2 1.5 1 0.5 0

0

0.2

0.4

0.6

0.8

1 t

1.2

1.8

2

8

Prbs Ch 13

March 18, 2002

(b) F(s) = ∫

∞ 0−

f (t)e

−st

P13-9

∞

dt = ∫ − 5e 1

−4t −st

e


∞

e −(s +4) t  dt = −5  s + 4 

1−

−4 5e −(s +4) −s 5e = =e s+4 s+4

0.1 0.09 0.08 0.07

f(t)

0.06 0.05 TextEnd 0.04 0.03 0.02 0.01 0

(c) F(s) = ∫

∞ 0−

f (t)e

−st

0

0.2

∞

0.4

dt = ∫ 5e

0.6

−4(t −1) −st

e

1

0.8

1 t

1.2

1.4

1.6

∞

e −(s+ 4)t  dt = −5e  s + 4  4

1.8

= e −s

1−

2

5 s+ 4

5 4.5 4 3.5

f(t)

3 2.5 TextEnd 2 1.5 1 0.5 0

(d) F(s) = ∫ (e) F(s) = ∫

∞ 0− ∞

0−

0

0.2

f (t)e −st dt = ∫

∞


∞

0−

0−

0.4

0.6

0.8

1 t

1.2

1.4

1.6

1.8

2

5e −4( t −1) (t)e −st dt = 5e 4

5e −4( t −1) (t − 1)e −st dt = 5e −s

9

Prbs Ch 13

March 18, 2002

P13-10


(f) F(s) = ∫

∞ 0−

f (t)e −st dt = 2 ∫

∞

0−

= 2∫

∞ 0−

[ u(t)u(1− t)]e −0.25t e −stdt = 2∫0 − 1

e −0.25t e −stdt − 2∫

∞

1−

e −0.25te −st dt =

e −0.25te −st dt =

(

2 1− e −(s +0.25) s + 0.25

(

2 1− e −s s + 0.25

)

)

2 1.8 1.6 1.4

f(t)

1.2 1 TextEnd 0.8 0.6 0.4 0.2 0

F(s) = ∫

∞ 0−

0

0.2

0.4

0.6

0.8

1 t

1.2

1.4

1.6

1.8

2

f (t)e −st dt =

S OLUTION 1 3 . 8 . (a) F(s) = ∫

∞


∞


∞


∞

0− ∞

−st

1

0−

(b) F(s) = ∫ (c) F(s) = ∫

∞ 0− ∞

(d) F(s) = ∫

0−

0− 0−

0−

f (t)e

dt = ∫

0−

∞ Ae t e −stdt = ∫ − Ae−(s − )t dt = 0

A s−

∞ Ae (t −1) u(t − 1)e −st dt = ∫ − Ae− e −(s − )t dt = e −s 1

A s−

Ae (t −1) (t)e −st dt = Ae− t −st

Ae e

1 e −(s− )t  A dt = − A  = (s − )  − s − 0

(1− e−(s − )) 10

Prbs Ch 13

March 18, 2002

P13-11


S OLUTION 1 3 . 9 . Consider the following in which q = at and t = q/a. ∞

L[ f (at)] =

∫−

f (at)e

−st

0

1 dt = a

∞

∫−

f (q)e −(s/a)q dq =

0

1  s F a  a

S OLUTION 1 3 . 1 0 . (a) e j t = cos t + j sin t e − j t = cos t − j sin t Add these equations and divide by 2 to obtain e j t + e− j t 2 Similarly, subtract the equations and divide by 2j to obtain cos( t) =

sin( t) =

e j t − e− j t 2j

Note that

[ ]

L ej t =

[

]

1 1 , L e− j t = s− j s+ j

(b)  e j t + e− j t  0.5 0.5 s L[cos t ] = L = + = 2 2 s+ j s + 2   s − j and L[sin t] =

−0.5 j 0.5 j + = 2 s− j s+ j s + 2

S OLUTION 1 3 . 1 1 . From the time frequency scaling property,

11

Prbs Ch 13

March 18, 2002

P13-12

L[sin( t) u(t)] =

1


1 2

 s +1  

= 2 s + 2

Using the time differentiation property, s d  L[ cos( t)u(t)] = L  (sin( t)u(t)) = sL[sin( t) u(t)]− sin(0) = 2  dt  s + 2 Finally using the multiplication by t property

L[− t cos( t)u(t)] =

d s  2 s2 = −  2  2 ds s + 2  s2 + 2 s2 + 2

(

)

Hence

( s2 + 2 ) − 2 2 (s 2 + 2 )

2 2 2 s L[sin( t)u(t) − t cos( t)u(t)] = 2 − = 2 s + 2 s2 + 2

(

)

s2

=

2 3

(s 2 + 2 )

2

S OLUTION 13.12. (a) 1 g1 (t) = At sin( t), f (t) = sin(t)u(t), F (s) = 2 s +1 By the frequency scaling property     2  1 1  = 1  L[ f ( t)] =  2  2 + 2  = s2 + 2 s s       + 1     By the multiplication by t property   d    −2 s  2 s L[t sin( t)u(t)] = −  2 = − =    2 2 ds  s + 2   s2 + 2  s2 + 2   By the linearity property

(

L[ At sin( t)] = A

)

(

)

2 s

(s 2 + 2 )

2

12

Prbs Ch 13

March 18, 2002

P13-13


(b) 1 g2 (t) = Ae at sin( t)u(t), f (t) = sin(t)u(t), F(s) = 2 s +1 By the frequency scaling property, L[sin( t)] = 2 s + 2 By the damping property

[

]

L e at sin( t) =

(s − a) 2 +

2

and by the linearity property L[g2 (t)] = A

(s − a) 2 +

2

(c) Before beginning, note that sin( t + ) = cos( )sin( t) + sin( )cos( t) Here g3 (t) = Aeat sin( t + )u(t) = cos( )Aeat sin( t)u(t) + sin( )Aeat cos( t)u(t) 1 f (t) = sin(t)u(t), F(s) = 2 s +1 From the linearity property and part (b), G3(s) = Acos( )

(s − a) + 2

2

+ L{sin( )Aeat cos( t)u(t)}

d sin(t) which implies dt s s L[cos(t)] = s{L[sin(t)]} − f (0) = 2 −0= 2 s +1 s +1

By the differentiation in the time domain property cos(t) =

By the frequency scaling property   s 2 1  1  s L[cos t ] =  =   2 2 s2 +  + 1  s

( )

 s  = 2  s2 + 2

Then by the linearity property

13

Prbs Ch 13

March 18, 2002

P13-14


s L[ Asin( )cos( t)] = A sin( ) 2 s + 2 Then by the frequency-shift property

[

]

s−a (s − a)2 + 2

L Asin( )e at cos( t) = Asin( )

[

]

G3(s) = L Aeat sin( t + ) = Acos( ) =

A

(s − a) 2 +

2

s−a + Asin( ) (s − a)2 + 2 (s − a) 2 + 2

( cos( ) + sin( )(s − a))

S OLUTION 13.13. (a) We are given that 1 f ( t) = sin( t) ⇒ F(s) = 2 s +1 And must find the transform of g1 (t) = At cos( )tu(t) By the time differentation property cos(t ) =

d sin(t) dt

( )

s s ⇒ L[cos(t )] = 2 − sin 0 − = 2 s +1 s +1

By the frequency scaling property with

L[cos( t )] =

1

 s   s 1   F  =  = 2 s + 1  

( )

2

(

s 2 s2

+

2

)

s = 2 s + 2

Using the multiplication-by-t property d  s  s2 + 2 − s(2s) s2 − 2 L[t cos( t )] = −  2 =− 2 = 2 2 ds  s + 2  s2 + 2 s + 2

(

)

(

)

Finally, by the linearity property L[ At cos( t )u(t)] =

A

(

2

− s2

(s 2 + )

)

2 2

14

Prbs Ch 13

March 18, 2002

P13-15


(b) Let g2 (t) = Ae at cos( t )u(t) . Recall from part (a) that s L[ Acos( t)] = A 2 s + 2 By the Frequency-shift property

[

]

L Aeat cos( t) = A

s− a

(s − a) 2 +

2

(c) From Trig identities, cos( t + ) = cos( t)cos( ) − sin( t)sin( ) Recall from part (b) that

[

]

L Aeat cos( t) = A

s− a

(s − a) 2 +

2

Hence

[

]

]

s− a

L Aeat cos( )cos( t) = Acos( )

s−a

( s − a)2 +

2

It follows that

[

L Aeat cos( t + ) = Acos( ) = A cos( )

(s − a) + 2

2

[

− A sin( )L e at sin( t)

]

 cos( )(s − a) − sin( )  − A sin( ) = A   (s − a) 2 + 2 (s − a) 2 + 2  (s − a) 2 + 2  s− a

S OLUTION 1 3 . 1 4 . (a) We are given that s L[cos(t )] = 2 s +1 which implies by the frequency scaling property that s L[cos( t )] = 2 s + 2 Using the multiplication-by-t property L[t cos( t )] = −

d  s  s2 + 2 − 2s2 s2 − 2 = − = 2 2 ds  s2 + 2  s2 + 2 s2 + 2

(

)

(

)

15

Prbs Ch 13

March 18, 2002

P13-16


(b) If g(t) = teat cos( t), then using part (a) and frequency shift property,

[

]

at

L te cos( t) =

(s − a) 2 −

[( s − a)

2

2

]

2 + 2

S OLUTION 1 3 . 1 5 . (a) With e at − e −at sinh(at) = 2 L[sinh(at )] =

1 1 1  1  s + a − ( s − a)  1  2a  a − =  =  2 = 2     2 2 2 2  s − a s + a 2  s − a  2  s − a  s − a2

(b) With e at + e −at cosh(at) = 2 L[cosh(at)] =

1 1 1  1  s − a + s + a  1  2s  s + =  2 =  2 = 2     2 2 2 s + a s − a  2  s − a  2  s − a  s − a2

S OLUTION 1 3 . 1 6 . (a) From Problem 15 a L[sinh(at )] = 2 s − a2 So that by the multiplication by t property L[t sinh(at )] = −

d  a  − a(2s) 2as =− =   2 2 2 2 ds  s − a  s2 − a 2 s2 − a 2

(

) (

)

(b) From Problem 15 s L[cosh(at)] = 2 s − a2 So that by the multiplication-by-t property

16

Prbs Ch 13

March 18, 2002

P13-17

(

)


s2 − a 2 − s(2s) −s2 + a 2 + 2s2 d  s  s2 + a2 L[t cosh(at)] = −  2 =− = = 2 2 2 ds  s − a 2  s2 − a2 s2 − a2 s2 − a 2

(

)

(

)

(

)

S OLUTION 13.17. Here s+2 s +1 Since g1 (t) = 5 f (t − 2), use the time shift and linearity properties to obtain F(s) =

(a)

s + 2  L[g1 (t)] = 5e −2s   s + 1  (b)

Since g2 (t) = 5e −2t f (t), use the frequency shift and linearity properties,  ( s + 2) + 2  s + 4  L[g2 (t)] = 5 = 5   s + 3   ( s + 2) + 1 

(c) From part (a),  s + 2 L[5 f (t − 2)] = 5e −2s   s + 1  Therefore, since g3 (t) = 5e −2t f (t − 2) , by the frequency shift property  (s + 2) + 2  s + 4  L[g3 (t)] = G1(s + 2) = 5e −2(s+ 2)  = 5e −2(s +2)    s + 3   (s + 2) + 1  (d)

Since g4 (t) = 5tf (t) , use the multiplication-by-t and linearity principle to obtain L[g4 (t)] = −

d  5(s + 2)  5[(s + 1)− (s + 2)] 5 =− =   ds  s + 1  (s + 1)2 (s + 1)2

S OLUTION 1 3 . 1 8 . In all parts s L[ f (t)u(t)] = F (s) = 2 s +4

17

Prbs Ch 13

(a)

(b) (c)

March 18, 2002

P13-18


e −Ts s L[ Af (t − T )u(t − T )] = A 2 s +4

(

L[ A(t − T ) f (t − T)u(t − T )] = Ae

(d)

)

( (

) )

 2  s2 − 4 d d  s   s + 4 − s(2s)  L[ Atf (t)u(t)] = − Af (s) = − A  2 = − A = A 2 2 2 ds ds  s + 4   s2 + 4  s +4   Note that the answer is simply a time shift of the function given in (b).

(

−sT

)

( s2 − 4 ) 2 (s 2 + 4 )

This function is that of part (a) multiplied by t. Hence, by the multiplication by t property,

(

)

 −Ts −Ts (s 2 + 4) − 2s × se −Ts  d  e −Ts s   e − Tse  L[ Atf (t − T )u(t − T )] = − A  2  = − A  2 ds  s + 4   s2 + 4   

(

)

 2   3 2  2  2s − ( 1− Ts ) (s + 4)   Ts + s + 4Ts − 4  = Ae−Ts  = Ae −Ts    2 2  s2 + 4   s2 + 4     

(

)

(

)

S OLUTION 1 3 . 1 9 . In all parts F(s) = L[ f (t)u(t)] = 2 s + 2 e −Ts

(a) By the time shift property, L[ Af (t − T )u(t − T )] = A 2 s + 2

(

)

Using the multiplication-by-t property, L[ Atf (t)u(t)] = − A

d d   (2s) 2 s [F (s)] = − A  2 2  = A =A 2 2 ds ds  s +  s2 + 2 s2 + 2

(

)

(

)

18

Prbs Ch 13

March 18, 2002

P13-19


© The answer here is an application of the time shift property to the answer of part (b). L[ A(t − T ) f (t − T)u(t − T )] = Ae−sT

2 s

(s 2 + 2 )

2

(d) The answer here uses the multiplication-by-t property applied to the answer of part (a).  d  e −Ts L[ Atf (t − T )u(t − T )] = − A ds  s2 + 2 

(

)

 = A  

 −Ts 2   Te (s + 2 ) + 2se −Ts    2  s2 + 2   

(

)

 2 2  Ts + 2s + T  = A e −Ts   2  s2 + 2   

(

)

Solution 13.20. (a) f (2t) = (2t) + (2t − 1) For the first term on the right, the peak occurs at t = 0 and 0+

∫−

0+

(2t)dt = 0.5 ∫

( )d = 0.5

−

0

0

under the transformation τ = 2t. For the second term, the function peaks at t = 0.5 and 0.5+

∫−

0+

(2t − 1)dt = 0.5 ∫

0.5

( )d = 0.5

−

0

under the transformation τ = 2t – 1. Hence f (2t) = (2t) + (2t − 1) = 0.5[ (t) + (t − 0.5)] Therefore, a = 0.5 = b. 19

Prbs Ch 13

March 18, 2002

P13-20


(b) (i) Here F(s) = L[ f (t)] = 1 + e −s . By the time scaling property L[ f (2t)] =

(

1 F (s /2) = 0.5 1+ e −0.5s 2

)

(b)-(ii) For this part,

(

L[ f (2t)] = 0.5L[ (t) + (t − 0.5)] = 0.5 1+ e −0.5s

)

S OLUTION 13.21. (a) L[v(t)] = 2L[g"(t)] − L[g'(t)] = 2s2F (s) − 2sg(0− ) − 2g'(0 − ) − sF(s) + g(0 − )

(

)

(2s − 1)(s + 1) s2

= 2s2 − s F (s) − 2sg(0− ) − 2g'(0 − ) + g(0− ) = (b)

L[v(t)] = 2L[ f "(t)] − L[ f '(t)] = 2s2F (s) − sF(s) − 2sf (0 − ) − 2 f '(0 − ) + f (0 − )

(

)

= 2s2 − s F (s) − 2s − 2 + 1 =

(2s − 1)(s + 1) s2

− 2s − 2 + 1

(c) −

0 t  G(s) 1  1 −   L[v(t)] = L[ g'( t)] − L ∫ g(q)dq = sG(s) − g(0 ) − − ∫ g(q)dq =  s −  G(s) s s s  −∞  −∞

1 1  1 s + 1 =  s −  3 = (s + 1)  2 − 4  s s s s  (d) −

0 t  F(s) 1 −   L[v(t)] = L[ f '( t)] − L ∫ f (q)dq = sF(s) − f (0 ) − − ∫ f (q)dq s s  −∞  −∞ −1 1 1  = (s + 1) 2 − 4  − 1− s s s 

assuming λ > 0. The expression is ill-defined if λ ≤ 0.

20

Prbs Ch 13

March 18, 2002

P13-21


S OLUTION 1 3 . 2 2 . (a) (i) If  s2 + 4  F(s) = L[ f (t)u(t)] = ln 2  = ln s2 + 4 − ln s2  s 

(

) ( )

Then by the multiplication-by-t property L[−tf (t)u(t)] = +

[ (

) ( )]

d ln s2 + 4 − ln s2 ds

2s 2s 2s 2 2s2 − 2s2 − 8 −8 = 2 − 2= 2 − = = 2 2 s +4 s s +4 s s s +4 ss +4

(

(ii)

)

(

)

Using the solution to (a)-(i), by the frequency shift property

[

]

L −te −2t f (t)u(t) =

(

−8

(s + 2) ( s + 2) + 4 2

−8 = 3 s + 6s + 16s + 16

)

=

(

−8

)

(s + 2) s2 + 4s + 8

(b) If G(s) =

(

−8

s s2 + 4

)

a partial fraction expansion may be employed G(s) =

K1 As + B −2 2 = + = + 2 2 s s s +4 s +4 s s2 + 4

(

−8

)

Hence, g(t) = [2cos(2t) − 2] u(t) and f (t) = −

g(t) 2 2 = − cos(2t) t t t

S OLUTION 13.23: Part (a)-(i): From table 13.2, the multiplication by t property implies that 21

Prbs Ch 13

March 18, 2002

L[−tf (t)u(t)] =

P13-22


d d  s + a d 1 1 −2a F(s) = ln  = (ln[s + a] − ln[s − a]) = − = 2  ds ds  s − a  ds s + a s − a s − a2

−2a Part (a)-(ii): Let us make use of the answer to part (a)-1. Let G(s) = L[−tf (t)u(t)]= 2 . s − a2 Then by the frequency shift property in table 13.2, L[−te −at f (t)u(t)] = L[e −at (− tf (t)u(t))] = G(s + a) =

−2a −2a = 2 2 s(s + 2a) (s + a) − a

(

)

 −2a  –1  1  –1  1  −at Part (b): g(t) = L–1[G(s)]= L–1  2 − e at u(t)  = L  s + a  − L  s − a  = e 2 s − a  More specifically,  e at − e −at  g(t) = e −at − e at u(t) = −2  u(t) = −2sinh(at) u(t) 2  

(

Hence f (t) =

)

g(t) 2sinh(at) = . −t t

S OLUTION 1 3 . 2 4 . (a)-(i) If  s + a F(s) = L[ f (t)u(t)] = ln = ln(s + a) − ln(s + b) s + b Then by the multiplication-by-t property L[−tf (t)u(t)] = +

d 1 1 s + b − (s + a) b−a ln (s + a) − ln( s + b)] = − = = [ ds s + a s + b (s + a)(s + b) (s + a)(s + b)

(ii) By the frequency-shift property

[

]

L −te −at f (t)u(t) =

b−a b− a = 2 (s + 2a)(s + a + b) s + s(3a + b) + 2a(a + b)

(b) If

22

Prbs Ch 13

March 18, 2002

P13-23

G(s) = L[−tf (t)u(t)] =


b−a (s + a)(s + b)

a partial fraction expansion may be employed

G(s) =

(

b−a K K 1 −1 = 1 + 2 = + (s + a)(s + b) s + a s + b s + a s + b

)

Hence, g(t) = e −at − e −bt u(t) and f (t) =

g(t) e −bt e −at = − −t t t

d S OLUTION 1 3 . 2 5 . The relationship is f (t) = g(t) or equivalently, g(t) = dt

t

∫

f (q)dq .

−∞

Now we have that f (t) = 6 (t) − 12 (t − 2) + 6 (t − 4) . Therefore, F(s) = 6 − 12e −2s + 6e −4s From the time integration property, F (s) 6 12e −2s 6e −4s G(s) = = − + s s s s

t

S OLUTION 1 3 . 2 6 . For 0 ≤ t < T1, we see that g(t) =

∫

f (q)dq . Thus one presupposes here that

−∞

d the relationship is f (t) = g(t) or equivalently, g(t) = dt

t

∫

f (q)dq . As such E = A – B and D = A –

−∞

B + C. Further, f (t) = A (t) − B (t − T1 ) + C (t − T2 ) which implies that F(s) = A − Be−T 1s + Ce −T 2 s Thus 23

Prbs Ch 13

March 18, 2002

P13-24

G(s) =


F (s) A Be−T1s Ce −T 2s = − + s s s s

S OLUTION 1 3 . 2 7 . (a) f (t) = u(t) + u(t − 1) ⇒

1 e −s 1 f (s) = + = 1− e −s s s s

(

)

(b) f (t) = u(t) + u(t − 1)− u(t − 3) ⇒

1 e −s e −3s 1 f (s) = + − = 1+ e −s − e −3s s s s s

(

)

(c)

f (t) = u(t) + u(t − 1)− 2u(t − 3) ⇒ F (s) =

1 e −s 2e −3s 1 + − = 1 + e −s − 2e −3s s s s s

(

)

(d) f (t) = 2u(t) − u(t − 2) − u(t − 3) ⇒ F(s) =

2 e −2s e −3s 1 − − = 2 − e −2s − e −3s s s s s

(

)

S OLUTION 1 3 . 2 8 . (a)

2 2e −s 2 f (t) = 2r(t) − 2r(t − 1) ⇒ F (s) = 2 − 2 = 2 1− e −s s s s

(b)

2 2e −s e −2s 1 f (t) = 2r(t) − 2r(t − 1)+ r(t − 2) ⇒ F (t) = 2 − 2 + 2 = 2 2 − 2e −s + e −2s s s s s

(

) (

)

(c) f (t) = 2r(t) − 2r(t − 1)− 2r(t − 2) + 2r(t − 3). It follows that 2 2e −s 2e −2s 2e −3s 2 F(s) = 2 − 2 − 2 + 2 = 2 1− e −s − e −2s + e −3s s s s s s

(

)

24

Prbs Ch 13

March 18, 2002

P13-25

S OLUTION 1 3 . 2 9 . (a) Here f (t) =


3 3 r(t) − 3r(t − 2) + r(t − 4) . Thus, 2 2

3 3e −2s 3e −4s 3 F(s) = 2 − 2 + = 2 1− 2e −2s + e −4s 2 2s s 2s 2s

(

)

V 2V V (b) Here f (t) = o r(t) − o r(t − T ) + o r(t − 2T ). Thus T T T V  1 2e −T e −2T  F(s) = o  2 − 2 + 2  T s s s  (c) Here f (t) = 2r(t − 1)− 4r(t − 2) + 4r(t − 4) − 2r(t − 5)

(

1 F(s) = 2 2e −s − 4e −2s + 4e −4s − 2e −5s s

)

S OLUTION 1 3 . 3 0 . (a) Here f (t) = 2r(t) − 2r(t − 1)− 2u(t − 4) implies 2 2e −s 2e −4s 2 F(s) = 2 − 2 − = 2 1− e −s − se −4s s s s s

(

)

(b) Here f (t) = 2u(t) − r(t − 2) + r(t − 4) implies

F(s) =

2 e −2s e −4s 2 − 2 + 2 = 2 s − e −2s − e −4s s s s s

(

)

S OLUTION 13.31. (a) Here f (t) = 2u(t) − r(t) + 2r(t − 2) − 2r(t − 4) − 2u(t − 4) 2 1 2e −2s 2e −4s 2e −4s Thus F(s) = − 2 + 2 − 2 − . s s s s s −1 1 e −2s e −2s (b) f (t) = − u(t) + r(t) − r(t − 2) − u(t − 2). Hence F(s) = + − 2 − . s s2 s s

25

Prbs Ch 13

March 18, 2002

P13-26


(c) f (t) = 2r(t) − 2r(t − 1)− 2u(t − 1) + 2u(t − 2) − 2r(t − 2) + 2r(t − 3). 2 2e −s 2e −s 2e −2s 2e −2s 2e −3s Hence F(s) = 2 − 2 − + − 2 + 2 . s s s s s s

26

2/23/02

page P14.1


PROBLEM SOLUTIONS CHAPTER 14 S OLUTION 1 4 . 1 . (a) 1 ) 2 2 2 Cs = Cs(RLCs + R) = RLC(s + 1 LC) = R(s + 1 LC) Z(s) = 1 Cs(RCs + LCs2 + 1) LC(s2 + R s + 1 ) s2 + R s + 1 R + Ls + Cs L LC L LC R(Ls +

(b) 1 ) LCs Ls Cs = R + Z(s) = R + = R + 1 C(LCs2 + 1) LCs2 + 1 Ls + Cs 1 1 2 RLCs2 + Ls + R RLC(s + RC s + LC ) = = 1 LCs2 + 1 LC(s2 + ) LC (Ls)(

Hence, 1 1   R s2 + s+  RC LC Z(s) = 1 s2 + LC

SOLUTION 14.2. (a) 80 ) Vs (s) s(800 + 16s) 800s + 4000 s Zin (s) = = = = 2 2 80 I s (s) 10 + 0.2s + s(0.2s + 10s + 80) s + 50s + 400 s (b) If is (t) = 3e −20t u(t) A then 3 Is (s) = s + 20 and 800s + 4000 3 2400s + 12,000 Vs(s) = Zin (s)Is (s) = 2 − = s + 50s + 400 s + 20 (s + 10)(s + 20)(s + 40) K1 K2 K3 = + + s + 10 s + 20 s + 40 Here 2400s + 12,000 +12,000 − 24,000 K1 = = = −40 (s + 20)(s + 40) s =−10 (10)(30) (10 + 0.2s)(

K2 =

2400s + 12,000 12,000 − 48,000 = = 180 (s + 10)(s + 40) s=− 20 (−10)(20)

2/23/02

page P14.2

K3 =

and for t > 0,

2400s + 12,000 12,000 − 96,000 = = −140 (s + 10)(s + 20) s =−40 (−30)(−20) 180 40 140 Vs(s) = − − s + 20 s + 10 s + 40


v s (t) = 180e −20t − 40e −10t − 140e −40t V

SOLUTION 14.3. (a) Yp(s) = Cs +

1 1 = 2 × 10 −3 s + = 2 × 10 −3 (s + 50) R 0.10

Then Z p(s) = and Zin (s) = 1.25s + and

500 s + 50

500 1.25s2 + 62.5s + 500 = s + 50 s + 50

I (s) 1 s + 50 0.80s + 40 Yin (s) = s = = = Vs (s) Zin (s) 1.25s2 + 62.5s + 500 s2 + 50s + 400

With v s (t) = 90e −40t u(t) , then Vs(s) = and

90 s + 40

90 0.80s + 40 ⋅ s + 40 (s + 10)(s + 40) 72s + 3600 K1 C1 C2 = = + + 2 s + 10 s + 40 (s + 40)2 (s + 10)(s + 40)

Is (s) = Vs (s)Yin (s) =

Here K1 = and with

72s + 3600 3600 − 720 2880 = = = 3.20 2 900 (s + 40) s =−10 (30) 2

72s + 3600 720 → p(−40) = = −24 s + 10 −30 (s + 10)(72) − (72s + 3600) p′(s) = (s + 10) 2 −30(72) − (−2880 + 3600) −2160 − 720 p′(−40) = = = −3.20 900 (−30)2 p(s) =

Then C2 =

p(−40) p′(−40) = −24 , C1 = = −3.20 , 0! 1!

2/23/02

page P14.3

Is (s) = and for t > 0


3.20 3.20 24 − + s + 10 s + 40 (s + 40)2

is (t) = 3.20e −10t − 3.20e −40t + 24te−40t A

SOLUTION 14.4. (a) Find Zin (s) vis Yin (s) Yin (s) = Cs +

1 1 200Cs + 10LCs 2 + 10 + 20 + Ls + = Ls + 20 10 10(Ls + 20)

10LCs2 + (200C + L)s + 30 = 10Ls + 200

and

Zin (s) = With C = 10−3 F and L = 0.05 H Zin (s) = (b) If is (t) = 0.3u(t) A, then

1 10Ls + 200 = Ω 2 Yin (s) 10LCs + (200C + L)s + 30

0.50s + 200 1000s + 4 × 10 5 = Ω 0.0005s2 + 0.25s + 30 s2 + 500s + 60,000 Is (s) =

0.30 s

and   s + 400  s + 400  Vin (s) = Zin (s)Is (s) = 300  2  = 300   s(s + 500s + 60,000)   s(s + 2000)(s + 300)  K K  K = 300 1 + 2 + 3   s 200 300  It follows that K1 = K2 =

s + 400 200 1 = =− s(s + 300) s =−200 (−200)(100) 100

K3 =

s + 400 100 1 = = s(s + 200) s =−300 (−300)(−100) 300

and

Thus

s + 400 400 1 = = (s + 200)(s + 300) s =0 200(300) 150

2/23/02

page P14.4


1 1   1  150  Vin (s) = 300 − 100 + 300  s + 200 s + 300   s   2 3 1 = − + s s + 200 s + 300

and for t > 0

vin (t) = 2 − 3e −200t + e −300t V

SOLUTION 14.5. Z(s) = and the network is “at rest” (a) If

s + 20 s + 40 20 s

vin (t) = 20u(t) → Vin (s) = then

V (s) 20  s + 40   s + 40  Iin (s) = in = = 20    Z(s) s s + 20  s(s + 20)  Using a partial function expansion  s + 40  K2  K Iin (s) = 20 = 20 1 +  s + 20   s  s(s + 20)  in which case K1 = Thus

s + 40 40 s + 40 20 = = 2, K 2 = = = −1 s + 20 s=0 20 s s =−20 −20 1  2 Iin (s) = 20 −  s s + 20 

and (b) Note that Then

in which case (c) Note that

iin (t) = 20(2 − e −20t )u(t) A 20 vin (t) = 20e −40t → Vin (s) = s + 40 V (s) 20  s + 40  20 Iin (s) = in = =   Z(s) s + 40 s + 20 s + 20 iin (t) = 20e −20t u(t) A vin (t) = 20e −20t → Vin (s) =

Then

20 s + 20

2/23/02

page P14.5

Iin (s) =


 s + 40  Vin(s) 20  s + 40  = = 20   Z (s) s + 20  s + 20   (s + 20)2 

Using a partial fraction expansion  s + 40   C1 C2  Iin (s) = 20 = 20 +     (s + 20)2   s + 20 (s + 20)2  Here

p(s) = s + 40 → p'( s) = 1 →

p(−20) = 20 p'(−20) = 1

and C1 =

p'(−20) p(−20) 20 = 1, C2 = = = 20 1! 0! 1

in which case  1 20  Iin (s) = 20 +   s + 20 (s + 20)2  so that

(

)

iin (t) = 20e −20t + 400te −20t u(t) A

SOLUTION 14.6. (a) Apply an arbitrary Iin (s) to the upper terminal of Fig. P14.6a. Assuming branch currents Ia(s) and Ib (s), it follows by KCL that Iin (s) = I a (s) + Ib (s) = 0.020sVa (s) + 0.005s Va(s) − sVa (s) = (0.020 + 0.005 − 0.015)sVa (s) = 0.010sVa (s) Hence, V (s) 1 100 Zin (s) = a = = Ω Iin (s) 0.010s s (b) Similarly apply an arbitrary Iin (s) to Fig P14.6b to obtain, in the s-domain, by KCL 1 V (s)   1 30  Iin (s) = 10sVin (s) + Vin (s) + 0.10sVin (s) − in  = 10s + + V (s) 50 4   50 4s in 

[

]

 2000s2 + 4s + 1500  =  Vin (s) 200s   Hence I (s) 10s2 + 0.02s + 7.50 Yin (s) = in = Vin (s) s (c) Here we apply Vin (s) to the input terminals of figure P14.6c. By KCL 400  1 200   Vin (s) = 10Iin (s) + 0.2sIin (s) + Iin (s) − Iin (s)  =  10 + 0.2s + I  s  2 s  in  in which case

2/23/02

page P14.6


V (s) 0.2s2 + 10s + 200 Zin (s) = in = Ω Iin (s) s

S OLUTION 1 4 . 7 . Writing two loop equations we obtain: Vin (s) = 100I1(s) + 200I 2 (s)

and

100  100I1(s) + 100 + I (s) = 0 s  2 In matrix form (dropping the s-dependence) 200   I1  Vin  100 100    =    100 100 + s   I2   0  Using Cramer's rule, 200  Vin det  100   0 100 +  100s + 100 1  s = I1 = × V 200  100 s  100s + 100  in 100  − 200  det100 100 + 100   s     s  Hence

V s −1 Zin (s) = in = −100 I1 s +1

SOLUTION 14.8. Working in the s-domain, apply KVL to the left side of the circuit to obtain 100 s Iin (s) + I (s) + V2 (s) s 100 in

Vin (s) = Now apply KCL to the right side to obtain

Iin (s) =

s 100 V2 (s) + V (s) 100 s 2

Thus  s2 + 104  Vin (s) =   Iin (s) + V2 (s)  100s  To find V2 (s) note that Iin (s) = implying that

s2 + 104 V (s) 100s 2

2/23/02

page P14.7


 100s  V2(s) =  2 I (s)  s + 104  in Thus  s2 + 10 4 100s  (s2 + 104 )2 + 104 s2 Vin (s) =  + 2  Iin (s) = Iin (s) s + 104  100s(s2 + 104 )  100s implying that V (s) s4 + 3 × 104 s2 + 108 Zin (s) = in = Ω Iin (s) 100s(s2 + 104 ) and Yin (s) =

1 100s(s2 + 104 ) = 4 S Zin (s) s + 3 × 104 s2 + 108

SOLUTION 14.9. Three mesh equations for the circuit R + Z2 (s) −Z2 (s) − R   I1 (s)  Vin (s)       2R + Z2 (s) − R   I 2 (s)  =  0   − Z2 (s)  − R −R 2R + Z1 (s)  I 3(s)   0  Solve for I1(s) via Cramer’s rule Vin − Z2 (s) −R    det  0 2R + Z2 (s) −R   0 Vin 3R 2 + 2R( Z1 + Z2 ) + Z1Z2 −R 2R + Z1(s)  I1 = =  R + Z2 (s) −Z2 (s) −R  d(s)   det −Z2 (s) 2R + Z2 (s) −R   − R −R 2R + Z1 (s) where d(s) = (R + Z2 ) 3R 2 + 2R( Z1 + Z2 ) + Z1Z2 + Z2 − Z2(2R + Z1 ) − R 2 − 2R 2Z2 + 2R 3

(

(

)

(

)

) (

= R 3 + 2R 2 Z1 + 2R 2 Z2 + 3RZ1Z2 Under the condition that Z1 (s)Z2 (s) = R 2 , we have Vin R 3 + 2R 2 Z1 + 2R 2 Z2 + 3RZ1Z2 4R 3 + 2R 2 (Z1 + Z2 ) Zin (s) = = = =R I1 3R 2 + 2R( Z1 + Z2 ) + Z1Z2 4R 2 + 2R (Z1 + Z2 )

SOLUTION 14.10. 1 (a) Yin (s) = Cs + implies a parallel RC circuit with values R and C respectively. R

)

2/23/02

(b) Yin (s) =

page P14.8


1 1 = 2s + which is a parallel RC circuit of values 4 Ω and 2 F respectively. Zin (s) 4

1 1 =1 + . Using the result of part (b), this circuit is a 1 Ω resistor in series 2s + 0.25 Yb(s) with the parallel RC of part (b). (c) Zin (s) = 1 +

2s + 8 4 1 = 2+ = 2+ 1 . Using the results of parts (b) and (c), this circuit is a 2 Ω s+2 s+2 0.25s + 2 resistor in series with a parallel combination of a 0.25 F capacitor and a 2 Ω resistor. (d) Zin (s) =

s+3 s+6 2 2 1 1 + = 2+ + = 2+ + . Using the above results, this s +1 s + 4 s +1 s + 4 0.5s + 1 2 0.5s + 1 0.5 circuit is a 2 Ω resistor in series with a parallel combination of a 0.5 F capacitor and a 2 Ω resistor which is in series with another parallel combination of a 0.5 F capacitor and a 0.5 Ω resistor. (e) Zin (s) =

SOLUTION 14.11. (a) Clearly this is an inductor of value L in series with a resistor of value R. (b) Inverting the admittance we have Zin (s) of the form of part (a). Hence the circuit is a 0.5 H inductor in series with a 10 Ω resistor. 1 1 (c) Yin (s) = 0.2 + = 0.2 + . Using the result of part (b), the circuit is 0.2 S resistor in 0.5s + 10 Zb(s) parallel with a series connection of a 0.5 H inductor and a 10 Ω resistor. 10s + 50 40 1 (d) Yin (s) = = 10 + = 10 + . This is similar to part (c). Hence the circuit is a s +1 s +1 0.025s + 0.025 10 S resistor in parallel with a series connection of a 25 mH inductor and a 0.025 Ω resistor. s+3 s+6 2 2 1 1 (e) Yin (s) = + = 2+ + = 2+ + . Hence, the circuit is a 2 S s +1 s + 4 s +1 s + 4 0.5s + 0.5 0.5s + 2 resistor in parallel with the series connection of a 0.5 H inductor and a 0.5 Ω resistor which in turn is in parallel with a 0.5 H inductor and 2 Ω resistor.

SOLTUION 14.12. 1 (a) Zin (s) = Ls + represents a series connection of an inductance L and a capacitance C. Cs 1 (b) Yin (s) = Cs + represents a parallel connection of an inductance L and a capacitance C. Ls 0.125s2 + 1 1 (c) Zin (s) = = 0.5s + which is a series connection of a 0.5 H inductor and a 0.25 F 0.25s 0.25s capacitor. 0.125s2 + 1 1 (d) Yin (s) = = 0.5s + which is a parallel connection of a 0.5 F capacitor and 0.25 H 0.25s 0.25s inductor.

2/23/02

(e) Zin (s) =

page P14.9


s2 + 1 0.25s2 + 1 1 4 1 + = 2s + + = 2s + which is a 2 H inductor in series with a 0.2 F s 0.25s s s 0.2s

capacitor. s2 + 1 0.25s 1 1 + = s+ + . This circuit is a 1 H inductor in series with a 1 F 2 s s s+ 1 0.25s + 1 0.25s capacitor which is in series with a parallel connection of a 1 F capacitor and a 0.25 H inductor. s2 + 1 0.25s 1 1 (g) Yin (s) = + = s+ + . This circuit is a 1 F capacitor in parallel with a 1 H 2 s s s+ 1 0.25s + 1 0.25s inductor which is in parallel with a series connection of a 1 H inductor and a 0.25 F capacitor. (f) Zin (s) =

SOLUTION 14.13. With L[v out (t)] = Vo(s) and L[v in (t)] = Vi (s) and vout (0− ) = 0, sV0 (s) + 25V0 (s) +

100 10 V0 (s) = 5Vi (s) − Vi (s) s s

which implies that  s2 + 25s + 100  5s − 10   V0 (s) = Vi (s) s s   The transfer function is

V (s) 5s − 10 5s − 10 H(s) = 0 = 2 = Vi (s) s + 25s + 100 (s + 5)(s + 20) −5t (a) If vin (t) = te u(t) V, then 1 Vi (s) = (s + 5)2 and 5s − 10 K1 C C2 C3 Vout (s) = = + 1 + + 3 2 s + 20 s + 5 (s + 5) (s + 20)(s + 5) (s + 5)3 5s − 10 −110 −110 22 K1 = = = = 3 3 −3375 675 (s + 5) s =−20 (−15) 5s − 10 p(s) = s + 20 (s + 20)(5) − (5s = 10) 110 p′(s) = = = 110(s + 20) −2 2 2 (s + 20) (s + 20) −220 p′′(s) = (s + 20)3 −25 − 10 −35 7 1575 p(−5) = = =− = − 15 15 3 675 110 110 330 p′(−5) = = = (−15)2 225 675

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page P14.10

p′′(−5) = − Then C3 =


220 220 44 =− =− 3 3375 675 (−15)

p(−5) 1575 p′(−5) 330 p′ ′(−5) 1  44  22 =− , C2 = = , C1 = = − = − 01 675 1! 675 2! 2 675 675

and Vout (s) =

1  22 22 330 1575  − + −   675  5 + 20 s + 5 (s + 5) 2 (s + 5) 3 

This yields vout (t) = (b) If vin (t) = u(t) V,

1  1575 2 −5t  22e −20t − 22e −5t + 330te −5t − t e  u(t) V  675  2  Vin (s) =

and

1 s

5s − 10 K K K3 = 1+ 2 + s(s + 5)(s + 20) s s + 5 s + 20 5s = 10 −10 1 K1 = = =− (s + 5)(s + 20) s=0 100 10

Vout (s) =

K2 =

5s − 10 −25 − 10 −35 7 = = = s(s + 20) s=− 5 (−5)(+15) −75 15

and K3 = Thus

5s − 10 −100 − 10 −110 −11 = = = s(s + 5) s =−20 (−20)(−15) 300 30

11 1 7 vout (t) =  e −5t − e −20t −  u(t) V 30 10   15 By virtue of linearity and time invariance, if vin (t) = [u(t) − u(t − 0.5)] V , 11 1 7 vout (t) =  e −5t − e −20t −  u(t) 30 10   15 11 −20(t −0.5) 1  7 −  e −5(t −0.5) − e −  u(t - 0.5) V 30 10  15

SOLUTION 14.14. Here vin (t) = cos(t) u(t) V and iout (t) = 2sin(t)u(t) A, in which case

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page P14.11


2 2

I (s) s + 1 2 H(s) = out = s = Vin (s) s 2 s +1

SOLUTION 14.15. 1 . Further, (s + 1)2 vout (t) = (1+ t − 0.5t 2 )e −t u(t) + sin(t)u(t) − cos(t)u(t) V in which case

Here vin (t) = te−t u(t) V which implies that Vin (s) =

Vout (s) =

1 1 1 1 s + + + 2 − 2 2 3 s + 1 (s + 1) (s + 1) s +1 s +1

(a) Hence

Simplifying

V (s)  1 1 1 s −1  2 H(s) = out = + + −  (s + 1) 2 3 2 Vin (s)  s + 1 (s + 1) (s + 1) s + 1  1 (s − 1)(s + 1)2 = (s + 1)+ 1 + − (s + 1) s2 + 1 s3 + 2s2 + 5s + 2 H(s) = (s + 1)(s2 + 1) 1 1 s +1 + 2 = 2 . Hence s s s 3 s + 2s2 + 5s + 2 5 2 4s Vout (s) = H (s)Vin (s) = = + 2− 2 2 2 s s s (s + 1) s +1

(b) If vin (t) = (1+ t)u(t) V, then Vin (s) =

implying that

vout (t) = [5 + 2t − 4cos(t)]u(t) V

SOLTUION 14.16. (a) By a voltage divider (Fig. P14.16a) Vout (s) =

Z4 (s) V (s) Z3(s) + Z4 (s) in

and H(s) = (b) In Fig. P14.16b,

Z4 (s) Z3(s) + Z4 (s)

Yin (s) = Y1(s) + Y2 (s)

and Vout (s) =

1 1 Iin (s) = I (s) Yin (s) Y1 (s) + Y2 (s) in

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page P14.12


Hence V (s) 1 H(s) = out = Iin (s) Y1 (s) + Y2 (s) (c) By current division,

Iout (s) =

1 Z3 (s) + Z4 (s) Y1 (s) + Y2 (s) +

1 Z3 (s) + Z4 (s)

Iin (s) =

1 [Y1(s) + Y2 (s)][Z3(s) + Z4 (s)] + 1 Iin (s)

Hence Vout (s) = Z4 (s)Iout (s) =

Z4 (s) [Y1(s) + Y2 (s)][Z3 (s) + Z4 (s)] + 1 Iin (s) .

and V Z4 (s) H(s) = out = Iin [Y1(s) + Y2 (s)] [Z 3(s) + Z4 (s)] + 1

V SOLUTION 14.17. With Vin (s) = Vi and Vout (s) = V0 , H(s) = 0 . By voltage division, Vi 1 −4 V 1 10 10 s + 10−3 H(s) = 0 = = = 1 Vi 103 + 0.1s + 2 s + 20 −4 −3 10 s + 10 400 20 20 (a) Vout (s) = = − ⇒ v out (t) = 20 − 20e −20t u(t) V. Plot omitted. s(s + 20) s s + 20 (b) If vin (t) = 40[ u(t) − u(t − 0.2)]V , then by linearity and time invariance

(

)

[

]

[

]

vout (t) = 20(1− e −20t )u(t) − 20 1− e −20(t −0.2) u(t − 0.2) V

(c) If vin (t) = 40[ u(t) + u(t − 0.2)]V , then by linearity and time invariance

vout (t) = 20(1− e −20t )u(t) + 201 − e −20(t −0.2) u(t − 0.2) V 40 (d) If vin (t) = 40e −20t u(t) V , Vi (s) = . Hence, (s + 20) 400 ⇒ vout (t) = 400te −20t u(t) V 2 (s + 20) 40 (e) If vin = 40te−20t u(t) V , then Vi (s) = . Hence, (s + 20)2 V0(s) = H (s)Vi (s) =

V0(s) = H (s)Vi (s) =

400 (s + 20) 3

⇒ v out (t) = 200t 2e −20t u(t) V

2/23/02

page P14.13


SOLUTION 14.18. (a) By voltage division

2s + 1 2s + 1 2s + 1 s s Vout (s) = 2 2s + 1 Vin (s) = 2s + 2 + 2s + 1 Vin (s) = 4s + 3 Vin (s) 2+ + s s s

Hence

V (s) 2s + 1 H(s) = out = Vin (s) 4s + 3

(b) With vin (t) = 8u(t) then

 2s + 1   8  16s + 8 Vout (s) = H (s)Vin (s) =  = 4s + 3  s  s(4s + 3)

Using MATLAB »n = [16 8]; d = [4 3 0]; »[r,p,k] = residue(n,d) r= 1.3333e+00 2.6667e+00 p= -7.5000e-01 0 k = [] Then (c) If vin (t) = 8sin(2 t)u(t) , then

8 4  vout (t) =  + e −0.75t u(t) V 3 3  2s + 1   16  Vout (s) = H (s)Vin (s) =    4s + 3  s2 + 4 

Using MATLAB, »ilaplace( (32*s+16)/((4*s+3)*(s^2+4)) ) ans = -32/73*exp(-3/4*t)+32/73*cos(2*t)+280/73*sin(2*t) Hence,

(

)

vout (t) = −0.43836e −0.75t + 0.43836cos(2t) + 3.8356sin(2t) u(t) V (d) With vin (t) = 8sin(8 t)u(t) 128s + 64  2s + 1   64  Vout (s) = H (s)Vin (s) = =    4s + 3  s2 + 64  (4 s + 3)(s2 + 64) 0.12391s + 31.907 0.12391 = − s + 0.75 s2 + 64 Using MATLAB »ilaplace((128*s + 64)/((4*s+3)*(s^2+64))) ans = -128/1033*exp(-3/4*t)+128/1033*cos(8*t)+4120/1033*sin(8*t)

2/23/02

page P14.14

» »128/1033 ans = 1.2391e-01 »4120/1033 ans = 3.9884e+00

(


)

vout (t) = 0.12391cos(8t) + 3.9884sin(8t) − 0.12391e −0.75t u(t) V

SOLUTION 14.19. V (s) With a source transformation Iin (s) = in . R (a) By current division, Cs Vin (s) LCs 2 Vin (s) s2 Vin (s) IC (s) = 1 = = 1 L 1 1 R R + Cs + LCs2 + s + 1 R s2 + s+ R Ls R RC LC Here IC (s) 1 s2 H(s) = = Vin (s) R s2 + 1 s + 1 RC LC 2 (b) With R = Ω, C = 0.5F and L = 1H , 3  3 s2 H(s) =  2  2  s + 3s + 2  1 If vin (t) = e −t u(t) V, then Vin (s) = . Hence s +1 IC (s) = H (s)Iin (s) =

 1  3 s2 3s2 K1 C1 C2 = = + +   2  s2 + 3s + 2  s + 1 2(s + 1)2 (s + 2) s + 2 s + 1 (s + 1)2

Using MATLAB, »n = [3 0 0]; d = conv([2 4],[1 2 1]); »[r,p,k] = residue(n,d) r= 6.0000e+00 -4.5000e+00 1.5000e+00 p= -2.0000e+00 -1.0000e+00 -1.0000e+00 k= [] »

2/23/02

page P14.15


6 4.5 1.5 − + s + 2 s + 1 (s + 1)2

IC (s) = and

9 3   iC (t) =  6e −2t − e −t + te−t  u(t) A 2 2

SOLUTION 14.20. (a) Make a source transformation: Vin (s) =

1 250 Iin (s) = I (s) Cs s in

By voltage division   10 2500    250  Vout (s) =  I (s)   s Iin (s)  =  1 2 250 250 1  in  + + s + 10  s s + 10s + 500 s s 20 20 50,000 = Iin (s) s s2 + 200s + 10,000

(

)

and V (s) 50,000 H(s) = out = 2 Iin (s) s(s + 200s + 10,000) (b) If iin (t) = (t) implies Iin (s) = 1. Using MATLAB »n = 50e3; d = [1 200 10e3 0]; »[r,p,k] = residue(n,d) r= -5 -500 5 p= -100 -100 0 k= [] Hence 50,000 5 5 500 Vout (s) = = − − 2 s s + 100 (s + 100) 2 s(s + 100) and vout (t) = 5 − 5e −100t − 500te −100t u(t) V This is the impulse response 0.1 (c) If Lin (t) = 100u(t) mA so that Iin (s) = . Therefore s 5000 Vout (s) = 2 s (s + 100) 2 In MATLAB,

(

)

2/23/02

page P14.16


»n = 5000; d = conv([1 0 0],[1 200 1e4]) d= 1 200 10000 0 0 »[r,p,k] = residue(n,d) r= 1.0000e-02 5.0000e-01 -1.0000e-02 5.0000e-01 p= -100 -100 0 0 k= [] Hence −0.01 0.5 0.01 0.5 Vout (s) = + 2 + + s s + 100 (s + 100) 2 s and

[

]

vout (t) = 0.01e −100t + 0.5te −100t − 0.01 + 0.5t u(t) (d) By superposition and time invariance, if iin (t) = 100[ u(t) + u(t − 1)] mA then the result of part (c) can be adjusted to

[

]

vout (t) = 0.01e −100t + 0.5te −100t − 0.01 + 0.5t u(t)

[

]

− 0.01e −100(t −1) + 0.5te−100(t −1) − 0.01 + 0.5(t − 1) u(t − 1) V

SOLUTION 14.21. For this problem change the 20 mH inductor to one of 0.3 H. (a) ( s + 200) (s + 400) 1 1 Yin = 1 + + = 15 0.3s + 90 0.1s + 10 15( s+ 100) (s + 300) and ( s + 100) (s + 300) H(s) = Iout = 1/15 = Iin Yin ( s+ 200) (s + 400) (b) If iin(t) = δ(t), then Iin(s) = 1 and Iout (s) = H(s) = Hence

( s + 100) (s + 300) = 1 - 50 - 150 ( s+ 200) (s + 400) s + 200 s + 400

iout(t) = δ(t) + (- 50 e-200t - 150 e-400 t) u(t) A

(c) We first find the response to iin(t) = 16u(t) mA. Here Iin(s) = 0.016/s and

2/23/02

page P14.17

Iout (s) = H(s)Iin (s) =


0.016( s + 100) (s + 300) 0.004 0.006 = 0.006 s + s + 200 + s + 400 s( s+ 200) (s + 400)

Hence

iout(t) = (6 + 4 e-200t + 6 e-400 t)u(t) mA By linearity and time invariance, the response to iin(t) = 16[ut) – u(t – 0.01)] mA is

iout(t) = (6 + 4 e-200t + 6 e-400 t)u(t) - (6 + 4 e-200(t - 0.01) + 6 e-400 (t - 0.01) )u(t - 0.01) mA A plot of iout(t) using MATLAB is given below. t= 0: 0.0005: 0.05; f1= (6 + 4*exp(-200*t) + 6*exp(-400*t)).*u(t); f2= (6 + 4*exp(-200*(t-0.01)) + 6*exp(-400*(t-0.01))).*u(t-0.01); iout= f1 - f2; plot(t, iout) grid ylabel('iout in mA') xlabel(' time in second') 20

iout in mA

15 10 5 0 -5 -10

0

0.01

0.02 0.03 time in second

0.04

REMARK : Notice that the resistor current is not continuous.

S OLUTION 1 4 . 2 2 . (a) First observe that the admittance of a parallel LC is YLC (s) = Cs + Using voltage division,

1 Ls

0.05

2/23/02

page P14.18

C1s +

1 Ls

C1s +


1 Ls

LC1s2 + 1 Vout (s) = 1 1 Vin (s) = 2 Vin (s) = L(C + C )s2 + 2 Vin (s) 1 2 C1s + + C2 s + C1 + C2 )s + ( Ls Ls Ls Finally 1 Vout (s) C1 LC1 s2 + 4 × 10 6 H(s) = = = 0.2 2 2 Vin (s) (C1 + C2 ) s2 + s + 1.6 × 106 L(C1 + C2 ) (b) Using MATLAB, »syms s t »ilaplace(0.2*(s^2+4e6)/(s^2+1.6e6)) ans = 1/5*Dirac(t)+120*10^(1/2)*sin(400*10^(1/2)*t) »120*10^(1/2) ans = 3.7947e+02 s2 +

−1 

s2 + 4 × 106  h(t) = 0.2L  2  = 0.2 (t) + 379.47sin(1264.9t)u(t) V  s + 1.6 × 10 6 

SOLUTION 14.23.

Then Z1(s) =

Y1(s) = C1s +

1 R1C1s + 1 = R1 R1

Y2(s) = C2 s +

1 R C s +1 = 2 2 R2 R2

R1 R2 and Z2(s) = . By voltage division, R1C1s + 1 R2C2s + 1   R2 R2   V (s)  R2C2s + 1 in R2C2 S + 1 Vout (s) = Vin (s) = R R C s + R + R R C s + R R1 R2 1 2 2 1 1 2 1 2 + R1C1s + 1 R2C1s + 1 (R1C1s + 1)(R2C2 s + 1) R2 (R1C1s + 1) = V (s) (C1 + C2 )R1R2s + R1 + R2 in

Thus the transfer function is: V (s) R2 (R1C1s + 1) H(s) = out = Vin (s) (C1 + C2 )R1R2s + R1 + R2 (b) If C1 = 0.5 F , C2 = 1.0 F and vin (t) = 10 u(t) V, then

2/23/02

page P14.19


 0.5R1R2s + R2   10 Vout (s) = H (s)Vin (s) =   1.5R1R2s + R1 + R2   s  R Moreover, with 1 = 4 so that R1 = 4R2 R2

1 1   20R22 (s + ) s+   2R22 s + R2   10 2R2 10  2R2  Vout (s) =  2 = =   5  3  s(s + 5 )   6R2 s + 5R2   s  6R22  s(s + )  6R2  6R2  

The partial fraction expansion is  1    s+  10 2R2  10  K1 K2 Vout (s) =  = +  5 3  3 s 5  s+  s s +   6R2 6R2     

    

Observe that K2 5 s+ 6R2

L−1

→ K 2e

−

5 t 6R2

and that it is required that − Thus

5 5 =− 6R2 3

R2 = 0.5 Ω, R1 = 4R1 = 2 Ω

and

       10 s + 1  10 0.6 0.4  2 4 / 3 Vout (s) =  =  + 5  = s + 5 5  3  s 3   s + s + s s +    3 3 3  Thus, 5  − t 4 vout (t) =  2 + e 3  u(t)   3   (c) If R1C1 = R2C2 = , then the transfer function is

H(s) = =

R2 (R1C1s + 1) R2 s + R2 = R1R2C1s + R1R2C2 s + R1 + R2 [ R2 (R1C1) + R1 (R2C2 )]s + R1 + R2 R2 ( s + 1) R2 = (R1R2 )( s + 1) R2 + R1

The zero-state response is

2/23/02

page P14.20

Vout (s) =


R2 10u(t) R2 + R1

(d) Using H(s) from part (c) with the requirement that R1C1 = R2C2 , then H(s) = With R2 = 106 Ω, then 10R2 = R1 + R2

R2 1 = R1 + R2 10

⇒ R1 = 9R2 = 9 M Ω. Since C2 = 5 × 10 −12 F, then

RC C1 = 2 2 = 0.556 × 10 −12 F R1

SOLUTION 14.24. V (s) V (s) (a) H(s) = out . Here Ib(s) = in . The parallel admittance at the right is Vin (s) 2000 YR (s) = Cs +

1 RCs + 1 = R R

so that ZR (s) = Then

R RCs + 1

R Vout (s) = − ZR (s) Ib (s) = − 2000(RCs + 1) (b) With V1(s) = Vin (s) − Vout (s) , then

1 1 Cs(Vout (s) − Vin (s)) + Vout (s) + [Vout (s) − 3Vin (s) + 3Vout (s)] = 0 s 2

Hence 3 1    Cs + V (s) = Cs + + 2 V (s) in   2 s  out and

V (s) s(Cs + 1.5) s(s + 0.75) H(s) = out = 2 = 2 Vin (s) Cs + 2s + 1 s + s + 0.5

(c) Transform the current source iin (t) into a voltage source. In the s-domain with Iin (s) = 2Vin (s) 1 1 Here ZC (s) = = which implies YC (s) = 2s . A single node equation yields Cs 2s 2 2 2 2 2  Vout (s) − Vin (s)] − Vout (s) + 2sVout (s) = − Vin (s) +  − + 2s Vout (s) = 0 [ 3 3 3 3 3

2/23/02

page P14.21


2 V (s) Vin (s) = 2sVout (s) implies Vout (s) = in 3 3s I (s) But Vin (s) = in in which case 2 I (s) Vout (s) = in 6s and V (s) 1 H(s) = out = Iin (s) 6s

SOLUTION 14.25. Here in the s-domain

V (s) − VC (s) Iin (s) = in 0.5s + 10

and with a node at VC (s) −Iin (s) + Hence,

sVC (s) 1 sVC (s) 1 V (s) − VC (s) + Iin (s) = 0 implies = Iin (s) = in 500 2 50 2 s + 20  s2 + 20s + 500   VC (s) = Vin (s) 500  

and the transfer function is V (s) 500 H(s) = C = 2 Vin (s) s + 20s + 500 4 5 With vin (s) = 4 5 u(t) V which implies Vin (s) = , and s 2000 5 K1 K K* VC (s) = = + + s(s + 10 − j20)(s + 10 + j20) s s + 10 − j20 s + 10 + j20 where K * designates the complex conjugate of K 2000 5 2000 5 K1 = 2 = = 8.944 500 s + 20s + 500 s =0 K=

2000 5 2000 5 = = −4.472 + j2236 = Jeij (−10 + j20)( j40) −(800 + j400) = 153.44 o . Then with K * = −4.472 − j2.236 = 5e − j 8.944 A + jB A − jB VC (s) = + + s s + 10 + j20 s + 10 − j20 =

where

Here

2000 5 s(s + 10 + j20) s=− (10− j 20)

2/23/02

page P14.22

A = −4.472, B = 2.236,


A2 + B2 = 5

and arc tan

B 2.236 1 = arc tan = arc tan = −153.44 o A −4.472 −2

With the help of Table 13.1. vC (t) = 8.944 + 10e −10t cos(20t + 153.44 o ) u(t) V

[

]

SOLUTION 14.26. In the s-domain we first find Vx (s) in terms of Vin (s) via voltage division: Z p (s) Vx (s) = V (s) 40 + Z p (s) in where (0.40s)(40) 40s Z p(s) = = 0.40s + 40 s + 100 Hence 40s 40s 0.5s Vx (s) = s + 100 40s Vin (s) = 80s + 4000 = s + 50 Vin (s) 40 + s + 100 and V (s) 2.5 IL (s) = x = V (s) 0.4s s x Then from the right hand side by another voltage division 10 10s 0.25s Vout (s) = 1000 0.25V x (s) = 0.25V x (s) = Vx (s) 10s + 1000 s + 100 + 10 s 20 20 (a) If vin (s) = 20(1− e −40t )u(t), then Vin (s) = − . Hence s s + 40 20   s   10 10   0.5s   20 Vx (s) =  − = −   s + 50  s s + 40   s + 50  s s + 40 and IL (s) = Hence

2.5 1  0.05 −0.25 0.2  2.5   1 V x (s) =  − = + +   s s + 50  s s + 40  s s + 40 s + 50

[

]

iL (t) = 0.05 + 0.2e −50t − 0.25e −40t u(t) A 20 20 − . From, part (a), it was found that s s + 40 0.25s 0.25s 0.5s s s 2.5   2.5 Vout (s) = V x (s) = × Vin (s) = × × − s + 100 s + 100 s + 50 s + 100 s + 50  s s + 40 

(b) Vin (s) =

2/23/02

page P14.23

=


100s −20/ 3 10 −10 /3 = + + (s + 40)( s + 50)(s + 100) (s + 40) ( s + 50) ( s + 100)

Thus, 10 20   Vout (t) = 10e −50t − e −100t − e −40t  u(t) V 3 3  

SOLUTION 14.27. In both parts (a) and (b), the op-ampis ideal. It will not draw current and the virtual groundprincipal requires that v + = v− = 0 (a) For a note at the inverting terminal with mode voltage v1(t) = 0 , KCL gives in the s-domain  Vin (s) V (s) V (s) 1 = −CsVout (s) − out implies in = −  Cs +  Vout (s) R1 R2 R1  R2  in which case  R Cs + 1 Vin (s) = − 2 V (s) R1  R2  out Then,     V (s) R  1 1 1  H(s) = out =− 2 =−    Vin (s) R1  R2Cs + 1 R1C  s + 1   R2C  To make H(s) = − make 1 = 20 and R1C If

20 s+4 1 1 = R2C 4

C = 1 F = 10−6 F 1 1 106 = 4 or R2 = = = 250 kΩ R2C 4C 4

and 1 2 106 = 20 or R1 = = = 50 kΩ R1C 20C 20 (b) With Fig. 14.27b in the s-domain and v1(t) at the inverting terminal, KCL gives V (s) V (s) C1sVin (s) + in = −C2sVout (s) − out R1 R2  R1C1s + 1  R2C2s + 1  R  Vin (s) = −   Vout (s) R2 1

2/23/02

page P14.24


Then V (s) R  R C s + 1 H(s) = out =− 2 1 1 Vin (s) R1  R2C2 s + 1 (c) If then

1  1   s+ s+  RRC  R1C1  C R1C1 =− 1 2 1  =− 1 1 R1RC2  s + C2  s + 1    R2C2  R2C2

    

H(s) = −5, C2 = 1 F and R2 = 1M Ω C1 = 5 and C1 = 5 F C2

The bracketed term must cancel and with Then with C1 = 5 F

R2C2 = 106 (10−6 ) = 1

R1C1 = 1 1 1 R1 = = = 200kΩ C1 5 × 10−6

(d) Using H(s) in part (b) 1   s+ C  R1C1  H(s) = − 1   C2  s + 1   R2C2  to obtain H(s) = −5 with C2 = 1 F

s +1 s+2

C1 = 5C2 = 5 × 10 6 F (5 F) 1 1 106 = 1 or R1 = = = 200kΩ R1C1 C1 5 1 1 106 = 2 or R2 = =− = 500kΩ R2C2 2C2 2

SOLUTION 14.28. Here, the op-amp will not draw current at the non-inverting terminal and the principal of the virtual ground demand that v1 = v 2 = Vin (s) For Fig. 14.28 in the s-domain with a node V1(s) taken at the inverting terminal Vin (s) Vin (s) − Vout (s) + =0 R1 Z p (s) Here

2/23/02

page P14.25


R2 Cs

R2 1 Z p(s) = 1 = = 1 + R2 R2Cs + 1 C(s + ) Cs R2C and 1 1  1  R + C(s + R C ) Vin (s) = C(s + R C )Vout (s)  1 2  2     1  1  + 1Vin (s) = R1C  s + V (s) R1C s +   R2C    R2C  out  and V (s) H(s) = out = Vin (s) Here

1 ) +1 s + R2C = 1 R1C(s + ) R2C

R1C(s +

1 1 + R2C R1C 1 s+ R1C

1 1 + =4 R2C R1C

and 1 =2 R1C 1 10 6 If C = 1 F then R1 = = = 500 kΩ 2C 2 1 and = 4 − 2 = 2 implies R2 = 500 kΩ . R2C

SOLUTION 14.29. For the non-inverting configuration in the s-domain, each of the two op-amps in cascade have a transfer function Z f (s) H(s) = − Zin (s) Then for the two op-amps  Z f 1(s)   Z f 2 (s)  Z f ,1(s)Z f ,2 (s) H0 (s) =  − =  −  Zin1 (s)   Zin,2  Zin,1(s)Zin,2 (s) For Fig. P14.29a in the s-domain Zin,1 = 25kΩ, Zin,2 = 50kΩ , Z f ,1 =

Hence for Fig. P14.29a

1 1   C s+  RC 

=

250,000 , and Z f ,2 = s+5

1 1   C s+  RC 

=

250,000 s + 2.5

2/23/02

page P14.26


 250,000   250,000  50    Ha (s) =  s + 2.5   s + 5.0  =  25,000   50,000  (s + 2.5)(s + 5)    If vin (t) = u(t), then Vin (s) =

1 and s Vout (s) =

Hence,

50 4 8 4 = − + s( s + 2.5)(s + 5) s s + 2.5 s + 5

[

]

vout (t) = 4 − 8e −2.5t + 4e −5t u(t) V

(b) For Fig. P14.29b in the s-domain, Zin,1 , Z f 1 , and Z f 2 are in part (a). However, 1 250,000 Zin,2 = = Cs s Thus  250,000   250,000  10s    s + 5   10   s  Hb(s) =  s + 2.5   250,000 = =  5 s + 2.5  s + 5   25,000    s + (s + 5)      s 2 1 With Vin (s) = , s  10s   1  20 4 4 Vout (s) =   = = −     (s + 2.5)(s + 5)   s  (s + 2.5)(s + 5) s + 2.5 s + 5 and vout (t) = 4e −2.5t − 4e −5t u(t) V

(

)

SOLUTION 14.30. (a) -(b). The subcircuit is an integrator, with

Vout(t) = - 1 s V1 (s) (c) This subcircuit is again an integrator, with

V1(t) = - 1 s V2 (s) (d) Applying KCL to the inverting input terminal of the top left op amp, we have or

G2Vout(s) + V3(s) + V 2(s) = 0 V2(s) = - G 2Vout(s) - V3(s)

2/23/02

page P14.27


(e) Applying KCL to the inverting input terminal of the bottom op amp, we have or

G3V1(s) + G1 Vin(s) + V 3(s) = 0 V3(s) = - G 3V1(s) - G1 Vin(s)

(f) The results of parts (b), (c) and (d) do not involve Vin. Therefore, , we can solve for V1, V2 and V3 in terms of Vout from these threes equations:

V1(s) = - sV out(s) and

V2(s) = - sV 1(s) = s 2Vout(s) V3(s) = - V 2(s) - G2Vout(s) = - s2 + G 2 Vout(s)

Substituting these relationships into the result of part (e), we obtain

- sG3Vout(s) + G1 Vin(s) - s2 + G 2 Vout(s) = 0 Therefore

G1 H(s) = Vout(s) = 2 Vin (s) s + G 3s + G2 SOLUTION 14.31. Use the parallel equivalent circuit model for the capacitor with the standard directions for voltage and current as given in figure 14.16. For the single node with vC (0− ) = 20 V, VC (s) VC (s) 1 1 1  s − 10 + 40 + 10 − 10 vC (0 ) = 0 implies  10 + 50 VC (s) = 10 (20) = 2 s Equivalently, (50s + 10)VC (s) = (s + 0.2)VC (s) = 20 or VC (s) = vC (t) = 20e −0.2t u(t) V

20 . Therefore, s + 0.2

SOLUTION 14.32. Using the equivalent model for the inductor in figure 14.19, we can compute the total admittance as 5 1 1 5 1 2s + 40 s + 20 Y(s) = + + = + = = 2s 40 10 2s 8 16s 8s Using current division,

2/23/02

page P14.28


0.1 −i (0) 0.8iL (0 − ) 1.6 Iout (s) = s + 20 × L =− =− s s + 20 s + 20 8s Thus

iout (t) = −1.6e −20t u(t) A

SOLUTION 14.33. Using the equivalent model for the inductor in figure 14.19 and for the capacitor using figure 14.16, we may combine the current sources to form an equivalent source (with C = 0.1 F) to obtain 1 i (0− ) 1 Ieq (s) = vC (0− ) − L = 0.2 − 10 s s Note that 1   2 s + 1 LCs 2 + 1  LC  Y(s) = Cs + = = C  LS Ls s     1 With C = 0.1 F and L = 0.4 H, = 25 and LC 1 s  10s Z(s) =  2  = 2  C s + 25 s + 25 Thus  10s   s − 5 2(s − 5) 2s 10 VC (s) = Z(s)Ieq (s) =  2 = 2 = 2 − 2     s + 25  5s s + 25 s + 25 s + 25 and vC (t) = (2cos(5t) − 2sin(5 t)) u(t) V

SOLUTION 14.34. Consider the equivalent circuit below:

Writing a single node equation we have,

2/23/02

page P14.29

0 = 0.5Vout − 0.2VR1 +

= 0.5Vout + 5 × 0.2 ×

1 (V + 0.8) 5 + 0.4s out 1 1 Vout + 0.8) + ( (V + 0.8) 5 + 0.4s 5 + 0.4s out

= 0.5Vout + Therefore

2 (V + 0.8) 5 + 0.4s out

Vout = and


−8 s + 22.5

vout (t) = −8e −22.5t u(t) V

SOLUTION 14.35. Redraw the circuit in the s-domain and use an equivalent circuit for the capacitor (figure 14.16) that accounts for the initial condition. By KCL

With R = 50 Ω and C = 0.02 F ,

VC (s) + CsVC (s) = Cv(0 − ) + Iin (s) R VC (s) + 0.02sVC (s) = 0.02vC (0 − ) + Iin (s) 50

or which is equivalent to

VC (s) + sVC (s) = vC (0− ) + 50Iin (s)

(s + 1)VC (s) = vC (0 − ) + 50Iin (s) (a) With vC (0− ) = 8 V and iin (t) = 40 (t) mA so that Iin (s) = 0.04 , (s + 1)VC (s) = 8 + 2 implies VC (s) = and

10 s +1

vC (t) = 10e −10t u(t) V

(b) With vC (0− ) = 1 V and iin (t) = 200e −t u(t) mA we have that Iin (s) =

0.2 . Thus s +1

v C (0 − ) 50 1 10 VC (s) = + Iin (s) = + (s + 1) (s + 1) (s + 1) (s + 1)2 and

SOLUTION 14.36.

(

)

vC (t) = e −t + 10te−t u(t) V

2/23/02

page P14.30


(a) By current division

H(s) = IL(s) = 1 Iin (s) Ls + 1 (b) The given data in Laplace transforms are:

Iin(s) = I0 and IL (s) = 15 - 3 + 3 = 75 s2 s2 s s+ 5 s2 s+ 5 Under the assumption of zero initial inductor current,

I0 I0/L 75 IL (s) = H(s) Iin (s) = H(s) = IL(s) = = = 2 2 2 Iin (s) Ls + 1 s s + 1/L s s s+ 5 Equating coefficients, we obtain the answers L = 1/5 = 0.2 H and I 0 = 75L =15 A (c) The s-domain equivalent is shown below.

Applying KVL to the right mesh, we have

LsIL(s) - L iL(0-) + 1 ×[IL(s) - 1] = 0 Solving for IL(s) from this equation, and equating it to 10/(s+ 5), we have IL(s) = L i L(0 ) + 1 = i L(0 ) + 1/L = 10 Ls+1 s + 1/L s+5

from which L= 0.2 H and iL (0-) = 5 A.

SOLUTION 14.37. (a) By inspection,

2/23/02

page P14.31


1 H(s) = IL(s) = = 0.5 Vin (s) 2s + 200 s + 100 Given vin(t) = 2u(t) V, then Vin(s) = 2/s , and

I L(s) =

1 = 0.01 - 0.01 s s s + 100 s + 100

in which case

i L(t) = 0.01 ( 1 - e-100t) A

Plots are omitted. (b) By linearity and time invariance,

i L(t) = 0.01 ( 1 - e-100t)u(t) - 0.01( 1 - e-100(t- 0.05))u(t - 0.05) A (c) Correction: (a) should be (c). With nonzero initial inducto current, the s-domain equivalent becomes:

Given iL(0-) = 0.01 A and vin(t) = 2e-200tu(t) V, then Vin(s) = 2/(s + 200) and

2 + 0.02 s + 200 0.01s + 3 I L(s) = = 2s + 200 ( s + 100)( s + 200) = in which case

0.02 - 0.01 s + 100 s + 200

iL( t) = 0.02e-100t - 0.01e-200t u(t) A (d) Correction . (b) should be (d). Given iL(0-) = 0.01 A and vin(t)= 2cos(200t)u(t) V, then Vin (s) = and

2s s2 + 40000

2/23/02

page P14.32

I L(s) =

s2


2s + 0.02 + 40000 0.01s2 + s + 400 = 2s + 200 ( s + 100)( s2 + 40000)

We use MATLAB to do the partial fraction expansion. n= [ 0.01 1 400]; d= conv([ 1 100], [ 1 0 40000]); [ r p k ] = residue (n,d) r= 0.0010 - 0.0020i 0.0010 + 0.0020i 0.0080 p= 1.0e+02 * 0.0000 + 2.0000i 0.0000 - 2.0000i -1.0000 From the above MATLAB output,

I L(s) =

0.008 + 0.001-j0.002 + 0.001+j0.002 = 0.008 + 0.002s + 0.8 s + 100 s - j200 s + j200 s + 100 s2 + 2002

From table 13.1, item 18,

iL( t) = 0.008e-100t + 0.002 cos(200t) + 0.004 sin(200t) u(t) A

2/23/02

page P14.1


SOLUTION 14.38. In the s-domain, we break the response up into the part due to the initial condition and the part due to the source with the initial condition set to zero. The transfer function with the initial condition set to zero is V (s) 1 Cs 1 RC 0.25 H(s) = C = = = Vin (s) R + 1 Cs s + 1 RC s + 0.25 Using the parallel equivalent circuit for the charged capacitor while setting the source voltage to zero, the capacitor voltage due only the initial condition is: vC (0 − ) VC,IC (s) = 1 CvC (0 ) = s + 0.25 + Cs R 1

Hence,

[

−

]

0.25 vC (0− ) VC (s) = V (s) + s + 0.25 in s + 0.25

and

V (s) − VC (s) 0.25  0.05v C (0 − ) 0.05s 0.05vC (0 − )  IC (s) = in = 0.05 1− V (s) − = V (s) − 20 s + 0.25 s + 0.25 in s + 0.25  s + 0.25  in for all inputs and initial conditions. 20 (a) If vin (t) = 20u(t) and vC (0− ) = 10 V, then Vin (s) = and s 5 10 20 10 VC (s) = + = − ⇒ v C (t) = 20 − 10e −0.25t u(t) V s(s + 0.25) s + 0.25 s s + 0.25 and 1 0.5 0.5 IC (s) = − = ⇒ iC (t) = 0.5e −0.25t u(t) A s + 0.25 s + 0.25 s + 0.25

(

(b) If vin (t) = 5e −0.25t u(t) V, then Vin (s) = VC (s) =

5 . Hence, s + 0.25

1.25 10 + 2 s + 0.25 (s + 0.25)

and IC (s) =

⇒ v C (t) = (10 + 1.25t)e −0.25t u(t) V

0.25s 0.5 −0.25 0.0625 − = − (s + 0.25) 2 s + 0.25 s + 0.25 (s + 0.25) 2

Hence iC (t) = − (0.25 + 0.0625t)e −0.25t u(t) A

)

2/23/02

page P14.2


10 9

Input and Capacitor voltages, V

8 7 6 5 4 3 TextEnd 2 1 0

0

5

10 Time in s

15

20

SOLUTION 14.39. The figure which accounts for the initial conditions is given below.

(a) For the zero-input response, the above circuit reduces to a parallel RLC driven by two current sources. Hence VC(s) equals the total current divided by the total admittance, i.e.,

2/23/02

page P14.3


i (0− ) i (0− ) Cv C (0 − ) + L svC (0 − ) + L 20s + 10 26.6 6.6 s = C = VC (s) = = − 2 4 1 1 1 1 s + 200 s + 50 s + 250s + 10 Cs + + s2 + s+ R Ls RC LC Hence vC(t) = [26.6e

–200t

–50t

– 6.6e

]u(t) V

(b) For the zero-state response, the current sources disappear. Executing a source transformation on the remaining voltage source, we obtain a current, I(s) = Vin(s)/(Ls), driving a parallel RLC circuit. Hence, the zero input response is Vin (s) Vin (s) 20000 2 0.6667 2.6667  1  Ls VC (s) = = = = + − 3 + 250s 2 + 104 s 1 1  LC  2 1 1 s s + 200 s + 50 s Cs + + s + s+ R Ls RC LC Hence

–200t

–50t

vC(t) = [2 + 0.6667e – 2.6667e ]u(t) V (c) By superposition, the complete response is the sum of the answers to (a) and (b). Hence vC(t) = [2 + 27.267e (d) By linearity and time-invariance, vC(t) = [2 + 0.6667e

–200t

–50t

– 2.6667e

–50t

– 9.2667e

]u(t) V

]u(t)

–200(t–0.01)

+ [4 + 1.3334e

–200t

–50(t–0.01)

– 5.3334e

]u(t – 0.01) V

2/23/02

page P14.4


SOLUTION 14.40. f (t) = Lin (t) = e −2t u(t)A [part (c)]

L (0− ) for the zero-input response, f (t) = Lin (t) = 2 [part (d)] s for the zero-input response, f (t) = Lin (t) = CvC (0)A [part (e)]

1 s2 + 2s + 1 (s + 1)2 (a) Yin (s) = 1 + + s + 1= = s s s Y1 (s) s +1 s(s + 1) s (b) Iout (s) = Iin (s) = 2 = = Iin (s) Yin (s) s + 2s + 1 (s + 1)2 s + 1 s and I (s) s H(s) = out = Iin (s) s + 1 1 (c) If iin (t) = e −2t u(t) A, then Iin (s) = , then s+2 s −1 2 Iout (s) = H (s)Iin (s) = = + (s + 1)(s + 2) s + 1 s + 2 which implies that the zero-state response is

(

)

iout (t) = 2e −2t − e −t u(t) A −

−

(d) If iL (0 ) = 2 A, vC (0 ) = 0, and iin (t) = 0. Using the parallel equivalent circuit for the inductor, figure 14.19, we have i(0− )  s   2  2 Iout (s) = H (s) = − =− ⇒ iout (t) = −2e −t u(t) A   s s +1 s s +1 (e) Use the parallel equivalent circuit for the capacitor, figure 14.16, to obtain by current division,

Iout (s) = − 1 s

1 +1 s

s +1 4 [CvC (0− )] = 4 2 = ⇒ iout (t) = 4e −t u(t) A s + 1 s + 2s + 1 +1 + s +1

(f) By superposition, the complete response is the sum of the answers to parts (c), (d) and (e).

SOLUTION 14.41. With vin (t) = 4u(t)V and vC (0− ) = 1 V, a single node equation at the front half of the circuit yields with CvC (0− ) = 1× 1= 1:

2/23/02

page P14.5


 s2 + 2s + 4  4 8 s+8  4   −2  +  2 + + s VC1(s) − 1= 0 ⇒  VC1 (s) = + 1 =  5 s s s s   or

s+8 VC1(s) = 2 s + 2s + 4

For the rear-half, represent the capacitor by a series equivalent circuit. Thus we can obtain an equivalent voltage source with value: V (0− ) 2(s + 8) 1 2s(s + 8)− (s2 + 2s + 4) Veq (s) = 2VC1 (s) − C = 2 − = s s + 2s + 4 s s s2 + 2s + 4

(

)

or equivalently Veq (s) =

s2 + 14s − 4

(

s s2 + 2s + 4

)

By a voltage division, 1 s

vC (0 − ) 2 s2 + 14s − 4 1 Vout (s) = 1 Veq (s) + = × + s s + 2 s s2 + 2s + 4 s + 0.5 s =

(

(

2s2 + 28s − 8 + (s + 2) s2 + 2s + 4

(

s( s + 2) s2 + 2s + 4

)

)

) = (s2 + 6s + 36) (s + 2)(s2 + 2s + 4)

Using MATLAB »num = [1 6 36]; »den = conv([1 2],[1 2 4]) den = 1 4 8 8 »[r,p,k] = residue(num,den) r= 7.0000e+00 -3.0000e+00 - 2.8868e+00i -3.0000e+00 + 2.8868e+00i p= -2.0000e+00 -1.0000e+00 + 1.7321e+00i -1.0000e+00 - 1.7321e+00i k= However, it would appear easier here to use ilaplace: »syms t s »ilaplace((s^2+6*s+36)/((s+2)*(s^2+2*s+4))) ans = 7*exp(-2*t)-6*exp(-t)*cos(3^(1/2)*t)+10/3*exp(-t)*3^(1/2)*sin(3^(1/2)*t)

2/23/02

» Hence

page P14.6

[


(

)

]

vout (t) = 7e −2t + e −t 3.334 3 sin( 3t − 6cos( 3t) u(t) V

SOLUTION 14.42. Using the series equivalent circuit (figure 14.17) for C1, we have I1K(s) =

Next, since vC 2 (0 − ) = 0 , we have

vC1 (0-)/s = - 0.25/s = -0.25 1 1000 + 50/s 1000s + 50 R+ sC 1

I (s) −0.25 × 500 −0 / 1 2 5 VC 2 (s) = 1k = = C2s s(1000s + 50) s(s + 0.05)

Finally, Vout (s) = -VC2 (s) =

0.125 = 2.5( 1 1 s s + 0.05 ) s(s + 0.05)

and vout (t) = 2.5( 1 - e -0.05t )u(t) V

SOLUTION 14.43. (a) It is preferable to use the series equivalent circuit (figure 14.17) for C1, and the parallel equivalent circuit (figure 14.16) for C2. (b) The current through the 2.5 kΩ resistor is given by -8×10-4 v (0-)/s - 2/s -2 I2.5K (s) = C1 = = = 2500 + 5000/s 2500s + 5000 s+2 R1 + 1 sC 1 Next, -8×10-4 1 1 -4 Vout (s) =VC2 (s) = I 2.5K (s) = × = 1 s + 2 0.0002s + 0.0002 (s + 2) (s + 1) sC 2 + R2 (c) Hence Vout (s) = and

-4 =-4 1 - 1 (s + 2) (s + 1) s + 1 s +2

(

)

vout (t) = 4 e −2t − e −t u(t) V

(d) SPICE plot omitted.

SOLUTION 14.44. (a) From voltage division,

2/23/02

page P14.7


R2 V Z R C R2 C1 s 2 2 2s + 1 H(s) = C2 = = = 2 Vin Z1 + Z2 R2 1 + R1 + R1 C1 R2 C2 s + ( R1 C1 +R2 C2 + R2 C1 )s + 1 C1 s R2 C2 s + 1 (b) If vin(t) = 15u(t) V, then Vin (s) = 15/s and VC2 = H(s)Vin =

s2

1.75s 7 ×15 = - 7 s s+ 0.25 s+ 4 + 4.25 s + 1

Hence

(

)

vC 2 (t) = 7 e −0.25t − e −4t u(t) V 15

vin and vc2

10

TextEnd 5

0

0

1

2

3

4 Time in s

5

6

(c) Using the series equivalent for C1, we have VC2 = H(s)

vC1 (0-) 1.75s = ×15 s 2 s + 4.25 s + 1 s

which is the same as result in part (b). Therefore

(

)

vC 2 (t) = 7 e −0.25t − e −4t u(t) V (d) Using the parallel equivalent for C2, we have

7

8

2/23/02

page P14.8


1 1 1 =2 ×15× = + 14 2s + 4 + 1 7 2s + 4 + s s + 0.25 s+4 7 7 2 + 1/s 7 7 2s + 1

VC2 (s) = C 2 vC2 (0-) Hence

(

)

vC 2 (t) = e −0.25t + 14e −4t u(t) V

(e) By linearity, the answer is the sum of parts (b), (c) and (d).

SOLUTION 14.45. (a) Z1 = R1 + L1 s and Z2 = IL1 =

R2 L2 s . From Ohm's law R2 + L2 s

(R2 + L2 s)Vin Vin = Vin = Z1 + Z2 R + L s + R2 L2 s (R1 + L1 s) (R2 + L2 s) + + R2 L2 s 1 1 R2 + L2 s

Using current division, we have IL2 =

R2 R2 Vin I = R2 + L2 s L1 (R1 + L1 s) (R2 + L2 s) + + R2 L2 s

Therefore R2 G1 H(s) = IL2 = = 2 Vin (R1 + L1 s) (R2 + L2 s) + + R2 L2 s G1 L1 G2 L2 s + (G 1 L1 + G 2 L2 + G 1 L2 )s + 1 With the given element values, H(s) =

2 2 = 2 + 4.25s + 1 4 7 4 7 7 2 s 2×1× × s + (2×1+ × + 2× )s + 1 7 8 7 8 8

(b) If vin(t) = 15u(t) V, then Vin (s) = 15/s and IL2 = H(s)Vin = Hence

2 32 2 ×15 = 30 s - s+ 0.25 - s+ 4 s 2 + 4.25 s + 1 s

(

)

iL 2 (t) = 30 − 32e −0.25t + e −4t u(t) A

Plot omitted. (c) Using the series equivalent for L1, we have IL2 = H(s) L 1 iL1(0-) = Therefore

2 8 ×1×15 = - 8 s + 0.25 s +4 s 2 + 4.25 s + 1

(

)

iL 2 (t) = 8 e −0.25t − e −4t u(t) A

2/23/02

page P14.9


(d) Using the parallel equivalent for L2, we have 7 ×15 L2 iL2(0-) 15( s + 2.25) 8 8 IL2(s) = = = = + 7 2 +4.25s + 1 R2 (R1 + L1 s) 1.75(0.5 + s) s + 0.25 s + 4 s 7 L2 s + s+ R2 + (R1 + L1 s) 8 1.75 + (0.5 + s) Hence iL2(t) = (8e-0.25t + 7 e-4t ) u(t) A (e) By linearity, the answer is the sum of parts (b), (c) and (d).

SOLUTION 14.46. (a) Using the result of problem 14.44(a) VC2 = R2 C1 s 2Iin R1 C1 R2 C2 s2 + ( R1 C1 +R2 C2 + R2 C1 )s + 1 Therefore H1 (s) = VC2 = Iin

=

=

1.75s s2 + 4.25 s + 1

3.5s s2 + 4.25 s + 1

(b) Using the result of problem 14.45(a), R2 2 H2 (s) = IL2 = = 2 VC2 (R1 + L1 s) (R2 + L2 s) + + R2 L2 s s + 4.25s + 1 (c) 3.5s 2 7s H(s) = IL2 =H 1 (s)H 2 (s) = × = Iin s2 + 4.25s + 1 s 2 + 4.25s + 1 (s2 + 4.25s + 1)2 (d) We first represent the initialized capacitor by the series equivalent (figure 14.17), and then apply a source transformation. From this circuit, by utilizing the expression derived in part (c), we have IL2 =sIL2 H(s) =H (s)H (s) = 7s 1 2 vC1 (0) 7.5 (s2 + 4.25s + 1)2 2s Therefore IL2(s) =

(s2

52.5 = 1.9911 + 3.7333 - 1.9911 + 3.7333 2 s+4 + 4.25s + 1) ( s + 4) 2 s + 0.25 ( s + 0.25)2

and iL2(t) = [(1.9911 + 3.7333t) e-4t +(- 1.9911 + 3.7333t) e-0.25t ] u(t) A Note: the book answer for part (d) should be divided by 2. (e) Since Iin(s) = 15/s, we have

2/23/02

page P14.10

IL2(s) = H(s)Iin (s) =

(s2


105 = 3.9822 + 7.4666 - 3.9822 + 7.4666 2 s+4 + 4.25s + 1) ( s + 4) 2 s + 0.25 ( s + 0.25)2

and iL2(t) = [(3.9822 + 7.4666t) e-4t +(- 3.9822 + 7.4666t) e-0.25t ] u(t) A

SOLUTION 14.47. (a) For this passive circuit, we may write the nodal equations by inspection. 0.8s + 2 + 10 s - 10 s

- 10 s

VC VR

1 + 10 s

= 2Vs1 -Is2

(b) Vs1= 3/s and Is2 = 3/s. We solve for VR by Cramer's rule to obtain 0.8s + 2 + 10 s - 10 s VR = 0.8s + 2 + 10 s

6 s - 3s - 10 s

- 10 s and

=

- 2.4s2 - 6s + 30 = - 4 + 1 s(0.8s2 + 10s + 30) s + 7.5 s

1 + 10 s

(

)

v R (t) = 1− 4e −7.5t u(t) V

(c) We represent the initialized capacitor by the parallel equivalent circuit ( figures 14.16) In this case the nodal equations becomes 0.8s + 2 + 10 s - 10 s

- 10 s 1 + 10 s

VC VR

6 + 2.4 s = - 3s

Solve for VR by Cramer's rule to obtain 0.8s + 2 + 10 s - 10 s VR = 0.8s + 2 + 10 s - 10 s

6 + 2.4 s - 3s - 10 s 1 + 10 s

2 = - 2.4s +18s + 30 = s1 - 16 + 12 2 s + 7.5 s + 5 s(0.8s + 10s + 30)

2/23/02

page P14.11


vR(t) = (1 - 16e-7.5t + 12e- 5t ) u(t) V

SOLUTION 14.48. (a) After performing the suggested source transformation, and representing the initialized capacitor and inductor by their series equivalent circuits, we can write two mesh equations by inspection: 0.5 + 1.25 s - 1.25 s

- 1.25 s 1 + 0.1s + 1.25 s

v (0) Vs1 - Cs Is1 = IL vC(0) s + LiL(0) + I s2

(b) With Vs1= 3/s, Is2 = 3/s, vC(0) = 0, and iL(0) = 3 A, the above mesh equation becomes 0.5 + 1.25 s - 1.25 s

- 1.25 s 1 + 0.1s + 1.25 s

vC(0) s

3 s Is1 = = IL vC(0) 0.3 + 3s s + LiL(0) + I s2 Vs1 -

Solve for IL(s) by Cramer's rule to obtain IL(s) =

0.15s 2 +1.875s + 7.5 = 2 + -3 + 4s 3 2 s + 7.5 s+5 0.05s + 0.625s + 1.875s

Therefore iL(t) = (4 + 2e-7.5t -3 e-5t ) u(t) A

SOLUTION 14.49. (a) Represent the initialized capacitors by their parallel equivalent circuits. (b) Write two nodals equation by inspection

0.001s + 0.4 - 0.2 - 0.2 0.001s + 0.4

VC1 = 0.2V in + 0.001vC1 (0-) = VC2 0.001v C2 (0-)

2.4 + 0.006 s 0.002

(c) Solve for VC2(s) by Cramer's rule to obtain 0.001s + 0.4 2.4/s + 0.006 2s 2 + 2×103 s + 48×104 - 0.2 0.002 0 -2 VC2 = = = + + + s4 0.001s + 0.4 - 0.2 2 4 s + 600 s + 200 2 s(s + 8×10 s + 12×10 ) - 0.2 0.001s + 0.4 (d)

2/23/02

page P14.12


vC2 (t) = (4 - 2e-200t ) u(t) V

SOLUTION 14.50. (a) Let VC denote the node voltage across the capacitor. By inspection the nodal equations in matrix form are: −1 /R 1+ 1 /R + 4s   VC   Vin  =  −1 /R 1 + 1/ R + 1 / ( s4)  Vout  Vin / ( 4s)  (b) By Cramer's rule, 1  1 + 1/ R + 4s det  −1 /R 1 / ( 4s) Vout (s) ( 4s + 1)(1 + 1/ R)  H(s) = = = 1+ 1 /R + 4s −1 /R Vin (s)   4s (1+ 1/ R + 4s)(1+ 4s / R + 4s) − 1/ R 2 det 1 + 1/ R + 1 / ( s4)   −1 /R

(

=

)

(4s + 1)(1 + 1/ R) 1 = 2 (1 + 8s + 16s )(1+ 1 /R) (1+ 4s)

Clearly, R does not affect the transfer function. The question is why? Note that the circuit can be redrawn as a balanced Wheatstone bridge circuit in which there is no voltage across R and no current through R. Hence R has no effect on the transfer function and on the impedance at the input. Hence R can be removed in the analysis of the circuit. In this case, the transfer function follows trivially by voltage division. (c) In view of the answer to (b), the impedance can be calculated with R removed. Hence 1  1+ (1+ 4s)  (1 + 4s)2 4s Zin (s) = = 2 =1 1  ( 1 + 4s ) 1+ + (1+ 4s)  4s Hence, the input impedance is a constant resistance and the network is called a constant resistance network. (d) The input is vin(t) = 10e–atu(t) V and R = 5 Ω. Find vout(t) for t ≥ 0 for the three cases, a = 0, 0.5, 0.25. (d) From part (b), for s ≠ 0.25, 0.25 10 2 . 5 / ( 0 . 2−5 a) 2 . 5 / ( 0 . 2−5 a) Vout (s) = × = − (s + 0.25) s + a s+ a (s + 0.25) which leads to  2.5  −at vout (t) =  e − e −0.25t u(t) V  0.25 − a For a = 0.25,

(

)

2/23/02

page P14.13

Vout (s) =

2.5 (s + 0.25)2


vout (t) = 2.5te−0.25t u(t) V

implying that

SOLUTION 14.51. (a) This bridged-T circuit was analyzed in problem 14.9. Here R = 1 Ω, Z1(s) = 0.25s and Z2(s) = 4/s. Since the condition Z1(s) Z2(s) = R2 is met, we have Zin(s)= 1Ω. (b) The s-domain equivalent circuit accounting for initial conditions is given below.

(c) Two nodal equations at Vc and Vout are:

VC - Vin + VC - Vout + 0.25sVC = 0.25 vC (0-) and

Vout - VC +

Vout - Vin + Vout = iL(0 ) s 0.25s

Writing these in matrix form, we have

0.25s + 2 -1

-1

VC

4 +2 s

Vout

Solving for Vout by Cramer's rule yields

Vin + 0.25vC (0-) = 4Vin + iL(0-) s s

2/23/02

page P14.14


0.5(s+ 8) Vout (s) = 4 Vin (s) + iL(0-) + 0.5s vC(0-) 2 s+4 (s+4) (s+4) 2 (d) Given vin(t) = 4u(t) - 3e-t u(t) V, then Vin (s) = 4s - 3 = s + 4 s+1 s( s + 1) and Vout (s) =

0.25(s+ 8) 4 + + 0.75s s(s + 1) (s+4) 2 (s+4) 2

Taking the inverse Laplace transform, we obtain, for t ≥ 0,

vout(t) = (4 - 4 e-t) + (0.25 e-4t + te-4t) + (0.75 e-4t - 3 te-4t) = 4 - 4 e-t + e -4t - 2te-4t

V

SOLUTION 14.52. A supernode is defined by drawing a curve to enclose the controlled voltage source. One node within the supernode has voltage Vout and the other has voltage V1Ω that is equal to V1Ω = -2I1 - Vout = -2 Vin - VC = - Vin + VC - Vout 2 Next, we write nodal equations at VC and the supernode: At node VC 0.5 (VC - Vin ) + 0.5sVC + 0.5 s (VC- Vout ) = 0 At the supernode 1 (V - V ) + 1 Vout + - Vin + VC + Vout = 0 2s out C 2 1 In matrix form, the nodal equations are: 0.5(s + 1 + 1/s) 1 - 0.5/s

-0.5/s 1.5 +0.5/s

VC = 0.5 V in Vout Vin

Solving by Cramer's rule yields 2( s 2 + 1.5) H(s) = Vout = 0.5s + 0.75/s = Vin 0.75s + 1+ 1.5/s 3s 2 + 4s + 6

SOLUTION 14.53. A supernode is defined by drawing a curve to enclose the controlled voltage source. One node within the supernode has voltage Vout and the other has voltage V1Ω which is equal to

2/23/02

page P14.15


V1Ω = -2I1 - Vout = -2 Vin - VC = - Vin + VC - Vout 2 Next, write nodal equations at VC and the supernode: At node VC s (VC- Vout ) = 0 0.5 (VC - Vin ) + (0.25s + 4s )V + C s2 + 16 At the supernode s (Vout - VC) + 1 Vout + - Vin + VC + Vout = 0 2 1 s 2 + 16 In matrix form, the nodal equation are: 0.25s + 0.5 +s/( s2 + 16)

-s/(s2 + 16)

1 - s/(s 2 + 16)

1.5 + s/(s2 + 16)

VC = 0.5 V in Vout Vin

Solving by Cramer's rule yields 0.25s 3 +5.5s H(s) = Vout = Vin 0.375s 3 + s 2 + 9s + 12

SOLUTION 14.54. Write nodal equations at V1 and V2: 0.5 (V1 - Vin ) + 0.125( V1 + 0.2V 2 ) + 0.1s(V1 - V2 ) = 0 and 0.1s(V2 - V1 ) + 1 V2 + V2 - 5V1 = 0 4 1 In matrix form, the nodal equations are: (0.1s + 0.625) (-0.1s - 5)

(-0.1s + 0.025) (0.1s + 1.25)

VC = 0.5 V in Vout 0

Solving by Cramer's rule yields 0.05s + 2.5 H(s) = Iout = V2 = Vin Vin - 0.31s + 0.9062

SOLUTION 14.55. (a) Simply replace each capacitor by the parallel form circuit model given in figure 14.16. (b) For this passive circuit, we can write the nodal equation by inspection.

2/23/02

page P14.16

0.5s +2 -1 0 -1 0.5s +2 -1 0 -1 0.5s +2

VC1 VC2 VC3


Vin + 0.5vC1 (0) = 0.5v C2 (0) 0.5v C3 (0)

Solving for VC3 by Cramer's rule yields VC3 (s) =

Vin (s) + 0.5vC1 (0) + 0.25s + 1 vC2 (0) +( 0.125s2 + s +1.5 )vC3 (0) 0.125s 3 + 1.5s2 + 5s + 4

(c) Substituting Vin(s) = 12/s, vC1(0) = 0, vC2(0)= 6, and vC3(0) = 2 into the above expression, we obtain VC3 (s) =

12/s + 6 0.25s + 1 + 2( 0.125s2 + s +1.5 ) 3 2 = 0.25s + 3.5s + 9s +12 3 2 3 0.125s + 1.5s + 5s + 4 s 0.125s + 1.5s2 + 5s + 4

Now use MATLAB to do the partial fraction expansion. n= [ 0.25 3.5 9 12]; d= [ 0.125 1.5 5 4 0]; [ r p k ] = residue (n,d) r= -2.5000 4.0000 -2.5000 3.0000 p= -6.8284 -4.0000 -1.1716 0 From the MATLAB output, we have, for t ≥ 0,

vC3(t) = 3 - 2.5 e-6.828t + 4e-4t - 2.5e-1.1716t V

SOLUTION 14.56. For this problem we utilize loop analysis with loops as indicated below.

2/23/02

page P14.17


In doing the following loop analysis, note that we will use gmVout = I3 and that due to our judicious choice of loops 1  Vout =  + Ls I1 Cs or equivalently,  1  0 =  + Ls I1 − Vout Cs For loop 1, 1   V1 =  2R + + Ls I1 + 2R I2 + RgmVout Cs For loop 2, V1 − V2 = 2R I1 + 7R I2 + 5RgmVout In Matrix form  1000  + 0.016s 0 −1  s  1000   I1   0  2 + + 0.016s 2 2   I 2  =  V1  s   2 7 10 Vout  V1 − V2     By Cramer's rule

2/23/02

page P14.18

 1000 + 0.016s 0  s  1000 det 2 + + 0.016s 2 s  2 7   Vout =  1000 + 0.016s  s  1000 + 0.016s det 2 + s  2  


   V1   V1 − V2   1000  − + 0.016s (5V1 + 2V2 )   s  = 6000 7000    0 −1 + 0.096s − 10 + + 0.112s   s s  2 2  7 10   0

Hence  1000  + 0.016s (5V1 + 2V2 )  s  s2 + 62500 Vout = = 2 + 625s + 62500 (5V1 + 2V2 ) 1000 s 10 + + 0.016s s The answers to (a) and (b) are clear at this point. (c) Using MATLAB »n = [21 0 21*62500]; »d = [1 625 62500 0]; »[r,p,k] = residue(n,d) r= 35 -35 21 p= -500 -125 0 k = [] Hence vout(t) = 21− 35e −125t + 35e −500t u(t ) V.

(

)

SOLUTION 14.57. (a) Replace the LC combination by a 1 V source after setting V1 and V2 to zero. We need to compute the current leaving the 1 V source which will be 1/Rth. Let the left node be denoted by Va and the right node by Vb. Also let G = 1/R. The nodal equations are by inspection 0  Va   gm + G  2G Va   ( gm + G) 2G    0 1.25G  V  =  −g + 0.25G ⇒  V  =    b   m   b   (− gm + 0.25G) 1.25G

2/23/02

page P14.19


Thus, the current leaving the 1 V source is I1V = G(1− Va ) + 0.25G(1− Vb ) = G(1− (gm + G) 2G) + 0.25G(1− (−gm + 0.25G) 1.25G) = 0.5G − 0.5gm + 0.25G + 0.2gm − 0.05G = 0.7G − 0.3gm Substituting G = 1/R we obtain 1 1 R 1 Rth = = = = = 10 Ω I1V 0.7G − 0.3gm 0.7 − 0.3gm R 0.7 − 0.3 × 2 (b) Replace the LC combination by a short circuit and compute Isc. This makes the controlled source zero. By inspection V V Isc = 1 + 2 2R 5R Thus V  V Voc = RthI sc = Rth  1 + 2  = 10(0.5V1 + 0.2V2 ) 2R 5R (c) By voltage division 1 + Ls ZLC s2 + 1 LC Vout = Voc = 1 Cs Voc = (5V1 + 2V2 ) Rth Rth + ZLC 2 + Ls + Rth s + s + 1 LC Cs L s2 + 62500 = 2 (5V1 + 2V2 ) s + 625s + 62500

SOLUTION 14.58. (a) The last equation is the constraint equation for the controlled floating voltage source. Hence, we have V1 − V2 − z0 (s)I 0 = 0 (b) By Cramer's rule,  1   1    R + Cs Iin   det 0 0 −1   1 0 − z0 (s)   − Iin Iin  = V2 = = 1 1 1  1         1  −2 + Cs − z0 (s)Cs + Cs  + Cs (2 + z0 (s)Cs)   R + Cs 1 R R R R   det 0 Cs −1   1 −1 −z0 (s)     (c) Here

2/23/02

page P14.20

V2 =

in which case

=

Iin 1  + Cs 2 + LCs 2 R 

(


)

2 . LC

SOLUTION 14.59. (a) Since the switch has been at position A for a very long time, the inductor looks like a short and iL(5-) = iL(5+) = 10/4 = 2.5 A. For t > 5, the switch moves to position B and the inductor current decays according to iL (t) = iL (5+ )e −(t −5)/ = 2.5e −(t −5)/0.1 = 2.5e −10(t −5) A (b) Note that iL(0-) = iL(0+) = 0. Hence 1 0.1  50 50  12.5 62.5 50 IL (s) = Vin (s) = − = − + 10s + 4 s + 0.4  s s + 0.5  s s + 0.4 s + 0.5 −0.4 t −0.5t Hence for 0 ≤ t ≤ 5s, iL (t) = 12.5 − 62.5e A. Here iL(5 ) = iL(5+) = 8.1458 A. + 50e For t > 5, the inductor decays with a time constant of 0.1 s. Thus iL (t) = 8.1458e −10(t −5)

SOLUTION 14.60. (a) Since the switch has been at position A for a very long time, the capacitor looks like an open and vC(5-) = vC(5+) = 40 V. For t > 5, the switch moves to position B and the capacitor voltage decays according to vC (t) = vC (5+ )e −(t −5)/ = 40e −(t −5)/2 = 40e −0.5(t −5) V (b) Note that vC(0-) = vC(0+) = 0. Hence 1 /Cs 12.5  50 50  7812.5 VC (s) = Vin (s) = − = 1 /Cs + 40 s + 12.5  s s + 12.5 s( s + 12.5)2 In MATLAB, »syms t s »ilaplace(7.8125e3/(s*(s+12.5)^2)) ans = 50-625*t*exp(-25/2*t)-50*exp(-25/2*t) Hence for 0 ≤ t ≤ 5s, vC (t) = 50 − 625te−12.5t − 50e −12.5t V. Here vC(5-) = vC(5+) = 50 V. For t > 5, the capacitor voltage decays with a time constant of 0.08 s. Thus vC (t) = 50e −12.5(t −5) V

2/23/02

page P14.21


SOLUTION 14.61. (a) Since the switch has been closed for a very long time, the capacitor looks like an open and vC(5-) = vC(5+) = 32 V. For t > 5, the switch opens and the capacitor voltage decays according to vC (t) = vC (5+ )e −(t −5)/ = 32e −(t −5)/0.4 = 32e −2.5(t −5) V (b) Note that vC(0-) = vC(0+) = 0 and vout = vC. Hence for 0 ≤ t ≤ 5, 1 Vin (s) 10 10  50 50  VC (s) = 1 = V (s) = − in 1 s + 12.5 s + 12.5  s s + 12.5  + + 0.002s 50 50 200 6250 = s(s + 12.5)2 In MATLAB, »syms t s »ilaplace(6250/(s*(s+12.5)^2)) ans = 40-500*t*exp(-25/2*t)-40*exp(-25/2*t) Hence for 0 ≤ t ≤ 5s, vC (t) = 40 − 500te−12.5t − 40e −12.5t V. Here vC(5-) = vC(5+) = 40 V. For t > 5, the capacitor voltage decays with a time constant of 0.4 s. Thus vC (t) = 40e −2.5(t −5) V

-

-

SOLUTION 14.62. (a) At 0-, vC( 0 ) = 0 and iL(0 ) = 50/10 = 5 A. (b) For this part, consider the equivalent circuit below.

By inspection, VC =

From table 13.1,

5 1   s 1+ 0.5s +  0.184s

=

5 0.5s2 + s +

1 0.184

=

10

(s + 1) 2 + π2

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page P14.22


vC (t) = 3.1831e −t sin(πt)u(t) V (c) Using MATLAB,

-

+

SOLUTION 14.63. (a) Since the switch has been closed for a long time, iL(1 ) = iL(1 ) = 30/0.8 = 37.5 A -

+

and vC(1 ) = vC(1 ) = 0. Represent the initialized inductor by its parallel equivalent circuit. Then VC (s) =

−37.5 1 −150 × = 1 s s2 + 4 Cs + Ls

Hence from table 13.1, vC (t') = −75sin(2 t') V

⇒ vC (t) = −75sin(2( t − 1 ) ) V for t > 1s.

(b) All initial conditions at t = 0 are zero. For 0 ≤ t ≤ 1s, 1 Vin (s) 5s 150s  1 1  VC (s) = 1 = V (s) = − in 2 2 1 0.8 s + 5s + 4 s + 5s + 4  s s + 2  + 0.25s + 0.8 s

Hence, for 0 ≤ t ≤ 1s,

300s  1  100 150 50 = 2 − +   = s +1 s + 2 s + 4 s + 5s + 4 s(s + 2)

vC (t) = 100e −t − 150e −2t + 50e −4t V Here, vC(1-) = vC(1+) = 17.403 V. Note, vin(1-) = 25.94 V. Next,

2/23/02

page P14.23

iC (1− ) = C

[


]

dvC  = 0.25 −100e −1 + 300e −2 − 200e −4 = 0.037378  dt  t =1

Thus in MATLAB »vin1 = 30*(1 - exp(-2)) vin1 = 2.5940e+01 »vc1 = 17.403 vc1 = 1.7403e+01 »ic1 = 0.25*(-100*exp(-1) + 300*exp(-2)-200*exp(-4)) ic1 = 3.7378e-02 »iL1 = (vin1 - vc1)/0.8 - ic1 iL1 = 1.0634e+01 Therefore, iL (1− ) = iL (1+ ) = 10.634 A. For t ≥ 1, we use the parallel equivalent circuit for both the inductor and the capacitor: + 4s  −10.634  −s 4s  −iL (1 ) +  VC (s) = e + Cv (1 ) = e −s 2 + 4.3507  C 2  s s s + 4  s +4 Therefore from table 13.1, for t > 1, vC (t) = −21.268sin (2(t − 1)) + 17.403cos (2(t − 1)) V Plots omitted.

SOLUTION 14.64. Here vC(0) = 0 for both capacitors. Part 1: 0 ≤ t ≤ 1s. 10 20 s  20 − 20  = Vout (s) = 10  s s + 2 s(s + 0.5)(s + 2) 20 + s From MATLAB »num = 20; »den = [1 2.5 1 0]; »[r,p,k] = residue(num,den) r= 6.6667e+00 -2.6667e+01 2.0000e+01 p= -2.0000e+00 -5.0000e-01 0 k = []

2/23/02

Therefore for 0 ≤ t ≤ 1,

page P14.24

(


)

vout (t) = 20 − 26.667e −0.5t + 6.667e −2t (u(t) − u(t − 1) )

Part 2. 1 ≤ t. Here the initial condition for the right-most capacitor is vout (1− ) = vout (1 + ) = 4.7281 V. As above, the left-most capacitor has zero value at t = 1s. Let us use the series equivalent circuit for the right capacitor. Then, −4.7281 1 −s −0.47281 IC (s) = e −s × = e 20 s s+2 10 + s Therefore, 4.7281  −4.7281 4.7281  −s  2.3641 2.3641 10 Vout (s) =  IC (s) + e −s = e −s  + =e  +  s  s  s + 2  s  s  s(s + 2) and for t ≥ 1, vout (t) = 2.3641+ 2.3641e −2(t −1) u(t − 1) V

(

)

SOLUTION 14.65. Assume the switch has been in position A for a long time. Both capacitors behave as open circuits and both capacitors have initial voltages at t = 0 of 10 V. For t ≥ 0, use the parallel equivalent circuits for both capacitors and write nodal equations. Let the left capacitor have voltage VCa. −0.01 0.005s + 0.03  VCa   0.005 × 10   0.05  = =  −0.01 0.0025s + 0.01 Vout   0.0025 × 10 0.025   By Cramer's rule,  0.005s + 0.03 0.05  det −0.01 0.025 s + 10 4 0 /3 10 / 3  Vout (s) = = 10 2 = − −0.01  0.005s + 0.03  s + 10s + 16 s + 2 s + 8 det  −0.01 0.0025s + 0.01  Therefore for t ≥ 0, vout (t) =

2s . 2 = 2 + s2 + 4 0.5s + s (b) Here, the initial condition is zero and SOLUTION 14.66. (a) Zin (s) = 2 +

1

40 −2t 10 −8t V e − e 3 3

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page P14.25


2s VC (s) =

2

10 s + 4 × 10 = 2 2s s s +s+ 4 2+ 2 s +4

»syms t s »ilaplace(10/(s^2+s+4)) ans = 4/3*exp(-1/2*t)*15^(1/2)*sin(1/2*15^(1/2)*t) Hence using MATLAB above or table 13.1 we have for 0 ≥ t ≤ 1.5s, vC (t) = 5.164e −0.5t sin(1.9365t) V and

vC (1.5− ) = vC (1.5 + ) = 0.57237 V (c) Use the parallel equivalent circuit for the left capacitor. The right capacitor has a zero initial voltage at t = 1.5. Hence, we do not use an equivalent circuit for the right capacitor. (d) Therefore e1.5sVC (s) =

s+4 s × 0.5 × 0.57237 = 0.57237 s2 + 10s + 16 0.5s + 2 + s+ 4 1

(e) In MATLAB »[r,p,k] = residue(0.57237*[1 4],[1 10 16]) r =V 3.8158e-01 1.9079e-01 p= -8 -2 k= [] Hence

(f) Finally

s+4  0.19079 0.38158  VC (s) = e −1.5s 0.57237 2 = e −1.5s  + s + 8   s+2 s + 10s + 16

[

]

vC (t) = 0.19079e −2(t −1.5) + 0.38158e −8(t −1.5) u(t − 1.5) V

R 16 = 8 V. 2R (b) For 0 ≤ t ≤ 1, v1(t) = 8 V and v 2 (t) = 8e −t /RC = 8e −0.6931t V. (c) v1(1− ) = 8 V and v 2 (1− ) = 8e −0.6931 = 4 V. SOLUTION 14.67. (a) v1(0 − ) = v1 (0+ ) = v 2 (0− ) = v2 (0 + ) =

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page P14.26


(d) From KVL, v1(1+ ) = v 2 (1 + ). From conservation of charge, 1× v1(0 − ) + 1× v 2 (0− ) = 12 = 2 × v1 (0+ ) = 2 × v 2 (0+ ) . Therefore v1(1+ ) = v 2 (1 + ) = 6 V. (e) This response represents a decay with time constant τ = 2R = 2.8854 s. Hence v1(t) = v 2 (t) = 6e −0.34657( t −1)u(t − 1) V It follows that

v1(3) = v 2 (3) = 6e −0.34657×2 = 3 V

(f) Both capacitor voltages change abruptly at t = 1.

SOLUTION 14.68. Label the current down through the first inductor as i1(t). (a) i1 (0− ) = i1(0+ ) = 1 A a n diout (0− ) = iout (0+ ) = 0 . (b) For 0 ≤ t we use a parallel equivalent for the first inductor. By current division 1 −1 −175 5 + 0.1s Iout (s) = × = 2 1 1 s s + 275s + 2500 0.057143 + + 0.35s 5 + 0.1s Use MATLAB to do the partial fraction expansion num = -175; den = [ 1 275 2500]; [ r, p, k] = residue (num, den) r= 0.6831 -0.6831 p= -265.5869 -9.4131 From the MATLAB output Iout (s) = Therefore ,

0.68313 + 0.68313 s+ 9.4131 s+ 265.59

(

)

iout (t) = 0.68313 e −265.59t − e −9.413t u(t) A

SOLUTION 14.69. (a) Here we use voltage division:

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page P14.27


1 V1(s) =

1 −6

Therefore, v1(0 + ) = 5 V. (b) Again use voltage division:

4 × 10

s

+

4 × 10−6 s 1 4 × 10

−6

s

+

×

1 1× 10−6 s

30 30 5 = = s 6s s

1 V1(s) =

1 5 × 10−6 s

Therefore, v1(0 + ) = 4.7059 V.

+

5 × 10 −6 s 1 1× 10 −6 s

+

1

40 80 = s 17s

×

2 × 10−6 s

SOLUTION 14.70. (a) Consider a mesh current I(s) in the usual direction and use the series equivalent circuit for each capacitor. Thus I(s) =

1 4 × 10 −6 s

+

1 1 4 × 10−6 s

+

1 1× 10 −6 s

 3 0.3 0.9 0.6 1.2 −6 −6  s − s − s − s  = 1.5 × 10 = 0.8 × 10

Therefore for t > 0,

1 0.3 0.8 × 10 −6 0.3 0.5 V1(s) = I(s) + = + = ⇒ v1(t) = 0.5 V s s s 4 × 10−6 s 4 × 10−6 s Similarly for t > 0 0.8 × 10 −6 0.9 1.1 V2(s) = + = ⇒ v 2 (t) = 1.1 V s s 4 × 10−6 s and 0.8 × 10−6 0.6 1.4 V3(s) = + = ⇒ v 3(t) = 1.4 V s s 1× 10−6 s (b) Consider a mesh current I(s) in the usual direction and use the series equivalent circuit for each capacitor. Thus I(s) =

1 5 × 10 −6 s

+

1 1 1× 10−6 s

+

1 2 × 10 −6 s

 4 0.3 0.9 0.6 2.2 −6 −6  s − s − s − s  = 1.7 × 10 = 1.2941× 10

Therefore for t > 0,

1 0.3 1.2941× 10 −6 0.3 0.55882 V1(s) = I(s) + = + = s s s 5 × 10−6 s 5 × 10 −6 s Similarly for t > 0

⇒ v1(t) = 0.55882 V

2/23/02

page P14.28

V2(s) =

1.2941× 10 −6 0.9 2.1941 + = ⇒ v 2 (t) = 2.1941 V s s 1× 10 −6 s

V3(s) =

1.2941× 10−6 0.6 1.2471 + = ⇒ v 3 (t) = 1.2471 V s s 2 × 10−6 s

and


SOLUTION 14.71. (a) For 0 < t < 2, the 150 mF capacitor is charged to 25 V. From conservation of charge, 0.15 × 25 = 0.15vC (2+ ) + 0.1vC (2 + ) = 0.25vC (2+ ) Therefore vC (2+ ) =

0.15 × 25 = 15 V 0.25

This voltage remains constant for t > 2 s. (b) For 0 < t < 2, the 150 mF capacitor is charged to 25 V. From conservation of charge, 0.15 × 25 + 0.1× 10 = 0.15v C (2 + ) + 0.1vC (2+ ) = 0.25v C (2+ ) Therefore vC (2+ ) =

0.15 × 25 + 0.1× 10 = 19 V 0.25

This voltage remains constant for t > 2 s.

SOLUTION 14.72. (a) Let the middle node have voltage Va(s). Then writing node equations  8s −2s  Va  1.144   Va  1  6 2 1.144 s 0.156 / s −2s 6s  V  =  0  ⇒ V  = 44  2 8  0  = 0.052 / s V    out     out       Thus for t > 0, vout (t) = 0.052 V. (b) Again define Va(s) as the middle node. Then 3 0.286 6 0.286 0.132 Va (s) = 1 10s × = × = 1 3 s 13 s s + + 10s 4s 10s Hence for t > 0, 1 0.132 0.132 0.044 Vout (s) = 1 4s 1 × = = ⇒ v out (t) = 0.044 V s 3s s + 2s 4s

2/23/02

page P14.29


SOLUTION 14.73. With switches in position A, the equivalent capacitance to the right of v2 is 4 mF. Therefore at t = 0+, by voltage division V1(s) = V2 (s) =

10 s

⇒ v1(t) = v 2 (t) = 10 V for 0 < t < 1.

Hence with the switches in position B, let us write a single node equation using the parallel equivalent circuit the initialized capacitors: 0.002sV2 (s) − 0.002 × 10 + 0.004sV2 (s) + 0.004 × 10 = 0.004s × Equivalently

11 s

0.006sV2 (s) = 0.044 − 0.02 = 0.024 ⇒ V2 (s) = 4 / s

Hence, for t > 1 s, v 2 (t) = 4 V and v1(t) = 7 V.

SOLUTION 14.74. (a) At t = 0+, the frequency domain equivalent circuit is given below.

To compute Ceq, we observe 1 Ceq = 1+ 1 1 1 = 2 mF + + 3 2 6 By voltage division Ceq s 20 10 0.002s 20 10 and V2 = × = × = Ceqs + 0.002s s s Ceq s + 0.002s s s Hence, for 0 < t <1, v1(t) = v2(t) = 10 V. (b) When the switch moves to B, the pertinent part of the equivalent frequency domain circuit is given below. V1 =

2/23/02

page P14.30


By superposition, 1 8 1 1 6 × + × 0.02 − × 0.01= 1 + 2 s 0.003s 0.003s s Hence, for 1 < t < 2, v1(t) = 6 V and by KVL, v2(t) = 2 V. V1 =

SOLUTION 14.75. When the switch in position A, the 2 µF capacitor is charged to –2 V. Hence, the charge on the top plate is C*vC = –4 µC. When the switch is moved to position B, due to the virtual ground, the 2 µF capacitor voltage is zero meaning it cannot retain any charge. Hence, assuming an ideal op amp, all charge moves to the 1 µF capacitor with –4 µC on the left plate. Hence, vout = −(–4 µC)/1 µF = 4 V.

SOLUTION 14.76. (a) vout(t) = 0 for 0 < t < 1 ms. Every time the switch moves to position A, the capacitor, C, charges to 8 V. When the switch moves to position B, because of the virtual ground, all charges moves to the kC capacitor. Hence with k = 1, vout(t) = – 8 V for 1 ms < t < 3 ms. For 3 ms < t < 5 ms, vout(t) = – 16 V. Repeating the pattern implies that for 5 ms < t < 7 ms, vout(t) = – 24 V, etc. See for example figure 14.51. (b) With k = 0.5, the voltages computed in part (a) double. (c) With k = 2, the voltages computed in part (a) are halved.

SOLUTION 14.77. (a) For this part consider the circuit below.

2/23/02

page P14.31


For the normalized design C = 1 F, G1 = 0.5 S, G2 = 2 S, and G3 = 1.5 S. After magnitude scaling with Km = 106, then C = 1 F , R1 = 2 MΩ, R2 = 5 0 0 Ω k , R3 = 6 6 6 . 7 Ω. k (b) For this part, consider the circuit below.

For the normalized design C = 1 F, G1 = 1 S, G3 = 2 S, G2 = 0.5 S, and G4 = 2.5 S. After magnitude scaling with Km = 106, then C = 1 F , R1 = 1 MΩ, R3 = 5 0 0 Ω k , R2 = 2 MΩ, R4 = 4 0 0 Ω k. (c) Now consider the circuit below.

2/23/02

page P14.32


For the normalized design C = 1 F, G1 = 0.25 S, G3 = 0.5 S, G2 = 0.75 S, G4 = 1 S, and G5 = 1 S. After magnitude scaling with Km = 106, then C = 1 F , R1 = 4 MΩ, R3 = 2 MΩ, R2 = 4 / 3 MΩ, R4 = 1 MΩ, R5 = 1 MΩ. (d) Finally, we consider the following circuit.

For the normalized design C = 1 F, G1 = 2 S, G3 = 1.5 S, G6 = 1 S, G2 = 0.5 S, G4 = 2 S, and G5 = 2 S. After magnitude scaling with Km = 106, then C = 1 F , R1 = 0 . 5 M Ω, R3 = 2 /3 MΩ , R6 = 1 MΩ, R2 = 2 MΩ , R4 = 0 . 5 M Ω, R5 = 0 . 5 M Ω.

2/23/02

page P14.33


SOLUTION 14.78. With V1 = Vout a prototype design is given by the topology below.

Using MATLAB »Km = 1e7; »Gin = 1; G1 = 1.405; G2 = 0.402; G3 = 0.942; »DG = 1.865; »Rinnew = Km/Gin Rinnew = 10000000 »R1new = Km/G1 R1new = 7.1174e+06 »R2new = Km/G2 R2new = 2.4876e+07 »R3new = Km/G3 R3new = 1.0616e+07 »DRnew = Km/DG DRnew = 5.3619e+06 Later, when we study frequency scaling, Km will be smaller and the filter will have a cutoff frequency in a more reasonable range.

SOLUTION TO 14.79. Note corrections to problem statement. W(0+) should be W(∞) in part (b) and in part (c) one should calculate W(0-) – W(∞). The frequency domain equivalent circuit is given by the figure below.

2/23/02

page P14.34


(a) To find the required time functions, we first find their Laplace equivalents. a b a−b − K1 s s R I(s) = = ≡ 1 1  1 1  s + p1 R+ + s + +   C1s C2s  RC1 RC2 

in which case i(t) = K1e − p1t u(t) where

K1 =

 1 a−b 1  and p1 =  . + R  RC1 RC2 

Further, K1  − I(s) a a C1 K  1 K 1 VC1(s) = + = − = a − 1  × + 1 × C1s s s s(s + p1)  C1 p1  s C1 p1 (s + p1) in which case   K  K vC1 (t) =   a − 1  + 1 e − p1t  u(t) C1 p1  C1 p1    Also, by symmetry, K1  I(s) b b C2 K  1 K 1 VC 2 (s) = + = + = b + 1  × − 1 × C2s s s s(s + p1)  C2 p1  s C2 p1 ( s + p1 ) in which case   K  K vC 2 (t) =  b + 1  − 1 e − p1t  u(t) C2 p1  C2 p1   (b) The total energy stored in the capacitors at time 0- is W (0− ) = Also at t = ∞,

Hence

1 1 C1a2 + C2b2 2 2

  K  K  vC1 (∞) =  a − 1  and vC 2 (∞) =  b + 1  .  C1 p1   C2 p1 

2/23/02

page P14.35


2

1  K  1  K  W (∞) = C1  a − 1  + C2 b + 1  2  C1 p1  2  C2 p1  (c)

∞

∫ Ri 0

∞

2

(t)dt = ∫

0

RK12e −2 p1t dt =

2

RK12 (a − b)2 (a − b)2 = = 2p1 2Rp1  1 1  2 +   C1 C2 

Observe that 2

1 1 1  K  1  K  W (0 ) − W (∞) = C1a 2 + C2b 2 − C1 a − 1  − C2  b + 1  2 2 2  C1 p1  2  C2 p1  −

=

2

aK1 bK1 1 K12 1 K12 (a − b)2 1 (a − b)2 (a − b)2 (a − b)2 − − − = − × = = p1 p1 2 C1 p12 2 C2 p12 Rp1 2 Rp1 2Rp1  1 1 2 +   C1 C2 

This indicates that the total energy lost between 0- and infinity is the energy dissipated in the resistor and the result is independent of the value of R. (d) When R → 0, a b − s s = a − b = C1C2 (a − b) lim I(s) = 1 1 1 1 (C1 +C2 ) R→ 0 + + C1s C2s C1 C2 Therefore C1C2 i(t) = (a − b) (t) (C1 +C2 ) Further −C2 a C1 C2 VC1(s) = (a − b) + = a+ b s(C1 +C2 ) s s(C1 +C2 ) s(C1 +C2 ) and C1 b C1 C2 VC 2 (s) = (a − b) + = a+ b s(C1 +C2 ) s s(C1 +C2 ) s(C1 +C2 ) Therefore  C1  C2 vC1 (t) = vC 2 (t) =  a+ b u(t)  (C1 +C2 ) (C1 +C2 ) 

SOLUTION TO 14.80. (a) From conservation of charge C1vC1(0 − ) + C2vC 2 (0− ) = C1v C1(0+ ) + C2vC 2 (0 + ) = (C1 + C2 )v C1(0 + ) = (C1 + C2 )v C 2 (0+ ) Therefore

2/23/02

page P14.36


C v (0− ) + C2v C 2 (0− ) vC1 (0+ ) = v C2 (0 + ) = 1 C1 (C1 + C2 ) (b) Inserting values into out answer for part (a) yields C1 vC1 (0+ ) = v C2 (0 + ) = = 0.5 V (C1 + C2 ) and the voltage remains the same for t > 0. (c) Before the switch is closed, the energy in C2 is zero and the energy in C1 is the total stored energy: 2 Wtot (0 − ) = WC1(0− ) = 0.5C1v C1 (0− ) = 0.5 J

After the switch is closed, 2 Wtot (0 + ) = WC1(0 + ) + WC 2 (0+ ) = 0.5C1vC1 (0 + ) + 0.5C2vC2 2 (0 + ) = 0.25 J (d) (i) Using the series equivalent circuit for C1, we have I(s) =

1

1 1 /R 1 −2t R × = ⇒ i(t) = e u(t) A 2 s s + 2 /R R R+ s

Thus VC 2 (s) = and VC1(s) = −

(

)

I(s) 1/ R 0.5 0.5 = = − ⇒ vC 2 (t) = 0.5 1− e −2t R u(t) V s s(s + 2 / R) s s + 2 /R

(

)

I(s) 1 −1/ R 1 0.5 0.5 + = + = + ⇒ vC1 (t) = 0 . 5 1+ e −2t R u(t) V s s s(s + 2 / R) s s s + 2 /R

(ii) The energy dissipated in the resistor is given by ∞ ∞ 1 −4t R e −4t R  1 WR (0,∞) = R ∫ i ( )d = ∫ e d =−  = J R 4  4 0 0 0 ∞

2

(iii) For all R,

∞ ∞ 1 −2t R e −2t R  1 Area under i(t) = ∫ i( )d = ∫ e d =−  = R 2  2 0 0 0 ∞

1 −2t R e u(t) has a decay that becomes infinitely fast and its magnitude (1/R) R approaches ∞. Thus we have infinite height, zero-width, but a finite area of 0.5. Thus as R → 0, i(t) → 0.5 (t) A. (We have avoided a more rigorous explanation as the above argument is more plausible to sophomores.). As R → 0, the exponential terms in the expressions for vC1 (t) and vC 2 (t) have infinitely fast decays and hence disappear from the expressions yielding the stated result. Further, as R → 0, i(t) =

2/23/02

page P14.37


SOLUTION TO 14.81. (a) Zin (s) =

2s + 4.5 1 = + 1 = Za (s) + Z b (s) (s+ 0.5)(s + 4) s+ 0.5 s+4 Ya (s) =

1 = s+ 0.5 Za (s)

Yb (s) =

1 =s+4 Zb (s)

From the above expressions, the RC circuit consists of a series connection of (a 1 farad capacitor in parallel with a 2 Ω resistor) and (a 1 F capacitor in parallel with a 0.25 Ω resistor). (b) Yin (s) =

12s + 440 10 + 2 = = Ya (s) + Yb (s) (s+ 120)(s + 20) s+ 120 s + 20 Za (s) =

1 = 0.1s + 12 Ya (s)

Zb (s) =

1 = 0.5s + 10 Yb (s)

From the above expressions, the RL circuit consists of a parallel connection of (0.1 H inductor in series with a 12 Ω resistor) and (a 0.5 H inductor in series with a 10 Ω resistor). (c) Following the hint, we have Yin (s) = 0.225s +0.075 = 0.1 + 0.125 s (s+ 0.2)(s + 0.5) s+ 0.2 s + 0.5 Hence Yin (s) =

0.1s + 0.125s = Ya (s) + Yb (s) s+ 0.2 s + 0.5

Za (s) =

1 = s + 0.2 = 10 + 1 Ya (s) 0.1s 0.5s

Zb (s) =

1 = s + 0.5 = 8 + 1 Yb (s) 0.125s 0.25s

From the above expressions, we see that each term in Yin (s) represents a series RC circuit. The RC circuit for Yin(s) consists of a parallel connection of (a 0.5 F capacitor in series with a 10 Ω resistor) and (a 0.25 F capacitor in series with a 8 Ω resistor). (d) CORRECTION: for part (d),change the second term to 2s/(s2 + 2).

2/23/02

page P14.38


Yin (s) = 0.5s + 2s = Ya (s) + Yb (s) s2 + 1 s2 + 2 Za (s) = Zb (s) =

1 = s 2 + 1 = 2s + 1 Ya (s) 0.5s 0.5s 1 = s 2 +2 = 0.5s + 1 s Yb (s) 2s

From the above expressions, we see that each term in Yin (s) represents a series LC circuit. The LC circuit for Yin(s) consists of a parallel connection of (a 0.5 F capacitor in series with a 2 H inductor) and (a 1 F capacitor in series with a 0.5 H inductor).

SOLUTION 14.82. CORRECTIONS TO PROBLEM STATEMENT: (i) v0(t), should read vout(t) and (ii) there should be a connection from the circuit inside the shaded box to the bottom line or reference node. (a) (i) 0 ≤ t < 1 ms. Since the capacitor voltage is initially zero and the switch is in position (a), a simple source transformation yields a Norton equivalent (seen by the capacitor) consisting of a 20 mA current source in parallel with 9.8039 kΩ resistor. Hence Vout (s) =

1 1 + 10−6 s 98039

×

.02 20 × 103 = s s(s + 102)

Using MATLAB n = .02*1e6; p1 = 1e6/Rth p1 = 102 »d = [1 p1 0]; »[r,p,k] = residue(n,d) r= -1.9608e+02 1.9608e+02 p= -102 0 k = []

(

)

Hence, for 0 ≤ t < 1 ms, vout (t) = 196.08 1− e −102t u(t) V. It follows that vout (1 ms) = 19.014 V. (ii) 1 ms ≤ t < 1.05 ms. The frequency domain equivalent circuit is given below.

2/23/02

page P14.39


Writing a single node equation yields 200 1 Vout − Vout − s + s + 10 −6 sV 19.014 × 10−6 = out 10000 10 1.2 × 10 5 Equivalently 19.014 + = (s + 100100)Vout or s

(

19.014s + 1.2 × 10 Vôut = s(s + 100100)

5

)

Therefore, vôut (t') = 1.1988 + 17.815e −100100t' u(t') , and for 1 ms ≤ t < 1.05 ms, vout (t) = vôut (t − 0.001) in which case

(

)

vout (t) = 1.1988 + 17.815e −100100(t −0.001) u(t − 0.001)

Again using MATLAB,

(b) Part 1: for t > 0 up to t1 which denotes the time when vout(t) reaches 80 V, i.e., the capacitor is charging. The frequency domain equivalent circuit is

2/23/02

page P14.40


Using our knowledge of part (a), this circuit simplifies to

Hence 0.02 3 s = 5s + 20 × 10 Vout (s) = 1 s(s + 102) 10 −6 s + 9803.9 5 × 10 −6 +

and from MATLAB »syms s t »ilaplace((5*s+20e3)/(s^2+102*s)) ans = 10000/51-9745/51*exp(-102*t) in which case vout (t) = 196.08 − 191.08e −102t u(t) V

(

)

From this expression, vout (t1 ) = 80 = 196.08 − 191.08e −102t1 V and t1 = log((80-10000/51)/(-9745/51))/(-102) t1 = 4.8864e-03 This part of the problem considers t1 ≤ t < t2, i.e., the capacitor is discharging where vout(t2) = 5. The equivalent frequency domain circuit is given below which is a slight modification of the circuit of (a)-(ii):

2/23/02

page P14.41


Making use of our knowledge of part (a)-(ii), we have 80s + 1.2 × 105 ˆ Vout = s(s + 100100)

(

)

in which case vôut (t') = 1.1988 + 78.801e −100100t ' u(t') V, and

(

)

vout (t) = 1.1988 + 78.801e −100100(t −t1) u(t − t1) Here t'2 = 3.0286 × 10 −5 s and t2 = t1 + t'2 = 4.9167 ms where t'2 is the duration of the discharge cycle. As a final point, note that the frequency of the sawtooth is 1/t2 = 203.39 Hz. Finally a plot is given below.

2/23/02

page P14.42


SOLUTION 14.83. CORRECTION: In example 14.10, page 560, delete the four minus signs in the equation for VC(s) and one more for vC(t). We use MATLAB instead of SPICE to solve this problem. Applying voltage division to the circuit of figure P14.83, we have 1 i (0-) Cs 1 108 VC (s) = LiL(0-) = L = 1 + Ls + R C s2 + R s + 1 s2 + 125s + 1.25 ×109 Cs L LC From table 13.1, item 18 vC (t) = 2828e-62.5t sin(35,355t) u(t) V A plot of vC(t) is given below with the vertical axis in V and the horizontal axis in seconds. 3000 2000 1000 0 -1000 -2000 -3000

0

0.2

0.4

0.6

0.8

1 x10-3

The waveform for the first few cycles is essentially the same as the example 14.10. Thus for the firs few cycles, the lossless circuit of example 14.10 is a good approximation to the more accurate circuit model of this problem. The effect of the presence of 100 Ω resistance is a slow decay (with respect to ms intervals) of the peak values.

1/25/02

P16-1


PROBLEM SOLUTIONS Solution 16.1. (a) By the definition of the convolution integral ∞

f 2(t) ∗ f 2 (t) =

∫

−∞

∞

f 2 (t − ) f 2 ( )d =

∞

∫ 2u(t −

)2u( )d = 4 ∫ u(t − )d

−∞

0

The integrand, u(t − ) , is nonzero only when ≤ t. This suggests that there are two regions of consideration: t < 0 and t ≥ 0. Case 1: t < 0. Here u(t − ) = 0 since is restricted to the interval [0,∞). Hence f 2(t) ∗ f 2 (t) = 0 , for t < 0.

Case 2: t ≥ 0.

∞

f 2(t) ∗ f 2 (t) = 4 In sum,

∫

−∞

t

u(t − )d = 4 ∫ d = 4t , for t ≥ 0. 0

 0, t < 0 f 2(t) ∗ f 2 (t) =  4t, t ≥ 0

(b) By the definition of the convolution integral ∞

f 2(t) ∗ f 3 (t) =

∫

−∞

∞

f 2 (t − ) f 3( )d =

∫ 2u(t −

)4e

−2

−∞

∞

u( )d = 8 ∫ e −2 u(t − )d 0

The integrand, u(t − ) , is nonzero only when ≤ t. This suggests that there are two cases to consider: t < 0 and t ≥ 0. Case 1: t < 0. Here u(t − ) = 0 since is restricted to the interval [0,∞). Hence f 2(t) ∗ f 3 (t) = 0, for t < 0. Case 2: t ≥ 0. t

]

t

f 2(t) ∗ f 3 (t) = 8∫ e −2 d = −4e −2 = 4(1− e −2t ) 0 0

In sum,

t<0 0, f 2(t) ∗ f 2 (t) =  −2t 4(1− e ), t ≥ 0

(c) By the definition of the convolution integral and the sifting property of the delta function ∞

f1(t) ∗ f 3(t) =

∫

−∞

∞

f1 (t − ) f 3( )d =

= 20e

−2

u( )

∫5

−∞

(t − )4e −2 u( )d =

] =t = 20e−2t u(t)

(d) By the definition of the convolution integral

1/25/02

P16-2

∞

f 3(t) ∗ f 3(t) =

∫

−∞

∞

f 3 (t − ) f 3( )d =

∫

4e

−2(t − )


u(t − )4e

−2

u( )d = 16e

−∞

−2t

∞

∫ u(t −

)d

0

The integrand, u(t − ) , is nonzero only when ≤ t. This suggests that there are two cases to consider: t < 0 and t ≥ 0. Case 1: t < 0. Here u(t − ) = 0 since is restricted to the interval [0,∞). Hence f 3(t) ∗ f 3(t) = 0 , for t < 0.

Case 2: t ≥ 0.

f 3(t) ∗ f 3(t) = 16e

−2t

t

∫d

= 16te −2t

0

In sum,

t<0 0, f 3(t) ∗ f 3(t) =  −2t 16te , t ≥ 0

(e) By the definition of the convolution integral and the sifting property of the delta function ∞

f1(t + 2) ∗ f 2 (t + 4) =

∫

−∞

∞

f1(t + 2 − ) f 2 ( + 4)d =

∫

5 (t + 2 − )2u( + 4)d =

−∞

= 10u( + 4)] =t +2 = 10u(t + 6)

(f) By the distributive property of convolution f 2(t) ∗ [ f 2 (t) + f 3(t)] = f 2 (t) ∗ f 2 (t) + f 2 (t) ∗ f 3 (t) Using the results of parts (a) and (b) the result follows immediately t<0 0, f 2(t) ∗ [ f 2 (t) + f 3(t)] =  −2t 4(1+ t − e ), t ≥ 0

Solution 16.2. (a) By definition ∞

f 3(t) =

∫

−∞

f1( ) f 2(t − )d =

∞

∫ K1u(

−∞

Here observe that u(t − − T2 ) = 0 for t − T2 < T1 and t − T2 ≥ T1 .

∞

− T1)K2 u(t − − T2 )d = K1K2 ∫ u(t − − T2 )d > t − T2 .

T1

Hence there are two cases to consider:

Case 1: t − T2 < T1 . Here u(t − − T2 ) = 0 , since is restricted to the interval [T1,∞) . f 3(t) = 0 Case 2: t − T2 ≥ T1 .

1/25/02

P16-3


t −T 2

∫

f 3(t) = K1K2

d = K1K2 (t − T2 − T1 )

T1

In sum,

t < T1 + T2 0, f 3(t) =  K1K 2 (t − T2 − T1), t ≥ T1 + T2

(b) By definition ∞

f 3(t) =

∫

−∞

∞

f1(t − ) f 2( )d =

∞

∫ K1u(t −

+ T1 )K 2u( + T2 )d = K1K2

−∞

Here observe that u(t − + T1) = 0 for t + T1 < −T2 and t + T1 ≥ −T2 .

> t + T1.

Case 1: t + T1 < −T2 . Here u(t − + T1) = 0 , since Case 2: t + T1 ≥ −T2 .

u(t − + T1 )d

Hence there are two cases to consider: is restricted to the interval [−T2 ,∞): f 3(t) = 0.

t +T 1

f 3(t) = K1K2 In sum,

∫

−T 2

∫

−T 2

d = K1K 2 (t + T1 + T2 )

t < −T1 − T2 0, f 3(t) =  K1K 2 (t − T2 − T1), t ≥ −T1 − T2

(c) By definition ∞

f 3(t) =

∫

−∞

∞

f1(t − ) f 2( )d =

The integrand is nonzero only when

∫ K1u(t −

−∞

)K 2e

−a

∞

u( )d = K1K 2 ∫ e −a u(t − )d 0

≤ t. Hence, there are two cases to consider: t < 0 and t ≥ 0.

Case 1: t < 0. f 3(t) = 0, for t < 0. Case 2: t ≥ 0. t

f 3(t) = K1K2 ∫ e 0

−a

e −a d = K1K2 −a

t

 KK  = 1 2 (1− e −at ). a  0

Therefore KK f 3(t) = 1 2 (1− e −at )u(t). for t ≥ 0. a (d) By definition ∞

f 3(t) =

∫

−∞

f1(t − ) f 2( )d =

∞

∫ K1u(t −

−∞

+ T1 )K 2e

−a

∞

u( )d = K1K 2 ∫ e −a u(t − + T1)d 0

1/25/02

P16-4


≤ t + T1 . Hence, there are two cases to consider: t + T1 < 0

The integrand is nonzero only when and t + T1 ≥ 0.

Case 1: t + T1 < 0. Here u(t − + T1) = 0 , since f 3(t) = 0. Case 2: t + T1 ≥ 0.

t +T 1

f 3(t) = K1K2

∫

e

is restricted to the interval [0,∞).

e −a d = K1K 2 −a

−a

0

Therefore

t +T1

   0

[

[

Hence

]

KK = 1 2 1− e −a(t +T 1) . a

]

KK f 3(t) = 1 2 1− e −a(t +T1 ) u(t + T1 ). a (e) By definition ∞

f 3(t) =

∫

−∞

∞

f1(t − ) f 2( )d =

∫ K1u(−t +

−∞

)K2e

−a

∞

u( )d = K1K2 ∫ e −a u(−t + )d 0

The integrand is nonzero only when ≥ t. Hence, there are two cases to consider: t ≤ 0 and t > 0. Case 1: t ≤ 0. Here u(− t + ) = 1, since ≥ 0. Hence ∞

f 3(t) = K1K2 ∫ e

−a

0

Case 2: t > 0. ∞

f 3(t) = K1K2 ∫ e

−a

t

∞

e −a d = K1K2 −a

 KK  = 1 2 , for t ≤ 0. a  0 ∞

e −a  KK d = K1K2  = 1 2 e −at , for t > 0. − a  a t

In sum, K1K 2 t≤0  a , f 3(t) = K K  1 2 e −at , t > 0  a

Solution 16.3. (a) By definition ∞

f 3(t) =

∫

−∞

f1( ) f 2(t − )d =

∞

∫ K1e

−∞

−a

u( )K 2e

∞

u(t − )d

= K1K2 ∫ e −at u(t − )d 0

The integrand, u(t − ) , is nonzero only when and t ≥ 0. Case 1: t < 0. Here u(t − ) = 0 , since Case 2: t ≥ 0.

−a(t − )

≥ t. Hence, there are two cases to consider: t < 0

≥ 0. Hence f 3(t) = 0, for t < 0.

1/25/02

P16-5

f 3(t) = K1K2e

−at

t

∫d


= K1K 2e −at t , for t ≥ 0.

0

In sum, Ic (s) = (b) By definition

Cs

1 Iin (s) Cs + R a=1 K =1

The integrand, u(t − ) , is nonzero only when and t ≥ 0. Case 1: t < 0. Here u(t − ) = 0 , since Case 2: t ≥ 0. f 3(t) = K1K2e

−bt

≥ t. Hence, there are two cases to consider: t < 0

≥ 0. Hence f 3(t) = 0, for t < 0. t =  1 e (b−a)t − 1 b − a

t

[

∫ e (b−a) d 0

Therefore, for t ≥ 0,

K1K 2e −bt tu(t)  f 3(t) = K1K 2 −at − e −bt  b − a e

[

]

]

if

a= b

if

a≠ b

if a = b if a ≠ b

(c) By replacing K1 = 50, K 2 = 20 and a = 10 in the formula of f 3(t) in part (a) the answer for part (i) is easily obtained as t<0 0, f 3(t) =  −10t t, t ≥ 0 1000e t

For part (ii) the parameters have the following values: vc (t) =

∫

]−∞ = 1,

t e −(t − ) d = e −t

−∞

K 2 = 0.2, a = 10 and b = 0.2. Using these values in the formula developed in part (b) for f 3(t) the answer follows immediately f 3(t) = 0.102 e −0.2t − e −10t u(t)

(

)

SOLUTION 16.4. (a) Using the impulse response theorem and the definition of the convolution integral the response of the system, y(t), can be computed as follows ∞

y(t) =

∫

−∞

∞

h(t − )v( )d =

∫ 2e −2(t − ) u(t − )[u(

− 1)− u( − 3)]d

−∞

Observing that u( − 1)− u( − 3) is nonzero only when 1 ≤ < 3 yields

1/25/02

P16-6


3

y(t) = 2∫ e 2( −t) u(t − )d 1

The integrand in the above equation is nonzero only when ≤ t. This suggests three regions of consideration: t < 1, 1 ≤ t ≤ 3, and 3 < t. Case 1: t < 1. Here u(t − ) = 0 , since is restricted to the interval [1,3]. Hence y(t) = 0, for t < 1. Case 2: 1 ≤ t ≤ 3. t

]1

t y(t) = 2∫ e 2( −t) d = e 2( −t) = 1− e 2(1−t ) , for 1 ≤ t ≤ 3 1

Case 3: 3 < t.

3

]1

3 y(t) = 2∫ e 2( −t) d = e 2( −t) = e 2(1−t ) (e 4 − 1), for 3 < t 1

In sum, 0, t<1  2(1−t ) y(t) = 1− e , 1≤ t ≤ 3 e 2(1−t ) (e 4 − 1), 3 < t  A picture of y(t) is sketched in the next figure.

(b) Using the impulse response theorem and the definition of the convolution integral the response of the system, y(t), can be computed as follows ∞

y(t) =

∫

−∞

∞

h(t − )v( )d =

∫ 2e −2(t − ) u(t −

)u(2 − t + )[ u( − 1)− u( − 3)]d

−∞

Observing that u( − 1)− u( − 3) is nonzero only when 1 ≤ < 3 yields 3

y(t) = 2∫ e 2( −t) u(t − )u(2 − t + )d 1

The integrand in the above equation is nonzero only when t − 2 ≤ ≤ t. This suggests four regions of consideration: t < 1, 1 ≤ t ≤ 3, 3 < t ≤ 5and 5 < t. Case 1: t < 1. Here u(t − ) = 0 , since is restricted to the interval [1,3]. Hence y(t) = 0, for t < 1. Case 2: 1 ≤ t ≤ 3. Here u(t − )u(2 − t + ) is nonzero only when 1 ≤ ≤ t. Therefore,

1/25/02

P16-7

t


]1

t y(t) = 2∫ e 2( −t) d = e 2( −t) = 1− e 2(1−t ) , for 1 ≤ t ≤ 3. 1

Case 3: 3 < t ≤ 5. Here u(t − )u(2 − t + ) is nonzero only when t − 2 ≤ ≤ 3. Therefore, 3

y(t) = 2

∫

t −2

]t −2 = e2(3−t) − e−4 , for 3 < t ≤ 5.

3 e 2( −t) d = e 2( −t)

Case 4: 5 < t. Here u(t − )u(2 − t + ) = 0 , since is restricted to the interval [1,3]. Therefore, y(t) = 0, for 5 < t. A picture of y(t) is sketched in the next figure.

(c) By the impulse response theorem, the zero-state response of the circuit y(t) is y(t) = h(t) ∗v(t) Using the definition of the convolution integral and the sifting property of delta function it follows that 0, t < 0 2, 0 ≤ t < 1  −2, 1≤ t < 2 y(t) = 2h(t) − 2h(t − 1) + h(t − 2) =  0, 2 ≤ t < 3 1, 3≤ t < 4  0, 4 ≤ t Using the waveform of h(t) given in figure P16.4, y(t) is sketched in the next picture.

1/25/02

P16-8

SOLUTION 16.5. (a) By definition

∞

f 4 (t) =


∞

∫

f1(t − ) f 2( )d =

−∞

∫

(t − − 2) ⋅2 ⋅ u( + 1)d

−∞

By the sifting property of delta function it follows that f 4 (t) = 2u( + 1)] =t −2 = 2u(t − 1). (b) By the definition of convolution and the sifting property of delta function ∞

f 5(t) =

∫

−∞

∞

f1(t − ) f 3( )d =

=e

−2

u( )

(c) By definition ∞

f 6(t) =

∫

−∞

(t − − 2)e −2 u( )d

∫

−∞ −2(t −2)

] =t −2 = e

u(t − 2)

∞

f 2 (t − ) f 3( )d =

∫ 2u(t −

+ 1)e

−2

−∞

∞

u( )d = 2 ∫ e −2 u(t − + 1)d 0

The integrand in the above equation is nonzero only when ≤ t + 1. This suggests two regions of consideration: t < −1and −1≤ t . Case 1: t < −1. Here u(t − + 1)= 0, since 0 ≤ . Hence f 6(t) = 0 , for t < −1. Case 2: −1≤ t . Here u(t − + 1) is nonzero only when ≤ t + 1. Therefore, t +1

f 6(t) = 2

∫

e −2 d = e −2

0

It follows that

[

]0

t +1

= 1− e −2(t +1)

]

f 6(t) = 1− e −2(t +1) u(t + 1)

(d) By the definition of convolution and the sifting property of delta function we have

1/25/02

P16-9

∞

f 7(t) =

∫

−∞

∞

f1 (t − ) f 3( − 2)d = =e

−2( −2)


u( − 2)


(t − − 2)e −2( −2) u( − 2)d =

∫

−∞

] =t −2 = e−2(t −4)u(t − 4) .

[

]

f 6(t) = 1− e −2(t +1) u(t + 1)

Here observe that u(t – τ) = 0 for τ > t. Hence, there are two cases to consider: t ≤ 0 and t > 0. Case 1: t ≤ 0

t e aτ  e at a y(t ) = K ∫ e τdτ = K K = a  a −∞ −∞ t

Case 2: t > 0

t

y(t) = K

e aτ dτ = K

−∞

(b) By definition y(t) = K

∫

e aτ dτ = K

−∞

∞

∫

0

0

∫

u(τ − t)e

−aτ

−∞

e aτ  K  = a  −∞ a

∞

− aτ

u(τ)dτ = K ∫ u(τ)e

dτ

t

Here observe that u(τ – t) = 0 for τ < t; hence the lower limit of integration is t. Also, because of the presence of u(τ) in the integrand, there are two cases to consider: t < 0 and t ≥ 0. Case 1: t < 0

−aτ  ∞

∞

e y(t) = K ∫ e −aτ dτ = K −a 0

Case 2: t ≥ 0

∞

y(t) = K ∫ e −aτ dτ = K t

K  = a  0

∞

e− aτ  K −at  = e − a  t a

SOLUTION 16.7. (a) Using the definition of the convolution integral and the sifting property of delta function, f 5(t)can be computed as below ∞

f 5(t) =

∫

−∞

∞

f 2 (t − ) f 4 ( )d = =e

−a(t − )

u(t − )

∫

e −a(t − ) u(t − ) ( − 4)d =

−∞

] =4 = e−a(t −4)u(t − 4)

A picture of f 5(t), for a = 1, is sketched in the next figure.

1/25/02

P16-10

(b) By definition

∞

f 6(t) =

∫

−∞


∞

f1 (t − ) f1( )d = K

2

∫

∞

u(t − )u( )d = K

−∞

2

∫ u(t −

)d

0

Since u(t − ) is nonzero only when ≤ t, there are two regions of consideration: t < 0and 0 ≤ t. Case 1: t < 0. Here u(t − ) = 0 , since ≥ 0. Hence f 6(t) = 0 , for t < 0. Case 2: 0 ≤ t. t

f 6(t) = K

2

∫d

= K 2t , for 0 ≤ t.

0

A picture of f 6(t) , for K = 1, is sketched in the next figure.

(c) By definition f 7(t) =

∞

∫

−∞

∞

f1 (t − ) f 2 ( )d =

∫ Ku(t −

)e

−∞

−a

∞

u( )d = K ∫ e −a u(t − )d 0

Since u(t − ) is nonzero only when ≤ t, there are two regions of consideration: t < 0and 0 ≤ t. Case 1: t < 0. Here u(t − ) = 0 , since ≥ 0. Therefore, f 7(t) = 0 , for t < 0. Case 2: 0 ≤ t. t

f 7(t) = K ∫ e −a d = 0

]

(

)

−K −a t K e = 1− e −at , for 0 ≤ t. a 0 a

1/25/02

P16-11


A picture of f 7(t) , for t < 0 and a = 1, is sketched in the next figure.

(d) By definition f8 (t) =

∞

∫

−∞

∞

f1(t − ) f 3( )d =

∫ Ku(t −

t

Case 2: 0 < t.

∫ ea

u(t − )d =

−∞

∫ ea

−∞

u(− )d = K

∫

e a u(t − )d

−∞

≤ t . This suggests two regions of consideration:

]

K a t K e = e at , for t ≤ 0. a −∞ a

0

f8 (t) = K

)e

−∞

The integrand, u (t − ) , is nonzero only when t ≤ 0 and 0 < t . Case 1: t ≤ 0 . f8 (t) = K

0

a

u(t − )d =

]

K a 0 K e = , for 0 < t. a −∞ a

In sum, K at a e , t≤ 0 f8 (t) = K  , 0< t a A picture of f8 (t), for K = 1 and a = 1, is sketched in the next figure.

1/25/02

P16-12


SOLUTION 16.8. (a) Using the current division formula Ic (s) =

Cs

1 Iin (s) R By Ohm’s law the Laplace transform of capacitor’s voltage 1 Vc (s) = I (s) Cs c Therefore the transfer function of the circuit Cs +

V (s) 1 1 H(s) = c = = 1 Iin (s) Cs + s+ 4 R Taking the inverse Laplace transform of H(s) yields the impulse response h(t) = e −4t u(t) . (b) By the impulse response theorem ∞

vc (t) = iin (t) ∗ h(t) =

∫

−∞

∞

iin (t − )h( )d = ∞

∫

3e −(t − ) u(t − )e −4 u( )d =

−∞

= 3∫ e −(t + 3 ) u(t − )d 0

The integrand is nonzero only when ≤ t. Therefore there are two regions of consideration: t < 0and 0 ≤ t. Case 1: t < 0. Here u(t − ) = 0 , since 0 ≤ . Hence vc (t) = 0 , for t < 0. Case 2: 0 ≤ t.

1/25/02

P16-13

t


]0

t vc (t) = 3∫ e −(t + 3 ) d = −e −(t + 3 ) = e −t − e −4t , for 0 ≤ t. 0

In sum,

(

)

vc (t) = e −t − e −4t u(t)V.

SOLUTION 16.9. (a) By voltage division formula Vout (s) =

1 Cs R+

1 Vin (s) Cs

Therefore the transfer function 1 Cs

V (s) 1 H(s) = out = = 1 Vin (s) R + s +1 Cs Taking the inverse Laplace transform of H(s) yields yields the impulse response h(t) = e −t u(t). By the impulse response theorem and the convolution definition ∞

vout (t) =

∫

−∞

∞

h(t − )vin ( )d = 0

=

∫

e

−(t − )

−∞

∫

[

]

e −(t − ) u(t − ) u(− ) + 2e −2 u( ) d =

−∞

∞

u(t − )d + 2∫ e −(t + ) u(t − )d 0

For both integrals the integrand is nonzero only when ≤ t. This suggests two regions of consideration: t < 0and 0 ≤ t. Case 1: t < 0. Here the second integral is zero since, for this integral, is restricted to [0,∞). t

vc (t) = Case 2: 0 ≤ t.

∫

]−∞ = 1, for t < 0.

t e −(t − ) d = e −t

−∞ 0

vc (t) =

]−∞

]0

∫

e

−(t − )

−∞

t

d + 2∫ e −(t + ) d 0

(

)

0 t = e −t − 2e −(t − ) = e −t − 2 e −2t − e −t = 3e −t − 2e −2t , for 0 ≤ t

(c) By the impulse response theorem and the definition of convolution

1/25/02

P16-14

∞

vout (t) = 0

=

∫

−∞

=e

∫

−∞

∞

h(t − )vin ( )d =

∫


e −(t − ) u(t − )e −a| |d

−∞ ∞ −(t − )

e −(t − ) u(t − )e a d + ∫ e

−t

0

0

∫e

(a+1)

u(t − )d + e

−∞

−t

u(t − )e −a d

∞

∫e

(1−a)

u(t − )d

0

For both integrals the integrand is nonzero only when ≤ t. This suggests two regions of consideration: t < 0and 0 ≤ t. Case 1: t < 0. Here the second integral is zero since, for this integral, is restricted to [0,∞). vout (t) = e Case 2: 0 ≤ t.

−t

t

∫

e

(a+1)

d =e

−t e

−∞

vout (t) = e

−t

(a+1)  t

 a + 1  −∞

0

∫

e

(a+1)

d +e

−t

−∞

e at = , for t < 0. a +1

t

∫e

(1−a)

d =

0 −t

0 t t e (a+1)  e −t (1−a) −t =e  +e ∫e d = + e ∫ e (1−a) d a + 1  a +1 −t

−∞

0

0

Here observe that a + 1 is nonzero since a > 0. For computing the second integral, in case 2, we need to distinguish two subcases: a = 1 and a ≠ 1. t e −t t if a = 1 −t (1−a) e ∫e d = 1 −at −t if a ≠ 1 1− a e − e 0 Therefore, for 0 ≤ t,  e −t + e −t t if a = 1  a + 1 vout (t) =  −t  e + 1 e −at − e −t if a ≠ 1  a + 1 1− a

(

)

(

)

SOLUTION 16.10. (a) The impulse response is obtained by taking the inverse Laplace transform of the transfer function h(t) = −2e −0.2t u(t) By the impulse response theorem the response y(t)equals

1/25/02

P16-15


∞

y(t) = h(t) ∗v(t) =

∫

h(t − )v( )d

−∞

Substituting v(t) = u(t + 1)− u(t − 1) in the above integral yields ∞

y(t) =

∫

h(t − )[u( + 1)− u( − 1)]d

−∞

Here observe that u( + 1)− u( − 1) is nonzero only when −1≤ ≤ 1. Hence 1

y(t) =

∫ h(t −

−1

The integrand is nonzero only when −1≤ t < 1 and 1 ≤ t. Case 1: t < −1. y(t) = 0 Case 2: −1≤ t < 1. t

1

)d = −2 ∫ e −0.2( t − ) u(t − )d −1

≤ t. This suggests three regions of consideration: t < −1,

[

]

y(t) = −2 ∫ e −0.2( t − ) d = 10 e −0.2(t +1) − 1 , for −1≤ t < 1. Case 3: 1 ≤ t.

−1 1

(

)

y(t) = −2 ∫ e −0.2( t − ) d = 10e −0.2t e −0.2 − e 0.2 , for 1 ≤ t. −1

(b) The transfer function of the leaky integrator (see equation 14.14 in the textbook) is given by 1 R1 H(s) = 1 Cs + R2 −

where R2 is the leakage resistance of the capacitor C and R1 is the resistance connected at the inverting input of the op amp. Equating the two expressions of H(s) we obtain that 1 − R1 −2 = 1 s + 0.2 Cs + R2 Matching the coefficients and taking into account that the smallest resistor is 10kΩ the following values are obtained: R1 = 10kΩ, R2 = 100kΩ and C = 5⋅10−5 F . (c) The impulse response is obtained by taking the inverse Laplace transform of the transfer function h(t) = Ke −at u(t) By the impulse response theorem the response y(t)equals

1/25/02

P16-16


∞

y(t) = h(t) ∗v(t) =

∫

h(t − )v( )d

−∞

Substituting v(t) = u(t + T ) − u(t − T ) in the above integral yields ∞

∫

y(t) =

h(t − )[u( + T ) − u( − T )]d

−∞

Here observe that u( + T ) − u( − T ) is nonzero only when −T ≤ ≤ T . Hence T

y(t) =

∫ h(t −

T

)d = K

−T

y(t) = K Case 3: T ≤ t.

∫

−T

∫

−T

In sum,

u(t − )d

−T

[

]

K e −a(t − ) d = 1− e −a(t +T ) , for −T ≤ t < T . a

T

y(t) = K

)

≤ t. This suggests three regions of consideration: t < −T ,

The integrand is nonzero only when −T ≤ t < T and T ≤ t. Case 1: t < −T . y(t) = 0 Case 2: −T ≤ t < T . t

∫ e −a(t −

(

)

K e −a(t − ) d = e −at e aT − e −aT , for T ≤ t. a

 0, t < −T K y(t) =  1− e −a(t +T ) , −T ≤ t < T a K e −at e aT − e −aT , T ≤ t  a

[

(

]

)

SOLUTION 16.11. (a) First observe, from figure P16.11(a), that f 2(t) = (−2t + 4)[u(t) − u(t − 2)] By definition ∞

f 3(t) =

∫

−∞

∞

f1(t − ) f 2( )d =

∫

−∞

2

4u(t − )(−2 + 4)[ u( ) − u( − 2)]d = −8∫ ( − 2)u(t − )d 0

The integrand is nonzero only when ≤ t. This suggests three regions of consideration: t < 0, 0 ≤ t < 2 and 2 ≤ t. Case 1: t < 0. Here u(t − ) = 0 due to the fact that is restricted to the interval [0,2]. Hence f 3(t) = 0, for t < 0.

1/25/02

P16-17


Case 2: 0 ≤ t < 2. t  2  f 3(t) = −8 ∫ ( − 2)d = −8 − 2   = −4 t 2 − 4t , for 0 ≤ t < 2.  2   0 0 t

Case 3: 2 ≤ t.

(

)

2  2  f 3(t) = −8 ∫ ( − 2)d = −8 − 2   = 16, for 2 ≤ t.  2   0 0 2

In sum, 0, t<0  2 f 3(t) = −4 t − 4t , 0 ≤ t < 2 16, 2≤ t  A picture of f 3(t) is sketched in the next figure.

(

)

(b) First observe, from figure P16.11(b), that f 2(t) = t[u(t) − u(t − 2)] + (4 − t)[u(t − 2) − u(t − 4)]

By definition

∞

f 3(t) =

∫

−∞

f1(t − ) f 2( )d

By replacing f1(t) and f 2(t) with their expressions we have ∞

f 3(t) = ∞

=

∫

∫

−∞

4u(t − ){ [u( ) − u( − 2)] + (4 − )[u( − 2) − u( − 4)]}d =

4u(t − ) [u( ) − u( − 2)]d +

−∞

∞

∫

4u(t − )(4 − )[u( − 2) − u( − 4) ]d =

−∞ 2

4

0

2

= 4 ∫ u(t − )d + 4 ∫ (4 − )u(t − )d

1/25/02

P16-18


The integrands are nonzero only when ≤ t. This suggests four regions of consideration: t < 0, 0 ≤ t < 2, 2 ≤ t < 4, and 4 ≤ t. Case 1: t < 0. Here u(t − ) = 0 due to the fact that is restricted to the interval [0,4]. Hence f 3(t) = 0, for t < 0. Case 2: 0 ≤ t < 2. here observe that the second integral is zero since, for this integral, to the interval [2,4]. Therefore

is restricted

t

f 3(t) = 4 ∫ d = 2t 2 , for 0 ≤ t < 2. 0

Case 3: 2 ≤ t < 4.

t

2   f 3(t) = 4 ∫ d + 4∫ (4 − )d = 8 + 4  4 −   = −2t 2 + 16t − 16, for 2 ≤ t < 4. 2    0 2 2 Case 4: 4 ≤ t. 2

t

4

2   f 3(t) = 4 ∫ d + 4∫ (4 − )d = 8 + 4  4 −   = 16, for 4 ≤ t. 2    0 2 2 2

4

In sum, 0, 2t 2,  f 3(t) =  2 −2t + 16t − 16, 16, A picture of f 3(t) is sketched in the next figure.

SOLUTION 16.12. (a) By voltage division,

t<0 0≤ t < 2 2≤ t < 4 4≤t

1/25/02

P16-19


2 Vout 1 2/3 2/3 s H(s) = = = = − 2 2 s + 2.5s + 1 (s + 0.5) (s + 2) Vin 2s + 5 + s Hence, the impulse response is 2 2 h(t) = e −0.5tu(t) − e −2tu(t) 3 3 (b) By definition h(t) * vin (t) =

∞

∫

h(t −τ )vin (τ) dτ =

−∞

20 3

∞

∫

e −0.5(t −τ)u(t −τ ) eτ u(−τ) dτ

−∞

20 − 3

∞

∫e

−2(t −τ)

u(t −τ ) eτ u(−τ) dτ

−∞

Case 1: t ≤ 0.

20 vout (t) = 3

t

∫

e

20 e dτ − 3

−0.5(t −τ) τ

−∞

t

∫e

20 e dτ = 3

−2(t− τ) τ

−∞

t

∫e

−0.5t 1.5τ

e

−∞

20 dτ − 3

t

∫e

−∞

20e −0.5t 1.5τ t 20e−2t 3τ t = e − e = 4.444e t − 2.222et = 2.222e t −∞ −∞ 4.5 9

[ ]

[ ]

Case 2: t > 0. vout (t) =

20e −0.5t 1.5τ 0 20e −2t 3τ 0 20e −0.5t 20e −2t e − e = − −∞ −∞ 4.5 9 4.5 9

[ ]

[ ]

SOLUTION 16.13. (a) The impulse response of the circuit has been computed in problem 16.12 h(t) =

2 −0.5t 2 e (t) − e −2t u(t) 3 3

By the impulse response theorem and the convolution definition ∞

vout (t) = h(t) ∗vin (t) =

∫

−∞

h(t − )vin ( )d =

−2t 3τ

e dτ

1/25/02

P16-20

20 = 3

∞

∫ [e −0.5(t −

)


]

− e −2(t − ) u(t − )e −| |d

−∞

The integrand is nonzero only when

≤ t. Hence t

∫ [e −0.5(t −

20 vout (t) = 3

)

]

− e −2(t − ) e −| |d

−∞

The existence of the function e −| | under the integral suggests two regions of consideration: t ≤ 0 and 0 < t. Case 1: t ≤ 0. t

∫ [e −0.5(t −

20 vout (t) = 3 20 −0.5t = e 3

−∞ t

)

]

− e −2(t − ) e d =

20 d − e −2t 3

∫e

1.5

−∞

[ ]

t

∫

e3 d =

−∞

[ ]

t 20 −0.5t 1.5 t 20 e e − e −2t e 3 = 4.5 −∞ 9 −∞ = 4.444e t − 2.222e t = 2.222e t , for t ≤ 0.

= Case 2 : 0 < t. 20 vout (t) = 3 20 = 3

0

∫ [e

−∞ 0

∫ [e

−0.5(t − )

−0.5t +1.5

−e

−2(t − )

−e

−2t + 3

−∞

]

t

[

]

20 e d + e −0.5(t − ) − e −2(t − ) e − d = ∫ 3

]

0

t

[

]

20 d + e −0.5t −0.5 − e −2t + d = 3 ∫ 0

0

t

 20  e −0.5t +1.5 e −2t + 3  20  e −0.5t −0.5 =  −  +  − e −2t +  = 3  1.5 3  3  −0.5  0 −∞ = 17.778e −0.5t − 20e −t + 4.444e −2t , for 0 < t.

In sum, 2.222e t , 0≤ t vout (t) =  −0.5t −t −2t 17.778e − 20e + 4.444e , 0 < t

SOLUTION 16.14. (a) The impulse response of the circuit is obtained by taking the inverse Laplace transform of H(s) h(t) = 2e −t − 2e −2t + 4e −4t u(t)

(

(b) The result follows from the following MATLAB code:

)

1/25/02

P16-21


>> p = [-1,-2,-4]; >> r = [2,-2,4]; >> k = 0; >> [n,d] = residue(r,p,k) n= 4

14

16

7

14

d= 1

8

Therefore, 4s2 + 14s + 16 H(s) = 3 s + 7s2 + 14s + 8 (c) By the impulse response theorem

[

y(t) = u(t) ∗ h(t) = u(t)* 2e −t u(t) − 2e −2t u(t) + 4e −4t u(t)

]

Using the distributive property of convolution we have

[

]

[

]

[

]

y(t) = u(t) ∗ 2e −t u(t) + u(t) ∗ −2e −2t u(t) + u(t) ∗ 4e −4t u(t) In problem P16.2(c) the following equation has been obtained

[K1u(t)] ∗ [K 2e −at u(t)] = K1aK2 (1− e −at )u(t) Using the above equation y(t)is immediately obtained

(

)

(

)

(

)

y(t) = 2 1− e −t u(t) − 1− e −2t u(t) + 1− e −4t u(t) =

(

)

= 2 − 2e −t + e −2t − e −4t u(t) (d) By the impulse response theorem

[

y(t) = f (t) ∗ h(t) = [8u(− t) − 8u(− t − T )] * 2e −t u(t) − 2e −2t u(t) + 4e −4t u(t)

]

Using the distributive property of convolution we have

[

]

y(t) = 8u(− t) ∗ 2e −t u(t) − 2e −2t u(t) + 4e −4t u(t) − We denote

[

−t

−8u(−t − T )* 2e u(t) − 2e −2t u(t) + 4e −4t u(t)

]

1/25/02

P16-22


[

y1(t) = 8u(−t) ∗ 2e −t u(t) − 2e −2t u(t) + 4e −4t u(t)

]

By the time invariance property it follows that y(t) = y1(t) − y1(t − T ) In order to compute y1(t) we will use the following equation which has been obtained in problem 16.2(e) K1K 2 t≤0  a , −at [K1u(−t)] ∗ K 2e u(t) = K1K 2 −at  e , t>0  a Therefore, t≤0 16, y1(t) =  −t −2t −4t + 8e , 0 < t 16e − 8e

[

]

The zero-state response to the input f (t) can now be computed 0, t≤0  −t −2t −4t y(t) = 16e − 8e + 8e − 16, 0< t≤T 16 e −t − e −(t −T ) − 8 e −2t − e −2(t −T ) + 8 e −4t − e −4(t −T ) , T < t 

[

] [

] [

]

SOLUTION 16.15. (a) Using the convolution theorem the transfer function of the cascade is H(s) = L [h(t)] = L[ h1(t) ∗ h2 (t) ∗ h3(t)] = L[ h1(t)] ⋅ L[ h2(t)] ⋅ L[ h3(t)] = H1 (s)⋅ H 2 (s)⋅ H 3(s) From table 13.1 1 s 10 H2 (s) = s+2 2 H3(s) = 2 s H1(s) =

Therefore, 20 H(s) = 3 s (s + 2) A partial fraction expansion of H(s) can be obtained using the residue command in MATLAB: >> num = [20]; >> den = [1 2 0 0 0]; >> [r,p,k] = residue(num,den)

1/25/02

P16-23


r= -2.5000 2.5000 -5.0000 10.0000 p= -2 0 0 0 k= [] Hence H(s) =

−2.5 2.5 −5 10 + + 2 + 3 s+2 s s s

Taking the inverse Laplace transform yields the impulse response of the cascade h(t) = −2.5e −2t u(t) + 2.5u(t) − 5tu(t) + 5t 2 u(t) (b) By the impulse response theorem and the convolution theorem, the Laplace transform of the step response of the cascade equals 20 1 20 Y(s) = H (s)⋅U (s) = 3 ⋅ = 4 s (s + 2) s s (s + 2) A partial fraction expansion of H(s) can be obtained using the residue command in MATLAB: >> num = [20]; >> den = [1 2 0 0 0 0]; >> [r,p,k] = residue(num,den) r= 1.2500 -1.2500 2.5000 -5.0000 10.0000 p= -2 0 0

1/25/02

P16-24


0 0 k= [] Hence Y(s) =

1.25 −1.25 2.5 −5 10 + + 2 + 3 + 4 s+2 s s s s

Taking the inverse Laplace transform yields the step response of the cascade y(t) = 1.25e −2t u(t) − 1.25u(t) + 2.5tu(t) − 2.5t 2 u(t) + 1.667t 3u(t) .

SOLUTION 16.16. (a) By the voltage division formula Vout (s) =

1 Cs

1 V (s) = Vin (s) i 1 CRs + 1 R+ Cs

Therefore, the transfer function of the circuit is V (s) 1 1 H(s) = out = = Vin (s) CRs + 1 s + 1 Taking the inverse Laplace transform yields the impulse response h(t) = e −t u(t). (b) From table 13.1 the Laplace transform of vin (t) is Vin (s) =

1 1 + s (s + 1)2

By the impulse response theorem and the convolution theorem it follows that Vout (s) = H (s)⋅Vin (s) =

1 1 1  1 1 = +  +  s + 1  s (s + 1)2  s(s + 1) (s + 1)3

A partial fraction expansion of Vout (s) is 1 1 1 Vout (s) = − + s s + 1 (s + 1)3 Taking the inverse Laplace transform yields

1/25/02

P16-25


vout (t) = u(t) − e −t u(t) − 0.5t 2e −t u(t). Using the time domain convolution method vout (t) can be computed as follows ∞

vout (t) = ∞

=

∫

∫

−∞

h(t − )vin ( )d =

[

]

e −(t − ) u(t − ) u( ) + e − u( ) d

−∞

From the experience earned by computing convolution integrals we know that the computation of the above integral requires more computational work than the Laplace transform method. More computations imply, of course, more sources of errors. From the solution of this problem we have seen that, in the case of the Laplace transform method, the computational burden consists in computing Laplace and inverse Laplace transforms. For a large class of functions these transforms can be found in tables(for example table 13.1). The only computation that we did, in the solution of this problem, was the partial fraction expansion of Vout (s) . (c) In this case the (unilateral) transform method cannot be used because vin (t) ≠ 0 for t < 0.

SOLUTION 16.17. (a) The impulse response can be obtained by taking the inverse Laplace transform of H(s) . Therefore h(t) = 8e −10t u(t) (b) From table 13.1 8 Vin (s) = 2 s + 16 By the impulse response theorem and the convolution theorem it follows that Vout (s) = H (s)⋅Vin (s) =

64 (s + 10)(s2 + 16)

Vout (s) can be further written as Vout (s) =

−0.5517s + 5.517 0.5517 s 4 0.5517 + = −0.5517 2 + 1.379 2 + 2 s + 10 s + 16 s + 16 s + 16 s + 10

The above expansion of Vout (s) can be obtained by using the technique of example 13.14, page 514. Taking the inverse Laplace transform yields vout (t) = −0.5517cos(4t)u(t) + 1.379sin(4 t)u(t) + 0.5517e −10t u(t).

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P16-26

(c) In this case

[


]

8 (s + 2)2 + 16

Vin (s) = L 2e −2t sin(4t)u(t) = Therefore Vout (s) = H (s)⋅Vin (s) =

[(s + 2)

64 2

]

+ 16 (s + 10)

Using again the technique of example 13.14, page 514, Vout (s) can be written as Vout (s) =

−0.8s + 4.8 0.8 s+2 4 0.8 + = −0.8 + 1.6 + 2 2 2 (s + 2) + 16 s + 10 (s + 2) + 16 (s + 2) + 16 s + 10

Taking the inverse Laplace transform yields vout (t) = −0.8e −2t cos(4t)u(t) + 1.6e −2t sin(4 t)u(t) + 0.8e −10t u(t) V. In this context the Laplace transform method is faster than the time domain convolution. This is due to the fact that vin (t) has a relatively complicated expression and the convolution integral will require more computational work than the Laplace transform method. (d) In this context the Laplace transform method cannot be used because time domain convolution method will be used to compute the response response thorem vout (t) = h(t) ∗vin (t) = 8e −10t u(t) ∗ [u(− t)]

[

vin (t) ≠ 0 for t < 0. The vout (t). By the impulse

]

Using the result of problem 16.2, part (e), it follows that t≤0 0.8, vout (t) =  −10t , 0
SOLUTION 16.18. (a) Replacing R1, R2 , C1 and C2 with their values the transfer function can be obtained s H(s) = 2 s +5s + 2 The only zero of H(s) is 0 and the poles of H(s) are –0.5 and –2. A partial fraction expansion of H(s) is: −0.167 0.667 H(s) = + s + 0.5 s + 2 The impulse response can be obtained by taking the inverse Laplace transform of H(s) h(t) = −0.167e −0.5t + 0.667e −2t

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P16-27


(b) vout (t) will be computed using the Laplace transform method. This approach is valid because h(t) and vin (t) are zero for t < 0. From table 13.1 the Lapace transform of vin (t) is 1 Vin (s) = (s + 2)2 By the impulse response theorem and the convolution theorem it follows that Vout (s) = H (s)⋅V (s) =

2s2

s 1 s = 4 2 3 + 5s + 2 (s + 2) 2s + 13s + 30s2 + 28s + 8

A partial fraction expansion of Vout (s) can be obtained using the residue command in MATLAB: >> a = [1 0]; >> b = [2 13 30 28 8]; >> [r,p,k] = residue(a,b) r= 0.0741 0.1111 0.6667 -0.0741 p= -2.0000 -2.0000 -2.0000 -0.5000 k= [] Therefore, Vout (s) =

−0.0741 0.0741 0.1111 0.6667 + + + s + 0.5 s+2 (s + 2) 2 (s + 2)3

Taking the inverse Laplace transform yields

[

]

vout (t) = −0.0741e −0.5t + 0.0741e −2t + 0.1111te−2t + 0.3333t 2e −2t u(t) V. One would prefer the time domain convolution method to compute vout (t), but the computations may require more work relatively to the Laplace transform method. (c) In this part vin (t) ≠ 0 for t < 0. Therefore the time domain convolution method will be used to compute vout (t). By the impulse response theorem ∞

=

vout (t) = h(t) ∗vin (t) =

∫ [−0.1667e −0.5(t −

−∞

)

]

+ 0.6667e −2(t − ) u(t − )e 2 u(− )d =

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P16-28

0

=

∫ [−0.1667e −0.5(t −

)


]

+ 0.6667e −2(t − ) u(t − )e 2 d

−∞

≤ t. This suggests two regions of

The integrand of the previous integral is nonzero only when consideration: t < 0 and 0 ≤ t. Case 1: t < 0. vout (t) = −0.1667e

Case 2: 0 ≤ t.

−0.5t

t

∫e

2.5

d + 0.6667e

−∞ 2.5  t

−2t

t

∫

e4 d =

−∞ 4 t

e e = −0.1667e −0.5t   + 0.6667e −2t   =  2.5  −∞  4  −∞ = 0.1e −2t , for t < 0. vout (t) = −0.1667e

−0.5t

0

∫e

2.5

d + 0.6667e

−∞ 2.5  0

−2t

0

∫

e4 d =

−∞ 4 0

e e  = −0.1667e −0.5t   + 0.6667e −2t   =  2.5  −∞  4  −∞ = −0.1667e −0.5t + 0.6667e −2t , for 0 ≤ t.

SOLUTION 16.19. Replacing Rand C with their values H(s) =

s−5 10 = 1− s+5 s+5

The zero-state response vout (t) will be computed using the time domain convolution method because vin (t) ≠ 0 for t < 0. The impulse response of the circuit is h(t) = (t) − 10e −5t u(t) By the impulse response theorem ∞

vout (t) = h(t) ∗vin (t) = ∞

∫

−∞

h( )vin (t − )d =

( ) − 10e −5 u( )]cos[10(t − [ −∞

= 10 ∫ ∞

= 10 ∫

−∞

∞

( )cos(t − )d − 10

∫ 10e −5 u(

)]d =

)cos[10(t − )]d

−∞

Using the sifting property of the delta function and expanding cos(t − )it follows that

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P16-29

∞

vout (t) = 10cos(t − ) − 100cos(10t) ∫ e 0

−5


∞

cos(10 )d − 100sin(10t) ∫ e −5 sin(10 )d 0

Using the definition of the Laplace transform we observe that ∞

∫ e −5 0

and

∞

∫ e −5 0

Therefore

s  cos(10 )d = L [cos(10t)u(t)]s=5 = 2 = 0.04  s + 100  s =5 10  sin(10 )d = L[sin(10t)u(t)]s=5 = 2 = 0.08  s + 100  s =5 vout (t) = 10cos(10t) − 4cos(10t) − 8sin(10t)

[

]

= 6cos(t) − 8sin(10t) = 10cos 10t + tan−1 ( 43 ) Notice that vout (t) and vin (t) have the same frequency and magnitude.

SOLUTION 16.20. (a) First notice that vin (t − T) = u(t). Therefore w(t) = u(t) and 1 W (s) = s (b) Using the properties of the Laplace transform it follows that w Vout (s) = H (s)⋅W (s) =

2 s(s + 2)

w A partial fraction expansion of Vout (s) is

1 1 w Vout (s) = − s s+2 Taking the inverse Laplace transform yields w vout (t) = u(t) − e −2t u(t) (b) Since vin (t) = w(t + T ) it follows, by the time invariance property, that

Therefore,

v w vout (t) = vout (t + T ). v vout (t) = u(t + T ) − e −2(t +T ) u(t + T ) V.

SOLUTION 16.21. (a) First observe from figure P16.21 that vin (t) = u(t + T ) − u(t − T ) From the definition of w(t) it follows that

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P16-30


w(t) = vin (t − T ) = u(t) − u(t − 2T )

Therefore

1 1 −2sT − e s s (a) By the impulse response theorem and the convolution theorem it follows that W (s) =

(

)

(

)

2 1  1 1− e −2sT =  − 1− e −2sT =  s(s + 2) s s+2 1 1 1  −2sT 1 = − − − e s s+2 s s + 2

w Vout (s) = H (s)⋅W (s) =

Taking the inverse Laplace transform and using the time shift property of the Laplace transform yields w vout (t) = 1− e −2t u(t) − 1− e −2(t −2T ) u(t − 2T ) Because vin (t) = w(t + T )

(

)

[

]

it follows, by the time invariance property, that

(

v w vout (t) = vout (t + T ) =

)

[

]

= 1− e −2(t +T ) u(t + T ) − 1− e −2(t −T ) u(t − T ) V.

SOLUTION 16.22. (a) The use of t = t + T1 in the problem statement means replace t by t + T1 . However, strictly speaking we should have used a statement of the form t = t' +T1 which is done in the proof below. By definition of the convolution and the property of commutivity,

f (t − T1) * g(t) =

∞    f (t − T −τ )g(τ) dτ = f (t' −τ)g(τ) dτ = [ f (t' ) * g(t' )]t' =t −T 1 ∫ 1 ∫  −∞  −∞  t' =t −T1 ∞

Observe that t = t' +T1 . Hence

[ f (t − T1 )* g(t)]t =t' +T1 = f (t' ) * g(t' ) Realizing that t and t' are simply dummy variables, we immediately obtain the result. From a systems perspective, this corresponds to the property of time-invariance where a shift of an input function by T1 yields a corresponding shift of the output function by T1 . (b) The steps in this part are similar to those of part (a).

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P16-31

f (t − T1) * g(t − T2 ) =

∞

∫


f (t − T1 −τ )g(τ − T2 ) dτ

−∞

Let λ = τ − T2 in which case τ = λ + T2 and dλ = dτ . Hence ∞  f (t − T1) * g(t − T2 ) = ∫ f (t − T1 − T2 − λ )g(λ)dλ =  ∫ f (t' −λ)g(λ )dλ    −∞  −∞  t' =t −T1 −T2 ∞

= [ f (t' )* g(t') ]t' =t −T −T 1

2

Since t = t' +T1 + T2 , and t and t' are dummy variables, we have

[ f (t − T1 )* g(t − T2 )]t =t' +T1 +T2 =

f (t' )* g(t')

and the result follows.

SOLUTION 16.23. From table 13.1, G(s) =

1 1 , H(s) = . s+2 (s + 2) 2

(a) Define p(t) = f(t – 2) = u(t). 1 Then, P(s) = . Consider s P(s)G(s) =

1 0.5 0.5 = − s(s + 2) s s+2

Hence p(t) * g(t) = 0.5u(t) − 0.5e −2tu(t) From problem 16.22 part (a),

[

f (t)* g(t) = [ p(t) * g(t)]t =t +2 = 0.5u(t) − 0.5e −2tu(t)

]t =t +2 = [0.5 − 0.5e−2(t +2) ]u(t + 2)

(b) Define p(t) = f(t – 2) = u(t). 1 Then, P(s) = . Consider s P(s)H(s) = Hence

1 0.25 0.25 0.5 2 = s − s+ 2− s(s + 2) (s + 2)2

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P16-32

(


)

p(t) * h(t) = 0.25 − 0.25e −2t − 0.5te−2t u(t) From problem 16.22 part (a),

[(

) ]t =t +2

f (t)* h(t) = [ p(t)* h(t)]t =t +2 = 0.25 − 0.25e −2t − 0.5te−2t u(t)

[

]

= 0.25 − 0.25e −2(t +2) − 0.5(t + 2)e −2(t +2) u(t + 2)

SOLUTION 16.24. (a) The pictures of f (t) and g(t) are sketched in the next figures

(b) Using the convolution theorem it follows that L[ f (t − 2 ) ∗ g(t − )] = L[ f (t − 2 )] ⋅L [g(t − )] From table 13.1 1 L[ f (t − 2 )] = L[u(t)] = s 1 L[ g(t − )] = L[sin(t)u(t)] = 2 s +1 Therefore 1 1 s L[ f (t − 2 ) ∗ g(t − )] = = − 2 2 s(s + 1) s s + 1 Taking the inverse Laplace transform yields f (t − 2 ) ∗ g(t − ) = u(t) − cos(t)u(t) = [1− cos(t)]u(t) Using the property give in problem 16.22(b) it follows that f (t) ∗ g(t) = [ f (t − 2 ) ∗ g(t − )]t =t +2 + = [1− cos(t + 3 )]u(t + 3 )

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P16-33

SOLUTION 16.25. Define


w(t) = vin (t − 2)

Hence

w(t) = u(t)

and, from table 13.1, W (s) =

1 s

From table 13.1 we also have that H1(s) = The impulse response of the cascade is

1 1 and H2 (s) = s +1 (s + 1)2 h(t) = h1 (t) ∗ h2 (t)

Hence the transfer function of the cascade is 1 (s + 1)3 w We denote by vout (t)the zero state response due to the input w(t). Hence, 1 w Vout (s) = H (s)⋅W (s) = s(s + 1)3 w A partial fraction expansion of Vout (s) is obtained using the residue command in MATLAB: H(s) = H1 (s)H 2 (s) =

>> a = [1]; >> b = [1 3 3 1 0]; >> [r,p,k] = residue(a,b) r= -1.0000 -1.0000 -1.0000 1.0000 p= -1.0000 -1.0000 -1.0000 0 k= [] Therefore w Vout (s) =

1 −1 −1 −1 + + + s s + 1 (s + 1)2 (s + 1)3

Taking the inverse Laplace transform yields

1/25/02

P16-34


[

]

w vout (t) = 1− e −t − te−t − 0.5t 2e −t u(t) V

Due to the fact that

vin (t) = w(t + 2) the time invariance property implies that w vout (t) = vout (t + 2) =

[

]

= 1− e −(t + 2) − (t + 2)e −(t +2) − 0.5(t + 2) 2 e −(t + 2) u(t + 2) V.

SOLUTION 16.26. (a) Using the sifting property of the delta function it follows that f 4 (t) = [ (t) + (t − 4)] ∗ f 2 (t) = f 2 (t) + f 2 (t − 4) The right-hand side of the above equation interprets as a graphical sum of shifted pictures of f 2(t) . A picture of f 4 (t) is sketched in the next figure. 4 2 0

0

1

2

3

4

5

6

7

(b) In order to compute the area beneath f 2(t − ) ⋅ f 2 ( ) four regions will be considered: t < 0, 0 ≤ t < 1, 1 ≤ t < 2 and 2 ≤ t. Step 1: t < 0. In this case f 2(t − ) ⋅ f 2 ( ) = 0 for all . Therefore f 2(t) ∗ f 2 (t) = 0 for t < 0. Step 2: 0 ≤ t < 1. In this case f 2(t − ) ⋅ f 2 ( ) = 16 for 0 ≤ ≤ t and is zero elsewhere. The area beneath f 2(t − ) ⋅ f 2 ( ) equals 16t . Therefore f 2(t) ∗ f 2 (t) = 16t for 0 ≤ t < 1. Step 3: 1 ≤ t < 2. In this case f 2(t − ) ⋅ f 2 ( ) = 16 for t − 1 < ≤ 1 and is zero elsewhere. Hence the area beneath f 2(t − ) ⋅ f 2 ( )equals f 2(t) ∗ f 2 (t) = 16(2 − t) for 1 ≤ t < 2. Step 4: 2 ≤ t. In this case f 2(t − ) ⋅ f 2 ( ) = 0 for all . Therefore f 2(t) ∗ f 2 (t) = 0 for2 ≤ t. A picture of f 5(t)is sketched in the next figure.

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P16-35


(c) In order to compute the area beneath f 2(t − ) ⋅ f 3( ) five regions will be considered: t < 0, 0 ≤ t < 1, 1 ≤ t < 2, 2 ≤ t < 3 and 3 ≤ t. Step 1: t < 0. In this case f 2(t − ) ⋅ f 3( ) = 0 for all . Therefore f 2(t) ∗ f 3 (t) = 0 for t < 0. Step 2 : 0 ≤ t < 1. In this case f 2(t − ) ⋅ f 3( ) = 8 for 0 ≤ ≤ t and is zero elsewhere. Therefore the area beneath f 2(t − ) ⋅ f 3( ) equals Step 3: 1 ≤ t < 2. Here

f 2(t) ∗ f 3 ( ) = 8t for 0 ≤ t < 1. 8, t − 1 < < 1  f 2(t − ) ⋅ f 3( ) = 24, 1 ≤ < t 0, otherwise 

Hence, the area beneath f 2(t − ) ⋅ f 3( ) equals f 2(t) ∗ f 3 ( ) = 81 [ − (t − 1)] + 24(t − 1) = 8(2t − 1) for 1 ≤ t < 2. Step 4: 2 ≤ t < 3. In this case f 2(t − ) ⋅ f 3( ) = 24 for t − 1 < < 2 and is zero otherwise. Hence, the area beneath f 2(t − ) ⋅ f 3( ) equals f 2(t) ∗ f 3 (t) = 24[2 − (t − 1)] = 24(3 − t) for 2 ≤ t < 3. Step 5: 3 ≤ t. Here f 2(t − ) ⋅ f 3( ) = 0 for all . Therefore f 2(t) ∗ f 3 (t) = 0 for t < 0. A picture of f 6(t) is sketched in the next figure.

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P16-36


SOLUTION 16.27. By the impulse response theorem, it follows that the response is y(t) = h(t) ∗ f (t) = = h(t) ∗ [ (t) − (t − 1)] Using the distributive property of convolution and the sifting property of delta function y(t) can be written as y(t) = h(t) − h(t − 1) The right-hand side of the above equation interprets as a graphical sum of (shifted) pictures of h(t). The pictures of h(t), h(t − 1) and y(t) are sketched in the next figures.

SOLUTION 16.28. (a) From the picture of f (t) and h(t) in figure P16.28 we observe that, in order to compute the area beneath h(t − ) f ( ) , we need to consider four cases: t < 0, 0 ≤ t < 4, 4 ≤ t < 8 and 8 ≤ t. Step 1: t < 0. Here h(t − ) f ( ) = 0 for all . Therefore the area beneath h(t − ) f ( ) equals zero and h(t) ∗ f (t) = 0 for t < 0. Step 2: 0 ≤ t < 4. In this case h(t − ) f ( ) = 1 for 0 ≤ ≤ t and is zero otherwise. Hence the area beneath h(t − ) f ( ) equals h(t) ∗ f (t) = t for 0 ≤ t < 4. Step 3: 4 ≤ t < 8. In this case 1, t − 4 < ≤ 4  h(t − ) f ( ) = 2, 4 ≤ ≤ t 0, otherwise  Therefore the area beneath h(t − ) f ( ) equals h(t) ∗ f (t) = [ 4 − (t − 4)] + 2(t − 4) = t for 4 ≤ t < 8.

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P16-37


Step 4: 8 ≤ t. Here h(t − ) f ( ) = 2 for t − 4 < ≤ t and is zero otherwise. Hence h(t) ∗ f (t) = 2[ t − (t − 4) = 8] for 8 ≤ t. A picture of y(t) is sketched in the next figure.

(b) The impulse response is

h(t) = u(t) − u(t − 4)

By the impulse response theorem ∞

y(t) = x(t) ∗ h(t) =

∫

∞

x( )h(t − )d =

−∞

∫

x( )[ u(t − ) − u(t − − 4) ]d

−∞

Here observe that u(t − ) − u(t − − 4) is nonzero only when t ≤ < t − 4. Therefore t −4

y(t) =

∫

x( )d

t

which interprets as the running area under x(t) over the interval [t − 4,t].

SOLUTION 16.29. The response, y(t), is obtained as indicated in the statement of the problem, by using the following MATLAB code: >> tstep = 1; >> vin = [1]; >> h = [0, 2, 3, 1, 1]; >> y = tstep*conv(vin, h); >> y = [0 y 0]; >> t = 0:tstep:tstep*(length(vin)+length(h)); >> plot(t,y) >> grid The response is plotted in the next figure.

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P16-38


SOLUTION 16.30. A picture of vin (t) sketched in the next figure.

In order to plot the response, y(t), the MATLAB code of problem 16.29 will be used with only one modification. Namely vin = [1, 1, 2, 2] as it can be observed from the picture of vin (t) with the time step tstep = 1. Therefore the MATLAB code is: >> tstep = 1; >> vin = [1, 1, 2, 2]; >> h = [0, 2, 3, 1, 1]; >> y = tstep*conv(vin,h); >> y = [0 y 0]; >> t = 0:tstep:tstep*(length(vin)+length(h)); >> plot(t,y) >> grid The response is plotted in the next figure.

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P16-39


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P16-1


SOLUTION 16.31. In order to compute the area beneath v(t − )h( ) seven regions will be considered: t < 0, 0 ≤ t < 1, 1 ≤ t < 2, 2 ≤ t < 3, 3 ≤ t < 4, 4 ≤ t < 5 and 5 ≤ t. Step 1: t < 0. For t in this region v(t − )h( ) = 0 for all . Hence y(t) = v(t) ∗ h(t) = 0 for t < 0. Step 2: 0 ≤ t < 1. In this case v(t − )h( ) = v0 × h0 for 0 ≤ ≤ t and is zero otherwise. Therefore the area beneath v(t − )h( ) equals y(t) = v(t) ∗ h(t) = v0 × h0 × t for 0 ≤ t < 1. Step 3: 1 ≤ t < 2. For t in this region we have v1× h0, v0 × h0,  v(t − )h( ) =  v0 × h1, 0, Therefore the area beneath v(t − )h( ) equals

0 ≤ ≤ t −1 t − 1< < 1 1≤ < t otherwise

y(t) = v(t) ∗ h(t) = v1× h0 × [(t − 1)− 0] + v0 × h0 × [1− (t − 1)] + v0 × h1× (t − 1)= = t × (v1× h0 − v0 × h0 + v0 × h1)− v1× h0 + 2 × v0 × h0 − v0 × h1, for 1 ≤ t < 2 Step 4: 2 ≤ t < 3. In this case

Hence, for 2 ≤ t < 3,

v1× h0, v1× h1,  v(t − )h( ) = v0 × h1, v0 × h2,  0,

t −2< <1 1≤ ≤ t − 1 t − 1< < 2 2≤ < t otherwise

y(t) = v(t) ∗ h(t) = = v1× h0 × [1− (t − 2)] + v1× h1× [(t − 1)− 1] + v0 × h1× [2 − (t − 1)] + v0 × h2 × (t − 2) = = t × (−v1× h0 + v1× h1− v0 × h1+ v0 × h2) + 3× v1× h0 − 2 × v1× h1 + 3 × v0 × h1− 2 × v0 × h2 Step 5: 3 ≤ t < 4. In this case v1× h1, t − 2 < < 2 v1× h2, 2 ≤ ≤ t − 1  v(t − )h( ) =  v0 × h2, t − 1< < 3 0, otherwise Hence, for 3 ≤ t < 4, y(t) = v(t) ∗ h(t) = = v1× h1× [2 − (t − 2)] + v1× h2 × [(t − 1)− 2] + v0 × h2 × [ 3− (t − 1)] = = t × (−v1× h1 + v1× h2 − v0 × h2) + 4 × v1× h1− 3 × v1× h2 + 4 × v0 × h2 Step 6: 4 ≤ t < 5. In this case v(t − )h( ) = v1× h2 for t − 2 < < 3 and is zero otherwise. Therefore y(t) = v(t) ∗ h(t) = v1× h2 × [ 3− (t − 2)] = v1× h2 × (5 − t) for 4 ≤ t < 5.

1/25/02

P16-2


Step 7: 5 ≤ t. For t in this region v(t − )h( ) = 0 for all . Hence y(t) = v(t) ∗ h(t) = 0 for 5 ≤ t. In sum, v0 × h0 × t, t × (v1 × h0 − v 0 × h0 + v 0 × h1) − v1 × h0 + 2 × v 0 × h0 − v0 × h1,  t × (−v1 × h0 + v1 × h1 − v0 × h1 + v0 × h2) + 3 × v1 × h0 − 2 × v1 × h1 + 3 × v 0 × h1 − 2 × v 0 × h2, y( t) =  t × (−v1 × h1 + v1 × h2 − v0 × h2) + 4 × v1 × h1 − 3 × v1 × h2 + 4 × v 0 × h2, v1 × h2 × (5 − t),  0,

0≤ t <1 1≤ t < 2 2≤t< 3 3≤t< 4 4≤t<5 otherwise

Hence, y1 = y(1) = v0 × h0 = 6 y 2 = y(2) = v0 × h1+ v1× h0 = 8 y 3 = y(3) = v0 × h2 + v1× h1 = −6 y 4 = y(4) = v1× h2 = 4 (b) Using the expressions of p(x) and q(x) it follows that p(x) × q(x) = x 3 × (v0 × h0) + x 2 × (v0 × h1+ v1× h0) + x × (v0 × h0 + v1× h1)+ v1× h2 We observe that the coefficients of p(x) × q(x) are exactly y1 , y 2 , y 3 and y 4 , respectively. Therefore r(x) = p(x) × q(x) .

SOLUTION 16.32. (a) This part will be solved using the techniques of convolution algebra. Therefore we can write f 3(t) as f 3(t) = f1(−1) (t) ∗ f 2(1)(t) Where the superscript (-1) means integration and the superscript (1) means differentiation. From figure P16.32 we observe that f1(t) = 4[ u(t) − u(t − 4) ] Hence f1(−1) (t) = 4[ tu(t) − (t − 4)u(t − 4) ] = = 4[r(t) − r(t − 4)] By inspection, from the same figure, we have f 2(1)(t) = 4[ (t) − 2 (t − 2) + 2 (t − 4) − 2 (t − 6) + (t − 8)] Using the sifting property of the delta function f 3(t) can be computed as follows f 3(t) = {4[ r(t) − r(t − 4) ]} ∗ {4[ (t) − 2 (t − 2) + 2 (t − 4) − 2 (t − 6) + (t − 8)]} = = 16[r(t) − 2r(t − 2) + r(t − 4) − r(t − 8) + 2r(t − 10) − r(t − 12)] A picture of f 3(t) is sketched in the next figure.

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P16-3


(b) Using the techniques of problem 16.31 and considering the time step tstep = 2, the polynomials p(x) , q(x) and r(x) can be associated with the functions f1(t), f 2(t) and f 3(t), respectively, as below: p(x) = 4x + 4 q(x) = 4 x 3 − 4 x 2 + 4 x − 4 r(x) = 32x 4 − 32 We need to verify that the equality

p(x) ⋅q(x) ⋅tstep = r(x)

holds. The equality indeed holds because p(x) ⋅q(x) ⋅tstep = 32(x + 1)(x 3 − x 2 + x − 1) = 32(x 4 − 1)= r(x). The results obtained in part (a) and part (b) coincide.

SOLUTION 16.33. (a) Using the techniques of convolution algebra f 3(t) can be written as f 3(t) = f1(−1) (t) ∗ f 2(1)(t) Where the superscript (-1) means integration and the superscript (1) means differentiation. From figure P16.33 we observe that f1(t) = 2[ u(t + 1)− u(t − 4)] Therefore f1(−1) (t) = 4[(t + 1)u(t + 1)− (t − 4)u(t − 4)] = = 4[r(t + 1)− r(t − 4)] = g(t) By inspection, from the same figure, we have f 2(1)(t) = 4 (t) − 8 (t − 2) + 6 (t − 4) − 2 (t − 6) Using the sifting property of the delta function f 3(t) can be computed as follows f 3(t) = 4g(t) − 8g(t − 2) + 6g(t − 4) − 2g(t − 6)

1/25/02

P16-4


This is plotted in MATLAB as follows: >> t = -2:0.01:14; >> g = 2*(t+1).*u(t+1)-2*(t-4).*u(t-4); >> g1 = 4*g; >> g2 = -8*( 2*(t-1).*u(t-1)-2*(t-6).*u(t-6) ); >> g3 = 6*( 2*(t-3).*u(t-3)-2*(t-8).*u(t-8) ); >> g4 = -2*( 2*(t-5).*u(t-5)-2*(t-10).*u(t-10) ); >> f3 = g1+g2+g3+g4; >> plot(t,f3); >> grid; A picture of f 3(t) is sketched in the next figure.

(b) To account for the fact that f1(t)is nonzero for negative t the following formula (see problem 16.22) f1(t) ∗ f 2 (t) = [ f1(t − 1)∗ f 2 (t)]t =t +1 will be used to compute f 3(t). Using a slightly modified version of the code of problem 16.31, we have >> f1 = [2, 2, 2, 2, 2]; >> f2 = [4, 4, -4, -4, 2, 2]; >> T = 1; >> tstep = T; >> f3 = tstep*conv(f1,f2); >> f3 = [0 f3 0]; >> t = -1:tstep:tstep*(length(f1)+length(f2))-1; >> plot(t,f3) >> grid

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P16-5


The results of parts (a) and (b) coincide.

SOLUTION 16.34. This problem is solved using the techniques of the convolution algebra with the graphical method left to the student. f3 (t) = f1(t) * f2 (t) = [ f1(t)]

(−1)

* [ f2 (t)]

(1)

where the superscript (-1) means integration and the superscript (1) means differentiation. By inspection,

[ f1(t )](−1) = 4tu(t) − 4(t − 6)u(t − 6) = g(t) and

[ f2 (t)](1) = 4δ(t) − 8δ(t − 2) + 8δ(t − 6) − 4δ(t − 8) Hence the response say y(t) satisfies f3 (t) = 4g(t) − 8g(t − 2) + 8g(t − 6) − 4g(t − 8) This is plotted in MATLAB as follows: »t=0:.05:20; »g = 4*t .* u(t) - 4*(t-6) .*u(t-6); »g1=4*g; »g2 = -8*(4*(t-2) .* u(t-2) - 4*(t-8) .*u(t-8)); »g3 = 8*(4*(t-6) .* u(t-6) - 4*(t-12) .*u(t-12)); »g4 = -4*(4*(t-8) .* u(t-8) - 4*(t-14) .*u(t-14)); »f3 = g1+g2+g3+g4; »plot(t,f3) »grid

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P16-6


40 30 20 10 0 -10 -20 -30 -40

0

2

4

6

8

(b) Using the code of problem 31, we have »f1 = [4 4 4]; »f2 = [4 -4 -4 4]; »T = 2; »tstep = T; »f3 = [0 conv(f1,f2)*tstep 0]; »t = 0: tstep : tstep* (length(f1) + length(f2)); »plot(t,f3) »grid

10

12

14

16

18

20

1/25/02

P16-7


40 30 20 10 0 -10 -20 -30 -40

0

2

4

6

8

10

12

14

The results of parts (a) and (b) coincide.

SOLUTION 16.35. (a) In order to compute the area beneath f1(t − )⋅ f1 ( ) four regions will be considered: t < 0, 0 ≤ t < 1, 1 ≤ t < 2, and 2 ≤ t. Step 1: t < 0. Here f1(t − )⋅ f1 ( ) = 0 for all . Hence f 3(t) = f1 (t) ∗ f1(t) = 0 for t < 0. Step 2: 0 ≤ t < 1. In this case f1(t − )⋅ f1 ( ) = 1 for 0 ≤ ≤ t and is zero otherwise. Therefore the area beneath f1(t − )⋅ f1 ( ) equals f 3(t) = t for 0 ≤ t < 1. Step 3: 1 ≤ t < 2. In this case f1(t − )⋅ f1 ( ) = 1 for t − 1 < < 1 and is zero otherwise. Therefore the area beneath f1(t − )⋅ f1 ( ) equals f 3(t) = 1− (t − 1)= 2 − t for 1 ≤ t < 2. Step 4: 2 ≤ t. Here f1(t − )⋅ f1 ( ) = 0 for all . Hence In sum,

f 3(t) = f1 (t) ∗ f1(t) = 0 for2 ≤ t. t, 0 ≤ t <1  f 3(t) = 2 − t, 1 ≤ t < 2 0, otherwise 

1/25/02

P16-8


(b) In order to compute the area beneath f1(t − )⋅ f 2 ( ) four regions will be considered: t < 0, 0 ≤ t < 1, 1 ≤ t < 2, and 2 ≤ t. Step 1: t < 0. Here f1(t − )⋅ f 2 ( ) = 0 for all . Hence f 4 (t) = f1 (t) ∗ f 2 (t) = 0 for t < 0. Step 2: 0 ≤ t < 1. In this case f1(t − )⋅ f 2 ( ) = for 0 ≤ ≤ t and is zero otherwise. Therefore the area beneath f1(t − )⋅ f 2 ( ) equals f 4 (t) = 0.5t 2 for 0 ≤ t < 1. Step 3: 1 ≤ t < 2. For t in this region f1(t − )⋅ f 2 ( ) = for t − 1 < < 1 and is zero otherwise. Therefore the area beneath f1(t − )⋅ f 2 ( ) equals f 4 (t) = 0.5 − 0.5t 2 for 1 ≤ t < 2. Step 4: 2 ≤ t. Here f1(t − )⋅ f 2 ( ) = 0 for all . Hence f 4 (t) = f1 (t) ∗ f1(t) = 0 for2 ≤ t. In sum, 0.5t 2, 0≤ t <1  2 f 4 (t) = 0.5 − 0.5t , 1≤ t < 2 0, otherwise  (c) In order to compute the area beneath f 2(t − ) ⋅ f 2 ( ) four regions will be considered: t < 0, 0 ≤ t < 1, 1 ≤ t < 2, and 2 ≤ t. Step 1: t < 0. Here f 2(t − ) ⋅ f 2 ( ) = 0 for all . Hence f 5(t) = f 2 (t) ∗ f 2 (t) = 0 for t < 0. Step 2: 0 ≤ t < 1. In this case f 2(t − ) ⋅ f 2 ( ) = (t − ) for 0 ≤ ≤ t and is zero otherwise. Therefore the area beneath f 2(t − ) ⋅ f 2 ( ) equals t

[

]0

t f 5(t) = ∫ (t − ) d = −0.3333 3 + 0.5t 2 =0.1667t 3 for 0 ≤ t < 1. 0

Step 3: 1 ≤ t < 2. For t in this region f 2(t − ) ⋅ f 2 ( ) = (t − ) for t − 1 < < 1 and is zero otherwise. Therefore the area beneath f 2(t − ) ⋅ f 2 ( ) equals 1

f 5(t) =

∫

t −1

[

]t −1

1 (t − ) d = −0.3333 3 + 0.5t 2

= −0.1667t 3 + t − 0.6667 for 1 ≤ t < 2. Step 4: 2 ≤ t. Here f 2(t − ) ⋅ f 2 ( ) = 0 for all . Hence

1/25/02

In sum,

P16-9


f 5(t) = f 2 (t) ∗ f 2 (t) = 0 for2 ≤ t. 0.1667t 3, 0 ≤ t <1  3 f 5(t) = −0.1667t + t − 0.6667, 1 ≤ t < 2 0, otherwise 

SOLUTION 16.36. In order to compute the area beneath f1(t − )⋅ f 2 ( ) five regions will be considered: t < 0, 0 ≤ t < 1, 1 ≤ t < 2, 2 ≤ t < 3and 3 ≤ t. Step 1: t < 0. Here f1(t − )⋅ f 2 ( ) = 0 for all . Hence f 3(t) = f1 (t) ∗ f 2 (t) = 0 for t < 0. Step 2: 0 ≤ t < 1. In this case f1(t − )⋅ f 2 ( ) = for 0 ≤ ≤ t and is zero otherwise. Therefore the area beneath f1(t − )⋅ f 2 ( ) equals f 3(t) = 0.5t 2 for 0 ≤ t < 1. Step 3: 1 ≤ t < 2. For t in this region

 , t − 1< < 1  f1(t − )⋅ f 2 ( ) = 2 − , 1≤ ≤ t 0, otherwise 

Therefore the area beneath f1(t − )⋅ f 2 ( ) equals

[

] [

f 3(t) = 0.5 − 0.5(t − 1)2 + 0.5 − 0.5(2 − t)2 2

= −t + 3t − 1.5 for 1 ≤ t < 2.

]

Step 4: 2 ≤ t < 3. For t in this region f1(t − )⋅ f 2 ( ) = 2 − for all t − 1 < < 2. Hence f 3(t) = f1 (t) ∗ f 2 (t) = 0.5 − 0.5(2 − t) 2 = −0.5t 2 + 2t − 1.5 for 2 ≤ t < 3. Step 5: 3 ≤ t. Here f1(t − )⋅ f 2 ( ) = 0 for all . Hence f 3(t) = f1 (t) ∗ f 2 (t) = 0 for 3 ≤ t. In sum, 0.5t 2,  2 − t + 3t − 1.5, f 3(t) =  2 −0.5t + 2t − 1.5, 0, 

SOLUTION 16.37.

0≤ t <1 1≤ t < 2 2≤ t < 3 otherwise

1/25/02

P16-10


(a) In order to compute the area beneath f1( )⋅ f 2 (t − ) three regions will be considered: t < 0, 0 ≤ t < 2, and 2 ≤ t. Step 1: t < 0. Here f1( )⋅ f 2 (t − ) = 0 for all . Hence f 3(t) = f1 (t) ∗ f 2 (t) = 0 for t < 0. Step 2: 0 ≤ t < 2. In this case f1( )⋅ f 2 (t − ) = 8 for 0 ≤ ≤ t and is zero otherwise. Therefore the area beneath f1( )⋅ f 2 (t − ) equals f 3(t) = 0.5(t ⋅ 8t) = 4t 2 for 0 ≤ t < 2. Step 3: 2 ≤ t. For t in this region

8 , 0 ≤ < 2  f1( )⋅ f 2 (t − ) = 16, 2 ≤ < t 0, otherwise  Therefore the area beneath f1( )⋅ f 2 (t − ) equals In sum,

f 3(t) = 16 + 16(t − 2) = 16(t − 1) for 2 ≤ t. 4t 2, 0≤ t < 2  f 3(t) = 16(t − 1), 2 ≤ t 0, otherwise 

A picture of f 3(t) is sketched in the next figure.

(b) From figure P16.37 we observe that f1(t) = 2tu(t) − 2(t − 2)u(t − 2) From table 13.1 and the time shift property of the Laplace transform it follows that

(

2 F1(s) = 2 1− e −2s s

)

1/25/02

P16-11

F2 (s) = By the convolution theorem


4 s

F3(s) = F1 (s)F2 (s)

Therefore

(

8 F3(s) = 3 1− e −2s s Taking the inverse Laplace transform yields

)

f 3(t) = 4t 2 u(t) − 4(t − 2) 2 u(t − 2) The results of parts (a) and (b) coincide.

SOLUTION 16.38. (a) In order to compute the area beneath f1(t − )⋅ f 2 ( ) six regions will be considered: t < 0, 0 ≤ t < 2, 2 ≤ t < 6, 6 ≤ t < 8, 8 ≤ t < 10, and 10 ≤ t. Step 1: t < 0. Here f1(t − )⋅ f 2 ( ) = 0 for all . Hence f 3(t) = f1 (t) ∗ f 2 (t) = 0 for t < 0. Step 2: 0 ≤ t < 2. In this case f1(t − )⋅ f 2 ( ) = 8(t − ) for 0 ≤ ≤ t and is zero otherwise. Therefore the area beneath f1( )⋅ f 2 (t − ) equals f 3(t) = 0.5(t ⋅ 8t) = 4t 2 for 0 ≤ t < 2. Step 3:2 ≤ t < 6. For t in this region

16, 0≤ < t −2  f1(t − )⋅ f 2 ( ) = 8(t − ), t − 2 ≤ < t 0, otherwise 

Therefore the area beneath f1(t − )⋅ f 2 ( ) equals f 3(t) = 16 + 16(t − 2) = 16(t − 1) for 2 ≤ t < 6. Step 4: 6 ≤ t < 8. For t in this region

16, 8(t − ),  f1(t − )⋅ f 2 ( ) =  −8(t − ), 0,

0 ≤ < t− 2 t− 2 ≤ < 6 6≤
Therefore, for 6 ≤ t < 8, the area beneath f1(t − )⋅ f 2 ( ) equals  8(t − 6)2  8(t − 6) 2 f 3(t) = 16(t − 2) + 16 − − 2 2  

1/25/02

P16-12


= −8(t − 6)2 + 16t − 16 = −8t 2 + 112t − 304 Step 5: 8 ≤ t < 10. For t in this region

0≤ <6 16, −16, 6 ≤ < t− 2  f1(t − )⋅ f 2 ( ) =  −8(t − ), t − 2 ≤ < 8 0, otherwise

Therefore, for 8 ≤ t < 10, the area beneath f1(t − )⋅ f 2 ( ) equals  8(t − 8) 2  f 3(t) = 96 − 16(t − 8) − 16 −  = 4t 2 − 80t + 464 2   Step 6: 10 ≤ t. For t in this region

Therefore, In sum,

16, 0 ≤ < 6  f1(t − ) f 2 ( ) = −16, 6 ≤ < 8 0, otherwise  f 3(t) = 96 − 32 = 64 for 10 ≤ t.

0, 4t 2,  16t − 16, f 3(t) =  2 −8t + 112t − 304, 4t 2 − 80t + 464,  64, A picture of f 3(t) is sketched in the next figure.

(b) From figure P16.38 we observe that

t<0 0≤ t < 2 2≤ t < 6 6≤ t < 8 8 ≤ t < 10 10 ≤ t

1/25/02

P16-13


f 2(t) = 4[u(t) − u(t − 6)] − 4[ u(t − 6)u − (t − 8)] = 4u(t) − 8u(t − 6) + 4u(t − 8) From table 13.1 and the time shift property of the Laplace transform it follows that 4 8 4 F2 (s) = − e −6s + e −8s s s s From problem 16.37 we have that 2 F1(s) = 2 1− e −2s s By the convolution theorem F3(s) = F1 (s)F2 (s) Therefore 8 8 16 24 8 F3(s) = 3 − 3 e −2s − 3 e −6s + 3 e −8s − 3 e −10s s s s s s

(

)

Taking the inverse Laplace transform yields f 3(t) = 4t 2 u(t) − 4(t − 2) 2 u(t − 2) − 8(t − 6)2 u(t − 6) +12(t − 8)2 u(t − 8) − 4(t − 10)2 u(t − 10) The results of parts (a) and (b) coincide.

SOLUTION 16.39. p(t) ∗ q(t) will be computed using the techniques of convolution algebra. Therefore we can write p(t) ∗ q(t) = p (−1)(t) ∗ q(1) (t) where the superscript (−1)means integration and the superscript (1)means differentiation. By inspection p(t) = (t + 4)[ u(t + 4) − u(t)] + (− t + 4)[u(t) − u(t − 4)] + 4[u(t − 4) − u(t − 8)] = = (t + 4)u(t + 4) + (−2t)u(t) + tu(t − 4) − 4u(t − 8) Therefore, p(−1) (t) = 0.5(t + 4)2 u(t + 4) − t 2u(t) + (0.5t 2 − 8)u(t − 4) − (4t − 32)u(t − 8) By inspection we also have q (1) (t) = 4 (t) By the sifting property of the delta function it follows that p(t) ∗ q(t) = 4 p (−1) (t) = = 2(t + 4) 2 u(t + 4) − 4t 2u(t) + 2(t 2 − 16)u(t − 4) − 16(t − 8)u(t − 8).

SOLUTION 16.40. (a) First observe that

1/25/02

P16-14


h(t) = 0.1u(t − 0.1) + 0.2u(t − 0.2) + 0.2u(t − 0.3) + 0.2u(t − 0.4) + 0.2u(t − 0.5) + +0.1u(t − 0.6) − 0.1u(t − 1)− 0.2u(t − 1.3)− 0.2u(t − 1.5) − −0.2u(t − 1.7) − 0.2u(t − 2) − 0.1u(t − 2.2) Due to the fact that h(t) is a linear combination of terms of the type Ku(t − T ), the convolution of h(t) and vin (t) reduces to a linear combination of terms of the following type: [K1u(t)] ∗ [K 2u(t − T )] . Using the definition of the convolution, the previous convolution product is computed below ∞

[K1u(t)] ∗ [K 2u(t − T )] = K1K 2 ∫

u( )u(t − − T )d =

−∞

t
vout (0) = 0 V.

At t = 0.5s

vout (0.5) = 16 V.

At t = 1s

vout (1) = 65 V.

At t = 1.5s

vout (1.5)= 106 V. (b) In this case vout (t) will be computed using the techniques of the convolution algebra. Hence we have (−1) vout (t) = vin (t) ∗ h(t) = v in (t) ∗ h (1) (t)

[

]

= 50t 2 u(t) ∗ [0.1 (t − 0.1) + 0.2 (t − 0.2) + 0.2 (t − 0.3) + 0.2 (t − 0.4) + +0.2 (t − 0.5) + 0.1 (t − 0.6) − 0.1 (t − 1)− 0.2 (t − 1.3)− −0.2 (t − 1.5) − 0.2 (t − 1.7)− 0.2 (t − 2) − 0.1 (t − 2.2)]

Using the sifting property of delta function it follows that vout (t ) = 5(t − 0.1) 2 u (t − 0.1) + 10(t − 0.2) 2 u (t − 0.2) + 10(t − 0.3) 2 u (t − 0.3) +

+10(t − 0.4) 2 u(t − 0.4) + 10(t − 0.5)2 u(t − 0.5) + 5(t − 0.6) 2 u(t − 0.6) − −5(t − 1)2 u(t − 1)− 10(t − 1.3)2 u(t − 1.3) − 10(t − 1.5)2 u(t − 1.5) − −10(t − 1.7) 2 u(t − 1.7) − 10(t − 2)2 u(t − 2) − 5(t − 2.2)2 u(t − 2.2) V.

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At t = 1s

P16-15


vout (1) = 22.25 V.

(c) From the expression of vout (t)obtained in part (a) we observe that vout (t)does not change after t = 2.2s . Therefore it is sufficient to compute vout (t)for t ≤ 2.2s . Hence vin (t) can be considered to be equal to vin (t) = 100[u(t) − u(t − 2.2)] V. Using the code of problem 16.31 we have >> vin = 100*ones(1,22); >> h = [0, 0.1, 0.3, 0.5, 0.7, 0.9, 1, 1, 1, 1, 0.9, 0.9, 0.9, 0.7, 0.7, 0.5, 0.5, 0.3, 0.3, … …0.3, 0.1, 0.1]; >> T = 0.1; >> tstep = T; >> y = tstep*conv(vin,h); >> y = [0 y 0]; >> t = 0:tstep:tstep*(length(h)+length(vin)); % After t = 2.2s vout(t) does not change >> t = t(1:length(h)+1); >> y = y(1:length(h)+1); >> plot(t,y) >> grid A picture of y(t) is sketched in the next figure.

Using the previous MATLAB code we the values of y(t) at the specified instants of time are: At t = 0s y(0) = 0 V

1/25/02

P16-16

At t = 0.5s At t = 1s At t = 1.5s


y(0.5) = 16 V y(1) = 65 V y(1.5)= 106 V

The results of parts (a) and (c) coincide.

SOLUTION 16.41. Using the MATLAB code of problem 16.31 we have: >> vin = [1]; >> h = [9, -6, 3, -2]; >> T = 1; >> tstep = T; >> y = tstep*conv(vin,h); >> y = [0 y 0]; >> t = 0:tstep:tstep*(length(h)+length(vin)); >> plot(t,y) >> grid The breakpoints in y(t) of the above figure are [9, -6, 3 –2] as expected because the polynomial associated with vin (t) is the constant 1 and the polynomial associated with h(t) is the polynomial 9x 3 − 6x 2 + 3x − 2, as it can be observed from figure P16.41.

SOLUTION 16.42. (a) Let vout ,40 (t) denote the response that has been obtained in problem 16.40, part (a), to the input 100u(t) . The expression of vout ,40 (t) is (see problem 16.40, part (a)): vout ,40 (t) = h(t) ∗ [100u(t)] = = 10(t − 0.1)u(t − 0.1)+ 20(t − 0.2)u(t − 0.2) + 20(t − 0.3)u(t − 0.3) + 20(t − 0.4)u(t − 0.4) +

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P16-17


+20(t − 0.5)u(t − 0.5) + 10(t − 0.6)u(t − 0.6) − 10(t − 1)u(t − 1)− 20(t − 1.3)u(t − 1.3) − −20(t − 1.5)u(t − 1.5) − 20(t − 1.7)u(t − 1.7) − 20(t − 2)u(t − 2) − 10(t − 2.2)u(t − 2.2) Using the distributive property of the convolution product and the time invariance property it follows that vout (t) = h(t) ∗vin (t) = h(t) ∗ [100u(t) − 100u(t − 0.2)] = = h(t) ∗ [100u(t)] − h(t) ∗ [100u(t − 0.2)] = = vout,40 (t) − vout,40 (t − 0.2) Using the above expression of vout ,40 (t) we have: vout ,40 (0) = 0 V and vout ,40 (−0.2) = 0V, vout ,40 (0.5) = 16 V and vout ,40 (0.3) = 4 V, vout ,40 (1) = 65 V and vout ,40 (0.8) = 45V, vout ,40 (1.5) = 106 V and vout ,40 (1.3)= 92V. At t = 0s At t = 0.5s At t = 1s At t = 1.5s

vout (0) = vout ,40 (0) − vout ,40 (−0.2) = 0 V. vout (0.5) = vout ,40 (0.5) − vout ,40 (0.3) = 12 V. vout (1) = vout ,40 (1) − vout ,40 (0.8) = 20 V. vout (1.5)= vout ,40 (1.5)− vout ,40 (1.3) = 14 V.

(b) In this case vout (t) will be computed using the techniques of convolution algebra. We have (−1) vout (t) = vin (t) ∗ h(t) = v in (t) ∗ h (1) (t) where the superscript (-1) means integration and the superscript (1) means differentiation. From figure P16.42 observe that vin (t) = 100t[u(t) − u(t − 0.5)] + 100(1− t)[u(t − 0.5) − u(t − 1)] = = 100tu(t) + 100(1− 2t)u(t − 0.5) + 100(t − 1)u(t − 1) Therefore vin(−1) (t) = 50t 2u(t) + (−100t 2 + 100t − 25)u(t − 0.5) + (50t 2 − 100t + 50)u(t − 1) = g(t) By the sifting property of the delta function we have (−1) vout (t) = vin (t) ∗ h (1)(t) = g(t) ∗ h (1) (t)

= g(t) ∗ [0.1 (t − 0.1) + 0.2 (t − 0.2) + 0.2 (t − 0.3) + 0.2 (t − 0.4) + +0.2 (t − 0.5) + 0.1 (t − 0.6) − 0.1 (t − 1)− 0.2 (t − 1.3)− −0.2 (t − 1.5) − 0.2 (t − 1.7)− 0.2 (t − 2) − 0.1 (t − 2.2)] = = 0.1g(t − 0.1)+ 0.2g(t − 0.2) + 0.2g(t − 0.3) + 0.2g(t − 0.4) + +0.2g(t − 0.5) + 0.1g(t − 0.6) − 0.1g(t − 1)− 0.2g(t − 1.3)−

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P16-18


−0.2g(t − 1.5) − 0.2g(t − 1.7) − 0.2g(t − 2) − 0.1g(t − 2.2) The values ofvout (t) at the specified instants of time can be computed using MATLAB. The results are: vout (0) = 0V vout (0.5) = 2.2V vout (1) = 17.85V vout (1.5)= 23.3V.

SOLUTION 16.43. Using the techniques of convolution algebra we have vout (t) = h(t) ∗vin (t) = h (−1) (t) ∗vin(1) (t) where the superscript (-1) means integration and the superscript (1) means differentiation. We have h (−1) (t) = 2(1− e −2t )u(t) and vin(1) (t) = (t) − (t − 1) Using the sifting property of the delta function it follows that vout (t) = 2(1− e −2t )u(t) ∗ [ (t) − (t − 1)] =

[

−2t

]

= 2(1− e )u(t) − 2(1− e −2(t −1) )u(t − 1) A picture of vout (t)is sketched in the next figure.

SOLUTION 16.44. (a) From table 13.1 it follows that the Laplace transform of vin (t) is 1 1 1 −s2 + 2 Vin (s) = − − = s s + 1 s + 2 s(s + 1)(s + 2)

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P16-19


And the Laplace transform of vout (t) is 1 2 1 1 −s2 + 2 Vout (s) = − − + = s s + 1 (s + 1)2 s + 2 s(s + 1)2(s + 2) Therefore the transfer function of the circuit is V (s) 1 H(s) = out = Vin (s) s + 1 A simple RC circuit that represents this transfer function is a series RC circuit with R = 1Ω and C = 1F . vout (t)is represented by the capacitor voltage and vin (t) is the source voltage. (b) The impulse response of the circuit is  1  −t h(t) = L−1[ H (s)] = L−1  = e u(t).  s + 1  (c) Assuming zero initial conditions we have

[

]

(1) vout (t) = vin (t) ∗ h(t) = v in (t) ∗ h (−1) (t) = (t) ∗ (1− e −t )u(t) = (1− e −t )u(t) V.

(d) Using the techniques of convolution algebra the zero-state response can be computed as (2) vout (t) = vin (t) ∗ h(t) = v in (t) ∗ h (−2) (t) where the superscript (2) means double differentiation and the superscript (-2) means double integration. First, from figure P16.44, observe that vin(1) (t) = [ u(t − 1)− u(t − 2)] + [ u(t − 3)− u(t − 4) ] Therefore vin(2)(t) = (t − 1)− (t − 2) + (t − 3) − (t − 4) h (−2) (t) is computed as the integral of h (−1) (t) . t

h

(−2)

(t) =

∫

t

h

(−1)

( )d =

−∞

∫

(1− e

−

t

)u( )d =

−∞

∫− (1− e −

)d = (t + e −t − 1)u(t)

0

The zero-state response can now be computed (2) vout (t) = vin (t) ∗ h (−2) (t) =

[

= [ (t − 1)− (t − 2) + (t − 3)− (t − 4) ] ∗ (t + e −t − 1)u(t) By the sifting property of the delta function it follows that

(

)

( ) +( t − 4 + e −(t −3) ) u(t − 3)− ( t − 5 + e −(t −4) ) u(t − 4) V.

vout (t) = t − 2 + e −(t −1) u(t − 1)− t − 3 + e −(t −2) u(t − 2)

]

1/25/02

P16-20


SOLUTION 16.45. Using the techniques of the convolution algebra we have y(t) = f (t) ∗ g(t) = f (2) (t) ∗ g(−2) (t) where t

(−1)

g

2

(t) =

∫− cos(

)d = sin( t)u(t)

0

and t

(−2)

g

(t) =

2

∫− sin(

)d = [1− cos( t)]u(t)

0

Differentiating f (t) twice leads to f (2) (t) = − (t) + 2 (t − 1)− 2 (t − 3)+ (t − 4) Therefore y(t) = f (2) (t) ∗ g(−2) (t) = = [− (t) + 2 (t − 1)− 2 (t − 3) + (t − 4)] ∗ {[1− cos( t)]u(t)} Using the sifting property of the delta function it follows that y(t) = −[1− cos( t)]u(t) + 2{1− cos[ (t − 1)]}u(t − 1) −2{1− cos[ (t − 3)]}u(t − 3) + {1− cos[ (t − 4)]}u(t − 4) Simplifying the expression of y(t) yields y(t) = −[1− cos( t)] ⋅ [u(t) − u(t − 4)] + 2 ⋅ [1 + cos( t)] ⋅ [ u(t − 1)− u(t − 3)] . A picture of y(t) is sketched in the next figure.

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P16-21


SOLUTION 16.46. (a) By the current division formula IC (s) =

s2 ⋅ I (s) = 2 + 1 ⋅ Iin (s) 1 in s Cs + Ls Cs

The transfer function of the circuit can be computed as below V (s) IC (s) ⋅ ZC (s) s2 1 s H(s) = C = = 2 ⋅ = 2 Iin (s) Iin (s) s +1 s s +1 (b) The impulse response is computed as the inverse Laplace transform of the transfer function h(t) = L−1[ H (s)] = cos(t)u(t). (c) Assuming zero initial conditions it follows, by the impulse response theorem, that vout (t) = iin (t) ∗ h(t) Using the techniques of the convolution algebra we have vout (t) = iin(2) (t) ∗ h (−2) (t) By inspection

(1) iin (t) = [u(t) − u(t − 2 )] + [ u(t − 4 ) − u(t − 6 )]

Therefore (2) iin (t) = (t) − (t − 2 ) + (t − 4 ) − (t − 6 )

And t

h

(−1)

(t) =

∫− cos(

)d = sin(t)u(t)

0

Hence t

h

(−2)

(t) =

∫− sin(

)d = [1− cos(t)]u(t)

0

Using the sifting property of the delta function we have vout (t) = [ (t) − (t − 2 ) + (t − 4 ) − (t − 6 )] ∗ {[1− cos(t)]u(t)} = [1− cos(t)]u(t) − [1− cos(t − 2 )]u(t − 2 ) +[1− cos(t − 4 )]u(t − 4 ) − [1− cos(t − 6 )]u(t − 6 ) = [1− cos(t)] ⋅ [u(t) − u(t − 2 ) + u(t − 4 ) − u(t − 6 )]

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P16-22


V. (d) A picture of vout (t) is sketched in the next figure.

SOLUTION 16.47. (a) The step response, vout (t), is computed using the convolution algebra techniques. We have (−1) vout (t) = h(t) ∗vin (t) = h (1)(t) ∗ vin (t) From figure P16.47 observe that h(t) = 2u(t) − u(t − 1)− 2u(t − 2) − u(t − 3)+ u(t − 5) + 2u(t − 6) + u(t − 7) − 2u(t − 8) Differentiating we have h (1) (t) = 2 (t) − (t − 1)− 2 (t − 2) − (t − 3) + (t − 5) + 2 (t − 6) + (t − 7) − 2 (t − 8) Since by integration it follows that

vin (t) = u(t), vin(−1) (t) = tu(t).

Using the sifting property of the delta function it follows that (−1) vout (t) = h (1) (t) ∗ vin (t) = [2 (t) − (t − 1)− 2 (t − 2) − (t − 3) + (t − 5) + 2 (t − 6) + (t − 7) − 2 (t − 8)] ∗ [tu(t)] = 2tu(t) − (t − 1)u(t − 1)− 2(t − 2)u(t − 2) − (t − 3)u(t − 3) + +(t − 5)u(t − 5) + 2(t − 6)u(t − 6) + (t − 7)u(t − 7) − 2(t − 8)u(t − 8)V.

A picture of the step response is sketched in the next figure.

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P16-23


(b) Using the convolution algebra techniques we have (−1) vout (t) = h(t) ∗vin (t) = h (1)(t) ∗ vin (t)

For computing vin(−1) (t) we have

t ∫ e d , t<0 t  e t , t < 0 vin(−1) (t) = ∫ e u(− )d = −∞ =  0 1, 0 ≤ t  −∞  ∫ e d , 0≤ t −∞

Using the sifting property of the delta function it follows that (−1) vout (t) = h (1) (t) ∗ vin (t) =

[

]

(−1) = [2 (t) − (t − 1)− 2 (t − 2) − (t − 3) + (t − 5) + 2 (t − 6) + (t − 7) − 2 (t − 8)] ∗ vin (t) =

Using the

(−1) (−1) (−1) (−1) = 2v in (t) − vin (t − 1)− 2vin (t − 2) − vin (t − 3) + (−1) (−1) (−1) +v in (t − 5) + 2vin(−1) (t − 6) + vin (t − 7) − 2v in (t − 8) (−1) expression of vin (t) computed above it follows that

vout (7.5) = 0.7869 V, vout (6.5) = 1.1603V, vout (5.5) = 0.2720V, vout (0.5) = 0.8848V.

SOLUTION 16.48. First observe, from figure P16.48(a), that h(t) = (1− t)⋅ [ u(t) − u(t − 1)] =

V.

1/25/02

P16-24


= (1− t)u(t) − (1− t)u(t − 1). Using the convolution algebra techniques it follows that the step response y(t) is y(t) = v(t) ∗ h(t) = v (1) (t) ∗ h (−1)(t) = = (t) ∗ h (−1) (t) = h (−1) (t) where t

h

(−1)

=

(t) =

[

∫

−∞

− 0.5

t

(1− )u( )d −

]

2 t 0 2

⋅ u(t) −

[

∫ (1−

−∞

)u( − 1)d =

]1

t − 0.5 2 ⋅ u(t − 1) =

= (t − 0.5t )u(t) − (t − 0.5t 2 − 0.5)u(t − 1). Hence the step response is y(t) = (t − 0.5t 2 )u(t) − (t − 0.5t 2 − 0.5)u(t − 1). Let y v (t) denote the zero-state response to the input v(t). From figure P16.48(b) we observe that v(t) = u(t) + u(t − 1)− 2u(t − 2) . Using the distributive property of the convolution it follows that y v (t) = h(t) ∗ v(t) = = h(t) ∗ [ u(t) + u(t − 1)− 2u(t − 2)] = = h(t) ∗ u(t) + h(t) ∗ u(t − 1)− 2 ⋅ h(t) ∗ u(t − 2). Due to the fact that y(t) = h(t) ∗ u(t) by the linearity and time invariance properties it follows that y v (t) = y(t) + y(t − 1)− 2y(t − 2) = = (t − 0.5t 2 )u(t) + (t − 1)u(t − 1)+ +(1.5t 2 − 8t + 12)u(t − 2) + (0.5t 2 − 3t + 4)u(t − 3).

SOLUTION 16.49. (a) By the voltage division formula it follows that 1 1 Vout (s) = 1 Cs ⋅Vin (s) = 2 + 1 ⋅Vin (s) LCs + Ls Cs Therefore the transfer function is V (s) 1 H(s) = out = Vin (s) LCs 2 + 1 Taking the inverse Laplace transform yields 1     1  1 −1 LC  = 1 sin 1 t u(t) . h(t) = L−1  = L   LC LC  LC   LCs 2 + 1   s2 + 1   LC  (b) The step response is computed as the convolution of the impulse response and the step function. vout (t) = h(t) ∗ u(t) Using the techniques of the convolution algebra it follows that

1/25/02

P16-25


vout (t) = h (−1) (t) ∗ u(1) (t) where the superscript (-1) means integration and the superscript (1) means differentiation. Taking the integral of h(t) we have t

h

(−1)

(t) =

1  1  sin  u( )d = LC  LC  −∞

∫

t

   1   1  = u(t)⋅  − cos   = 1− cos t ⋅ u(t) .  LC   0   LC    Therefore vout (t) = h (−1) (t) ∗ u(1) (t) =    1  = 1− cos t  ⋅ u(t) ∗ (t) =  LC        1  = 1− cos t ⋅ u(t) V.  LC    T (c) We denote by vout (t) the output to the rectangular pulse in figure P16.49(b). Observe, from figure P16.49(b), that 1 vin (t) = ⋅ [u(t) − u(t − T )] V. T By linearity and time invariance it follows that 1 T vout (t) = ⋅ [v out (t) − vout (t − T )] T where vout (t)is the step response obtained in part (b). Therefore T vout (t) =

  1  1 t  ⋅ u(t) − 1− cos LC  LC   2 LC

1 2

=

  1  t  ⋅ u(t) − u(t − 2 1− cos LC  LC  

1 2

  1  t − 2   ⋅ u(t − 2 1− cos  LC 

[

T A picture of vout (t), for L = 1H and C = 1F , is sketched in the next figure.

LC

]

LC )

1/25/02

P16-26


SOLUTION 16.50. The impulse response of the configuration in the figure P16.50 is h(t) = h1 (t) ∗ [h2 (t) + h3(t)] ∗ h4 (t) Due to the fact that h4 (t) = 2 (t) , the sifting property of the delta function can be applied and it follows that h(t) = 2⋅ h1 (t) ∗ [h2 (t) + h3(t)] Using the techniques of the convolution algebra we can further write h(t) = 2⋅ h1(1) (t) ∗ [h2 (t) + h3(t)] h(t) =

2⋅ h1(1) (t) ∗

We have

∫

=

h3(−1) (t) =

[

2e −2 u( )d = u(t) ⋅ −e −2

−∞ t

∫

=

h2(−1) (t) + h3(−1)(t)

]

h1(1) (t) = u (1)(t) = (t) t

h2(−1) (t)

[

(−1)

[

8e −4 u( )d = u(t)⋅ −2e −4

−∞

]0 = (1− e−2t )u(t) t

]0 = 2(1− e−4t )u(t) t

Substituting the expressions of h1(−1) (t) , h2(−1) (t) and h3(−1) (t) in the expression of h(t), and using the sifting property of the delta function we have

(

)

(

)

h(t) = 2 1− e −2t u(t) + 4 1− e −4t u(t). A picture of h(t) is sketched in the next figure.

1/25/02

P16-27


SOLUTION 16.51. Observe first that h2(t) , h3(t) and h4 (t) have the same expressions as in the problem 16.50. In problem 16.50 the following convolution has been computed

(

)

(

)

u(t) ∗ [ h2 (t) + h3(t)] ∗ h4 (t) = 2 1− e −2t u(t) + 4 1− e −4t u(t) Using the time shift property of the convolution (see problem 16.22, part (a)) it follows that

(

)

(

)

u(t − 2) ∗ [ h2 (t) + h3(t)] ∗ h4 (t) = 2 1− e −2(t −2) u(t − 2) + 4 1− e −4( t −2) u(t − 2) Using the above expressions and the distributive property of the convolution it follows that the overall impulse response can be computed as below h(t) = h1 (t) ∗ [h2 (t) + h3(t)] ∗ h4 (t) = [u(t) − u(t − 2)] ∗ [h2 (t) + h3 (t)] ∗ h4 (t) = u(t) ∗ [h2 (t) + h3 (t)] ∗ h4 (t) − u(t − 2)∗ [ h2 (t) + h3(t)] ∗ h4 (t) =

(

)

(

)

(

)

(

)

= 21 − e −2t u(t) + 4 1− e −4t u(t) − 2 1− e −2(t −2) u(t − 2) + 4 1− e −4(t −2) u(t − 2).

A picture of h(t) is sketched in the next figure.

1/25/02

P16-28


SOLUTION 16.52. The overall impulse response of the configuration is h(t) = h1 (t) ∗ [h2 (t) + h3(t)] ∗ h4 (t) Using the distributive property of the convolution we have h(t) = h1 (t) ∗ h2 (t) ∗ h4 (t) + h1(t) ∗ h3 (t) ∗ h4 (t) Replacing the expressions for h2(t) and h3(t), and using the sifting property of the delta function it follows that h(t) = 2⋅ h1 (t) ∗ (t) ∗ h4 (t) − 2⋅ h1 (t) ∗ (t − 2) ∗ h4 (t) = 2⋅ h1(t) ∗ h4 (t) − 2⋅ [ h1(t) ∗ h4 (t)]t =t −2 Using the techniques of the convolution algebra it follows that h1(t) ∗ h4 (t) = h1(1)(t) ∗ h4(−1) (t) where the superscript (1) means differentiation and the superscript (-1) means integration. We have h1(1) (t) = u (1)(t) = (t) and t

h4(−1) (t) Therefore

=

∫

−∞

t

−

[

2e u( )d = 2u(t)∫ e − d = 2u(t) −e − 0

[

]

]0 = 2(1− e−t )u(t) t

h1(t) ∗ h4 (t) = (t) ∗ 2(1− e −t )u(t) = 2(1− e −t )u(t)

1/25/02

P16-29


Replacing the above expression into the expression of h(t) it follows that

[

h(t) = 4(1− e −t )u(t) − 4 (1− e −t )u(t)

]t =t −2 = 4(1− e−t )u(t) − 41[ − e−(t −2)]u(t − 2).

A picture of h(t) is sketched in the next picture.

SOLUTION 16.53. (a) The overall impulse response of the configuration in figure P16.53 is h(t) = [ h1(t) + h2 (t)] ∗ h3 (t) By the distributive property of convolution it follows that h(t) = h1 (t) ∗ h3(t) + h2 (t) ∗ h3(t) = (t) ∗cos( t)u(t) + (t − 1)∗cos( t)u(t) By the sifting property of the delta function it follows that h(t) = cos( t)u(t) + cos[ (t − 1)]u(t − 1). A picture of h(t) is sketched in the next figure.

1/25/02

P16-30


(b) The response y(t) is computed as y(t) = h(t) ∗ u(t) = {cos( t)u(t) + cos[ (t − 1)]u(t − 1)} ∗ u(t) Using the distributive property and the time shift property of convolution(see problem 16.22, part (a) ) we have y(t) = [cos( t)u(t)] ∗ u(t) + {cos[ (t − 1)]u(t − 1)} ∗ u(t) = = [cos( t)u(t)] ∗ u(t) + {[cos( t)u(t)] ∗ u(t)}t =t −1 Using the techniques of convolution algebra, the convolution cos( t)u(t) ∗ u(t) is computed as

[cos(

t)u(t)] ∗ u(t) = [cos( t)u(t)](−1) ∗ u (1)(t)

where the superscript (-1) means integration and the superscript (1) means differentiation. We have

[cos(

t)u(t)]

t

(−1)

=

∫ cos(

t

)u( )d

−∞

Therefore

[cos(

)d = u(t)

1

[sin( t)]t0

=

sin( t)

0

sin( t)  sin( t)  t)u(t)] ∗ u(t) =  u(t) ∗ (t) = u(t)  

Hence the step response is y(t) =

= u(t)∫ cos(

sin( t)

u(t) +

sin[ (t − 1)]

u(t − 1) =

sin( t)

[ u(t) − u(t − 1)] .

SOLUTION 16.54. (a) The overall impulse response of the configuration in figure P16.53 is

u(t) .

1/25/02

P16-31


h(t) = [ h1(t) + h2 (t)] ∗ h3 (t) By the distributive property of convolution it follows that h(t) = h1 (t) ∗ h3(t) + h2 (t) ∗ h3(t) = (t) ∗cos( t)u(t) + (t − 3) ∗cos( t)u(t) By the sifting property of the delta function it follows that h(t) = cos( t)u(t) + cos[ (t − 3)]u(t − 3). A picture of h(t) is sketched in the next figure.

(b) The response y(t) is computed as y(t) = h(t) ∗ u(t) = {cos( t)u(t) + cos[ (t − 3)]u(t − 3)} ∗ u(t) Using the distributive property and the time shift property of convolution(see problem 16.22, part (a) we have y(t) = [cos( t)u(t)] ∗ u(t) + {cos[ (t − 3)]u(t − 3)} ∗ u(t) = [cos( t)u(t)] ∗ u(t) + {[cos( t)u(t)] ∗ u(t)}t =t −3 Using the techniques of convolution algebra, the convolution cos( t)u(t) ∗ u(t) is computed as

[cos(

t)u(t)] ∗ u(t) = [cos( t)u(t)](−1) ∗ u (1)(t)


[cos(

t)u(t)]

t

(−1)

=

∫ cos(

−∞

Therefore

t

)u( )d = u(t)∫ cos( 0

)d = u(t)

1

[sin( t)]t0

=

sin( t)

u(t) .

1/25/02

P16-32

[cos(




sin( t)

u(t) +

sin[ (t − 3)]

u(t − 3) =

sin( t)

[ u(t) − u(t − 3)] .

SOLUTION 16.55. (a) The overall impulse response of the configuration in figure P16.55 is h(t) = [ h1(t) + h2 (t) + h3(t) + h4 (t)] ∗ h5(t) By the distributive property of convolution it follows that h(t) = h1 (t) ∗ h5(t) + h2 (t) ∗ h5(t) + h3 (t) ∗ h5 (t) + h4 (t) ∗ h5(t) = = (t) ∗cos( t)u(t) + (t − 1)∗cos( t)u(t) − − (t − 3)∗ cos( t)u(t) − (t − 4) ∗cos( t)u(t) By the sifting property of the delta function it follows that h(t) = cos( t)u(t) + cos[ (t − 1)]u(t − 1)− − cos[ (t − 3)]u(t − 3)− cos[ (t − 4) ]u(t − 4) . A picture of h(t) is sketched in the next figure.

(b) The response y(t) is computed as y(t) = h(t) ∗ u(t) = {cos( t)u(t) + cos[ (t − 1)]u(t − 1) − cos[ (t − 3)]u(t − 3)− cos[ (t − 4) ]u(t − 4)} ∗ u(t)

1/25/02

P16-33


Using the distributive property and the time shift property of convolution (see problem 16.22, part (a)) we have y(t) = [cos( t)u(t)] ∗ u(t) + {cos[ (t − 1)]u(t − 1)} ∗ u(t) − − {cos[ (t − 3)]u(t − 3)} ∗ u(t) − {cos[ (t − 4)]u(t − 4)} ∗ u(t) = = [cos( t)u(t)] ∗ u(t) + {[cos( t)u(t)] ∗ u(t)}t =t −1 − − {[cos( t)u(t)] ∗ u(t)}t =t −3 − {[cos( t)u(t)] ∗ u(t)}t =t −4

Using the techniques of convolution algebra, the convolution cos( t)u(t) ∗ u(t) is computed as

[cos(

t)u(t)] ∗ u(t) = [cos( t)u(t)](−1) ∗ u (1)(t)


[cos(

t)u(t)]

t

(−1)

=

∫ cos(

t

)u( )d

−∞

= u(t)∫ cos(

)d = u(t)

1

[sin( t)]t0

=

sin( t)

u(t) .

0

Therefore

[cos(



sin( t)

u(t) +

sin[ (t − 1)] =

sin( t)

u(t − 1) −

u(t − 3)−

sin[ (t − 4)]

[ u(t) − u(t − 1)+ u(t − 3) − u(t − 4)] .


∞

∫

h(t) ∗ f (t) =

1

we have

h(t − ) f ( )d

−∞

By making a change of variable

= t−

∞

h(t) ∗ f (t) = By definition

sin[ (t − 3)]

∫

−∞

∞

h( 1) f (t − 1)d 1 =

∫

−∞

f (t − 1 )h( 1 )d 1

∞

f (t) ∗ h(t) =

∫

−∞

f (t − )h( )d

From the above expressions we observe that h(t) ∗ f (t) = f (t) ∗ h(t) because and 1 are only variables of integration.

u(t − 4)

1/25/02

P16-34


(b) Using the definition of convolution we have ∞  ∞ ∞   [h(t) ∗ f (t)] ∗ g(t) =  ∫ h( ) f (t − )d  ∗ g(t) = ∫  ∫ h(  −∞   −∞   −∞ ∞ ∞  = ∫  ∫ [ h( ) f (t − 1 − )g( 1 )]d −∞  −∞

  ) f (t − 1 − )d  g( 1 )d 1 =    d 1 

Changing the order of integration we have  [h(t) ∗ f (t)] ∗ g(t) = ∫ h( −∞  

∞  ) ∫ f (t − 1 − )g( 1 )d 1 d  −∞  

∞

By the definition of the convolution we have

∞

f (t) ∗ g(t) =

∫

f (t − )g( )d

−∞

Therefore

∞

[ f (t) ∗ g(t)]t =t − Hence

∞

[h(t) ∗ f (t)] ∗ g(t) = ∫

−∞

=

∫

f (t − − )g( )d

−∞

{h( )⋅ [ f (t) ∗ g(t)]t =t − }d

By the definition of the convolution product we have h(t) ∗ [ f (t) ∗ g(t)] = Therefore

∞

∫ {h( )⋅ [ f (t) ∗ g(t)]t =t − }d

−∞

[h(t) ∗ f (t)] ∗ g(t) = h(t) ∗ [ f (t) ∗ g(t)] .

Thus the associative property of convolution is proved.

SOLUTION 16.57. We have f (t) ⋅ h(t) =

∞

∑

f (t) (t − kT )

k=0

For some nonnegative k , the Laplace transform of f (t) (t − kT) is L[ f (t) (t − kT)] = f (kT)e −skT

1/25/02

P16-35


by the sifting property of the delta function. Therefore we have ∞  ∞  L[ f (t)h(t)] = L  ∑ f (t) (t − kT)  = = ∑ f (kT )e −skT  k =0  k =0

Using the notation z = e sT we have L[ f (t)h(t)] =

∞

∑ f (kT)z −k .

k =0

SOLUTION 16.58. (a) By the voltage division formula we have Vout (s) =

1 Cs

1 ⋅Vin (s) R+ Cs

Therefore the transfer function of the circuit is 1 Vout (s) 1 1 H(s) = = Cs1 = = . Vin (s) R + CRs + 1 2s + 1 Cs Taking the inverse Laplace transform yields  1   1  -1  2  −0.5t h(t) = L-1  = = L u(t).   = 0.5e  1  2s + 1  s +   2 (b) By the impulse response theorem it follows that vout (t) = h(t) ∗vin (t) =

[

]

= 0.5e −0.5t u(t) ∗ [ (t) + (t − 1) + (t − 2) + K]

Using the sifting property of the delta function it follows that vout (t) = 0.5e −0.5t u(t) + 0.5e −0.5(t −1) u(t − 1) + 0.5e −0.5(t −2) u(t − 2) + K V. Therefore for 0 < t < 1

vout (t) = 0.5e −0.5t V

because only u(t) is nonzero for 0 < t < 1. (c) From the expression of vout (t)obtained in the part (b) it follows that

1/25/02

P16-36


vout (t) = 0.5e −0.5t + 0.5e −0.5(t −1) V, for 1 < t < 2 because only u(t) and u(t − 1) are nonzero for 1 < t < 2. (d) For t in the interval (4,5) , only u(t) , u(t − 1), u(t − 2) , u(t − 3) and u(t − 4) are nonzero. Therefore vout (t) = 0.5e −0.5t + 0.5e −0.5(t −1) + 0.5e −0.5(t −2) + 0.5e −0.5(t −3) + 0.5e −0.5(t −4) V. The above expression can be written as

(

) (

) (

) (

2 3  vout (t) = 0.5e −0.5(t −4) 1+ e −0.5 + e −0.5 + e −0.5 + e −0.5 

(e) Using the sum formula for geometric series with

(

) (

Therefore

(

1− e −0.5

)

) + (e ) + (e ) = 1− e−0.5 −0.5 5 1− e ( ) V. (t) = 0.5e −0.5(t −4) ⋅

1 + e −0.5 + e

vout

= e −0.5 we have

−0.5 2

−0.5 3

−0.5 4

)

4

 V. 

5

1− e −0.5

vout (t) = 8.6189⋅ e −0.5t V.

(f) Using the expression of vout (t)obtained in part (b) it follows that, for n < t < n + 1, vout (t) = 0.5e −0.5t + 0.5e −0.5(t −1) + 0.5e −0.5(t −2) + K + 0.5e −0.5(t −n) = 2 n  = 0.5e −0.5(t −n) 1+ e −0.5 + e −0.5 + K + e −0.5  V.  

(

) (

Using the sum formula for geometric series we have

(

(g) For large n, e −0.5

)

)

(

(

)

n+1

1− e −0.5 −0.5(t −n) vout (t) = 0.5e ⋅ 1− e −0.5 n+1

)

V.

≅ 0. Therefore, for large n we have vout (t) ≅ e −0.5(t −n) ⋅

0.5 V for n < t < n + 1. 1− e −0.5

A picture of vout (t), for large n, is sketched in the next figure.

1/25/02

P16-37


SOLUTION 16.59. By the impulse response theorem we have

[

]

vout (t) = h(t) ∗vin (t) = 0.5e −0.5t u(t) ∗ [ (t) + (t + 1)+ (t + 2) + K] V. Using the sifting property of the delta function it follows that vout (t) = 0.5e −0.5t u(t) + 0.5e −0.5(t +1)u(t + 1) + 0.5e −0.5(t + 2) u(t + 2) + K V. For 0 < t we have

vout (t) = 0.5e −0.5t + 0.5e −0.5(t +1) + 0.5e −0.5(t + 2) + K = 2   = 0.5e −0.5t 1+ e −0.5 + e −0.5 + K V.  

(

) (

)

Using the sum formula for geometric series( for n = ∞ ) we have vout (t) = 0.5e −0.5t ⋅

1 V for 0 < t. 1− e −0.5

Simplifying the expression of vout (t) it follows that vout (t) = 1.2707⋅ e −0.5t V for 0 < t.

SOLUTION 16.60. First we plot for reader convenience vin (t) and its staircase approximation.

1/25/02

P16-38


vin and its Staircase Approximation 2.8 2.6 2.4 2.2 2 1.8 1.6 1.4 1.2 1 0

0.1

0.2

0.3

0.4

0.5

0.6

The transfer function of the circuit of figure 16.58 is H(s) =

0.7

0.8

0.9

1

1 in which case 2s + 1

h(t) = 0.5e −0.5t u(t). Because we only want the output for 0 ≤ t ≤ 2, we only need h(t) for 0 ≤ t ≤ 2 s. Hence we need to generate staircase approximations to both vin (t) and h(t) as follows: t = 0:0.05:2; vin = exp(t .^2) .* (u(t) - u(t - 1)); h = 0.5*exp( - 0.5*t) .* (u(t) - u(t - 2)); T = 0.05; tstep = T; y = [0 conv(vin,h)*tstep 0]; t = 0:tstep:tstep*(length(vin)+length(h)); % For plotting through time 2 s we set t=t(1:41); y = y(1:41); plot(t,y) grid

1/25/02

P16-39


Convolution of vin and h

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

5/15/01

P17-1

R.A. DeCarlo & P.M. Lin

CHAPTER 17 P ROBLEM SOLUTIONS

SOLUTION PROBLEM 17.11. Case 1: suppose R1 > R2 . From example 17.3, page 696, if L and C are connected as indicated in part (a), then Z1 can be made real and larger than R2 . This means we can solve the problem at least for Z1 . Specifically, consider the figure below

From example 17.3, at a specified frequency, ωr, for which Z1 is real, then L, C, and R2 must satisfy, 1 − LC

ωr =

R22 L2

(1)

Further from example 17,3, at ωr, Z1( jω r ) =

L R2C

We require that Z1( jω r ) = R1 in which case L (2) C It is necessary to solve equations (1) and (2) simultaneously for L and C. From (2), L = R1R2 C . R1R2 =

Substituting into the square of (1) yields ω 2r =

1 R22 − R1R2C 2 R12 R22 C2

Hence C= It follows that

1 ω r R1

R1 − R2 R2

5/15/01

P17-2

 1 L = R1R2   ω r R1

R1 − R2  1 =  R2  ω r


R2 (R1 − R2 )

Observe that since R1 > R2 , both C and L are real, i.e., exist. Please note that this connection would not result in real values of C and L had R1 < R2 . If we can now show that Z2 = R 2 , then parts (a) and (b) are valid for this case. By direct computation 1 R1 R1 Z2 (jω r ) = jω r L + = jω L + = j R ( R − R ) + r 2 1 2 1 R − R2 jω r CR1 + 1 jω r C + j 1 +1 R1 R2

= j R2 ( R1 − R2 ) +

R1 − R2 R2 R1 − R2 = j R ( R − R ) + R − jR = R2 2 1 2 2 2 R1 − R2 R 2 1+ R2

R1 − jR1

Thus, (a) and (b) are true for the case R1 > R2 . We can also arrive at the conclusion that Z2 = R2 using maximum power transfer concepts. Since Z1 is constructed so that Z1 = R1 , we have set up the conditions for maximum power transfer of a V-source in series with R1 to the "load" Z 1 . Since the LC coupling network is lossless, whatever average power is received by the network to the right of R1 , will be dissipated by R2 . Therefore maximum power is transferred to the load R2 . Looking back from R2 , it must be that R2 sees a Thevenin resistance Z2 = R2 since it is known that there is a non-zero R1 . Case 2, R 1 < R 2 . Now consider the configuration

5/15/01

P17-3


Interchanging the subscripts of 1 and 2 in case 1 produces the derivation for this case. Combining cases 1 and 2 using the text notation produces C=

1 ωRlarge

Rlarge − Rsmall Rsmall

L=

SOLUTION PROBLEM 17.16. (a) »C = 0.1e-6; »R = 510; »wr = 2*pi*1.56e3; »%From equation 17.4 in example 17.3 which analyzes »%this particular circuit, »%L must satisfy »% wr^2*L^2 - L/C + R^2 = 0. »% Therefore » »L = roots([wr^2 -1/C R^2]) L= 5.3133e-02 5.0952e-02 (b) From HW problem 17.9,

(ω r )2 Hence, in MATLAB, »Linv = C*(wr^2 + 1/(R^2*C^2)) Linv = 4.8054e+01 »L = 1/Linv L = 2.0810e-02

=

(

1 R R − Rsmall ω small large

1 1 − 2 2 LC R C

)

5/15/01

P17-4


SOLUTION PROBLEM 17.22. (Correction: change RL in figure P17.22 to R.) We solve this problem using phasors rather than the Laplace transform approach as it is simpler. Consider

As such, since by definition ω 0 = 1

LC , the resonant frequency,

VC = VL = Zin ( jω 0 )Iin = RIin = RIm Hence vC (t) = vL (t) = RI m cos(ω 0t) and iL (t) =

RIm sin(ω 0 t). ω0 L

(a) By direct computation wC (t) =

1 2 1 2 CvC (t) = CR2 Im cos2 (ω 0 t) 2 2

(b) By direct computation w L (t) =

2 1 2 R2 Im LiL (t) = sin 2 (ω 0t) 2 2 2ω0 L

(c) Also by direct computation, since ω 20 = 1 LC ,

wC (t) + wL (t) = =

2 1 R2 Im 2 2 CR2 Im cos2 (ω 0 t) + 2 sin (ω 0t) 2 2ω 0 L

 1 1 2 2  2 1 2 R ImC  cos (ω 0t) + 2 sin 2 (ω 0t) = R2 Im C 2  ω 0 LC  2

(d) The energy dissipated in the resistance in one period is T

w R (0,T ) = R ∫

0

iR2 (t) dt =

T

R∫

0

iin2 (t) dt =

2 RIm

T

∫ cos 0

2

(ω 0t)dt

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P17-5

T

=


T

2 2 2 2π RIm RIm RIm RIm dt + cos(2ω t) dt = T = 0 2 ∫ 2 ∫ 2 ω0 0 0

(e) Finally

2π

maximum energy stored 0.5R2 I 2mC = 2π = ω 0 RC = Q total energy lost per period RI 2m π ω0

by equation 17.13.

SOLUTION PROBLEM 17.27. 1 s IL 1 1 L (a) H1(s) = = = = R 1 Vs Zin (s) R + Ls + 1 s2 + s + Cs L LC s 1 1 × 1 1  1  VC L Cs LC H2 (s) = H1(s) × =  IL ×  = = = R 1 R 1 2 2 Cs Vs Cs Vs s + s + s + s+ L LC L LC (b) H1(s) is precisely of the form of equation 17.18 with a single zero at the origin. It follows that ω 2p =

1 R and 2σ p = . Hence, from equation 17.19, ω m = ω p = LC L

Bω = 2σ p =

ωp R , and Qcir = Qp = = L 2σ p

1 1 , H1( jω m ) = , LC R

1 L 1 L × = . LC R R C

(c) With s = jω, 2  1   LC 

2

H2 ( jω) =

2

 1 −ω 2  +  R ω   LC  L  2

2

=

1

(1− LCω2 )

2

+ ( RCω)2

Instead of maximizing H2 ( jω) we minimize its reciprocal, through differentiation. Let

5/15/01

P17-6

f (C) =

1 H2 ( jω)


(

2 2 = 1 − LCω

)

2

+ ( RCω )2

Then

(

)

f '(C) = 2C(ωR)2 + 2 ω 2 LC − 1 ω 2 L = 0 implies that         L 1  1 1 = 1   C= 2 =  Lω 2  1 + 1  R (ωL)2 Lω 2  R2  2 2 + 1  Q2   coil  Lω  If the coil has high Q, then 1 C≅ Lω 2 in which case V Lω H2 ( jω) = C ≅ = Qcoil Vs R Therefore VC max ≅ Qcoil Vs

SOLUTION PROBLEM 17.31. Here

a + jω H( jω) 2 = K 2 (ω p −ω 2 ) + j2σ pω

2

= K2

( )

a2 +ω 2 2 2 2 2 2 2 2 ≡K f ω (ω p −ω ) + 4σ pω

(a) Differentiating with respect to ω 2 and setting the derivative equal to zero yields

0=

( )=

d f ω2 dω 2

 d  a2 +ω 2   dω 2  (ω 2p −ω 2 )2 + 4σ 2pω 2 

5/15/01

P17-7

(


)(

a 2 +ω 2 −2(ω 2p −ω 2 ) + 4σ 2p 1 = 2 2 2 2 2− 2 2 2 2 2 2 (ω p −ω ) + 4σ pω (ω p −ω ) + 4σ pω

[

]

)

Given that the denominator of the first term is non-zero, this is equivalent to

(

)(

2 2

2 2

0 = (ω 2p −ω 2 )2 + 4σ2p ω2 − a 2 +ω 2 −2(ω 2p −ω 2 ) + 4σ2p

( 2)

= ω

2

4

(

2

+ω p − 2ω pω + 4σ pω + 2 a +ω

2

)

)(ω 2p −ω 2 ) − 4(a 2 +ω 2 )σ2p

( )2 +ω 4p − 2ω 2pω 2 + 4σ 2pω 2 + 2a 2ω 2p − 2a2 ω 2 + 2ω2p ω2 − 2(ω 2 )2 − 4a2 σ2p − 4σ 2pω 2

= ω2

( )2 − 2a 2ω 2 +ω 4p + 2a2ω 2p − 4a 2σ 2p

= − ω2

Hence

(ω 2 )2 + 2a 2ω 2 + a 4 =ω 4p + 2a2 ω 2p − 4a2 σ2p + a 4 where we have added a 4 to produce perfect squares, i.e.,

(ω 2 + a2 )2 = (ω 2p + a2 )2 − (2aσ p )2 This implies that 2

2

ω = −a ±

(ω 2p + a2 )2 − (2aσ p )2

Thus, to achieve a real positive solution we obtain,

2

ωm = − a +

(ω 2p + a 2 )2 − (2aσ p )2

(b) Here, as in part (a), the arithmetic is simpler if we deal with 1/H(s) rather than H(s). Specifically s2 + 2σ ps +ω 2p s2 + 2σ p (s + a) − 2σ pa +ω 2p 2σ p s2 − 2σ pa +ω 2p 1 = = = + H(s) K(s + a) K(s + a) K K(s + a)

5/15/01

P17-8


The problem asks for a non-zero value of ω. For zero phase shift of H(jω), Im[H(jω)] = 0 Im[H(jω)] > 0. This necessarily requires that the imaginary part of the above expression at s = jω must be zero. Thus

(

)

 −ω 2 − 2σ pa +ω 2p (a − jω)   −ω 2 − 2σ pa +ω 2p   1    = Im  Im   = Im  2 2   H( jω) K( jω + a)   K(ω + a )       −ω 2 − 2σ pa +ω 2p )(−ω) ( = =0 K(ω 2 + a 2 )

The solutions to this are dc, i.e., ω = 0, and ω (for zero phase shift) = ω 2p − 2σ pa Note that such a frequency may not exist if the quantity under the radical is negative.

SOLUTION PROBLEM 17.40. (a) Compute the transfer function: Rs s + V 1 1 1 L H(s) = out = = = × 1 1 Iin Yin(s) C R  1 1   Rs + Cs + s2 +  s + s+ + 1 Ls + Rs Rp   L R pC   Rp  LC s + 1.333 × 103 = 8 × 105 2 s + 2000s + 1.0006ω 20 where ω 20 = 1.6 × 10 9 and ω 2p = 1.0006ω 20 ≅ 1.60089 × 109 and ω p = 4.001× 10 4 rad/sec. ωp R (b) Qp = = 20 which is high Qp . Further a = s = 1.333 × 103 <<ω p = 40.011109 × 103 . 2σ p L (c) Equation 17.28,

5/15/01

P17-9

2

ωm = − a +


(ω 2p + a 2 )2 − (2aσ p )2

gives the exact value of ωm. In MATLAB, »wmsqrd = -a^2 + sqrt((wp^2+a^2)^2 - (twosig*a)^2) wmsqrd = 1.600886670362673e+09 »wm = sqrt(wmsqrd) wm = 4.001108184444246e+04 Clearly, ωm approximates ωp . Finally, H( jω m ) ≅ H( jω p ) = 8 × 10 5

jω p + 1.333 × 10 3

8 × 105 1.333 × 103 = 1− j 2000 ωp −ω 2p + j2000ω p +ω 2p

≅ 400 Ω (d) From equation 17.30, Bω ≅ 2σ p = 2000 rad/sec, ω1 ≅ ω m − 0.5Bω = 39.011 × 103 rad/sec, and ω 2 ≅ω m + 0.5Bω = 41.011 × 103 rad/sec. (e) From equation 17.30,

Qcir ≅ Qp =

1 ≅ Qcir Hence,

Rs 1 + L RpC ω0

=

ωp 2σ p

=

 Rs  1  R + 1 LC  p   Rs 1   L + R C  p 

≅

ω0

 Rs 1   L + R C  p 

Qcoil + Qcap Rs 1 1 1 + = + = Lω 0 Rp Cω 0 Qcoil Qcap Qcoil Qcap

5/15/01

P17-10

Qcir ≅


QcoilQcap Qcoil + Qcap

SOLUTION PROBLEM 17.65. Here we consider the equivalent circuit valid for t > 0. Note that -

+

-

+

iL(0 ) = iL(0 ) = E/Rs and vC(0 ) = vC(0 ) = 0. Hence

It follows that 1 E 1 E VC (s) = 1 1 × = × 1 1 CRs + + Cs Rs s s2 + s+ R Ls RC LC Complex roots occur when 2 2  1  − 4 =  1  − 4ω 2 < 0 0  RC  LC  RC 

Equivalently, ω 0 RC >

1 2

+

as was to be shown. Further, since vC(0 ) = 0, the general form of the capacitor voltage for constant excitation is (as per chapter 10) vC (t) = e−σt ( Acos(ω d t) + Bsin(ω d t)) = Be−σt sin(ω dt) From the characteristic equation, the complex roots are −1 1  1  2 −ω 0 1 −σ ± jω d = ±j − = ± jω 1− 0 2RC LC 2RC  2Q 4Q2 Here

5/15/01

P17-11

a=σ=

ω0 2Q

ω d =ω 0 1−


1 4Q2

Letting B = Vm, for the appropriate value of Vm, we obtain the desired result. 1 2Q 2Q (b) If Q is large, ω d ≅ω 0 . The Vm e −at will drop to 1/e of its peak value in t = = ≅ a ω0 ωd 2π seconds. The period of oscillation of the damped sinusoid is . Therefore the number of cycles ωd contained in this interval is 2Q ωd Q 2π = π ωd

5/31/01 Mag Crt Probs

P18-1

© R. A. DeCarlo & P.M. Lin

CHAPTER 18 P ROBLEM SOLUTIONS

SOLUTION PROBLEM 18.34. There is a correction to this problem: set M = 3 H. (a) The stored energy at t = 0 is: W(0) = 0.5L1i12 (0) + 0.5L2i22 (0) + Mi1 (0)i2 (0) = 8 J »L1 = 10; L2 = 2; M = 3; i10 = 1; i20 = -3; »W0 = 0.5*L1*i10^2 + 0.5*L2*i20^2 + M*i10*i20 W0 = 5 (b) Writing two differential mesh equations we obtain di di di di L1 1 + M 2 + R1i1 = 10 1 + 3 2 + i1 = 0 dt dt dt dt and di2 di di di + M 1 + R2i2 = 2 2 + 3 1 + i2 = 0 dt dt dt dt

L2

Taking the Laplace transform of these equations yields and

10sI1 − 10i1 (0) + 3sI2 − 3i2 (0) + I1 = (10s + 1)I1 + 3sI2 − 10i1(0) − 3i2 (0) = 0 2sI2 − 2i2 (0) + 3sI1 − 3i1(0) + I2 = (2s + 1)I2 + 3sI1 − 3i1(0) − 2i2 (0) = 0

Putting these equations in matrix form yields 3s   I1  10i1(0) + 3i2 (0)   1  (10s + 1) = =  3s (2s + 1)  I2   3i1(0) + 2i2 (0)   −3  Solving yields 3s  −1  1  −3s   1   I1  (10s + 1) 1  2s + 1 =  I  =  3s     (2s + 1)  −3 11s2 + 12s + 1  −3s 10s + 1  −3  2  =

1  11s + 1  1 1  =   (11s + 1)(s + 1)  −33s − 3 s + 1 −3

Therefore, by inspection, i1(t) = e −tu(t) A and i2 (t) = −3e −tu(t) A Remark: normally, i1(t) and i2(t) would have two exponential terms present. Because of the special choice of initial conditions, a pole cancelled out. (c) From equation 18.24 with the lower limit changed to zero and the upper limit changed to ∞, we have


P18-2

W(0,∞) = ∫

∞

0


(v1i1 + v2i2 ) dt = 0.5L1i12 (∞) + 0.5L2i22 (∞) + Mi1(∞)i2 (∞) −0.5L1i12 (0) + 0.5L2i22 (0) + Mi1(0)i2 (0)

From part (b) all currents at t = ∞ are zero, hence W(0,∞) = −0.5L1i12 (0) + 0.5L2i22 (0) + Mi1(0)i2 (0) = −5 J The result of part (a) indicates that the initial store energy is 5 J. The result of part (c) indicates that the energy returned to the circuit is also 5 J, i.e., the total energy accumulated in the inductors over [0,∞) is –5 J. Hence 5 J is dissipated in the resistors. ∞

Remark: the interested student might computer the integral

∫ (R1i1 (t) + R2i2 (t))dt , the actual energy 2

0

dissipated in the resistors over [0,∞), and show that this is 5 J.

SOLUTION PROBLEM 18.35. (a) »L1 = 4; L2 = 9; M = 3; »I1 = 2; I2 = -3; »W = 0.5*L1*I1^2 + 0.5*L2*I2^2 - M*I1*I2 W= 6.6500e+01 (b) »K = 0.5*L1*I1^2 K= 8 »% Minimize (over I2) K + 0.5*9*I2^2 - 3*2*I2 »% Take Derivative and set to zero; then solve for I2. »% Derivative is: 9*I2 – 6 = 0 »% The result is I2 = 2/3 A. »I2 = 2/3; »Wmin = 0.5*L1*I1^2 + 0.5*L2*I2^2 - M*I1*I2 Wmin = 6 (c)

2


(d) »L1 = 4; L2 = 9; M = 3; »k = M/sqrt(L1*L2) k = 5.0e-01

SOLUTION PROBLEM 18.37. »k = 0.5; »L1 = 9; L2 = 4;L3 = 1; »M = 0.5*sqrt(L1*L2) M= 3 »Lcpld = L1 + L2 + 2*M Lcpld = 19 »Leq = Lcpld + L3 Leq = 20 »Imax = 2; »Wmax = 0.5*Leq*Imax^2 Wmax = 40 J

SOLUTION PROBLEM 18.41. »RL = 100; Rs = 300e3; R = 10e3; »m = 20; n = 5; (a) »Z2 = RL*m^2 Z2 = 40000 »Req1 = Z2*R/(Z2+R) Req1 = 8000 »Z1 = Req1*n^2 Z1 = 200000

P18-3



P18-4


(b) »%Gv1 = v1/vin »%v1 = [Z1/(Rs + Z1)]vin »Gv1 = Z1/(Rs+Z1) Gv1 = 4.0000e-01 »% Gv2 = v2/vin »% Gv2 = v2/v1 * v1/vin = (1/n) * G1 »Gv2 = G1/n Gv2 = 8.0000e-02 »% Gv3 = v3/vin »Gv3 = -Gv2/m Gv3 = -4.0000e-03 (c) »% Gi2 = i2/iin »Gi2 = n*R/(R+Z2) Gi2 = 1 »% Gi3 = i3/iin = i3/i2 * i2/iin = –m*Gi2 »Gi3 = -m*Gi2 Gi3 = -20

SOLUTION PROBLEM 18.55. (a) The parameters in the circuit of figure P18.55b are given by equations in figure 18.22b. Specifically, since k = M L1 L2 = 0.16 3.5 × 0.008 = 0.95618 »M = 0.16; L1 = 3.5; L2 = 0.008; »k=M/sqrt(L1*L2) k= 9.5618e-01 »La = (1 - k^2)*L1 La = 3.0000e-01 »Lb = k^2 * L1 Lb = 3.2000e+00 »N =M/L2 N= 20 (b) »R = 500; »w = 2*pi*60; »Vseff = 110; »Zin = R + j*La*w + j*Lb*w Zin = 5.0000e+02 + 1.3195e+03i »Iseff = Vseff/Zin Iseff = 2.7624e-02 - 7.2899e-02i »Pave = R*abs(Iseff)^2


P18-5

Pave = 3.0387e+00 (c) »Zin2 = R + j*La*w Zin2 = 5.0000e+02 + 1.1310e+02i »Iseff2 = Vseff/Zin2 Iseff2 = 2.0929e-01 - 4.7341e-02i »Is2mag = abs(Iseff2) Is2mag = 2.1458e-01 » % The current in the secondary is (in A): »Isecmag = Ismag*N Isecmag = 4.2916e+00 (d) »% Our first step is to compute the reflected impedance: »Zrefl = 100*N^2 Zrefl = 40000 »% We now compute the impedance of the parallel combination »% of Lb and Zrefl denoted Zpar »Zpar = 1/(1/Zrefl + 1/(j*w*Lb)) Zpar = 3.6350e+01 + 1.2053e+03i »% We now compute the input impedance: »Zin = R+j*w*La + Zpar Zin = 5.3635e+02 + 1.3184e+03i »% Now we compute the voltage across the primary of the »% ideal transformer, by voltage division: »Vpar = Vseff*Zpar/Zin Vpar = 8.7342e+01 + 3.2500e+01i »% Now we compute the voltage across the load: »Vload = Vpar/N Vload = 4.3671e+00 + 1.6250e+00i »Vloadmag = abs(Vload) Vloadmag = 4.6596e+00 »Iloadmag = Vloadmag/100 Iloadmag = 4.6596e-02



P18-6


SOLUTION PROBLEM 18.65. (a) The equivalent circuit accounting for initial conditions is given below:

(b) From the definition of coupled inductors 0.1472s  I1'   V1   0.6s = V   0.1472s 0.1472s  '   2    I2  (c) Hence

 I1'   0.6s 0.1472s −1  V1  15  0.1472 '=   =   I2   0.1472s 0.1472s  V2  s  −0.1472

−0.1472   V1  0.6   V2 

(d) Writing nodal equations we obtain, Vs   2 −1  V1   I1   2 −1  V1  1  i1 (0− )   I1'  +    0  =  −1 2  V  +  I  =  −1 2  V  + s       2   2    2 i2 (0 − )  I'2  (e) Now we substitute our result of part (c): Vs   2 −1  V1  1  −2 1  2.208 −2.208  V1   0  =  −1 2  V  + s  −2 + s  −2.208 9  V2      2    which simplifies to

Vs  1  −2 1  2s + 2.208  0  − s  −2 = s  −(s + 2.208)     

−(s + 2.208)  V1  2s + 9   V2 

or equivalently  2s + 2.208 −(s + 2.208)  V1  sVs + 2 = −(s + 2.208) 2s + 9  V2   2   (f) Solving these equations we obtain:


P18-7


−1  V1   2s + 2.208 −(s + 2.208)  sVs + 2  V  =  −(s + 2.208) 2s + 9   2   2 

=

(s + 2.208)  sVs + 2  1  2s + 9    3s2 + 18s + 15 (s + 2.208) 2s + 2.208  2 

(g) If vs (t) = 10u(t) V, then (s + 2.208) 12  V1  1  2s + 9 1 26s + 112.42  V  = 2  (s + 2.208) 2s + 2.208   2  = 2    2  3s + 18s + 15     3s + 18s + 15 16s + 30.912  From MATLAB »[r,p,k]=residue([16 30.912], [3 18 15]) r= 4.0907e+00 1.2427e+00 p= -5 -1 k= [] Therefore,

(

)

v2 (t) = 4.0907e −5t + 1.2427e −t u(t) V

SOLUTION PROBLEM 18.67. The solution to this problem is based upon the following: (i) Leq = L1 + L2 + 2M for series aiding connection (see example 18.4) and (ii) k = M L1 L2 (a definition), and (iii) k = 1 (an assumption). (a) Given L1 = L2 = L and k = 1, Leq = L1 + L2 + 2M = L + L + 2k L2 = 4L . Hence, when the number of turns is doubled, the inductance is quadrupled. (b) For this part, let us first consider L2 which has 2N turns. We can view L2 as two coils of N turns each connected in series aiding with coupling coefficient k = 1. Hence, according to part (a), the inductance of L2 is four times that of L1 which only has N turns. Hence, Leq = L1 + L2 + 2M = L + 4L + 2k 4L2 = 9L 2

Observe that the coil has 3N turns yielding an inductance of 9L = 3 L. (c) Suppose coil 1 and coil 2 consist of one turn each. Here the total number of turns is 2N where N = 1 turn. Suppose further that L1 = L. From part (a), Leq = 4L = (2)2 L . Now suppose coil one consists of one turn and coil 2 consists of M turns. We assume here as an induction hypothesis that


P18-8


Leq = ( M + 1)2 L We must show that if coil 2 has (M+1)N turns then, Leq = ( M + 2)2 L Our first step is to compute the equivalent inductance of coil 2. However, coil 2 consists of a single turn coupled to an M-turn coil, which by the induction hypothesis means that L2 = ( M + 1)2 L Thus coil 1 in a series aiding connection with L2 leads to Leq = L1 + L2 + 2M = L + (M + 1)2 L + 2 (M + 1)2 L2 = L + ( M + 1)2 L + 2( M + 1) L

[

]

= L ( M + 1)2 + 2( M + 1) + 1 = L[( M + 1) + 1] = ( M + 2)2 L 2

Given this relationship, if coil 1 consists of N1 turns, and one turn has an inductance L, then L1 = ( N1 ) L . Similarly, L2 = ( N2 ) L , and M = k L1 L2 = L1L2 = immediately follows that L1 : L2 : M = N12 : N22 : (N1N2) 2

2

(N1)2 ( N2 )2 L2

= N1N 2 L . It

SOLUTION PROBLEM 18.70. (a) k = M / L1L2 »M = 1.5; L1 = 1.5; L2 = 6; »k = M/sqrt(L1*L2) k = 5.0000e-01 (b) For this part and the remaining parts consider the following equivalent circuit where the coupled coils have been replaced by the model of figure 18.22(b).

»La = (1 - k^2)*L1 La = 1.1250e+00 »Lb = k^2 * L1 Lb = 3.7500e-01


P18-9


»% The turns ratio is M:L2, i.e., »1.5/6 ans = 2.5000e-01 »% Therefore the turns ratio is 1:4. It follows that »Rb = 200*(1/4)^2 Rb = 1.2500e+01 To compute Z(s) we have, Z(s) = 20 +

1 RL s 12.5s 1 + Las + b b = 1.125s + 20 + + 100 Cs Cs Lbs + Rb s+ 3

Hence,    j12.5ω j 1 416.67ω  Z( jω) = j1.125ω + 20 + − = 20 + + j 1.125ω − +   4 100 Cω Cω 10 10 4  2  2 jω + ω +  ω +  3  9 9  12.5ω 2

(c) For this part we need to make the imaginary part of Z(jw) real. To this end: »K1 = 12.5*100/3 K1 = 4.1667e+02 »w = 1333; »K2 = 1.125*w + K1*w/(w^2 + 1e4/9) K2 = 1.4999e+03 »C = 1/(K2*w) C = 5.0015e-07 Hence, we take C = 5 µF. (d) At resonance, we have 12.5ω 2r Z( jω r ) = Z( j1333) = 20 + = 32.5 Ω 10 4 2 ωr + 9 and Zb (jω r ) =

j12.5ω r j16,662 = 100 33.33 + j1333 = 12.492 + j0.31238 jωr + 3

By voltage division Vout ( jω r ) Vout Vb 4 Zb ( jω r ) = × = × = 1.5379 + j0.038456 Vs ( jω r ) Vb Vs 1 Z( jω r ) »Zb = j*12.5*w/(j*w + 100/3) Zb = 1.2492e+01 + 3.1238e-01i »Zwr=20 + 12.5*w^2/(w^2 + 1e4/9) Zwr =


P18-10


3.2492e+01 »Gv = 4*Zb/Zwr Gv = 1.5379e+00 + 3.8456e-02i »MagGv = abs(Gv) MagGv = 1.5384e+00 »AngGv = angle(Gv)*180/pi AngGv = 1.4325e+00

SOLUTION PROBLEM 18.71. (a) Following the hint we apply a source transformation to obtain

Writing two mesh equations we obtain the following matrix form of the mesh equations: ks s + β + 1 s  ks s +β+ 1 

  I1   Iin s = s  I2   0 

Solving for I2 yields  s + β + 1 s Iin s det ks 0  − kIin  I2 = = ks s + β + 1 s  (s + β + 1 s)2 − (ks )2 det ks s + β + 1 s  To find H(s) we have V −I s ks ks H(s) = out = 2 = = 2 2 Iin Iin ( s + β + 1 s) − ( ks) (s + ks + β + 1 s)(s − ks + β + 1 s) =

[(1+ k)s

ks 2

][

]

+ β s + 1 (1− k)s 2 + βs + 1


P18-11

(b) R = 0.02 Ω, b = R = 0.02 Ω, Q = 50, and k = 0.01, 0.02, and 0.04. »beta = 0.02; k1 = 0.01; k2 = 0.02; k3 = 0.04; »p11 = roots([(1+k1) beta 1]); »p12 = roots([(1-k1) beta 1]); »p1 = [p11;p12] p1 = -9.9010e-03 + 9.9499e-01i -9.9010e-03 - 9.9499e-01i -1.0101e-02 + 1.0050e+00i -1.0101e-02 - 1.0050e+00i »p21 = roots([(1+k2) beta 1]); »p22 = roots([(1-k2) beta 1]); »p2 = [p21;p22] p2 = -9.8039e-03 + 9.9010e-01i -9.8039e-03 - 9.9010e-01i -1.0204e-02 + 1.0101e+00i -1.0204e-02 - 1.0101e+00i »p31 = roots([(1+k3) beta 1]); »p32 = roots([(1-k3) beta 1]); »p3 = [p31;p32] p3 = -9.6154e-03 + 9.8053e-01i -9.6154e-03 - 9.8053e-01i -1.0417e-02 + 1.0206e+00i -1.0417e-02 - 1.0206e+00i (c) »f = 0.14:.0001:.18; »n1 = [k1/(1-k1)^2 0]; n2 = [k2/(1-k2)^2 0]; n3 = [k3/(1-k3)^2 0]; »h1 = freqs(n1, poly(p1), 2*pi*f); »h2 = freqs(n2, poly(p2), 2*pi*f); »h3 = freqs(n3, poly(p3), 2*pi*f); »plot(f, abs(h1), f, abs(h2), f, abs(h3)) »grid



P18-12


(d) An inspection of [f, abs(h2)] (i.e., a tabulation of the values) in part (c) indicates that fpeak = 0.15865 Hz and Hpeak = 26.148 Ω. The frequency scale factor Kf is defined according to: »Kf = 455e3/fpeak Kf = 2.8679e+06 Further, »Km = Kf*2.35e-3 Km = 6.7397e+03 »Lnew = Km*1/Kf Lnew = 2.3500e-03 »Cnew = 1/(Km*Kf) Cnew = 5.1736e-11 »Rnew = Km*0.02 Rnew = 1.3479e+02 The 3 dB down value of h2 is Hpeak/sqrt(2). Hence


P18-13


»Hmax = max(abs(h2)) Hmax = 2.6148e+01 »H3db = Hmax/sqrt(2) H3db = 1.8490e+01 Again, inspecting the tabulated values indicates that the 3 dB frequencies are: f1 = 0.1569 Hz f2 = 0.1614 Hz. Finally »Bf = f2 - f1 Bf = 4.5000e-03 Hz »Bfnew = Kf*Bf Bfnew = 1.2906e+04 Hz

and

(e) For this part we redo part (a) with R, L, and C as literals. Writing two mesh equations we obtain the following matrix form of the mesh equations:  Ls + R + 1 Cs  kLs 

kLs   I1   Iin Cs = Ls + R + 1 Cs   I2   0 

Solving for I2 yields  Ls + R + 1 Cs Iin Cs det  kLs 0  − kL C  I2 = = I Ls + R + 1 Cs kLs   ( Ls + R + 1 Cs)2 − (kLs)2 in det kLs Ls + R + 1 Cs  To find H(s) we have V − I Cs kL C 2s kL C2 s H(s) = out = 2 = = Iin Iin ( Ls + R + 1 Cs)2 − (kLs)2 ( Ls + kLs + R + 1 Cs)( Ls − kLs + R + 1 Cs) ks kLs LC2 = = 2 2 LC(1+ k)s + CRs + 1 LC(1− k)s + CRs + 1  (1+ k)s2 + R s + 1   (1− k)s2 + R s + 1   L LC  L LC 

(

)(

=

)

kω 20 s

C

 2 ω 2  2 ω 2  (1+ k)s + 0 s +ω 0   (1− k)s + 0 s +ω 0   Q  Q 

Evaluating this expression at s = jω0 , yields


P18-14

H( jω 0 ) =


jkω 30

C

2  2+ 2   −(1− k)ω2 + jω 0 +ω 2  −(1+ k)ω +ω 0 0  0 0  Q Q   

jω 20

− jk

− jk Cω 0 Cω 0 = = 1 2  j  j1  k −  k +  k + Q2  Q  Q Therefore H( jω 0 ) =

1 k × Cω 0 k 2 + 1 Q2

(f) To solve this part we differentiate and set equal to zero as follows:       d H (jω 0 )    1 d  k 1 1 2k 2 = × × −  2 =  =0 2 1 1 dk Cω 0 dk  k + Cω 0  k 2 +    1  2  Q2   Q2  k + 2   Q   It follows that 2k 2 k2

1 =1 + 2 Q

Hence k2 =

1 Q2

or k = 1/Q. With this value of k, 1 1/ Q Q × 2 = Cω 0 2Cω0 Q2 At ω0 , the magnitude of the transfer function increases with increasing k, reaching a peak at k = 1/Q and then decreases with a further increase in k as born out in the plots of part (c). H( jω 0 ) max =

Prbs Chap 19, 1/7/02

P19-1

CHAPTER 19 PROBLEM SOLUTIONS SOLUTION PROBLEM 19.1. Refer to figure 19.3. »Vs = 100; ZL = 20; Rs = 1e3; »beta = 149; »Zinbox = (beta + 1)*ZL Zinbox = 3000 »% By voltage division »V1 = Vs*Zinbox/(Zinbox + Rs) V1 = 75 »% To obtain the power delivered by the source »I1 = Vs/(Zinbox +Rs) I1 = 2.5000e-02 »Psource = I1*Vs Psource = 2.5000e+00

SOLUTION PROBLEM 19.2. Refer to figure 19.4. »Vs = 100; Z1 = 30e3; Rs = 50; beta = 149; »Zboxin = Z1/(beta+1) Zboxin = 200 »V1 = Vs*Zboxin/(Zboxin + Rs) V1 = 80 »Psource = Vs^2/(Rs + Zboxin) Psource = 40 SOLUTION PROBLEM 19.3. Refer to figure 19.5. (a) »C = 0.1e-3; vc0 = 10; Z1 = 300; Z2 = 1e3; »Z3 = 1e3; gm = 9e-3; » »Zin = Z1 + (1 + gm*Z1)*Z2 Zin = 4.0000e+03 (b) »tau = Zin*C



P19-2


tau = 4.0000e-01 Hence, vC (t) = vC (0)e −t / = 10e −2.5t V.

SOLUTION PROBLEM 19.4. (a) First observe that since no current can flow into the secondary we have Voc = aV pri = aRI in = 800 Vrms Now Zth =

1 + a2 R = 640 − j 360 Ω j C

(b) ZL = (Zth ) = 640 + j360 Ω. »Voc = 800; Rth = 640; »Pmax = Voc^2/(4*Rth) Pmax = 250 *

(c) By inspection the circuit is a 640 Ω resistor in series with a 3.6 H inductor.

SOLUTION PROBLEM 19.5. Because the output is open circuited, no current flows into the secondary of the transformer, hence voc = v sec + sin(3t)u(t) = 2v pri + sin(3t)u(t) = [cos(3t) + sin(3t)]u(t). Additionally Zth (s) =

10  2.5  20 + s + 4 + 4 + 0.25s + 9 = + 2s + 40 s s s

SOLUTION PROBLEM 19.6. Using Cramer's rule,


P19-3

I1 =

V1  det 0  0


1 − a  0.5 0  0 0.5 0.25V1 1 = = V 1.5a 1.5a 6a 1

Therefore Rin = 6a Ω. 2 V1,eff 100 To compute the average power, V1,eff = 10 V. Hence Pave = = watts. Rin 6a

SOLUTION PROBLEM 19.7. As per the hint, we write loop equations as follows: −Vout   s + 1 −1 0   Iout        10 s  =  −1 6 −2   I2  −40 s  0 −2 4   I3  Using equation 19.6,   0.2 0.1 −1  0.2 0.1  10 s  −Vout =  s + 1− [−1 0] Iout + [−1 0]         0.1 0.3  0   0.1 0.3 −40 s 2 = (s + 0.8)Iout + s Therefore Vout = (s + 0.8)(−Iout ) −

2 = Zth (− Iout ) + Voc s

2 i.e., Zth = s + 0.8, and Voc = − . s

SOLUTION PROBLEM 19.8. (a) Let the node voltages from left to right be V1 , V2 , and Vout . Also inject a current I3 into node 3. Writing nodal equations by inspection we have:


P19-4


Iin   1.5 −1 −0.25   V1       2 −0.5  0  =  −1   V2   I 3   −0.25 −0.5 0.0625s + 0.75  Vout  Using equation 19.11, we have

(

)

−1 −1  Iin  I3 = W22 − W21W11 W12 Vout + W21W11 0    0.5  0.25  0.5   Iin  1 1 =  0.0625s + 0.75 − [0.25 0.5] Vout − [0.25 0.5]         0.5 0.75  0.5  0.5 0.75  0 

Thus I3 = (0.0625s + 0.375)Vout − 0.5Iin 1 16 Therefore Isc = −I 3]V =0 = 0.5Iin . Further Zth (s) = = Ω. out 0.0625s + 0.375 s + 6 16I sc 8Iin (b) Vout (s) = Zth (s)I sc = = . By inspection, the impulse response is s+6 s+6 vout ,imp (t) = 8e −6t u(t) V. Further, from MATLAB »n = 8; d = [1 6 0]; »[r,p,k] = residue(n,d) r= -1.3334e+00 1.3334e+00 p= -6 0 k= [] Hence the step response is: vout ,step (t) =

(

)

4 1− e −6t u(t) V 3

SOLUTION PROBLEM 19.9. (a) Consider the following figure:


P19-5


Let Iout enter the output terminal and I1 and I2 be the currents entering the primary and secondary of the transformer respectively. It follows that

Iin =

bVout (b − 1)Vout (2b − 1)Vout + I1 + = + I1 R R R

which implies that I1 = Iin −

(2b − 1)Vout R

Further, Iout =

(1− b)Vout (1− b)Vout + I2 = − bI1 R R

Therefore Iout =

(1− b)Vout (1− b)Vout (1− b)Vout (2b − 1)Vout   + I2 = − bI1 = − b Iin −  R R R R =

2b 2 − 2b + 1 Vout − bIin R

Equivalently Vout =

2b 2

R bR Iout + 2 Iin − 2b + 1 2b − 2b + 1

Therefore Zth = (b) Isc = b∠45o , Zth =

2b2

2b2

R bR , Voc = 2 Iin , Isc = bIin − 2b + 1 2b − 2b + 1

R . − 2b + 1


(c) B = 10 =

P19-6

1 ZthC

1 R

=

=

2b 2 − 2b + 1

C


2b2 − 2b + 1 5 = . Hence C = 0.02 F. Further, RC 25C

1 1 = . Therefore L = 2 H. LC 0.02L V V 1 2 100s (d) H(s) = out = out = 2 2 = = 1 Iin 0.5I sc 2b − 2b + 1 1 s2 + 10s + 25 + + Cs 0.2 + 2s + 0.02s R Ls 2 0

= 25 =

2 s − 10 and Iin (s) = × 2 . Hence, 2 s + 100 Vout (s) =

(s2

50 2 s(s − 10) + 10s + 25)(s2 + 100)

To compute vout (t) in MATLAB, %partial fraction expansion of Vout/√2 = = n(s)/d(s) n =50*[ 1 -10 0]; d = conv([ 1 0 100], [ 1 10 25]); [r ,p, k] = residue(n,d) %numerator polynomial of combined complex pole terms num = [1 -p(1)]*r(2) +[1 -p(2)]*r(1) Output from MATLAB r= 2.8000 + 0.4000i 2.8000 - 0.4000i -5.6000 30.0000 p= -0.0000 +10.0000i -0.0000 -10.0000i -5.0000 -5.0000 num = 5.6000 -8.0000 Laplace transform of ouput Vout (s) =

−5.6 2 30 2 (5.6s − 8) 2 + + 2 s+5 (s + 5) s2 + 100


P19-7


Taking the inverse Laplace transform using table 13.1 on page 515:

[

]

vout (t) = 30 2te −5t − 5.6 2e −5t + 5.6 2 cos(10t) − 0.8 2 sin(10t) u(t) V The steady state part consists of the cosine and sine terms only. Since the parallel RLC acts like a band pass circuit and the peak value occurs at 5 rad/s, one expects the magnitude of the steady state output to be much smaller at 100 rad/s.

SOLUTION PROBLEM 19.10. (a) Write two nodal equations by inspection:  I1  Y1 + Y3 −Y3 + gm   V1   y11 = I  =  −Y Y2 + Y3  V2   y 21  2  3

y12  V1  y 22 V2 

(b) When port-2 is shorted, y11 is the input admittance. Therefore

Zin =

1 1 = y11 Y1 + Y3

and since V2 = 0, I2 = y 21V1 =

−KY3 s

SOLUTION PROBLEM 19.11. (a) By inspection −Y3  V1   y11  I1   Y1 + Y3 I  =  −Y + g   =   2  3 m Y2 + Y3  V2   y 21

y12   V1  y 22  V2 

Clearly, Y3 = − y12 . Then, Y1 = y11 − Y3 = y11 + y12 and Y2 = y22 − Y3 = y 22 + y12 . Finally, gm = y 21 + Y3 = y 21 − y12 . (b) Recall that


P19-8


 I1  Y1 + Y3 −Y3 + gm   V1   y11 = I  =  −Y Y2 + Y3  V2   y 21  2  3

y12  V1  y 22 V2 

Clearly, Y3 = − y 21 . Then, Y1 = y11 − Y3 = y11 + y 21 and Y2 = y22 − Y3 = y 22 + y 21 . Finally, gm = y12 + Y3 = y12 − y 21 .

SOLUTION PROBLEM 19.12. (a) By definition of coupled inductors V1   L1s Ms  I1   I1   L2 1 V  =  Ms L s  I  ⇒  I  =   2  2  2   2  L1L2 − M 2 s − M

(

)

− M   V1  L1  V2 

Hence, the y-parameters are:

(

 L2  L1L2 − M 2 s  − M 1

)

− M L1 

(b) By definition of coupled inductors V1   L1s − Ms  I1   I1   L2 1 V  = − Ms L s  I  ⇒  I  =   2  2  2   2  L1L2 − M 2 s  M

(

)

M   V1  L1  V2 

Hence, the y-parameters are:

(

L2  L1L2 − M 2 s  M 1

)

M L1 

If the coupling coefficient is 1, L1L2 = M 2 and the y-parameters do not exist since the determinant of the z-parameter matrix is zero.

SOLUTION PROBLEM 19.13. Let I2' denote the current entering the dotted terminal of the secondary of the coupled inductors. Then using the result of problem 12a,


P19-9

 I1   L2 1 I '  =   2  L1L2 − M 2 s  − M

(

)


− M   V1  1  4 −1  V1  = L1  V2  6s  −1 1  V2 

−1 1 8 2 7 1 From the given circuit I2 = I2' + 2I1 = V1 + V2 + V1 − V2 = V1 − V2 . Therefore 6s 6s 6s 6s 6s 6s the y-parameter matrix is: 1  4 −1 6s  7 −1

SOLUTION PROBLEM 19.14. I  (a) By definition and the properties of the ideal transformer y11 = 1  = G1 and V1  V =0 2 I  G +G y22 = 2  = 1 2 2 . Additionally, since the circuit is obviously reciprocal, V2  V =0 a 1 I  G y12 (= y21) = 1  = 1. V2  V =0 a 1 (b) V 2 reflected to the primary side, denoted by Vˆ1 , is −2K a Vˆ1 = 2 s +4 Hence I1 =

2G1K a s2 + 4

To compute I2 , we reflect the parallel of G1 and G2 to the secondary of the ideal transformer. Hence the impedance in parallel with V 2 , denoted Zsec, is a2 Zsec = G1 + G2 Therefore,


P19-1 0

I2 =


2K( G1 + G2 ) a2 V2 = Zsec s2 + 4

SOLUTION PROBLEM 19.15. (a) By definition I  1  1 y11 = 1  = = S  V1 V =0 Zin V =0 9 2 2 where 1  Zin = 6 + 12/ / × (320/ /80) = 9 Ω 16 Similarly, by definition I  1  3 y 22 = 2  = = S  V2 V =0 Zout V =0 400 1 1 where Zout = 80 + 320/ /16 ( × (6 //12) ) =

400 Ω 3

SOLUTION PROBLEM 19.16. Write nodal equations:  I1   2s + 2    I 2  =  −1  0   −2s

−1 −2s V1    s + 2 −s  V2  −s 5s  V3 

Using the matrix partitioning method, we obtain the 2-port y parameters  V1   I1   2s + 2 −1  1 −2s −  [−2s − s]   I  =   −1  s + 2  5s  −s   V2   2    1.2s + 2 −0.4s − 1  V1   y11 =   =  −0.4s − 1 0.8s + 2  V2   y 21

y12   V1  y 22  V2 


P19-1 1


SOLUTION PROBLEM 19.17. (a) The ideal transformer yields the constraints I1 = −aIˆ2 and Vˆ2 = aV1 The three resistors have nodal equations − Iˆ2   2 −1 Vˆ2   =     I2   −1 2  V2  Substituting the first two equations into the last one, we obtain  I1   2a 2 I  =   2   −a

−a  V1   y11   =  2  V2   y 21

y12   V1  y 22  V2 

(b) 2 y12 y 21 2 a Yin = y11 − = 2a − = 1.75a2 S y 22 + YL 4

Zin =

1 4 = 2 Ω Yin 7a

and

Gv = V2 = -y21 = a V1 y22 + y L1 4

SOLUTION PROBLEM 19.18. This problem is solved in MATLAB. Part (a) »% Parameter specification »Ys = 1e-3;YL = 1e-3; »y11=4e-3; y12 = -0.1e-3; »y21 = 50e-3; y22 = 1e-3; »% Calculation of input admittance and impedance »Yin = y11 - y12*y21/(y22 + YL) Yin = 6.5000e-03 »Zin = 1/Yin Zin = 1.5385e+02 »% Calculation of output admittance


P19-1 2


»Yout = y22 - y12*y21/(y11 + Ys) Yout = 2.0000e-03 Part (b) »% Calculation of voltage gain »Gv = (Ys/(Ys + Yin))*(-y21/(y22 + YL)) Gv = -3.3333e+00 Part (c) »V2 = Gv*10 V2 = -3.3333e+01 Therefore, v2 (t) = −33.333u(t) V. Finally, »PL = V2^2/1e3 PL = 1.1111e+00

SOLUTION PROBLEM 19.19. (a)

V1 = Z in = Z in = 0.5 ⇒ Z in = 10 Ω or 0.1 S Vs Z in + Z s Z in + 10

Now

Yin = y 11 - y12y21 = y11 - 0.02×2 = 0.1 y22 +Y L 0.2 + 0.1 Solving for y11 yields y11 = 0.2333 S. (b)

v2 = - y 21 = -2 v1 y22 +YL 0.2 +0.1

= - 6.667

Hence and

v2(t) = - 6.667 v1(t) = - 3.333 vs(t) = - 33.33 u(t) V 2 2 PL = V2 = 33.33 = 111.11 W

RL

10

SOLUTION PROBLEM 19.20. (a) Writing a node equation at port 1 and mesh equation at port 2, we obtain by inspection

I1 = 2V 1 + 3I 2 V2 = 2V 1 + 2I 2 Rearranging in matrix form , we have


P19-1 3

1 -3 I1 0 2

Therefore

I1 I2 (b) Yin = y 11 -

=

=

I2

-1

1.5

V1

-1

0.5

V2


2 0

V1

-2 1

V2

=

y11 y21

y12 y22

V1 V2

y12y21 = -1 + 1.5×1 = 1 S y22 +Y L 0.5 + 0.25

(c) Here we compute Y out seen looking into port-2, i.e.,

Yout = y 22 - y12y21 = 0.5 + 1.5×1 = -1 S y11 +Y s -1 + 0 From current division,

-1 i2(t) = Yout i s(t) = ×5u(t) = 6.667u(t) A Yout +YL -1 + 0.25 Finally from Ohm's law

v1(t) = - 0.5 ×3i2(t) = - 0.5 ×3×6.667u(t) = - 10u(t) V REMARK: Because the current source sees a negative resistance, the circuit is unstable as it stands.

SOLUTION PROBLEM 19.21. (a) With port 2 shorted, the Laplace transform of the given data are: 1 1 1 4 −1 I1(s) = , V1(s) = − = , I 2 (s) = s s s + 4 s(s + 4) s+ 3 Hence I y11 = 1 = V1 and

1s s+ 4 = 4 4 s(s + 4)

I −1 (s + 3) − s(s + 4) y 21 = 2 = = 4 V1 4(s + 3) s(s + 4)

Next, with port-2 terminated in a 1-Ω reistor, the Laplace transform of the given data are:


P19-1 4


Z L=YL = 1 I 1(s) = 1/s, V1(s) = 1 - 1 + 1 = 5s + 16 s s+4 (s+4)2 s(s + 4) 2 I2 = - 1 s +7 Now

I 2 = - V2 = - V 2 = y21V1 ZL y22 +YL Solving for y22, and using y21 expression found earlier, we obtain

y22 = y 21V1/I 2 - Y L= - s(s + 4) × 5s + 16 ×s + 7 - - 1 4 (s + 3) s(s + 4) 2 -1 2 = s + 23s + 64 4(s + 3)(s+ 4) Finally, to compute y12, we use the defining equation for y-parameters, i.e., from which we obtain

I1 = y11V1 + y 12V2

1 - s + 4 × 5s + 16 s 4 s(s + 4 ) 2 y12 = I1 - y 11V1 = = - (s + 7) V2 1 4 ( s + 4) s+7 (b) Given YL =1 S, the input admittance is

Yin = y11 - y12y21 y22 +Y L

- (s + 7) ×- s(s + 4) 2 = s + 4 - 4 ( s + 4) 4(s + 3) = (s + 4) 4 5(s+3.2) s2 + 23s + 64 +1 4(s + 3)(s+ 4)

and

Zin = 1 = 5(s+3.2) Yin (s + 4)2


P19-1 5


(c) For this part, we use phasors to do the sinusoidal steady state analysis: ω = 10 rad/s and I 1 = 1. Also,

V 1 = Z = 5(j10+3.2) = 0.1973 - j0.4072 in I1 (j10 + 4)2 and

V 2 = -y21 V1 y22 +YL

- j10(j10 + 4) 4(j10 + 3) = = -0.2873+ j0.1787 -100 + 230j + 64 +1 4(j + 3)(j+ 4)

Thus

V 2 = V2 ×V 1 I1 V 1 I1

= (-0.2873+ j0.1787)(0.1973 - j0.4072)

from which we obtain in Ohms,

V2 I1

= -0.2873+ j0.1787 ×0.1973 - j0.4072 = 0.8192

SOLUTION PROBLEM 19.22. Looking into port-1, the admittance is:

and

2×2 Yport 1 (s) = y11 - y12y21 = 0 + = 40 s y22 +Y L 0 + 0.1s Zport 1 (s) = s 40

Therefore

Hence

2 2 Zin (s) = 1000 + s = s + 200 s 40 40s

Zin ( j ) = The imaginary part is zero when

4 × 10 4 − 2 4 × 10 4 − 2 =−j j40 40

= 200 rad/s. Hence, the resonant frequency is 200 rad/s.

SOLUTION PROBLEM 19.23. (a) By writing nodal equations for the boxed 2-port, we have by inspection (note passive circuit in which y21 = y12 ):


P19-1 6

I1

=

I2

3s + 2

-2s -2

V1

-2s -2

3s + 3

V2


=

y11 y21

V1

y12 y22

V2

(b) In Ohms, ZL = s+1; Zs = 2. Therefore, 2 (2s +2) 2 Yin = y 11 - y12y21 = 3s+2 = 2s + 3s + 1 = 2s + 1 y22 +Y L 3s + 3 + s + 1 s+1

and

V1 = Z in 1 1 = = = 1 Vs Z in + Z s 1 + Z s Yin 1 + 2×(2s + 1) 4s + 3 -y21 = V2 = 2s +2 = 0.5 y + y 22 L V1 3s + 3 + s + 1

Thus,

V2 = V1 ×V2 = 1 × 0.5 = 1 = 1 2 Vs Vs V1 4s + 3 8s + 14s + 6 8s+ 6 (c) The impulse reponse is  1  −1  0.125  −0.75t h(t) = L−1{H (s)} = L−1  u(t) = L   = 0.125e 8s + 6 s + 0.75     For the step response,

(

)

 H (s)  −1  0.125  1 −0.75t v 2 (t) = L−1 u(t) V = L   = 1− e s s(s + 0.75) 6     (d) We must compute the complete Laplace transform and invert. Here

Vs(s) = 12.75 × 2 s2 + 4

and

3.1875 V2(s) = H(s)V2(s) = 0.125 ×12.75× 2 = 2 s+ 0.75 s + 4 (s + 0.75) (s2 + 4) We use MATLAB to compute the partial fraction expansion n=3.1875; d= conv([1 0.75], [ 1 0 4]); [r p k ] = residue (n,d) r= -0.3493 - 0.1310i -0.3493 + 0.1310i 0.6986 p= -0.0000 + 2.0000i


P19-1 7


-0.0000 - 2.0000i -0.7500 Hence, after combining the two complex terms, we obtain

V2(s) = 0.6986 + - 0.6986s + 0.524 s + 0.75 s2 + 4 From Table 13.1, the steady state response is

v2,ss (t) = [-0.6986cos(2t) + 0.262sin(2t)]u(t) V and the transient response is

v2,tran(t) = 0.6986e-0.75t u(t) V

SOLUTION PROBLEM 19.24. (a) Using the y-parameters of stage 2,

Yin2 = y 11 - y12y21 = 500 + 0.1×75000 = 501.8 µS y22 +Y L 185 +4000 The load for stage 1 is the parallel combination of Zin2 and the 2 kΩ resistor. Hence, using the yparameters of stage 1, we obtain

Yin1

4 y y 0.5×24×10 12 21 = y 11 = 2000 + = 2108.9 µS y22 +Y L 100 +501.8 + 500

(b) We compute the following voltage gains:

V1 = Z in1 1 1 = = = 0.9499 Vs Z in1 + Z s 1 + Z s Yin1 1 + 25×2108.9×10-6  V2  − y 21 −0.24 = = = −217.8  V  y 22 + YL,stage1 (100 + 500 + 501.8) × 10 −6 1 stage1 and  V2  − y 21 −0.075 = = = −17.92  V  y 22 + YL,stage2 (185 + 4000) × 10 −6 1 stage2 Finally Gv is the product of the three gains calculated above

Gv = 0.9499×(- 217.8) ×(-17.92) = 3708.2


P19-1 8


SOLUTION PROBLEM 19.25. (a) With the switch in position A, the load to the 2-port is YL = Cs = 0.25s. Hence, V2 − y 21 − y 21 = = V1 y22 + YL y 22 + 0.25s (b) From the given data, V1 (s) = –2/s and hence, V2(s) =

− y 21 −1 s −2 8 −2 1 1 V1 (s) = × = = + + 2 y 22 + 0.25s −1 s + 0.25s s s s+2 s−2 s(s − 4)

Hence, for t ≥ 0,

v2(t) = ( -2 + e-2t + e 2t) u(t) V (c) The circuit is not stable in the time interval 0 to 1 s, because the transfer function has a pole in the right half plane. (d) v2(1-) = -2 + e-2 + e 2 = 5.524 V (e) Replace the charged capacitor by the parallel combination of an admittance of 0.25s and a current source of value 0.25×8.524 (in accordance with figure 14.16 (f) Z1 (s) = b2 × 1 = 4 Ω. (g) For t ≥ 1s, the capacitor is discharging through a 4-Ω equivalent resistance, with a time constant 0.25×4 = 1 s, and an initial voltage v2 (1-) = 5.542 V. Hence v 2 (t) = 5.524e −(t −1)u(t − 1)V and by the ideal transformer voltage ratio property, v 3(t) = 2.762e −(t −1)u(t − 1)V

SOLUTION PROBLEM 19.26. (a) By definition of coupled inductors V1   L1s Ms  I1  V  =  Ms L s  I   2  2  2  Hence, the z-parameters are:


P19-1 9


L1s Ms   Ms L s  2  (b) By definition of coupled inductors V1   L1s − Ms  I1  V  = − Ms L s  I   2  2  2  Hence, the y-parameters are:  L1s − Ms − Ms L s   2  The z-parameters exist independent of the values of M, L1 , and L2 .

SOLUTION PROBLEM 19.27. (a) By definition and the properties of the ideal transformer V  V  z11 = 1  = R1 + R2 and z22 = 2  = a2 R2 . Additionally, since the circuit is obviously I1  I =0 I2  I =0 2

1

V  reciprocal, z21 (= z12 ) = 2  = aR2 . I1  I =0 2 (b) The input impedance is given by the formula z12z21 a 2R22 1 Zin = z11 − = R1 + R2 − 2 = R1 + 2 R2 z22 + ZL a R2 + R2 a +1 (c) If port-1 is open circuited, I1 = 0. Hence,

2aKR V1 = z12 I2 = 2 2 s +4

and V2 = z22I 2 =

2a 2KR2 s2 + 4

SOLUTION PROBLEM 19.28. (a) For this part consider the figure below:


P19-2 0


With port 2 open, it follows that Zpr = 320/16 = 20 Ω and V 20 ×12 z11 = 1  =6 + =13.5 Ω I1  I =0 20 + 12 2 With port 1 open and I2 injected into port 2, we have Zsec = 12*16 = 192 Ω and Vpr  V −V  −I (320 / /192 )  z12 = 1  = = sec  = 2 = −30 Ω   I2  I = 0 I2  I =0 4I2  I =0 4I2  I =0 1 1 1 1 With port 2 open and I1 injected into port 1, we have Zpr = 20 Ω and −4Vpr  V  V  −4I1 (12 //20 )  z21 = 2  = sec  = = = −30 Ω   I1  I =0 I1  I =0 I1  I =0 I1  I2 =0 2 2 2 With port 1 open, it follows that Zsec = 12*16 = 192 Ω and V  320 ×192 z22 = 2  = 80 + = 200 Ω I2  I =0 320 + 192 1 −300 2000 (b) V1 (s) = z12 I2 (s) = 2 and V2 (s) = z 22 I2 (s) = 2 . s +4 s +4


P19-2 1


SOLUTION PROBLEM 19.29. (a) Writing two mesh equations we have by inspection, V1   Z1 + Z3 V  =  Z + r  2  3 m (b) V1 = z11I1 =

Z3   I1   z11 = Z2 + Z3   I2   z21

z12   I1  z22  I 2 

K ( Z1 + Z3 ) K (Z3 + rm ) and V2 = z21I1 = s s

SOLUTION PROBLEM 19.30. From the result of problem 29,  Z1 + Z3 Z + r  3 m

Z3   z11 = Z2 + Z3   z21

z12  z22 

Therefore, Z3 = z12 , Z1 = z11 − Z3 = z11 − z12 , Z2 = z22 − Z3 = z22 − z12 , and rm = z21 − Z3 = z21 − z12 .

SOLUTION PROBLEM 19.31. (a) The z-parameters can be computed by inspection (first write the z-parameters of the passive part of the network, i.e., with the dependent source ignored; then add the effect of the dependent source to the resulting equations.) As such, using loop equations, 10   10 5 + s 10 + s  10  0.5s + 1 s + 1 Z =  10 Ω 10  =  s + 1  10 +  s  1  s s 10 10 × (s + 1) 5s + 10 10(s + 1) (5s + 10)(2s + 1)− 10(s + 1) 5s + 10 s (b) Zin = − s = − = s  10(s + 1)  s s(2s + 1) 2s(s + 0.5) + 10   s 5s2 + 7.5s 5s + 7.5 = = s(s + 0.5) s + 0.5 V (s) 10 s + 0.5 0.6667 1.3334 (c) I1(s) = 1 = × = + . Hence Zin (s) s 5s + 7.5 s s + 1.5


P19-2 2


2 4  i1 (t) =  + e −1.5t  u(t) A 3 3

SOLUTION PROBLEM 19.32. (a) To find the resonant frequency we first find Zin (s) = z11 −

z12 z21 z z 10z11 10 10 10s = z11 − 11 11 = = = = 2 2 10 z22 + Z L z11 + 10 z11 +10 1 + s + 25 s + s + 25 z11 1 + s

This is of the form of equation 17.18 with K = 10. Here according to equation 17.19f, ω r = ω p = 25 = 5 rad/s (b) To find Q we use equation 17.19e, i.e.,

Q = Qp =

ωp 2σ p

=

5 =5 1

(c) Using MATLAB we obtain the frequency response plots below:


P19-2 3


10 9 8

Magnitude of Zin Ω

7 6 5 4 TextEnd 3 2 1 0

0

2

4

6


14

16

18

20


P19-2 4


100 80 60

Phase of Zin degrees

40 20 0 -20 TextEnd -40 -60 -80 -100

0

2

4

6


 z11 SOLUTION PROBLEM 19.33. (a) Since Z =  z21 not include Zs , it follows that Zin = z11 −

14

16

18

20

z12   2 −3 = Ω. Assuming that Zin does z22   30 4 

z12z21 90 = 2+ = 5.75 Ω z22 + ZL 24

Assuming that Zout does not include the parallel connection of Z L, then z z 90 Zout = z22 − 12 21 = 4 + = 34 Ω z11 + Zs 3 (b) From equation 19.27  ZL   z21  GV = Gv2Gv1 =   .  z22 + ZL   Zin + ZS 


P19-2 5


Thus in MATLAB »Gv = (20/(4 + 20))*(30/(5.75+1)) Gv = 3.7037e+00 »% Therefore »v2 = Gv*30 v2 = 1.1111e+02 Hence, v 2 (t) = 111.11u(t) V. The power absorbed by ZL is therefore (in watts): »PZL = v2^2/20 PZL = 6.1728e+02

SOLUTION PROBLEM 19.34. (a) Before determining b, it is necessary to compute the impedance seen at the secondary of the 2-port. Here, z z 2 3R02 Zout = z22 − 12 21 = 33 2 R0 − = 33 2R0 − 2R0 = 32 2 R0 z11 + Zs R0 + R0 Thus b must be chosen so that the impedance reflected to the secondary of the transformer is 2 Ω, i.e., 32 2R0 2= ⇒ b = 4 R0 b2 (b) First do a source transformation on the front end of the two port to obtain v s (t) = R0 2 cos(2t) = 32 2 cos(2t) V Therefore, Vs,eff = 32 V. Also note that the impedance looking into the primary of the two port is z z 2 3R02 Zin = z11 − 12 21 = R0 − = 31.015 Ω z22 + ZL 33 2 R0 + 32 2R0 Hence  ZL   z21  GV = Gv2Gv1 =  = 0.5  z22 + ZL   Zin + ZS  Consider the following MATLAB calculations: »a=2; R0 = 16; »Zin = a*R0-2*a^3 * R0^2/(65*a^2 *R0)


P19-2 6


Zin = 3.1015e+01 »ZL = 32*a^2 * R0 ZL = 2048 »Zs = a*R0 Zs = 32 »z22 = 33*a^2*R0 z22 = 2112 »z21 = 2*a*R0 z21 = 64 »Gv = (ZL/(z22+ZL))*(z21/(Zin+Zs)) Gv = 5.0000e-01 »V1eff = 32; »V2eff = Gv*V1eff V2eff = 16 »b = 4*a*sqrt(R0) b = 32 »VLoadeff=V2eff/b VLoadeff = 5.0000e-01 »Pmax = VLoadeff^2/2 Pmax = 1.2500e-01 Hence, max power transferred to the load is 125 mW.

SOLUTION PROBLEM 19.35. (a) Using the usual formula

Z(s) = z11 −

z12 z21 1000 × (−1000) =0− = 0.01s z22 + ZL 0 + 108 s

Yes, the input to the 2-port looks like a 0.01 H inductor. (b) Under the given conditions, the circuit reduces to parallel RLC with Req = 50 kΩ, L = 0.01 H, and C = 100 pF. Therefore, (c) For this circuit

m

=

1 1 = 106 rad/s and B = = 2 × 105 rad/s. R C LC eq


P19-2 7


1 Vin sC Vin 106 s V1(s) = 1 × = 1 1 1 × R1 = s2 + 2 × 10 6 s + 1012 Vin + + Cs R1 s2 + s+ Req Ls ReqC LC For the impulse response, Vin = 1 and from MATLAB »n = [1e6 0];d = [1 2e6 1e12]; »[r,p,k] = residue(n,d) r= 1.0000e+06 -1.0000e+12 p= -1000000 -1000000 k= [] Therefore, the impulse response is:

(

)

6 h(t) = 106 − 1012 t e −10 t u(t) V

1000 −1  0 −0.001  0 (d) y =  = S  0  0  −1000  0.001

(e) V2(s) =

− y 21 −0.001 106 s −1011 V1(s) = × = y 22 + YL 0 + 10−8 s s2 + 2 × 10 6 s + 1012 (s + 10 6 )2

Hence 6 v 2 (t) = −1011 te−10 t u(t) V.

SOLUTION PROBLEM 19.36.


P19-2 8


z z 1.2075 × 63.75 (a) Zin 2 = z11 − 12 21 = 62.582 − = 1.7778 kΩ z22 + ZL 1.25 + 0.016 Because z12 for stage 1 is zero, z12z21 Zin = z11 − = z = 2 kΩ z22 + Zin 2 / /2 11 V Zin (b) Gv1 = 1 = = 0.96386 . Let ZL1 = Zin 2 //2 = 0.94118 kΩ. Then Vs Zin + 75 V ZL1 z V ZL z Gv2 = 2 = × 21 = −22.472. Finally Gv3 = out = × 21 = 0.45319. V1 ZL1 + z22 Zin V2 ZL + z22 Zin 2 Thus Gv = Gv1Gv 2Gv 3 = −9.816

SOLUTION PROBLEM 19.37. (a) By inspection via mesh standard equations  L1s Ms z=    Ms L2s (b) Utilizing the properties of an ideal transformer, V  z11 = 1  = L1s I1  I =0 2

V  k 2 L2 z22 = 2  = L s + (1− k 2 )L2s = L2s I2  I =0 L1 1 1

k L2 V  z12 = 1  = L s = k L1L2 s = Ms I2  I =0 L1 1 1 Finally, k L2 V  z21 = 2  = L s = k L1L2 s = Ms I1  I =0 L1 1 2 (c) Utilizing the properties of an ideal transformer,


P19-2 9

V  z22 = 2  = L2 s I2  I =0 1


V  k 2 L1 z11 = 1  = L s + (1− k 2 )L1s = L1s I1  I =0 L2 2 2

k L1 V  z12 = 1  = L s = k L1L2 s = Ms I2  I =0 L2 2 1 Finally, k L1 V  z21 = 2  = L s = k L1L2 s = Ms I1  I =0 L2 2 2 10−2 (d) For this circuit k = 1 and the turns ratio a = = 0.1. Under this condition the given 1 circuit reduces to a current source of value Is = Vin 5000 driving a parallel RLC with R = 5000 Ω, L = L1 = 0.01 H, the capacitance reflected to the primary of value C = 10−8 F. Therefore m

=

1 1 = 105 rad/s and B = = 2 × 104 rad/s. ReqC LC

Finally, at w = wm, the circuit is resonant and Vin appears across the primary of the transformer. V This voltage is then stepped up by a factor of 10. Therefore out = 10. Vin max

SOLUTION PROBLEM 19.38. (a) By inspection V1   Z1 1   I1   h11 I  =  −1 Y  V  =  h  2  2   2   21

h12   I1  h22  V2 

(b) By inspection V1 = V2 − Z1I 2 ⇒ V1 + Z1I2 = V2 and I1 = Y2V1 − I2

⇒ Y2V1 − I 2 = I1


P19-3 0


In matrix form, V1   0 1  I1  V1   1 I  =  1 0 V  ⇒ I  = Y  2   2   2  2

Z1  −1  0 1  I1  −1  1 0 V2 

Thus V1   Z1 1   I1  1 I  = Z Y + 1  −1 Y  V   2 1 2  2  2 

SOLUTION PROBLEM 19.39. (a) From problem 19.38b we have

V1   Z1 1   I1  1 I  = Z Y + 1  −1 Y  V  =  2 1 2  2  2 

1  Cs 1  1 + 1  −1 RCs 

 1  I  RCs  I1  1  R 1 =     1  V2  RCs + 1  −RCs Cs  V2  R

(b) This part is a cascade of part (a) and an ideal transformer. Label the voltage and current at the port 2 of N as Vˆ and Iˆ . From the properties of the ideal transformer, V = nVˆ and 1

2

2

2

2

I2 = Iˆ2 n . Hence RCs   I1  RCs  I1  V1   V1   R 1  R 1 Iˆ  = RCs + 1  −RCs Cs  Vˆ  ⇒  nI  = RCs + 1 −RCs Cs  V n   2   2   2   2  Therefore RCs n   I1  V1  1  R I  = RCs + 1  − RCs n Cs n 2  V   2   2 

From table 19.1, if h22 =

Cs n 2(RCs + 1)

≠ 0, then the z-parameters exist and if h11 =

the y-parameters exist, i.e., if C ≠ 0 and R ≠ 0 respectively.

R ≠ 0, RCs + 1

Prbs Chap 19, 1/13/02

P19-1


CHAPTER 19 PROBLEM SOLUTIONS SOLUTION PROBLEM 19.40. (a) Let Z1 = R and Z2 = 1/Cs or Y2 = Cs. From problem 38, Z1 1   R 1  h= =   −1 Y2  −1 Cs (b) This part is a cascade of an ideal transformer and part (a). Label the voltage and current at the port 1 of N as Vˆ and Iˆ . From the properties of the ideal transformer, V = −bVˆ and 1

1

1

1

1

I1 = − Iˆ1 b . Hence Vˆ1   R 1   Iˆ1  −V1  =    ⇒   I 2   −1 Cs V2   I2

b   R 1  −bI1   = −1 Cs V     2 

Therefore V1   b2 R −b   I1    I  =  Cs V2   2   b From table 19.1, if h22 = Cs ≠ 0, then the z-parameters exist and if h11 = b 2R ≠ 0, the yparameters exist, i.e., if C ≠ 0 and R ≠ 0 (assuming reasonably that b ≠ 0) respectively.

SOLUTION PROBLEM 19.41. For this solution we apply the definition of h-parameters: by inspection V 1 h11 = 1 = I1 V =0 2 + 2s 2 I 2V − 2sV1 (2 − 2s)V1 1− s h21 = 2 = 1 = = I1 V =0 I1 (2 + 2s)V1 V =0 s + 1 V2 =0 2 2 1 s +1 When I1 = 0, then I2 = 2V1 + 2V1 = 4V1 and V2 = 0.5(2V1) + (2V1 ) = V. 2s s 1

Prbs Chap 19, 1/13/02

P19-2


V s I 4V 4s Therefore, h12 = 1 = and h22 = 2 = 1 = . V2 I =0 s + 1 V2 I =0 V2 I =0 s + 1 1 1 1

SOLUTION PROBLEM 19.42. (a) In MATLAB »h11 = 250; h12 = 0.025; h21 = 12.5; h22 = 2.25e-3; »Zs = 1e3; ZL = 500; »YL = 1/ZL YL = 2.0000e-03 »Zin = h11 - h12*h21/(h22 + (1/ZL)) Zin = 1.7647e+02 »Yout = h22 - h12*h21/(h11 + Zs) Yout = 2.0000e-03 »Zout = 1/Yout Zout = 500 (b) »% Gv1 = V1/Vs »Gv1 = Zin/(Zin + Zs) Gv1 = 1.5000e-01 »% Gv2 = V2/V1 »Gv2 = -h21/(Zin*(h22 + YL)) Gv2 = -1.6667e+01 »Gv = Gv1*Gv2 Gv = -2.5000e+00 (c) Given the above, the Thevenin equivalent seen by the capacitor is Voc = −2.5Vin and Rth = 500 Ω. In MATLAB »Zth = ZL*Zout/(ZL + Zout) Zth = 250 »Vin = 10; »Voc = -2.5*Vin; »w = 400;

Prbs Chap 19, 1/13/02

P19-3


»Zc = 1/(j*w*10e-6) Zc = 0 - 2.5000e+02i »Vc = Voc*Zc/(Zth + Zc) Vc = -1.2500e+01 + 1.2500e+01i»V2mag = abs(Vc) »V2mag = abs(Vc) V2mag = 1.7678e+01 »V2ang = angle(Vc)*180/pi V2ang = 135 From above, v 2 (t) = 17.678 2 cos(400t + 135o ) V Therefore »Pave = V2mag^2/500 Pave = 6.2500e-01

SOLUTION PROBLEM 19.43. (a) Using the h-parameters of stage 2

Zin2 = h 11 - h12h21 = 1000+ 0.966×51 = 4000 Ω h22 +Y L 0.0008 + 1/64 The load for stage 1 is the parallel combination of Zin2 and the 3 kΩ resistance. However, because h12 = 0, the input impedance is unaffected by the load, and hence for stage 1,

Zin1 = h 11 = 2000 Ω (b) For stage 1, because h12 = 0, the output impedance is unaffected by the source impedance. Thus,

Yout1 = h 22 = 0.05 ×10-3 S, Zout1 =1/Y out1 = 20 ×103 Ω

Prbs Chap 19, 1/13/02

P19-4


For stage 2, the source impedance is the parallel combination of Zout1 and the 3 kΩ resistance. Thus

Zs2 = 20000×3000 =2608.7 Ω 20000 + 3000

and

0.966×51 Yout2 = h 22 - h12h21 = 0.0008+ = 0.0145 S h11 +Z s2 1000 + 2608.7 Z out2 =1/Yout2 = 69.19 Ω (c)

V1 = Z in1 2000 = = 0.5 Vs Z in1 + Z s 2000 + 2000 The load of stage 1 is the parallel combination of Zin2 and Zm. Thus

YL1 = Y in2 + 1/3000 = 5.834 ×10-4 S Hence

ZL1 = 1/ YL1 = 1714 Ω

and

V2 V1

stage 1

-50 = 1 × -h 21 = 1 × = - 39.46 Z in1 h22 + Y L1 2000 (0.05 + 0.5834) ×10-3

For stage 2, the load is 64 Ω. Hence

V2 V1

stage 2

51 = 1 × -h 21 = 1 × = 0.7762 Z in2 h22 + Y L2 4000 (0.8 + 1000/64) ×10-3

Finally, the overall voltage gain is the product of the three gains calculated above

Vout = 0.5 ×(- 39.46) ×(0.7762) = - 15.32 Vs (d) The input circuit consists of a series connection of Vs, Zs, C and Zin1. The remainder of the circuit is resistive and has no effect on the frequency response. The magnitude response is of the high pass type with f 3dB =

1 1 = = 39.789 Hz 2π(Rs + Rin1)C 2π(2000 + 2000)10−6

SOLUTION PROBLEM 19.44. To meet the required matching, we must have

Prbs Chap 19, 1/13/02

P19-5

Zout = Z out2 = Z L = 64 = This requires that


1 Yout2

Yout2 = h 22 - h12h21 = 0.0008+ 0.966×51 = 1 h11 +Z s2 1000 +Z s2 64 Solving for Zs2, we obtain Zs2 = 2323.2 Ω. Now for stage 2, Z s2 is the parallel combination of Zout1 = 20 kΩ and Zm:

Zs2 = 20000×Z m = 2323.2 Ω 20000 +Z m from which Zm = 2628.5 Ω. With this new value of Zm, we repeat the calculations of problem 19.43 to obtain Zin = 2000 Ω and Vout/Vs = -14 .26. Details follow. Using the h parameters of stage 2

Zin2 = h 11 - h12h21 = 1000+ 0.966×51 = 4000 Ω h22 +Y L 0.0008 + 1/64 For stage 1, because h12 = 0, the input impedance is not affected by the load and

Zin1 = h 11 = 2000 Ω. The voltage gains of the various stages are: V1 = Z in1 2000 = = 0.5 Vs Z in1 + Z s 2000 + 2000 The load of stage 1 is the parallel combination of Zin2 and Zm. Thus

YL1 = Y in2 + 1/2628.5 = 6.3044 ×10-4 S

Hence

V2 V1

stage 1

-50 = 1 × -h 21 = 1 × = - 36.74 Z in1 h22 + Y L1 2000 (0.05 + 0.63044) ×10-3


V2 V1

stage 2

51 = 1 × -h 21 = 1 × = 0.7762 Z in2 h22 + Y L2 4000 (0.8 + 1000/64) ×10-3


Vout = 0.5 ×(- 36.74) ×(0.7762) = - 14.26 Vs

Prbs Chap 19, 1/13/02

P19-6


SOLUTION PROBLEM 19.45. (a) Using the h-parameters of stage 2

Zin2 = h 11 - h12h21 = 500 + 0.966×51 = 2000 Ω h22 +Y L 0.0016 + 1/32 The load for stage 1 is the parallel combination of Zin2 and the 1.5 kΩ resistance. However, because h12 = 0, the input impedance is unaffected by the load, and Zin1 = h 11 = 1000 Ω. (b) For stage 1, because h12 =0, the output impedance is unaffected by the source impedance, and

Yout1 = h 22 = 0.1 ×10-3 S, Zout1 =1/Y out1 =104 Ω For stage 2, the source impedance is the parallel combination of Zout1 and the 1.5 kΩ resistor 10 4 × 1.5 × 10 3 Zs2 = 4 = 1304.3 Ω 10 + 1.5 × 10 3

and

Yout2 = h 22 - h12h21 = 0.0016+ 0.966×51 = 0.0289 S h11 +Z s2 500 + 1304.3 Z out2 =1/Yout2 = 34.6 Ω (c)

V1 = Z in1 1000 = = 0.5 Vs Z in1 + Z s 1000 + 1000 The load of stage 1 is the parallel combination of Zin2 and Zm. Thus

YL1 = Y in2 + 1/1500 = 1.1667 ×10-3 S

Hence

ZL1 = 1/ YL1 = 857.1 Ω V2 V1

stage 1

-50 = 1 × -h 21 = 1 × = - 39.47 Z in1 h22 + Y L1 1000 (0.1 + 1.1667) ×10-3


Prbs Chap 19, 1/13/02

V2 V1

P19-7

stage 2


51 = 1 × -h 21 = 1 × = 0.7762 Z in2 h22 + Y L2 2000 0.0016 + 1/32


Vout = 0.5 ×(- 39.47) ×(0.7762) = - 15.32 Vs (d) The input circuit consists of a series connection of Vs, Zs, C and Zin1. The remainder of the circuit is resistive and has no effect on the frequency response. The magnitude response is of the highpass type with 1 1 f 3dB = = = 79.58 Hz 2π(Rs + Rin1)C 2π(1000 + 1000)10−6

SOLUTION PROBLEM 19.46. (a) Since the currents through YL and h22 are the same, h22 = YL. (b) From current division, I2 =

YL h I ⇒ YL + h22 21 1

I2 Y h = L 21 . I1 YL + h22

I Y h (c) 150 = 2 = L 21 = 0.5h21 ⇒ h21 = 300. I1 YL + h22 V  −1 (d) h12 = 1  = = −0.5 . V2  I =0 2 1 I1 Zs 9 × 10 3 (e) = = = 0.9 ⇒ Zin = 1000 Ω. Given this quantity, Is Zs + Zin 9 × 10 3 + Zin h11 = Zin +

h12h21 150 = 1000 − = 400 Ω. h22 + YL 0.25

SOLUTION PROBLEM 19.47. Recall that V1   h11 I  = h  2   21

h12   I1  h22  V2 

Prbs Chap 19, 1/13/02

P19-8


V  (a) From this expression and specification 1, h12 = 1  =0. V2  I =0 1 (b) From the formula for Y out (equation 19.50), specification 2, and the result of part (a), we have Yout =

1 Zout

=

1 h h = h22 − 12 21 = h22 800 h11 + Z s

1 Hence, h22 = = 1.25 mS. 800 For maximum power transfer from amplifier to the load, Zout = 800 = b2 Z L = 8b 2 Therefore, b = 10. (c) and (d) Observe that Zin = h11 −

h12 h21 = h11 h22 + YL b 2

From specification 3 and voltage division, V1 24 Z in = = Vs 25 Zin + 40 Equivalently, Vs 25 40 = = 1+ V1 24 Zin Hence h11 = Z in = 40 × 24 = 960 Ω. (e) From equation 19.51, V − h21 −h21 −h Gv2 = 2 = −100 = = = 21 2 −3 −3 V1 2.4 Zin h22 + YL b 960 1.25 ×10 +1.25 × 10

(

Hence, h21 = 240 .

)

(

)

Prbs Chap 19, 1/13/02

P19-9


V2 (f) The power delivered to the load is PL = 2 and the power delivered to the amplifier is 800 V2 V2 Pamp = 1 = 1 . Therefore the power gain is Zin 960 2

PL 96  V2  4 =   = 1.2 ×10 Pamp 80  V1 

SOLUTION PROBLEM 19.48. Recall that V1   h11 I  = h  2   21

h12   I1  h22  V2 

V  (a) From this expression and specification 1, h12 = 1  = 0.01. V2  I =0 1 (b) For maximum power transfer from amplifier to the load, Zout = 800 = b2 Z L = 8b 2 Therefore, b = 10. Now we find Zin. From specification 3 and voltage division, V1 24 Z in = = Vs 25 Zin + 40

(1)

Equivalently, Vs 25 40 = =1 + V1 24 Zin

⇒ Zin = 40 × 24 = 960 Ω

Using the formula for Zin we have the following equation

Zin = 960 = h11 −

0.01h21 h22 + 1.25 × 10 −3

⇒ h11 = 960 + 0.01

h21 h22 + 1.25 × 10 −3

(2)

Prbs Chap 19, 1/13/02

P19-10


But from the given specs, V2 − h21 −h21 = −100 = = V1 Zin h22 + YL b2 960 h22 + 1.25 × 10 −3

(

)

(

)

which implies that

(h22

h21

)

+ 1.25 × 10−3

= 960 × 10 2

(3)

Substituting (3) into (2) allows us to solve for h11: h11 = 960 + 0.01× 960 × 102 = 1920 Ω Now let us rewrite equation (3) as: h21 − 960 × 10 2 h22 = 960 × 102 × 1.25 × 10 −3 = 120

(4)

Also Yout = 1.25 × 10−3 = h22 −

h12h21 0.01h21 = h22 − h11 + Zs 1960

Equivalently, 1960 × 1.25 × 10 −3 = 2.45 = 1960h22 − 0.01h21 Solving equations (4) and (5) simultaneously in MATLAB yields »A = [1 -960e2;-0.01 1960] A= 1.0000e+00 -9.6000e+04 -1.0000e-02 1.9600e+03 »b = [120; 2.45] b= 1.2000e+02 2.4500e+00 »x = A\b x=

(5)

Prbs Chap 19, 1/13/02

P19-11


4.7040e+02 3.6500e-03 »h21 = x(1); h22 = x(2); h = [1920 0.01;h21 h22] h= 1.9200e+03 1.0000e-02 4.7040e+02 3.6500e-03 We can verify these results as well as compute the overall amplifier gain using the following mfile: % two-port analysis in terms of h-parameters function [zin, zout] =twoport(h, zL, zs) ['twoport analysis using h-parameters'] h11= h(1,1); h12=h(1,2); h21=h(2,1); h22=h(2,2); zin = h11 - h12*h21/(h22+ 1/zL) yout= h22 - h12*h21/(h11+zs); zout= 1/yout v1tovs= zin/(zin+zs) v2tov1= -h21/(zin*(h22+1/zL)) v2tovs= v1tovs*v2tov1

»twoporth(h,ZL,Zs) ans = twoport analysis using h-parameters zin = 960 zout = 8.0000e+02 v1tovs = 9.6000e-01 v2tov1 = -100 v2tovs = -96

Prbs Chap 19, 1/13/02

P19-12


Hence the overall voltage gain is VL/Vs = –96/10 = –96 because of the transformer. Finally to compute power gains, »Vs = 1; Vin = 24/25; »VL = -9.6; »Pin = Vin^2/960 Pin = 9.6000e-04 »Pload = VL^2/8 Pload = 1.1520e+01 »Pgain = Pload/Pin Pgain = 12000

SOLUTION PROBLEM 19.49. −C sVπ + gmVπ −C s + gm I (a) h21 = 2 = = 1 I1 V =0  1  Cπ + C s + 2 + Cπ + C s Vπ  Rπ  Rπ 

(

(

)

)

    1  Rx + 1  I1  + Cπ + C s  Rπ  V1 1 (b) h11 = = = Rx + 1 I1 V =0 I1 + Cπ + C s 2 Rπ

(

)

(

)

(c) Under the condition that I1 = 0, V1 = Vπ. Using voltage division from V2 to Vπ:

Prbs Chap 19, 1/13/02

P19-13


    1  1   + Cπ s C s C s R  V V h12 = 1 = π= 1 π = = 1 1 1 V2 I =0 V2 + + Cπ s + C s Cπ + C s + 1 1 C s Rπ + Cπ s Rπ Rπ

(

I (d) h22 = 2 V

2

)

 1   1  gmVπ +  + Cπ s Vπ C s Cπ s + gm +   Rπ   Rπ  = = 1 1 I 1 =0 Cπ + C s + Cπ + C s + Rπ Rπ Vπ C s

(

)

(

)

SOLUTION PROBLEM 19.50: (a) Recall t-parameter relationship: V1   t11 I  = t  1   21

t12   V2  t22   −I2 

For the given network,V2 = ZL (−I 2 ) and V1 = t11V2 + t12 (−I 2 ) = t11ZL (− I2 ) + t12 (− I2 ) = (t11ZL + t12 )(−I 2 ) Further, I1 = t21V2 + t22 (− I2 ) = (t21ZL + t22 )(−I 2 ) Hence, V (t Z + t )(−I 2 ) Zin = 1 = 11 L 12 = I1 (t21ZL + t22 )(−I 2 )

t11ZL + t12 t21ZL + t 22

(b) For the output impedance relationship, from the t-parameter relationships V1 = Zs(− I1 ) = t11V2 + t12 (−I2 ) = − Zs (t21V2 + t 22 (− I2 )) Grouping V2 and I2 terms together on separate sides of the equation implies Thus

(t 21Zs + t11)V2 = (t22 Zs + t12 )I 2

Prbs Chap 19, 1/13/02

P19-14

V Zout = 2 = I2


t22Zs + t12 t21Zs + t11

SOLUTION PROBLEM 19.51: 45. From the z-parameter relationships V1 = z11I1 + z12 I2

and

⇒ V1 − z11I1 = z12 I2 = −z12 (− I2 )

V2 = z21I1 + z22I 2 ⇒ z21I1 = V2 + z22(− I2 )

These two equations in matrix form are:

Solving for [V1

I1 ] yields

1 −z11  V1  0 − z12   V2  0 z   I  = 1 z   −I   21   1   22   2 

T

V1  1  z21 I  = z  0  1 21 

z11  0 − z12   V2  1  z11 = 1  1 z22  − I2  z21  1

∆z   V2  z22  − I2 

SOLUTION PROBLEM 19.52: For figure 19.52a, by inspection V1   1 Z1   V2   I  =  0 1   −I   1   2  Therefore T is as indicated in the problem. For figure 19.52b, by inspection V1   1  I  = Y  1  2

0   V2  1  − I2 

Therefore T is as indicated in the problem.

SOLUTION PROBLEM 19.53: Here we use the results of problem 19.52: (a) For figure (a) 1 Tnew =  0 (b) For figure (b)

Z1   1 1  Y2

0 1+ Z1Y2 = 1  Y2

Z1  1 

Prbs Chap 19, 1/13/02

P19-15

1 Tnew =  Y2


0  1 Z1   1 Z1  =    1  0 1  Y2 1 + Z1Y2 

SOLUTION PROBLEM 19.54: By the properties of an ideal transformer, V1 = nV2

I1 =

−1  1 I 2 =   (− I2 ) n n

Therefore V1   n 0  V2   I  =  0 1 n  −I   1   2  with the t-parameters given by the 2x2 matrix.

SOLUTION PROBLEM 19.55: This problem uses the results of the previous two problems. (a) 1+ Z1Y2 Z1   n 0  n (1+ Z1Y2 ) Z1 n  T= =  1   0 1 n   nY2 1n   Y2 (b) n 0  1 + Z1Y2 T=  0 1 n   Y2

Z1  n (1+ Z1Y2 ) nZ1  =  1   Y2 n 1 n

SOLUTION PROBLEM 19.56: (a) 0.25 0   8 4   2 1  T= = 4   2 5   8 20  0 (b)  1 0  8 T=  1 R 1  2 (c) 1+ Z1Y2 T=  Y2

4  8 4  =  5  2 + 8 R 5 + 4 R 

Z1   8 4  1+ 0.25s 0.5s 8 4   8 + 3s 4 + 3.5s = = 1   2 5   0.5 1  2 5   6 7 

SOLUTION PROBLEM 19.57: For each 2-port of the form of figure P19.53a, we have that the tparameters are given by

Prbs Chap 19, 1/13/02

P19-16


1+ Z1Y2 Z1  T= 1   Y2 The given network consists of three such sections in cascade whose t-parameters are respectiely, 1 + s2 T1 =   s

1+ 1× 0.5s 1 1 + 0.5s 1 s  , T2 =  = , and 1  0.5s 1 1  0.5s 1    1 2  1 + 2s × 4 s 2s 1+ 2 s 2s T3 =  =  1  1  s 1  s 1  4   4 

Observe that 1 3 3 2 1  2  2 s + 2 s + 2 s + 1 s + s + 1 T1T2 =   1 2 3  s + s s +1   2 2  and the overall t-parameters are 1 5 3 4  3 9 2 3 4 3 2  4 s + 4 s + s + 4 s + 4 s + 1 s + 3s + 2s + 3s + 1 T = T1T2T3 =   1 4 3 3 3 2 7  s + s + s + s s3 + 3s2 + s + 1   4 4 4 4 

SOLUTION PROBLEM 19.58: This poblem is done primarily in MATLAB. Part (a) % The following code solves part (a) of the problem. % Parameter Specification t11= 0.895+j*0.022; t22= t11; t12= 40 + j*180; % t21= (t11*t22 -1)/t12; % The above formula follows because it is a reciprocal network. % The actual value is specified. t21= -2.6175e-05+j*1.1023e-03; t=[t11 t12; t21 t22] % Part (a) calculations vr= 115200 ir = 361 vs= t11*vr +t12*ir magvs=abs(vs) angvs= angle(vs)*180/pi

Prbs Chap 19, 1/13/02

P19-17

is=t21*vr+t22*ir magis= abs(is) angis= angle(is)*180/pi pscomp=vs*conj(is) ps=real(pscomp) pr=real(vr*conj(ir)) eff= pr/ps pf= ps/abs(pscomp) ploss= ps- pr The MATLAB output is as follows: T= 8.9500e-01 + 2.2000e-02i 4.0000e+01 + 1.8000e+02i -2.6175e-05 + 1.1023e-03i 8.9500e-01 + 2.2000e-02i vr = 115200 ir = 361 vs = 1.1754e+05 + 6.7514e+04i magvs = 1.3555e+05 angvs = 2.9872e+01 is = 3.2008e+02 + 1.3493e+02i magis = 3.4736e+02 angis = 2.2857e+01 pscomp = 4.6733e+07 + 5.7501e+06i ps = 4.6733e+07 pr = 41587200 eff = 8.8989e-01 pf = 9.9252e-01 ploss = 5.1458e+06 Part (b) % The following code solves part (b) of the problem. zL=500; zin= (t11*zL + t12)/( t21*zL + t22) yin = 1/zin vsnew=134000; iin= yin*vsnew psnew= vsnew^2*real(yin) m= inv(t) v2=m(1,1)*vsnew +m(1,2) *iin magv2=abs(v2) iload= m(2,1) *vsnew + m(2,2)* iin magild = abs(iload) % Check value of rload rload=abs(v2)/abs(iload) The MATLAB output for part (b) is:


Prbs Chap 19, 1/13/02

P19-18


zin = 4.8759e+02 - 1.0031e+02i yin = 1.9676e-03 + 4.0478e-04i iin = 2.6366e+02 + 5.4241e+01i psnew = 3.5331e+07 % m is the inverse of T-matrix. m= 8.9500e-01 + 2.2000e-02i -4.0000e+01 - 1.8000e+02i 2.6175e-05 - 1.1023e-03i 8.9500e-01 + 2.2000e-02i v2 = 1.1915e+05 − 4.6681e+04i magv2 = 1.2796e+05 iload = 2.3829e+02 - 9.3362e+01i magild = 2.5593e+02 rload = 5.0000e+02

SOLUTION PROBLEM 19.59. From the given information, the circuit is linear and reciprocal. (a) Here i2 (t) is the integral of i1 (t). Therefore, the new v1(t) is the integral of the old v 2 (t). The result for t ≥ 0 is: v1(t) = 3.005 − 3e −t + e −t[0.00865sin(500t) − 0.005cos(500t)]

[

= 3.005 − 3e −t + e −t −2 ⋅10 −5 cos(500t − π /6) + 0.01sin(500t − π /6) (b) From the problem statement

 π cos (s + 1) + 250  6

]

3 +5 s +1 ( s + 1)2 + (500)2 From reciprocity, z12 (s) = z21(s). For steady state analysis, we use phasors to obtain z21 (s) =

V1 = z12 ( j500)I2 =

 π cos (1+ j500) + 250 6

3 +5 1+ j500 (1+ j500) 2 + (500) 2

v1ss (t) = 2.505cos(500t − 30.15o ) V

SOLUTION PROBLEM 19.60. Writing loop equations we have: (i) For the left loop, V1 − aV2 − 3(I1 + I3 ) = 0 ⇒ V1 − aV2 = 3I1 + 3I3 (ii) For the right loop,

Prbs Chap 19, 1/13/02

P19-19


V2 + bI1 − (I 2 − I 3) = 0 ⇒ V2 = −bI1 + I2 − I3 (iii) For the middle loop, bI1 + (I3 − I 2 ) + I3 + 3(I1 + I3 ) = 0 ⇒ 0 = (b + 3)I1 − I 2 + 5I 3 Writing the first two equations in matrix form yields I  1 −a  V1   3 0 3  1  0 1  V  =  −b 1 −1 I 2    2      I3 

whose solution is V1  1 V  = 0  2  Hence

I  a  3 0 3   1   3 − ab I = 1  −b 1 −1  2   −b  I3 

I  a 3 − a  1  I 1 −1   2   I3 

V1   3− ab a   I1   3 − a + I V  =  −b 1   I2   −1  3  2 

From the third equation I3 = [−0.2b − 0.6

 I1  0.2]  I 2 

Thus V1    3 − ab V  =   −b  2  

a  3− a + [−0.2b − 0.6 1  −1 

  I1  0.2]     I2 

V1  −0.6b − 0.8ab + 0.6a + 1.2 0.6 + 0.8a  I1  V  =  −0.8b + 0.6 0.8   I2   2  For reciprocity,

z12 = z21 ⇒ a = −b

SOLUTION PROBLEM 19.61. (a) and (b) together: Observe that and

V1 − 0.5V1 − I1 = V2 ⇒ I1 = 0.5V1 − V2 ⇒ V1 = 2I1 + 2V2

The h-parameters are

I1 + I2 = 0.5(V2 − aI1 ) ⇒ I 2 = −(0.5a + 1)I1 + 0.5V2 2 2   H=  −(0.5a + 1) 0.5 

Prbs Chap 19, 1/13/02

P19-20

Reciprocity requires that Thus


h12 = −h21 ⇒ 2 = 0.5a + 1 ⇒ a = 2 2 2 H=  −2 0.5

SOLUTION PROBLEM 19.62. There are 2 corrections in the problem statement concerning the second set of expressions: (1) v1(t) = 2e-t - 1.5e -1.5t V (2) i2(t) = 0.5 e-1.5t A For both parts, recall, the y-parameters:  I1   y11 y12  V1  I  =  y y V   2   21 22  2  (a) Part-1: From the first set of given data (V2 = 0) 1 0.5 0.5 0.5 0.5(s + 2) I1 = , I 2 = , V1 = + = 2 s s +1 s + 1 (s + 1) (s + 1)2 Hence, I  2(s + 1)2 I  s +1 y11 = 1  = , y 21 = 2  = V1 V =0 s(s + 2) V1  V =0 s + 2 2 2 (a) Part-2: From the second set of given data (ZL = 1 Ω, I1 = 1 s , etc.), we have I2 =

0.5 2 1.5 0.5(s + 3) , V1 = − = s + 1.5 s + 1 s + 1.5 (s + 1)(s + 1.5)

For a terminated 2-port, I2 = Therefore

−V2 y 21 y 21 = −V2 = V1 = V ZL y 22 + YL y 22 + 1 1

V s +1 0.5(s + 3) s + 1.5 (s + 3) 1 y 22 = y 21 1 − 1 = × × − 1= −1 = I2 s + 2 (s + 1)(s + 1.5) 0.5 (s + 2) s+2

Also, I1 = y11V1 + y12V2 ⇒

y12 = −

I −y V I −y V y12 = 1 11 1 = 1 11 1 . Hence V2 −I 2

s + 1.5  1 2(s + 1)2 0.5(s + 3)  (2s + 3)(s + 2) 2(s + 1)(s + 3) × − =− +  0.5 s(s + 2) s(s + 2)  s s(s + 2) (s + 1)(s + 1.5) 

Prbs Chap 19, 1/13/02

P19-21

=


1 (s + 2)

In conclusion  2(s + 1)2  y11 y12   s(s + 2) y y  =  s + 1  21 22   s+2 

1   2(s + 1)2  (s + 2)  = 1  s 1  (s + 2)  s + 1   (s + 2) 

 1  1

(b) This is a straightforward application of the conversion table 19.1.

SOLUTION PROBLEM 19.63. (a) Consider figure (a). Write two mesh equations: 2 2 2  V1 = 16I1 + (I1 + I 2 ) + 4I1 =  20 +  I1 + I2 s s s 2 2 2   V2 = sI2 + (I1 + I 2 ) + 4I1 =  4 +  I1 +  s +  I2 s s s Therefore

1 20s + 2 2  Z=  s  4s + 2 s2 + 2 

Taking the inverse yields the y-parameters Y=

1  z22 ∆Z  −z21

 s2 + 2 −z12  1 −2  =    z11  20s2 + 2s + 32  −(4s + 2) 20s + 2 

where ∆Z =

(20s + 2)(s2 + 2)− 2(4 s + 2) 20s3 + 2s2 + 32s = = 20s2 + 2s + 32 s s

Finally, the h-parameters are given as  1 y11 H= − z21 z22

z12 z22  1  20s2 + 2s + 32 =  1 z22  s2 + 2  −(4 s + 2)

2  s

(b) Now consider figure (b). z11 is V1 when I2 = 0. But if I2 = 0, then because of the ideal transformer I1 = 0, meaning that the ratio is not defined. Hence the z-parameters do not exist. To find the h-parameters, observe that because of the ideal transformer, I2 = −0.5I1, I1 = −2I2 , and V pri = 0.5Vsec . Writing a mesh equation at the right mesh first we obtain V2 = RI 2 + Vsec + (I1 + I 2 )R = Vsec

Prbs Chap 19, 1/13/02

P19-22


Now writing a mesh equation on the left we have V1 = RI1 + V pri + (I1 + I 2 )R = 1.5RI1 + V pri = 1.5RI1 + 0.5V2 Therefore 1.5R 0.5  H=   −0.5 0  To obtain the y-parameters we use table 19.1: Y=

1  1 − h12  1  4 −2 = h11  h21 ∆h  6R  −2 1 

Note: the det[Y] = 0 implying again that the z-parameters do not exist. (c) For this network we consider it as a cascade (left to right) of an ideal transformer, the middle network of a transformer and an inductor across the top, and finally another ideal transformer. The t-parameters of these two ports are respectively: a 0  A B  1 b 0 T1 =  , T2 =  , T3 =      0 1 a C D  0 b To find T2 , we replace the mutually coupled inductors by the pi-equivalent circuit of figure ∆ 18.25c where L1 = 4 H, L2 = 9 H, M = k L1L2 = 3 H, and ∆ = 27. Thus Lleft = = 4.5 L2 − M ∆ ∆ H, Lright = = 27 H, and Ltop = = 9 H. Notice that the external 9 H inductor is in L1 − M M parallel with Ltop leading to Lpar = 4.5 H. The y-matrix of this new pi-network is by inspection: −1 4.5  1  2 4.5 1 12 −6  Ymid =  =  s −1 4.5 1 27 + 1 4.5  27s −6 7  To compute the t-parameters we have  − y 22  y 21 T2 =  −∆y   y 21

−1  y 21   7 6 4.5s − y11  = 8 27s 2  y 21 

Therefore a 0   7 6 4.5s 1 b 0  7a 6b 4.5abs T = T1T2T3 =    =   0 1 a  8 27s 2   0 b  8 27abs 2b a  From table 19.1, we obtain

Prbs Chap 19, 1/13/02

P19-23

 63a2  0.5a / b  H=  , Z = s 16 2  27ab −0.5a / b 4 27b    8  a 2 2.25s


27ab    12b2 −6ab 1 8 , Y =   27b 2  27a2b 2s  −6ab 7a2   4 

SOLUTION PROBLEM 19.64. (a) The defining equation for the g-parameters is:  I1   g11 V  =  g  2   21

g12  V1  g12   I2 

Because of the load impedance Z L, we have V2 = −ZL I 2 . Hence substituting for V2 in the second equation yields − g21V1 V2 = g21V1 + g22 I2 = − ZL I 2 ⇒ I 2 = g22 + ZL Substituting this equation into I1 = g11V1 + g12 I2 we obtain  g V   g g  I1 = g11V1 − g12  21 1  =  g11 − 12 21  V1  g22 + ZL   g22 + ZL  Therefore, g g Yin = g11 − 12 21 g22 + ZL (b) Because of the source impedance Zs , we have V1 = −Zs I1 or I1 = −YsV1 . Hence substituting for V1 in the first equation yields −g12 I2 I1 = g11V1 + g12 I2 = −YsV1 ⇒ V1 = g11 + Ys Substituting this equation into V2 = g21V1 + g22 I2 we obtain  −g I   g g  g g V2 = g21 12 2  + g22I 2 =  g22 − 12 21  I2 ⇒ Zout = g22 − 12 21  g11 + Ys   g11 + Ys  g11 + Ys (c)

Ys (g22 + ZL ) V Zin Ys Ys G1 = 1 = = = = Vs Zs + Zin Ys + Yin Y + g − g12g21 ( g22 + ZL )( g11 + Ys ) − g12g21 s 11 g22 + ZL

(d) Refer to figure P19.64, where V2 =

ZL V g Z g21V1 ⇒ G2 = 2 = 21 L . g22 + ZL V1 g22 + ZL

Prbs Chap 19, 1/13/02

P19-24


Ys (g22 + ZL ) g Z g21Ys ZL (e) Gv = G1G2 = × 21 L = (g22 + ZL )( g11 + Ys ) − g12 g21 g22 + ZL (g22 + ZL )(g11 + Ys ) − g12g21

SOLUTION PROBLEM 19.65. Recall  I1   g11 V  =  g  2   21

g12  V1  g22   I2 

For the given circuit V2 =

1 1 1 (I2 − 4V2 ) + V1 ⇒ V2 = V1 + I 2 4 2 8

Also V2 = V1 −

1 (I − sV1 ) 4 1

This implies that 1  1 1 I1 = (4 + s)V1 − 4V2 = (4 + s)V1 − 4  V1 + I2  = (2 + s)V1 − I2 2 8 2 Thus the g-parameter matrix is: 1  2 + s − 2  G= 1 1    2 8 

SOLUTION PROBLEM 19.66. We first convert the y-parameters to t-parameters using table 19.1:  − y 22  y 21 YN1 → TN 1 =  −∆y   y 21

−1  y 21   −0.1 −0.5  − y11  = −3.82 −25.1 y 21 

where ∆y = 50.2 × 0.2 − 2 × 1.2 = 7.64 . Now we convert the h-parameters to t-parameters using table 19.1:  −∆h  h HN 2 → TN 2 =  −h21  21  h21

−h11  h21  1  56 13 11.2 2.6 −1  = 5  2 1  =  0.4 0.2 h21 

where ∆h = 13 × 2 + 6 × 5 = 56 . To obtain the cascaded t-parameters we compute

Prbs Chap 19, 1/13/02

P19-25

Tcas = TN1TN 2 =


−1.8   −1.32 −0.36  1  −6.6 =   5 −264.12 −74.76  −52.824 −14.952

Thus, 603 1 t21ZL + t22 5 = 40.2 S Yin = = = 15 Zin t11ZL + t12 5 To obtain the voltage gain, we first convert the t-parameters back to y-parameters (table 19.1) and then use the derived voltage gain formula:  −1.32 Tcas =   −52.824 Hence,

−0.36  2   41.533 → Ycas =   S −14.952  2.7778 3.6667 

V − y 21 −2.7778 Gv = L = = = −0.66667 V1 y 22 + YL 3.6667 + 0.5

Alternately, one could consider the load as a 2-port, compute its t-parameters, construct 1 the overall t-parameters as a cascade of three networks, and then use Gv = . t11

SOLUTION PROBLEM 19.67. (a)

Since only one C or L , weh have in general: H(s) =

have that a = 0. Therefore, H(s) =

b c

K = d s+ c s+ c

Rth where Rth is the Thevenin resistance seen by the energy Rth C L storage element, we refer the reader to problem 19.69 which provides a general derivation with H(∞) arbitrary; hence this problem is the special case of H(∞) = 0. (b) To prove that

c

=

1

as + b a . Since H(∞) = 0 = , we cs + d c

or

(c) Here Rth = 2000/ /(50 + 1000/ /200)) = 195.49 Ω. Hence c

=

1 Rth C

= 25.58 × 10 6 rad/s

Prbs Chap 19, 1/13/02

P19-26


SOLUTION PROBLEM 19.68. (a) Because there is only one energy storage element, it follows that the most general form of as + b b the transfer function is: H(s) = . Since H(0) = 0 = , we have b = 0. Further, since the cs + d d transfer function is high pass, c ≠ 0, and  a s c as Ks H(s) = = = cs + d  d s + c s+  c 1 R (b) To prove that c = or th where Rth is the Thevenin resistance seen by the energy Rth C L storage element, we refer the reader to problem 19.69 which provides a general derivation with H(0) arbitrary; hence this problem is the special case of H(0) = 0. (c) Let us apply the result of example 19.1 to that part of the circuit to the right of the 1 kΩ resistor and call the associated resistance Zin. Here Z1 = ∞, Z2 = 2 kΩ, ZL = 100 Ω, and beta = 50. Hence Zin = (beta + 1) ZL = 5.1 kΩ

Thus Hence,

Rth = 200 + 1000 //(2000 + 5100) = 1076.5 Ω

c

=

1 Rth C

=

1 = 464.45 rad/s 1076.54 × 2 × 10−6

SOLUTION PROBLEM 19.69. In this problem we assume (i) a single input single output system and that linear circuit seen by the energy storage element has a Thevenin equivalent or a Norton equivalent. For simplicity we will presume the existence of a Thevenin equivalent. (a) The Thevenin equivalent seen by the dynamic element L or C consists of Zth(s) in series with Voc(s). Since the remainder network seen by L or C is non-dynamic (resistive, resistive with dependent sources and ideal op amps, etc), we have Zth(s) = Rth and Voc(s) = Ko×Input(s), R th and Ko being real constants.

Prbs Chap 19, 1/13/02

P19-27


By voltage division, VL =

Ls K 0s Voc = R × Input(s) Ls + Rth s + th L

(1)

and K0 1 RthC VC = 1 Cs Voc = 1 × Input(s) + Rth s+ Cs RthC

(2)

After VL(s) or V C(s) has been determined, we can find the Laplace transform of any other output (voltage or current) using the voltage source substitution theorem (chapter 6) and linearity (chapter 5): Ouput(s) = K1 × Input(s) + K 2 (VL (s)

or

VC (s))

(3)

For the case of VL(s),   K0 s   Ouput(s) =  K1 + K2 R  × Input(s)  s + th   L 

(4)

and for the case of VC(s), K0    RthC  Ouput(s) =  K1 + K2 1  × Input(s)  s+   Rth C 

(5)

For either case, the transfer function H(s) has the form H(s) =

Ouput(s) K3s + K4 = Input(s) s +ω c

(6)

Prbs Chap 19, 1/13/02

P19-28


R 1 where ω c = th or with K3 and K4 real constants. L RthC It remains to give K3 and K4 some physical interpretations. In (6), let s = ∞, we have K3s + K4  = K3  s + ωc  s=∞

H(∞) =

On the other hand, letting s = 0 in (6) produces H(0) =

K3 s + K 4  K = 4  s +ω c  s=0 ω c

Substituting K3 and K4 into (6), we obtain the desired result H(s) =

Ouput(s) H (∞)s +ω c H(0) = Input(s) s +ω c

(b) When s = ∞, the impedance of C is ZC (∞) = and the impedance of L is

(7)

1 =0 Cs  s=∞

ZL (∞) = Ls]s=∞ = ∞

Therefore in calculating H(∞), we may replace C by a short circuit and L by an open circuit. On the other hand, when s = 0, the impedance of C is ZC (0) =

1 =∞ Cs  s=0

and the iimpedance of L is ZL (0) = Ls ]s=0 = 0 Therefore in calculating H(0), we may replace C by an open circuit and L by a short circuit.

Prbs Chap 19, 1/13/02

P19-29


(c) For figure P19.69a, by inspection Rth = 3//(2 + 4) = 3//6 = 2 Ω ωc = 1/(Rth C) = 1/(2 × 0.5) = 1 H(∞) = 4/(2 + 4) = 2/3 H(0) = 4/(2 + 3 + 4) = 4/9 Therefore 2 2 4 2 s +  H(∞)s + ωc H(0) 3 s + 9 3  3 H(s) = = = s +ω c s+1 s +1 For figure P19.69b Rth = 3//(2 + 4) = 3//6 = 2 Ω ωc = Rth/ L = 2/2 =1 H(∞) = 4/(2 + 4) = 2/3 H(0) = 4/(2 + 3 + 4) = 4/9 Therefore 4 2 4 s + (s + 1.5) H(∞)s + ωc H(0) 9 3 9 H(s) = = = s +ω c s+1 s +1

SOLUTION PROBLEM 19.70. According to problem 19.68, the transfer function of the circuit is

H(s) =

Ks s+ c

1 where c = and Rth is the Thevenin resistance seen by the storage element C. To find Rth Rth C we make use of figure 19.4 and the associated formula. The details are in the MATLAB code below: »R1 = 200*1e3/1200 R1 = 1.6667e+02 »Z1 = R1+2e3 Z1 = 2.1667e+03 »Z3 = 100;

Prbs Chap 19, 1/13/02

P19-30


»beta = -50; »Zout = Z1/(1+beta) Zout = -4.4218e+01 »Rth = 2000 + Zout Rth = 1.9558e+03 »wc = 1/(Rth*2e-6) wc = 2.5565e+02 Hence c

=

1 = 255.65 rad/s 1955.8 × 2 × 10−6

SOLUTION PROBLEM 19.71. (a) Except for the terminating resistor, let the other element branches of the circuit be given by Ζ1 =

1 1 2s 1 , Y2 = = 2 + 1, and Z3 = 1 3s1 s 4s 2s + 2s

Consider the circuit

Here from problem 19.53a and from 19.52 a we obtain

Prbs Chap 19, 1/13/02

P19-31

2   V1  1+ 23  I  =  4s + 1  1  2s  4s2 + 1


 1  V3  1   V3  3s    and   = −I 3  −I 3  0 1  

1  V  2 s  − I  1  2 

which implies that

(16s2 + 6)  V2 

 V1  1  4s2 + 5/ 3 I  = 2   1  4s + 1  2s 

  −I  3s  2 2 4s + 3 

(b) GV (s) =

− y 21 y 22 + y L

(c) 5  3s 4s2 + 2  1 3s(4s + 1) t11 3 s(12s + 5) y 21 = − =− , y = = = , and YL = 1 22 t t12 16s2 + 6 16s2 + 6 16s2 + 6 12 2

(d) Hence, 3s(4s2 + 1) GV (s) =

2 − y 21 = 16s2 + 6 y 22 + y L s(12s + s)

16s2 + 6

= +1

3s(4 s2 + 1) 12s3 + 16s2 + 5s + 6

SOLUTION PROBLEM 19.72. For part (a), treating each capacitor as a short circuit yields the equivalent circuit below.

Prbs Chap 19, 1/13/02

P19-32


»h11T = 4.2e3; h12T = 0; h21T = 150; h22T = 0.1e-3; »hT = [h11T, h12T;h21T h22T]; »yT = htoy(hT) yT = 2.3810e-04 0 3.5714e-02 1.0000e-04 By inspection, the y11 parameter of the overall two-port consists of the sum of y11T plus the conductances of the two front end resistors. Also, the y22 parameter of the overall two port is y22T plus the conductance of the 4.7k Ω resistor. Hence, »y = yT + [1/1e4+1/1e5 0;0 1/4700] y= 3.4810e-04 0 3.5714e-02 3.1277e-04 Hence, the overall h-parameters are: »h = ytoh(y) h= 2.8728e+03 0 1.0260e+02 3.1277e-04 (b) »yout = h(2,2) - (h(1,2)*h(2,1)/(h(1,1)+50)) yout =

Prbs Chap 19, 1/13/02

P19-33


3.1277e-04 »Zout = 1/yout Zout = 3.1973e+03 »a = sqrt(8/Zout) a= 5.0021e-02 (c) To compute the gain we first need Zin . From equation 19.49, since h12 = 0, Zin = h11. » Zin = h(1,1) Zin = 2.8728e+03 From equation 19.51, »Gv2 = -h(2,1)/(Zin*(h(2,2)+h(2,2))) Gv2 = -5.7094e+01 From equation 19.52, »Gv1 = Zin/(Zin + 50) Gv1 = 9.8289e-01 To compute VL/Vs we use: »Gv = Gv1*Gv2*a Gv = -2.8071e+00 To compute VL/V1 we use: »Gvv = Gv2*a Gvv =

Prbs Chap 19, 1/13/02

P19-34


-2.8559e+00 (d) For these calculations, we assume Vs is normalized to 1 V. Since we are computing gains, we may do this without loss of generality. »Vs = 1; »Is = 1/(50 + Zin) Is = 3.4214e-04 Now we compute the normalized power delivered by the voltage source. »Psnorm = Vs*Is Psnorm = 3.4214e-04 Next we compute the normalized power absorbed by the load. »VL = Gv*1 VL = -2.8071e+00 »PLnorm = VL^2/8 PLnorm = 9.8496e-01 Next, the power gain from source to load is: »GpLs = PLnorm/Psnorm GpLs = 2.8788e+03 Further, we compute the power gain from input to the two port to the load as follows: »V1 = Vs*Zin/(Zin + 50) V1 =

Prbs Chap 19, 1/13/02

P19-35


9.8289e-01 »P1 = V1*Is P1 = 3.3629e-04 »GpL1 = PLnorm/P1 GpL1 = 2.9289e+03 (e) SPICE Simulation Because the frequency response is flat for freqency above 800 Hz, we only plotted up to 1.6k Hz. The circuit diagram reflects the load back to the primary of the ideal transformer. In general, this is not possible. Hence for a SPICE simulation, it is necessary to use one of the models given in Figure 18.15 consisting of two controlled sources. For this example, this is not necessary. Note however that the actual output voltage is 0.05 times the values on the graph given below. This simulation assumes a 1 V source voltage and the parameter of GVCCS is 0.035714. Notice that in this problem

Prbs Chap 19, 1/13/02

P19-36


Prb 19.72-Small Signal AC-3 +200.000

+400.000

+600.000

+800.000

Frequency +1.000k

+1.200k

+1.400k

(Hz)

+1.600k

+60.000

+50.000

+40.000

+30.000

+20.000

+10.000

+0.000e+000

MAG(V(IVM2))

(f) For this part, we change 100 µF to 10 µF. The resulting plot shows degradation of the low end frequency response. Prb 19.72-Small Signal AC-4 +200.000 +60.000

+50.000

+40.000

+30.000

+20.000

+10.000

+0.000e+000

MAG(V(IVM2))

+400.000

+600.000

+800.000

Frequency +1.000k

+1.200k

+1.400k

(Hz)

+1.600k

SOLUTIONS PROBLEMS CHAPTER 20 USEFUL MATLAB M-FILES FOR USE IN THE SOLUTION TO PROBLEMS IN THIS CHAPTER. Program 1: converts y-parameters to t-parameters % convert y parameters to t parameters function [t,t11,t12,t21,t22] = ytot(y) y11=y(1,1); y12=y(1,2); y21=y(2,1); y22=y(2,2); deltay= y11*y22-y12*y21; t11=-y22/y21; t12 = -1/y21; t21= -deltay/y21; t22= -y11/y21; t= [ t11 t12; t21 t22]; Program 2: Computes Zin, Zout, and gains using two port h-parameters. »% two-port analysis in terms of h-parameters »function [zin, zout] =twoport(h, zL, zs) »['twoport analysis using h-parameters'] »h11= h(1,1); h12=h(1,2); h21=h(2,1); h22=h(2,2); »zin = h11 - h12*h21/(h22+ 1/zL) »yout= h22 - h12*h21/(h11+zs); »zout= 1/yout »v1tovs= zin/(zin+zs) »v2tov1= -h21/(zin*(h22+1/zL)) »v2tovs= v1tovs*v2tov1 Program 3: Computes Zin, Zout, and gains using two port y-parameters. % two-port analysis in terms of y-parameters function [zin, zout] =twoporty(y, zL, zs) ['twoport analysis using y-parameters'] y11= y(1,1); y12=y(1,2); y21=y(2,1); y22=y(2,2); yin = y11 - y12*y21/(y22+ 1/zL) zin= 1/yin yout= y22 - y12*y21/(y11+1/zs) zout= 1/yout v1tovs= zin/(zin+zs) v2tov1= -y21/(y22+1/zL) v2tovs= v1tovs*v2tov1 Program 4: Computes Zin, Zout, and gains using two port t-parameters. % two-port analysis in terms of t-parameters function [zin, zout] =twoportt(t, zL, zs)

['analysis of terminated twoport using t-parameters'] t11= t(1,1); t12=t(1,2); t21=t(2,1); t22=t(2,2); zin= (t11*zL + t12)/(t21*zL + t22) zout= (t22*zs + t12)/(t21*zs + t11) v2tov1= zL/(t11*zL + t12) v1tovs= zin/(zin+zs) v2tovs= v2tov1*v1tovs Program 5: converts z-parameters to t-parameters %converting z to t paramters (same sormulas as %converting t to z parameters) function [t,t11,t12,t21,t22] = ztot(z) z11=z(1,1); z12=z(1,2); z21=z(2,1); z22=z(2,2); deltaz= z11*z22 - z12*z21; t11= z11/z21; t12= deltaz/z21; t21= 1/z21; t22= z22/z21; t= [ t11 t12; t21 t22];

SOLUTION 20.1. For network a, the z-parameters are by inspection R1 + R3 Za =   R3

R3  R2 + R3 

Similarly, for network b, the z-parameters are the same: R3  R1 + R3 Zb =  R2 + R3   R3 The interconnection of networks a and b conforms to figure 20.2b, which is a series interconnection. Hence, the new overall z-parameters are  R1 + R3 Znew = Za + Zb = 2   R3

R3  R2 + R3 

SOLUTION 20.2. For networks a and b, the y-parameters are by inspection  3 −2  Ya = Yb =  S −2 4  Hence, their z-parameters are the inverse of the y-parameter matrix: 1 4 Za = Zb =  8 2

2 Ω 3

The interconnection of networks a and b conforms to figure 20.2b, which is a series interconnection. Hence, the new overall z-parameters are Znew = Za + Zb =

1 4 2  Ω 4  2 3

SOLUTION 20.3. For network a consisting of the single inductor, s s Za =  Ω s s For network b, we have Zb = [Yb ]

−1

=

1 1 2 2

1  0.5 0.5  Ω = 4   1 2 

The interconnection of networks a and b conforms to figure 20.2b, which is a series interconnection. Hence, the new overall z-parameters are  s + 0.5 s + 0.5  Ω Znew = Za + Zb =  s + 2   s +1 SOLUTION 20.4. For network a consisting of the single capacitor, 1 1 1 Ω Za =  s 1 1 The interconnection of networks a and b conforms to figure 20.2b, which is a series interconnection. Hence, the new overall z-parameters are 1 1 + 0.5s 1+ 0.5s Ω Znew = Za + Zb =  s  1+ s 1 + 2s 

SOLUTION 20.5. For network a, the y-parameters are by inspection  3 −2  Ya =  S −2 3  Hence, their z-parameters are the inverse of the y-parameter matrix: Za =

1 3 2  0.6 0.4  Ω = 5 2 3  0.4 0.6

The interconnection of networks a and b conforms to figure 20.2b, which is a series interconnection. Hence, the new overall z-parameters are 1.6 Znew = Za + Zb =  0.4

0.9  Ω −0.4 

SOLUTION 20.6. For networks Na and Nb, the y-parameters are:  0.7 −0.2 YNa = YNb =  S  −0.2 0.7  Hence, their z-parameters are the inverse of the y-parameter matrix: ZNa = ZNb =

20  0.7 0.2 1 14 4  Ω = 9  0.2 0.7 9  4 14 

The network Na* has the same z-parameters as Na and continues to act as a two when series interconnected to another 2-port because of the transformer. Hence, the interconnection of networks Na* and Nb forms a valid series interconnection in which cas Znew = ZNa * + ZNb =

2 14 4  Ω 9  4 14 

SOLUTION 20.7. For network Nb, the y-parameters are the same as in problem 20.6, i.e.,  0.7 −0.2  1 14 4  Ω YNb =  S and ZNb =   9  4 14  −0.2 0.7 

For network Na, consider the purely resistive part without the transformer. The y-parameters of this part are half the y-parameters of Nb, i.e., 0.35 −0.1 YR =  S −0.1 0.35 In MATLAB we use the m-file which converts y-parameters to t-parameters: »y = [0.35 -0.1;-0.1 0.35]; »t = ytot(y) t= 3.5000e+00 1.0000e+01 1.1250e+00 3.5000e+00

From figure 19.34, the transformer has to parameters with a = 2 given by 1 a 0  0.5 0  ttrans =  =    0 a  0 2  »ttrans = [0.5 0;0 2]; »tNa = ttrans*t tNa = 1.7500e+00 5.0000e+00 2.2500e+00 7.0000e+00 » »zNa = ttoz(tNa) zNa = 7.7778e-01 4.4444e-01 4.4444e-01 3.1111e+00 » »zNb = [14 4;4 14]/9; »znew = zNa + zNb znew = 2.3333e+00 8.8889e-01 8.8889e-01 4.6667e+00 »

SOLUTION 20.8. This problem is identical numerically to problem 20.6. Here however the isolation transformer is on the right hand side which makes no difference to the interconnection. Therefore, Znew = ZNa * + ZNb = SOLUTION 20.9. Figure 20.4 is

2 14 4  Ω 9  4 14 

Using the values in figure 20.3, we obtain the following mesh equation matrix V1   4 + R1    V2  =  2  0   − R1

2 4 + R2 R2

−R1   I1    R2   I 2  R1 + R2   I 3 

For I3 to be zero, the third equation implies that R1I1 = R2I2. Therefore,  R 2R  V1 = (4 + R1) I1 + 2 1 I1 =  4 + R1 + 1  I1 R2  R2  Similarly  R (4 + R2 )R1  V2 = 2I1 + ( 4 + R2 ) 1 I1 + =  2 +  I1 R2  R2 Therefore  2R1   4 + R1 +   R2  V1 (4R2 + R1R2 + 2R1) = 42 = 7 = = V2  ( 4 + R2 )R1  (2R2 + R1R2 + 4R1) 48 8 2 +   R2  as was to be shown. V 7 7 Now suppose, 1 = or equivalently V1 = V2 . With specific values V2 8 8 V1   7 8 10 2 −6  I1         V2  = V2  1  =  2 7 3   I 2   0   0  −6 3 9   I3  In MATLAB »Z = [10 2 -6;2 7 3;-6 3 9]; »b = [7/8 1 0]'; »I = inv(Z)*b I= 6.2500e-02 1.2500e-01

2.7756e-17 Thus  I1  0.0625      I 2  = V2  0.124   I3   0  SOLUTION 20.10 (a) Write two loop equations assuming I1 and I2 are the usual port currents. Here V1 = Z1I1 + V13 + Z3( I1 + I2 ) and

V2 = Z2 I2 + V23 + Z3(I1 + I 2 )

which in matrix form are

V1   Z1 + Z3 V  =  Z  2  3

Z3   I1  V13  + Z2 + Z3   I 2  V23

However, V13   I1  V  = Z A  I   23   2 Therefore V1   Z1 + Z3 V  =  Z  2  3

Z3   I1   I1    Z1 + Z3 + ZA   =      Z2 + Z3 I 2   I2    Z 3

Z3    I1  + Z A    Z2 + Z3   I 2 

(b) The procedure of part (a) is repeated to produce the same result. REMARK: this problem states that the two networks are equivalent two ports. Thus the configurations can be used interchangeably.

SOLUTION 20.11. Because of the isolation transformers, the overall Z-parameters are the sum of the three component Z-parameters. For Na, 2 1 Za =  Ω 1 2 For Nb,

For Nc,

8 1 Zb =  Ω 1 5 2 1 Zc = Za =  Ω 1 2

Therefore 12 3 Z = Za + Zb + Zc =  Ω  3 9 SOLUTION 20.12. (a) Given the Z-parameters of N, then YN =

ZN−1

−1  7 2  3 −2  = =  S 10 3 −10 7 

The 1-F capacitor considered as a two port has y-parameters  s − s YC =  S − s s  Therefore, the parallel connection of the two ports has y-parameters −(s + 2)  s −s  3 −2   s + 3 S Y = YN + YC =  + =   s + 7  −s s  −10 7  −(s + 10) (b) For this part, the same reasoning applies with s replaced by 1/s. SOLUTION 20.13. The isolation transformer allows for the valid parallel connection of N *a and Nb in the sense that the overall y-parameters are the sum of the individual y-parameters. Further, because the ideal transformer is 1:1 with the standard dot locations, the y-parameters of N *a are those of N a . Further, the y-parameters of Nb are the same as those of N a as the circuits are simple vertical flips of each other. Therefore  0.7 −0.2 Ya* = Ya =   = Yb −0.2 0.7  Hence, the overall y-parameters are:  0.7 −0.2  1.4 −0.4  Y = 2× =  S −0.2 0.7  −0.4 1.4  True-False: FALSE because the connection does not conform to figure 20.2a.

SOLUTION 20.14. From the previous example, the y-parameters of Nb are

 0.7 −0.2  Yb =  S −0.2 0.7  The resistance values of the resistive part of Na are twice those of Nb. Hence, the y-parameters of the resistive part are half those of Nb, i.e.,  I1'   V '  0.35 −0.1 V1'    = Ya,R  1  =    S I 2  V2  −0.1 0.35 V2  We obtain the y-parameters of Na by considering the effect of the transformer on these yparameters. Observe that I1 = 2I1' and V1 = 0.5V1' Hence  I1   0.7 −0.2  V1'   0.7 −0.2  2V1  1.4 −0.2  V1  I  =  −0.1 0.35   = −0.1 0.35   V  =  −0.2 0.35 V   2   V2    2    2  Thus

It follows that

SOLUTION 20.15. (a)

 1.4 −0.2  Ya =  S −0.2 0.35   2.1 −0.4  Y = Ya + Yb =  S −0.4 1.05 

Prob 20.15-Small Signal AC-10 +100.000m

+200.000m

Frequency

+300.000m

+1.000

+800.000m

+600.000m

+400.000m

+200.000m

+0.000e+000

MAG(V(IVM))

(b)

C0 15.9n

C1 15.9n

R1 500

R 1k

R0 1k

V 0

IVm C 31.8n

+400.000m

(Hz)

+500.000m

Prob 20.15-Small Signal AC-11 +10.000k

+20.000k

Frequency +30.000k

+1.000

+800.000m

+600.000m

+400.000m

+200.000m

+0.000e+000

MAG(V(IVM))

SOLUTION 20.16. (a) For network Na, the y-parameters by inspection are: 0  G1 + j C1 0   2 + j10.21 mS Ya =  =  G0   95 0.07143  gm For network Nb, the y-parameters by inspection are:  1 −1  1 −1 Yb = G f + j C2  = (0.8333 + j1.021)   mS  −1 1   −1 1 

(

Therefore

)

2.8333 + j11.23 −0.8333− j1.021 Y = Ya + Yb =   mS 94.167 − j1.021 0.9048 + j1.021 

(b), (c), and (d). Here we use the MATLAB m-file for two port analysis in terms of yparameters:

(Hz)

»zs = 50; zL = 50; »twoporty(Y, zL, zs) ans = twoport analysis using y-parameters yin = 6.8501e-03 + 1.5594e-02i zin = 2.3614e+01 - 5.3756e+01i yout = 5.3617e-03 + 3.0023e-03i zout = 1.4199e+02 - 7.9506e+01i v1tovs = 5.5701e-01 - 3.2349e-01i v2tov1 = -4.4915e+00 + 2.6821e-01i v2tovs = -2.4150e+00 + 1.6023e+00i »

SOLUTION 20.17. The t-parameters of the LR circuit follow from problem 19.53 with Z1 = Ls = s Ω and Z2 = 0.5 Ω:

Therefore

1+ 2s s TLR =  1  2  5 + 2s 2 + s Tcascade = T * TLR =    5 + 2s 2 + s

SOLUTION 20.18. This problem can be solved in many ways. Here we emphasize the cascade nature of the two ports. »% The y-parameters of Nb are: »Yb = [8 2;20 6]; »% The z-parameters of Na are: »Za = [0.75 -0.25;-2.5 1]; »% The t-parameters of Na are: »Ta = ztot(Za) Ta = -3.0000e-01 -5.0000e-02 -4.0000e-01 -4.0000e-01 »% The t-parameters of Nb are: »Tb = ytot(Yb) Tb =

-3.0000e-01 -5.0000e-02 -4.0000e-01 -4.0000e-01 »% The t-parameters of the cascaded two port are: » »Tab = Ta*Tb Tab = 1.1000e-01 3.5000e-02 2.8000e-01 1.8000e-01 »% Doing a t-parameter analysis we obtain: »twoportt(Tab,0.25,0.5) ans = analysis of terminated twoport using t-parameters zin = 2.5000e-01 zout = 5.0000e-01 v2tov1 = 4 v1tovs = 3.3333e-01 v2tovs = 1.3333e+00 ans = 2.5000e-01 »

SOLUTION 20.19. Use the h-parameter analysis m-file. »H1 = [1e3 0.001;50 6e-5]; H2 = [1e3 0.99;-5 8e-4]; »zL = 100; zs = 2e3; zm = 10e3; »% For part (a), the common collector stage: »twoporth(H2,zL,0) ans = twoport analysis using h-parameters zin = 1.4583e+03 zout = 1.7391e+02 v1tovs = 1 v2tov1 = 3.1746e-01 v2tovs = 3.1746e-01

Remark: in the above, v2 is Vout and v1 is vm; zout has no signifigance. »% We now compute the load on the first stage. »zin2 = 1.4583e+03; »zLm = zm*zin2/(zin2 + zm) zLm = 1.2727e+03 »% zLm is the load impedance on stage 1. »% For part (a), the common emitter stage: »twoporth(H1,zLm,zs) ans = twoport analysis using h-parameters zin = 9.4088e+02 zout = 2.3077e+04 v1tovs = 3.1993e-01 v2tov1 = -6.2835e+01 v2tovs = -2.0103e+01 Remark: in the above, v2 is vm and v1 is Vin; zout is the output impedance of stage 1. Conclusion: the input impedance to stage 1 is 940.88 Ω and the input impedance to stage 2 is 1.4583 kΩ. (b) From the above output and remark, Vm/Vin = -62.835. Further Vout/Vm = 0.31746. (c) From above Vm/Vs = -20.103. Therefore Vout/Vs = Vout/Vm * Vm/Vs = -6.38. (d) »zout1 = 2.3077e+04; »zs2 = zm*zout1/(zm + zout1) zs2 = 6.9768e+03 »twoporth(H2,zL,zs2) ans = twoport analysis using h-parameters zin = 1.4583e+03 zout = 7.0395e+02 v1tovs =

1.7289e-01 v2tov1 = 3.1746e-01 v2tovs = 5.4885e-02 Conclusion: Zout of amplifier is 704 Ω. SOLUTION 20.20 (a) Using MATLAB »na = 1.1514; »nb = 3.4012; »Zlprime = nb^2*75 + j*1042.94 ZLprime = 8.6761e+02 + 1.0429e+03i »Zsprime = 75/na^2 + j*30 Zsprime = 5.6573e+01 + 3.0000e+01i (b) Since the h-parameters of the transistors are given, we can again use MATLAB and the mfile twoporth defined earlier. Hence: »h = [ 60-j*50 0.01; -j*2 0.0005+j*0.0004]; »[Zin, Zout] = twoporth(h,ZLprime, Zsprime) Zin = 5.6569e+01 - 3.0000e+01i Zout = 8.6763e+02 - 1.0429e+03i (c) Observe that Zin = 5.6569e+01 - 3.0000e+01i and Zsprime = 5.6573e+01 + 3.0000e+01i, which are clearly conjugates of each other. Further, Zout = 8.6763e+02 - 1.0429e+03i and ZLprime = 8.6761e+02 + 1.0429e+03i, which are also conjugates of each other. Hence maximum power is transferred into and out of the transistor. (d) For this part, we change all cascaded two ports to t-parameters. Specifically, t0 = [1 75;0 1]; t1 = [na 0; 0 1/na]; t2 = [ 1 j*30; 0 1]; t3 = htot(h); t4 = [ 1 j*1042.9; 0 1]; t5 = [nb 0; 0 1/nb]; t6 = [1 0;1/75 1]; t = t0*t1*t2*t3*t4*t5*t6

t= 2.2176e-01 - 2.5959e-01i 8.3160e+00 - 9.7351e+00i 1.4785e-03 - 1.7307e-03i 6.6577e-02 - 7.4415e-02i »gain = 1/t(1,1) gain = 1.9024e+00 + 2.2270e+00i »gainmag = abs(gain) gainmag = 2.9289e+00 »gainangle = angle(gain)*180/pi gainangle = 4.9494e+01 In this case, Vout/Vs = gain = 1.9024e+00 + 2.2270e+00i = 2.9289∠49.494

o

SOLUTION 20.21. (a) »Y2N = [1 0;20.1 0]*1e-3; »Y10k = [1 -1;-1 1]*1e-4; »Yshade = Y2N + Y10k Yshade = 1.1000e-03 -1.0000e-04 2.0000e-02 1.0000e-04 » (b) This is a series connection of two 2-ports. Hence we first convert the answer of part (a) to zparameters. »Y2N = [1 0;20.1 0]*1e-3; »Y10k = [1 -1;-1 1]*1e-4; »Yshade = Y2N + Y10k Yshade = 1.1000e-03 -1.0000e-04 2.0000e-02 1.0000e-04 »Zshade = inv(Yshade) Zshade = 4.7393e+01 4.7393e+01 -9.4787e+03 5.2133e+02 »Z1k = [1 1;1 1]*1e3; »Zdashed = Zshade + Z1k Zdashed = 1.0474e+03 1.0474e+03 -8.4787e+03 1.5213e+03 » (c) »twoportz(Zdashed,1e12,1e3)

ans = twoport analysis using z-parameters zin = 1.0474e+03 zout = 5.8588e+03 v1tovs = 5.1157e-01 v2tov1 = -8.0950e+00 v2tovs = -4.1412e+00 Conclusion: Vout/Vs = -4.1412.

SOLUTION 20.22. »Z = [3 1;5 2]*1e3; »Y = inv(Z) Y= 2.0000e-03 -1.0000e-03 -5.0000e-03 3.0000e-03 »% Consider the parallel connection of Y with the 1 kΩ resistor »Y1 = Y + [1 -1;-1 1]*1e-3 Y1 = 3.0000e-03 -2.0000e-03 -6.0000e-03 4.0000e-03 » »% Now consider Y/Z in parallel with 1 kΩ connected between B and C »Y2 = Y + [0 0;0 1]*1e-3 Y2 = 2.0000e-03 -1.0000e-03 -5.0000e-03 4.0000e-03 » »% This combination is in series with another 1 kΩ resistor »% Hence we need to compute Z2 first »Z2 = inv(Y2) Z2 = 1.3333e+03 3.3333e+02 1.6667e+03 6.6667e+02 »% Now we compute the series combo of Z2 and the 1 kΩ resistor »Z3 = Z2 + [1 1;1 1]*1e3 Z3 = 2.3333e+03 1.3333e+03 2.6667e+03 1.6667e+03 »

» »% We now convert Y1 and Z3 to t-parameters and then multiply »% together to obtain the overall t-parameters » »T1 = ytot(Y1) T1 = 6.6667e-01 1.6667e+02 0 5.0000e-01 »T3 = ztot(Z3) T3 = 8.7500e-01 1.2500e+02 3.7500e-04 6.2500e-01 »T = T1*T3 T= 6.4583e-01 1.8750e+02 1.8750e-04 3.1250e-01 »Y = ttoy(T) Y= 1.6667e-03 -8.8889e-04 -5.3333e-03 3.4444e-03 » SOLUTION 20.23. Refer to figure 20.13 for all problems. (a) By inspection,  I1   6 −4 −2   V1       I 2  =  −4 4 + 5s −5s  V2   I3   −2 −5s 2 + 5s V3  The 3x3 coefficient matrix is the desired Yind(s). (b) Writing the nodal equation by inspection yields 0 −0.25 −0.25  V1   I1   0.5 I   0 1 −0.5 −0.5  V2   2 =   I3   −0.25 −0.5 1.25 −0.5  V3        0   −0.25 −0.5 −0.5 1.25 V4  Using MATLAB, »W11=[0.5, 0 -0.25;0 1 -0.5;-0.25 -0.5 1.25] W11 = 5.0000e-01 0 -2.5000e-01 0 1.0000e+00 -5.0000e-01 -2.5000e-01 -5.0000e-01 1.2500e+00 »W12 = [-0.25 -0.5 -0.5]'

W12 = -2.5000e-01 -5.0000e-01 -5.0000e-01 »W21 = [-0.25 -0.5 -0.5] W21 = -2.5000e-01 -5.0000e-01 -5.0000e-01 »W22 = 1.25 W22 = 1.2500e+00 »Yind = W11 - W12*inv(W22)*W21 Yind = 4.5000e-01 -1.0000e-01 -3.5000e-01 -1.0000e-01 8.0000e-01 -7.0000e-01 -3.5000e-01 -7.0000e-01 1.0500e+00 » (c) Again, writing the nodal equation by inspection yields 0 0 −5  V1   I1   5 I   0 6 −7 1  V2   2 =   I3   0 −4 8 −4  V3        0   −5 −2 −1 8 V4  Again, using MATLAB, »W11 = [5 0 0;0 6 -7;0 -4 8] W11 = 5 0 0 0 6 -7 0 -4 8 »W12 = [-5 1 -4]' W12 = -5 1 -4 »W21 = [-5 -2 -1] W21 = -5 -2 -1 »W22 = 8; »Yind = W11 - W12*inv(W22)*W21 Yind = 1.8750e+00 -1.2500e+00 -6.2500e-01 6.2500e-01 6.2500e+00 -6.8750e+00 -2.5000e+00 -5.0000e+00 7.5000e+00 »

(d) This part is similar to part (a) as it does not require the method of matrix partitioning. By inspection, IG  (CGD + CGS )s −CGD s −CGS s  VG       CGD s − gm  VD  I D  =  −CGD s + gm  I S   −CGS s − gm 0 CGS s + gm   VS  The 3x3 coefficient matrix is the desired Yind(s). SOLUTION 20.24. (a) With regard to the given information, the associated indefinite admittance matrix is the coefficient matrix in the following nodal equation given reference to figure 20.13: IG   0 0 0  VG       I D  =  gm 0 − gm  VD   I S  − gm 0 gm   VS  We use property 5 to compute the remaining answers: (a)-(a): YGD is as given.  gm 0 (a)-(b): YSD =  S −gm 0 0   0 (a)-(c): YGS =  S  −gm gm   0 gm  (a)-(d): YDG =  S 0 0  0 − gm  (a)-(e): YDS =  S 0 gm   gm − gm  (a)-(e): YSG =  S 0  0 (b) Transmission from port-1 to port-2 occurs when the 2-1 entry of the 2-port y-matrix is nonzero. Hence, the following all have the desired transmission: YGD , YSD , YGS . SOLUTION 20.25. Using the zero-sum properties of the rows and columns, we have by inspection:  40 2 −42   40 2 −42     Yind =  ? ? −50  =  30 20 −50 S  ? −22 92  −70 −22 92 

 8 ? ?   8 −2 −6      SOLUTION 20.26. (a) Yind =  −2 9 −7 =  −2 9 −7  S, where the second equality arises  −6 ? ?   −6 −7 13  because a purely resistive network has a symmetric indefinite admittance matrix. (b) Construct a common ground 2-port with input terminal C, common terminal A, and output terminal B: 13 −7  YCB =  S −7 9  Therefore, since the output terminal B is shorted, YL = ∞, i.e., y y Yin = y11 − 12 21 = 13 S y 22 + YL Hence Zin =

1 Ω 13

(c) Construct a common ground 2-port with input terminal A, common terminal B, and output terminal C:  8 −6 YAC =  S  −6 13  Hence, the required voltage gain is V − y 21 − y 21 6 GV = 2 = = = V1 y 22 + YL y 22 13 Therefore, V2 =

6 V. 13

SOLUTION 20.27. (a) Using the zero-sum property of the rows and columns of an indefinite admittance matrix we can write down by inspection (in mS)  IB   1 −1 0   VB        IE  = −100 100.1 −0.1  VE   IC   99 99.1 0.1   VC 

(b) The y-parameters (also in mS) of the common emitter configuration are easily computed as 0   VBE   IB   1  I  = 99 0.1  V   C    CE  Hence in MATLAB »Y = [1 0;99 0.1]; »Z = inv(Y) Z= 1.0000e+00 0 -9.9000e+02 1.0000e+01 where Z is in kΩ. It follows that VCE = −990I B + 10IC where IB and IC are in mA and VCE is in volts.

SOLUTION 20.28. Expanding the given y-parameter matrix into an indefinite admittance matrix yields IG  VG   0.2 + j2.5 −0.01− j0.65 −0.19 − j1.85  VG         0.05 + j0.8 −3.15 − j0.15  VD  I D  = Yind VD  =  3.1− j0.65  I S   VS   −3.3 − j1.85 −0.04 − j0.15 3.34 + j2   VS  By inspection, with G as the common terminal, S as the input terminal, and D as the output terminal, we obtain, −0.04 − j0.15   3.34 + j2 Ynew =   mS  −3.15 − j0.15 0.05 + j0.8  SOLUTION 20.29. Here we use the zero-sum properties of the columns and rows to complete the indefinite admittance matrix:  −s ? ?   −s −2s − 1 3s + 1     Yind =  s − 1 2 ?  = s − 1 2 −s − 1 S  1 2s − 1 −2s  1 2s − 1 −2s 

 −s −2s − 1 In figure (b), the top 2-port has y top =  S and the bottom 2-port has y-parameters 2  s − 1  −2s 2s − 1 S. Since these 2-ports are connected in parallel, the overall 2-port yybot =  2   −s − 1 parameters are  −s −2s − 1  −2s 2s − 1  −3s −2  S y = y top + y bot =  + = 2   −s − 1 2   −2 4  s − 1 SOLUTION 20.30. From problem 28, IG  VG   0.2 + j2.5 −0.01− j0.65 −0.19 − j1.85  VG         0.05 + j0.8 −3.15 − j0.15  VD  I D  = Yind VD  =  3.1− j0.65  I S   VS   −3.3 − j1.85 −0.04 − j0.15 3.34 + j2   VS  Hence

 0.2 + j2.5 −0.19 − j1.85 S yGS =  3.34 + j2  −3.3− j1.85

SOLUTION 20.31. (a) Here, by inspection we can compute the indefinite admittance matrix as the coefficient matrix of the following nodal equations: I A  V A  Y1 + Y2 + 2 −Y2      −Y2  IB  = Yind VB  =  Y2 − 2 IC  VC   −Y1 − 2Y2 2Y2 (b)

(c)

−Y1 − 2 V A    2  VB  Y1  VC 

Y1 + Y2 + 2 −Y2  y AB =  −Y2   Y2 − 2 Y1 + Y2 + 2 −Y1 − 2  y AC =  Y1   −Y1 − 2Y2

SOLUTION 20.32. (a) Here, by inspection we can compute the indefinite admittance matrix as the coefficient matrix of the following nodal equations:

I A  V A   Y1 + Y2 −Y2 −Y1  VA         −Y3 − gm  VB   IB  = Yind VB  =  −Y2 + gm Y2 + Y3 IC  VC   −Y1 − gm −Y3 Y1 + Y3 + gm  VC  (b)

−Y2   Y1 + Y2 y AB =    −Y2 + gm Y2 + Y3 

(c)

−Y1  Y1 + Y2  y AC =    −Y1 − gm Y1 + Y3 + gm 

SOLUTION 20.33. (a) Here the nodal equation matrix is: 0 0 −8  VA  I A  VA   8 I  W       B  =  11 W12  VB  =  0 2 + 2s −2s − 10 8  VB  IC  W 21 W22  VC   0 −2s 16 + 2s −16  VC         −2 −6 16 VD   0  VD  −8 where VD is the internal node voltage. Using the method of matrix partitioning, −1 Yind = W11 − W12W22 W 21

8 0 0   −8    1   = 0 2 + 2s −2s − 10  −  8 [−8 −2 −6] 16 0 −2s −16 16 + 2s 

8 0 0  −0.5 4 −1 −3        = 0 2 + 2s −2s − 10  −  0.5 [−8 −2 −6] =  4 3 + 2s −2s − 7  S 0 −2s −8 −2s − 2 10 + 2s  16 + 2s   −1  (b) When C is grounded, 4 y AB =  4 (c) When B is grounded

−1  1  3 + 2s 1  S and zAB = Ω  3 + 2s 8s + 16  −4 4  −3  4 y AC =  S  −8 10 + 2s

SOLUTION 20.34. (a) Writing the usual node equations we have,

−s 0 −0.5   VA  I A  VA   s + 0.5 I  W    s + 0.5 0 −0.5   VB   B  =  11 W12  VB  =  −s IC  W 21 W22  VC   0 0 0.5s −0.5s   VC         −0.5 −0.5s 0.5s + 1 VD   0  VD   −0.5 Using the method of matrix partitioning, −1 Yind = W11 − W12W22 W 21

 s + 0.5 −s 0  1   0.5   =  −s s + 0.5 0  − 1 [1 1 s] s+2   0  s 0 0.5s

 s + 0.5 −s 0  1 1 s    0.5   =  −s s + 0.5 0 − 1 1 s  s+2  0  s s s2  0 0.5s s2 + 2.5s + 0.5 1  2 =  − s − 2s − 0.5 s+2 −0.5s 

−s2 − 2s − 0.5 −0.5s  s2 + 2.5s + 0.5 −0.5s S −0.5s s 

(b) Here y AB =

1  s2 + 2.5s + 0.5 − s2 − 2s − 0.5   S s + 2  −s2 − 2s − 0.5 s2 + 2.5s + 0.5 

SOLUTION 20.35. (a) Rule 1: Consider the two networks NA (3 external nodes) and NB (4 external nodes) given below:

The two indefinite admittance matrices are

G1 + G2  YindN A =  −G2  −G1

−G2 G2 + G3 −G3

G1 + G2 −G1   −G  2 −G3  and YindN B =  −G1  G1 + G3   0

−G2 −G1 G2 + G3 −G3 −G3 G1 + G3 0 0

0 0 0  0

Observe that YindN B can be obtained from YindN A by adding a column of zeros and a row of zeros to form a 4x4 matrix. (b) Rule 2: Consider two networks NA and NB and a third network NC which combines elements of NA and NB as given below:

The corresponding indefinite admittance matrices are:  6 −2 −4   5 −3    YindN A = −2 3 −1 , YindN B =  −3 5 −4 −1 5  −2 −2 Clearly, YindN C = YindN A (c) Rule 3: Consider the 3-terminal network

−2  11 −5 −6    −2  , and YindN C = −5 8 −3 . −6 −3 9  4  + YindN B

with indefinite admittance matrix  5 −3 −2   YindN A = −3 5 −2 S −2 −2 4  If we move node 3 inside to form a 2-terminal network and labeled as NB,

then from nodal analysis we have  I1   5 −3 −2  V1       I 2  =  −3 5 −2 V2   0   −2 −2 4  V3  Using the method of matrix partitioning,  5 −3 1 −2 4 YindN B =  −  [−2 −2] =   −3 5  4 −2  −4

−4  S 4 

This computation is the one given by the formula in the problem. To see that this is correct, we observe that the internal simplification of NB leads to the following:

SOLUTION 20.36. Part (a) »Yinda= [1/2 -1/4 0 -1/4 0; -1/4 1/2 0 0 -1/4; ... 0 0 0 0 0; -1/4 0 0 1/2 -1/4; 0 -1/4 0 -1/4 1/2] Yinda = 5.0000e-01 -2.5000e-01 -2.5000e-01 5.0000e-01 0 0 -2.5000e-01 0 0 -2.5000e-01

0 -2.5000e-01 0 0 0 -2.5000e-01 0 0 0 0 5.0000e-01 -2.5000e-01 0 -2.5000e-01 5.0000e-01

»Yindb= [0 0 0 0 0; 0 0 0 0 0; 0 0 3/4 -1/2 -1/4; ... 0 0 -1/2 5/8 -1/8; 0 0 -1/4 -1/8 3/8] Yindb = 0 0 0 0 0

0 0 0 0 0 0 0 0 0 7.5000e-01 -5.0000e-01 -2.5000e-01 0 -5.0000e-01 6.2500e-01 -1.2500e-01 0 -2.5000e-01 -1.2500e-01 3.7500e-01

Part (b) »Yind = Yinda + Yindb Yind = 5.0000e-01 -2.5000e-01 0 -2.5000e-01 0

-2.5000e-01 0 5.0000e-01 0 0 7.5000e-01 0 -5.0000e-01 -2.5000e-01 -2.5000e-01

-2.5000e-01 0 -5.0000e-01 1.1250e+00 -3.7500e-01

0 -2.5000e-01 -2.5000e-01 -3.7500e-01 8.7500e-01

% To suppress nodes 4 and 5 we use the partitioned matrix formula as follows: »W11= [Yind(1:3, 1:3)] W11 =

5.0000e-01 -2.5000e-01 -2.5000e-01 5.0000e-01 0 0 7.5000e-01

0 0

»W12=[Yind(1:3, 4:5)] W12 = -2.5000e-01 0 0 -2.5000e-01 -5.0000e-01 -2.5000e-01 »W21= [Yind(4:5, 1:3)] W21 = -2.5000e-01 0 -5.0000e-01 0 -2.5000e-01 -2.5000e-01 »W22= [Yind(4:5, 4:5)] W22 = 1.1250e+00 -3.7500e-01 -3.7500e-01 8.7500e-01 »Yind123 = W11 - W12*inv(W22)*W21 Yind123 = 4.3519e-01 -2.7778e-01 -1.5741e-01 -2.7778e-01 4.1667e-01 -1.3889e-01 -1.5741e-01 -1.3889e-01 2.9630e-01 (d) For the required Y-matrix we delete row and column 3 to obtain »Ysc = Yind123(1:2,1:2) Ysc = 4.3519e-01 -2.7778e-01 -2.7778e-01 4.1667e-01 »Zoc = inv(Ysc) Zoc = 4.0000e+00 2.6667e+00 2.6667e+00 4.1778e+00

SOLUTION 20.37. Since complex roots must occur in conjugate pairs, we will only check jω0.

(

)

0 = p( j 0 ) = − ja3 30 − a2 20 + ja1 0 + a0 = a0 − a2 20 + j a1 0 − a3 03 Both real and imaginary parts must be zero, i.e.,

(

)

a0 − a2 02 = 0 and a1 0 − a3 30 = 0 a1 − a3 02 = 0

From the first equation, a2 20 = a0 . The second equation above must be true for arbitrary ω0 which implies that a1 = a3 02 . Equivalently a1a2 = a3a2 02 = a3a0 . Conclusion: this condition leads to imaginary complex roots. (b) Given the above condition, what are the resulting imaginary roots of the polynomial? Since the polynomial is cubic, we can assume a3 ≠ 0. In this case, a 0 = p(s) = s3 + 2 s2 + a3

 a1 a s + 0 = s s2 + a3 a3 

a1  + a3 

 a2 2 a1a2 s + 2 = s s2 + a3  a3

a1  a2  2 a1  + s +  a3 a3  a3 

 a  a  = s2 + 1   s + 2   a3   a3  Therefore, the roots are: s = ± j

a1 a , − 2. a3 a3

SOLUTION 20.38. (a) The four 2-port equations arising from the interconnection are: V1 = V1a − V1b

I1 = I1a = −I1b

V2 = V2a = V2b

I2 = I2a + I2b

Thus V1 = V1a − V1b = (h11a I1a + h12aV2a ) − (h11b I1b + h12bV2b ) = ( h11a + h11b ) I1 + ( h12a − h12b )V2 and I2 = I2a + I2b = ( h21a I1a + h22aV2a ) + (h21b I1b + h22bV2b ) = ( h21a − h21b )I1 + (h22a + h22b )V2 This proves that the series-parallel connection has the required h-parameters. (b) The four 2-port equations arising from the interconnection are: V2 = V2a − V2b

I2 = I2a = −I 2b

V1 = V1a = V1b

I1 = I1a + I1b

Thus, I1 = I1a + I1b = (g11aV1a + g12a I2a ) + (g11bV1b + g12b I2b ) = ( g11a + g11b )V1 + (g12a − g12b ) I2 and V2 = V2a − V2b = ( g21aV1a + g22a I 2a ) − (g21bV1b + g22b I2b ) = ( g21a − g21b )V1 + ( g22a + g22b ) I2 This proves that the parallel-series connection has the required g-parameters.

SOLUTION 20.39. (a) Refer here to Na in figure P20.39b. With reference to figure 19.28b, h11 = 11 kΩ, h12 = 0, 5

-5

h21 = 95.9, and h22 = 1/10 = 10 S. Similarly, by inspection with reference to equation 19.33, -5

h11 = 90||10 = 9 kΩ, h12 = 0.1 (reverse voltage division), h21 = –0.1, and h22 = 10 S. (b) By problem 20.38, part (a), h11 = h11a + h11b = 20 kΩ h21 = h21a – h21b = 96

h12 = h12a – h12b = –0.1 -5

h22 = h22a + h22b = 2×10 S

V h h I h h (c) Recall from chapter 19 that Zin = 1 = h11 − 12 21 and Yout = 2 = h22 − 12 21 in I1 h22 + YL V2 h11 + Zs which case Zout is the reciprocal. Using our MATLAB script, we have h = [20e3 -0.1; 96 0.02e-3 ]; zL= 1e8; zs= 5e3; »twoporth(h,zL,zs) ans = twoport analysis using h-parameters zin = 4.9976e+05 zout = 2.4752e+03 v1tovs = 9.9009e-01 v2tov1 = -9.5998e+00 v2tovs = -9.5047e+00 REMARK: We have used the following m-file code for "twoporth": »% two-port analysis in terms of h-parameters »function [zin, zout] =twoport(h, zL, zs) »['twoport analysis using h-parameters'] »h11= h(1,1); h12=h(1,2); h21=h(2,1); h22=h(2,2); »zin = h11 - h12*h21/(h22+ 1/zL) »yout= h22 - h12*h21/(h11+zs);

»zout= 1/yout »v1tovs= zin/(zin+zs) »v2tov1= -h21/(zin*(h22+1/zL)) »v2tovs= v1tovs*v2tov1

SOLUTION 20.40. (a) The y-parameters for Na are:

−1  7 4 1  7 −4  S y AB =  =  33 −4 7   4 7

Hence the indefinite admittance matrix for Na is:  7 −4 −3 1   YindN a = −4 7 −3 S  33 −3 −3 6  Let us consider the associate 2-port with port A grounded and B as the new port 1 input. The problem is then solved by computing the input impedance with port 2 open circuited. Hence, the new y-parameters are: y BC =

1  7 −3 S 33 −3 6 

y y 1  9 5.5 1 Thus Yin = y11 − 12 21 = 7 − = = S. Hence, Zin = 6 Ω is the unique reading. y 22 33  6 33 6 (b) The answer is not unique as demonstrated in part (c). (c) The following two networks have the given Z-parameters, but the meter reading for N1 is 4 Ω but for N2 it is 2 Ω.

SOLUTIONS CHAPTER 21 PROBLEMS

SOLUTION TO 21.1. (a) Low pass (b) High pass

SOLUTION TO 21.2.

SOLUTION TO 21.3. (a) »n = 0.65378; »d = [1 0.80381643 0.82306043]; »w = 0:0.005:2; »h = freqs(n,d,w); »plot(w, 20*log10(abs(h))) »grid »xlabel('Frequency rads/s') »ylabel('dB Gain') »

0

dB Gain

-5

TextEnd

-10

-15

0

(b) »poles = roots(d) poles = -4.0191e-01 + 8.1335e-01i -4.0191e-01 - 8.1335e-01i (c) »% Poles of new transfer function »wp = 2*pi*750' wp = 4.7124e+03 »wp = 2*pi*750; »polesnew = poles*wp polesnew = -1.8939e+03 + 3.8328e+03i -1.8939e+03 - 3.8328e+03i »% All zeros remain at infinity.

0.5

1 Frequency rads/s

1.5

2

Further H(s) = HNLP (s

7 ( p)2 2.2207 × 10 ) = = p s2 + 0.80381643 p s + 0.82306043( p )2 s2 + 3.7879 × 103 s + 1.8277 × 10 7

1 SOLUTION TO 21.4. (a) The 2nd order normalized LP transfer function is HNLP (s) = 2 . This s + 2s + 1 must be frequency scaled by Kf = 1000π. Hence, (K f )2

6

9.8696 × 10 H(s) = HNLP (s K f ) = 2 = 2 2 s + K f 2s + (K f ) s + 4.4429 × 103 s + 9.8696 × 106 (b) Using MATLAB, »n = (1000*pi)^2; »d = [1 sqrt(2)*pi*1e3 (1000*pi)^2]; »w = 0:1:2*pi*1500; »h = freqs(n,d,w); »plot(w/(2*pi),abs(h)) »grid »xlabel('Frequency in Hz') »ylabel('Magnitude') »plot(w/(2*pi),20*log10(abs(h))) »grid »xlabel('Frequency in Hz') »ylabel('Magnitude in dB')

1 0.9 0.8

Magnitude

0.7 0.6 0.5 TextEnd 0.4 0.3 0.2 0.1

0

500

1000

1500

Frequency in Hz

0 -2 -4

Magnitude in dB

-6 -8 -10 -12 TextEnd

-14 -16 -18 -20

0

500

1000 Frequency in Hz

(c) »n n = 9.8696e+06 »d d = 1.0000e+00 4.4429e+03 9.8696e+06 »w = j*2000*pi; »mag = abs(n/(w^2 + d(2)*w + d(3))) mag = 2.4254e-01

1500

SOLUTION TO 21.5. (a) ∈max is that value of e that places the magnitude response curve through Amax at ω = ωp. Therefore 2n   p  2 Amax = 10log 10 H ( j p) = 10log 10  1+ ∈max    = 10log 10 1+ ∈2max   p   

(

2

)

Therefore ∈2max = 10 0.1Amax − 1 which upon a square root yields the final answer. (b) Similarly, ∈min puts the magnitude response curve through the Amin spec. Hence 2n     2 s Amin = 10log10 H ( j s ) = 10log 10 1+ ∈min        p  2

Therefore ∈2min =

100.1A min − 1  s    p

2n

which is equivalent to the required formula.

SOLUTION TO 21.6. The relationship of ε and ωc is given by the formula:

c min

=

p 1n

(∈max )

= 2n

p

10 0.1Amax − 1

≤ω≤

=

p 1 (∈) n

. Further, ∈max

and ∈min in putting the c max . Hence, from the solution to problem

in putting the magnitude response curve through the Amax spec produces magnitude response curve through the Amin spec produces 5,

c

c max

=

p

(∈min )1 n

c min ,

= 2n

s

10 0.1Amin − 1

SOLUTION TO 21.7. (a) From above material, the second order Butterworth NLP transfer function is 1 HNLP2 (s) = 2 s + 2s + 1 and from tables, the third order is

1 HNLP3 (s) = 3 2 s + 2s + 2s + 1 (b) n1 = 1; d1 = [1 sqrt(2) 1]; n2 = 1; d2 = [1 2 2 1]; w = 0:.01:5; h1 = freqs(n1,d1,w); h2 = freqs(n2,d2,w); plot(w,abs(h1)) grid xlabel('Normalized Frequency') ylabel('Magnitude') hold plot(w,abs(h2),'r') hold off

1 0.9 0.8

Magnitude

0.7 0.6 0.5 0.4

TextEnd

0.3 0.2 0.1 0

0

1

2 3 Normalized Frequency

4

Notice how the 3rd order filter has a sharper transition to zero. (c) »% The simplest way to obtain the step response is as follows: » »syms s t »StepResp1 = ilaplace(1/(s^3 + sqrt(2)*s^2 + s)) StepResp1 = 1-exp(-1/2*2^(1/2)*t)*cos(1/2*2^(1/2)*t)-exp(-1/2*2^(1/2)*t)*sin(1/2*2^(1/2)*t) »StepResp2 = ilaplace(1/(s^4 + 2*s^3 + 2*s^2 + s))

5

StepResp2 = 1-exp(-t)-2/3*exp(-1/2*t)*3^(1/2)*sin(1/2*3^(1/2)*t) » Thus the step response of the second order Butterworth normalized LP filter is: v(t) = u(t) − e −0.70711t [cos(0.70711t) − sin(0.70711t)]u(t) and that of the third order Butterworth normalized LP filter is:

(

v(t) = 1− e

−t

) u(t) − 1.1547e−0.5t sin(0.86603t)u(t)

SOLUTION TO 21.8. fp = 100; fs = 1200;Amax = 0.3; Amin = 35; n = buttord(fp,fs,Amax,Amin,'s') emax = sqrt(10^(0.1*Amax) - 1) emin = sqrt(10^(0.1*Amin) - 1)/(fs/fp)^n fcmin = fp/((10^(0.1*Amax)-1)^(1/(2*n))) fcmax = fs/((10^(0.1*Amin)-1)^(1/(2*n))) wcmin = 2*pi*fcmin wcmax = 2*pi*fcmax wc = wcmin; fc = fcmin; [z,p,k] = buttap(n) % Numerators are each 1. Denominators are the polynomials d1 = poly(p(1:2)) d2 = poly(p(3)) zplane(p) grid pause znew = z*wc pnew = p*wc knew = k*wc^n f = 0:fc/50:1.2*fs; h = freqs(knew*poly(znew),poly(pnew),2*pi*f); plot(f,abs(h)) grid xlabel('Frequency in Hz') ylabel('Gain magnitude') pause plot(f,20*log10(abs(h))) xlabel('Frequency in Hz') ylabel('Gain in dB')

grid n= 3 emax = 2.6743e-01 emin = 3.2538e-02 fcmin = 1.5521e+02 fcmax = 3.1324e+02 wcmin = 9.7524e+02 wcmax = 1.9681e+03 z = [] p= -5.0000e-01 + 8.6603e-01i -5.0000e-01 - 8.6603e-01i -1.0000e+00 k= 1 % Numerators are each 1. Denominators are the polynomials d1 = 1.0000e+00 1.0000e+00 1.0000e+00 d2 = 1 1

1 0.8 0.6

Imaginary part

0.4 0.2 0 -0.2 -0.4 -0.6 -0.8 -1 -1

znew = [] pnew = -4.8762e+02 + 8.4458e+02i -4.8762e+02 - 8.4458e+02i -9.7524e+02 knew = 9.2753e+08

-0.5

0 Real part

0.5

1

1 0.9 0.8

Gain magnitude

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

0

500

1000

1500

1000

1500

Frequency in Hz

0

-10

Gain in dB

-20

-30

-40

-50

-60

0

500 Frequency in Hz

SOLUTION TO 21.9.

fp = 100; fs = 1200;Amax = 0.3; Amin = 35; n = buttord(fp,fs,Amax,Amin,'s'); emax = sqrt(10^(0.1*Amax) - 1); emin = sqrt(10^(0.1*Amin) - 1)/(fs/fp)^n; fcmin = fp/((10^(0.1*Amax)-1)^(1/(2*n))); fcmax = fs/((10^(0.1*Amin)-1)^(1/(2*n))); wcmin = 2*pi*fcmin; wcmax = 2*pi*fcmax; [z,p,k] = buttap(n); wc = wcmax; fc = fcmax; znew = z*wc pnew = p*wc knew = k*wc^n f = 0:fc/50:1.2*fs; h = freqs(knew*poly(znew),poly(pnew),2*pi*f); plot(f,abs(h)) grid xlabel('Frequency in Hz') ylabel('Gain magnitude') pause plot(f,20*log10(abs(h))) xlabel('Frequency in Hz') ylabel('Gain in dB') grid znew = [] pnew = -9.8406e+02 + 1.7044e+03i -9.8406e+02 - 1.7044e+03i -1.9681e+03 knew = 7.6235e+09

1 0.9 0.8

Gain magnitude

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

0

500

1000

1500

1000

1500

Frequency in Hz

5 0 -5

Gain in dB

-10 -15 -20 -25 -30 -35 -40

0

500 Frequency in Hz

SOLUTION TO 21.10.

fp = 75; fs = 450;Amax = 1; Amin = 45; n = buttord(fp,fs,Amax,Amin,'s') emax = sqrt(10^(0.1*Amax) - 1) emin = sqrt(10^(0.1*Amin) - 1)/(fs/fp)^n fcmin = fp/((10^(0.1*Amax)-1)^(1/(2*n))) fcmax = fs/((10^(0.1*Amin)-1)^(1/(2*n))) wcmin = 2*pi*fcmin wcmax = 2*pi*fcmax [z,p,k] = buttap(n) d1 = poly(p(1:2)) d2 = poly(p(3:4)) zplane(p) grid pause wc = wcmin; fc = fcmin; znew = z*wc pnew = p*wc knew = k*wc^n W = 0:0.01:fs/fp; h = freqs(k*poly(z),poly(p),W); plot(W*wc/(2*pi),abs(h)) grid xlabel('Frequency in Hz') ylabel('Gain magnitude') pause plot(W*wc/(2*pi),20*log10(abs(h))) xlabel('Frequency in Hz') ylabel('Gain in dB') grid n= 4 emax = 5.0885e-01 emin = 1.3721e-01 fcmin = 8.8800e+01 fcmax = 1.2323e+02 wcmin = 5.5795e+02 wcmax = 7.7427e+02 z = [] p= -3.8268e-01 + 9.2388e-01i -3.8268e-01 - 9.2388e-01i -9.2388e-01 + 3.8268e-01i -9.2388e-01 - 3.8268e-01i k= 1

d1 = 1.0000e+00 7.6537e-01 1.0000e+00 d2 = 1.0000e+00 1.8478e+00 1.0000e+00 znew = [] pnew = -2.1352e+02 + 5.1548e+02i -2.1352e+02 - 5.1548e+02i -5.1548e+02 + 2.1352e+02i -5.1548e+02 - 2.1352e+02i knew = 9.6912e+10

1 0.8 0.6 Imaginary part

0.4 0.2 0 -0.2 -0.4 -0.6 -0.8 -1 -1

-0.5

0 Real part

0.5

1

1 0.9 0.8

Gain magnitude

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

0

100

200

300 400 Frequency in Hz

500

600

0

100

200

300 400 Frequency in Hz

500

600

0 -10

Gain in dB

-20 -30 -40 -50 -60 -70

SOLUTION TO 21.11. fp = 75; fs = 450;Amax = 1; Amin = 45;

n = buttord(fp,fs,Amax,Amin,'s'); % The order mfile may not be available in the student edition. emax = sqrt(10^(0.1*Amax) - 1); emin = sqrt(10^(0.1*Amin) - 1)/(fs/fp)^n; fcmin = fp/((10^(0.1*Amax)-1)^(1/(2*n))); fcmax = fs/((10^(0.1*Amin)-1)^(1/(2*n))); wcmin = 2*pi*fcmin; wcmax = 2*pi*fcmax; [z,p,k] = buttap(n); wc = wcmax; fc = fcmax znew = z*wc pnew = p*wc knew = k*wc^n W = 0:0.01:fs/fp; h = freqs(k*poly(z),poly(p),W); plot(W*wc/(2*pi),abs(h)) grid xlabel('Frequency in Hz') ylabel('Gain magnitude') pause plot(W*wc/(2*pi),20*log10(abs(h))) xlabel('Frequency in Hz') ylabel('Gain in dB') grid fc = 1.2323e+02 znew = [] pnew = -2.9630e+02 + 7.1533e+02i -2.9630e+02 - 7.1533e+02i -7.1533e+02 + 2.9630e+02i -7.1533e+02 - 2.9630e+02i knew = 3.5940e+11

1 0.9 0.8

Gain magnitude

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

0

100

200

300 400 500 Frequency in Hz

600

700

800

0

100

200

300 400 500 Frequency in Hz

600

700

800

0 -10

Gain in dB

-20 -30 -40 -50 -60 -70

SOLUTION TO 21.12. Here we require that

C + C2 2 C1 + C2 + L 2 s3 + 2s2 + 2s + 1= s3 + 1 s + s+ LC1C2 LC1C2 C1C2 Thus

C + C2 L LC1 + LC2 LC1C2 = 2 ⇒ C1 + C2 + L = 4 and 2 = 1 × = L 2 C1C2

in which case, LC1 + LC2 = 4 . Therefore, L(C1 + C2 + L) = 4L = 4 + L2 . Equivalently L2 − 4L + 4 = (L − 2)(L − 2) = 0 Hence L = 2 H is the only solution. Thus C1C2 = 1 ⇒ C1 =

1 1 ⇒ + C2 + 2 = 4 or equivalently C2 C2

C22 − 2C2 + 1 = (C2 − 1)(C2 − 1) = 0 which implies that C1 = C2 = 1 F is the only solution, as was to be shown.

SOLUTION TO 21.13. (a) By voltage division

(b) With R = 1 Ω,

1 1 LC H(s) = Cs + G = 1 1 1 2 Ls + s + s+ RC LC Cs + G 1 1 LC H(s) = = 2 1 1 s + 2s + 1 s2 + s+ RC LC

requires that C = 1/ 2 F and L = (c) »fc = 1000; »wc = 2*pi*fc wc = 6.2832e+03 »Kf = wc; »Km = 1000; »C = 1/sqrt(2); »L = sqrt(2); »Cnew = C/(Kf*Km) Cnew = 1.1254e-07 »Lnew = L*Km/Kf Lnew = 2.2508e-01 (d)

2 H.

»Km = C/(wc*1e-6) Km = 1.1254e+02 »Kf = wc; »Rnew = Km Rnew = 1.1254e+02 »Lnew = L*Km/wc Lnew = 2.5330e-02

SOLUTION TO 21.14. (a) By voltage division

(b) With R = 1 Ω,

1 1 Cs LC H(s) = 1 = 2 Rs 1 Ls + Rs + s + s+ L LC Cs 1 1 LC H(s) = = 2 R 1 s + 2s + 1 s2 + s s + L LC

requires that L = 1/ 2 F and C = (c) »L = 1/sqrt(2); C = 1/L; »Km = 10; »Kf = 2*pi*500; »Rnew = 10; »Cnew = C/(Km*Kf) Cnew = 4.5016e-05 »Lnew = Km*L/Kf Lnew = 2.2508e-03 (d) »Km = C/(1e-6*Kf) Km = 4.5016e+02 »Lnew = L*Km/Kf Lnew = 1.0132e-01 »Cnew = C/(Km*Kf) Cnew = 1.0000e-06

SOLUTION TO 21.15.

2 H.

(a) Since 1 = 2/(LC), L = 2/C. Since (1/L + 1/C) = (C/2 + 1/C) = sqrt(2), we have that C is a root of 2 the quadratic 0.5C – sqrt(2)C + 1 = 0. Hence »v = [0.5 -sqrt(2) 1]; »r = roots(v) r= 1.4142e+00 1.4142e+00 »C = r(1) C = 1.4142e+00 »L = 2/C L = 1.4142e+00 (b) »Km = 1e3; »Kf = 2*pi*3500; »Cnew = C/(Km*Kf) Cnew = 6.4308e-08 »Lnew = L*Km/Kf Lnew = 6.4308e-02 (c) »Km = C/(Kf*10e-9) Km = 6.4308e+03 »Cnew = C/(Km*Kf) Cnew = 1.0000e-08 »Lnew = L*Km/Kf Lnew = 4.1356e-01 »Rs = Km Rs = 6.4308e+03 »RL = Rs RL = 6.4308e+03

SOLUTION TO 21.16. (a) Let G = RL. Then by voltage division 1 1 Vout Cs + G LC H(s) = = = 1 Vin  1 R 1 + Rs RL 2 s Ls + Rs + Cs + G s +  CR + L  s + LC L

(b) Since 1 = 1.25/(LC), L = 1.25/C. Since (Rs/L + 1/RLC) = (2C/1.25 + 1/8C) = sqrt(2), we have that 2

C is a root of the quadratic (16/1.25)C – 8sqrt(2)C + 1 = 0. Hence »C = roots([16/1.25 -8*sqrt(2) 1]) C= 7.8427e-01 9.9615e-02 »L = 1.25 ./C L= 1.5938e+00 1.2548e+01 (c) »Km = 1e3; »Kf = 2*pi*5e3 Kf = 3.1416e+04 »Cnew = C/(Km*Kf) Cnew = 2.4964e-08 3.1708e-09 »Lnew = L*Km/Kf Lnew = 5.0734e-02 3.9942e-01

SOLUTION TO 21.17. (a) Define G = 1/Rs, execute two source transformations, and apply voltage division to obtain 1 V 1 G Rs LC H(s) = out = × = 1 Vin + Ls + 1 Cs + G s2 +  1 + 1  s + 1 + 1 Rs  R C L  Cs + G LC s 2

(b) Since L = 1.5/C, the values of C are the roots of the quadratic, (2/1.5)C – 2sqrt(2)C + 1 = 0. Hence »C = roots([(2/1.5) -2*sqrt(2) 1]) C= 1.6730e+00 4.4829e-01 »L = 1.5 ./C L= 8.9658e-01 3.3461e+00 »Km = 2e3; »Kf = 2*pi*5e3; »Cnew = C/(Km*Kf)

Cnew = 2.6627e-08 7.1347e-09 »Lnew = Km*L/Kf Lnew = 5.7078e-02 2.1302e-01 »Rsnew = 2*Km Rsnew = 4000

SOLUTION TO 21.18. (a) H( jω) =

K  ω j  +1  ωp

2

1

=

=

K 2

 ω ω  +1  p

The 3 dB down frequency, ωc, occurs when 2

 ωc   ω  +1  p

1 2

Equivalently ωc = ω p (b) h(t) = Kω 2pte

−ω p t

2 − 1 = 0.64359ω p = 6.4359 × 10 4 rad/sec.

u(t) since H(s) =

K   s + 1 ω   p 

2

=

Kω 2p

( s + ωp )

2

.

Further, Kω 2p Kω p 1 K K H(s) = = − − 2 s s s+ ω 2 s s + ωp s + ωp p

(

Hence, the step response is K 1 − e

)

−ω p t

(

− ωpte

−ω p t 

 u(t)

)

SOLUTION TO 21.19. (a) Using voltage division, 1 V V H(s) = C × out = Vin VC

Cs + 1 + L1s +

1 L2 s + 1 1 Cs +

1 L2s + 1

×

1 L2 s + 1

1 L1L2C = 1 1 2 L + L2 + C s3 +  +  s2 + 1 s+ L1L2C  L1 L2  L1L2C (b) Matching coefficients in 1 L1L2C 1 = 3 2   1 1 2 L + L2 + C s + 2s + 2s + 1 s3 +  +  s2 + 1 s+ L1L2C  L1 L2  L1L2C

yields C =

2 L + L2 + C and 1 = L1L2 L1L2C

L1 + L2 + 2

2 L1L2

= 2; equivalently, L1 + L2 +

2 = 4. Further, L1L2

1 1  L1 + L2 2  L +  = L L = 2 implies that L1 + L2 = 2L1L2 implies 2L1L2 + L = 4 . This requires that L2 L1 2 1 1 2 L1L2 = 1 and from earlier equations that L1 + L2 = 2L1L2 = 2 which forces L1 = L2 = 1 H and C = 2 F. The idea is to match the denominator coefficients and thus the dc gain is 0.5 instead of the desired 1. A transformer or some amplifier device is needed to increase the gain to 1.

(c) Using MATLAB: »Km = 1000; »Kf = 2*pi*20e3; »Lnew = Km/Kf Lnew = 7.9577e-03 »Cnew = 2/(Km*Kf) Cnew = 1.5915e-08

» Hence, L1new = L2new = 7.96 mH and C = 15.9 nF. (d) SPICE simulation

R 1K

L 7.96m

L0 7.96m

V 0

C 15.9n

IVm R0 1K

Prob 21.19-Small Signal AC-1 +10.000k

+500.000m

+400.000m

+300.000m

+200.000m

+100.000m

+0.000e+000 MAG(V(IVM))

+20.000k

+30.000k

Frequency +40.000k

(Hz)

SOLUTION TO 21.20. (a) From figure P21.19a Vin − V1 = I1, Rs in which case

V2 = Vout

V1 = z11I1 + z12 × 0 = Vin − Rs I1

Also V2 = z21I1 + z22 × 0 = Vout This implies that Vin = (z11 + Rs )I1 and Vout = z21I1 . Finally we conclude that Vout z21 = Vin z11 + Rs (b) Now from figure P21.20b, we have V1 = Vin and V2 = Vout = −RL I2 . This implies that V I2 = − out = y21V1 + y 22V2 RL Thus

Vout − y 21 − y21 = = 1 Vin y 22 + GL y22 + RL

Consider here that 1 1 Vout z 1 2s s = 21 = 2 = = 1  Vin z11 + 1 s + 2s + 1  1   s s+ + 2 +      s 2s 2 Hence z21 =

1  s 1  and z11 =  +  . This leads to the circuit  2 2s 2s

with L = ( c)

1 H and C = 2 F. 2

Similarly, 1 −y 21 2s = y 22 + 1  s 1    +1 +  2 2s

implies y 21 = −

1 and y 22 = 2s

s 1 . + 2s 2s

This yields the same circuit as above with L = 2 H and C = (d)

Here K m = 1000 and K f = 5000rad / s

(d-i)

For (b),

1 F. 2

1 10 3 1 L= × = = 0.1414 H 3 5 2 2 5 × 10 and C= 2×

1 5⋅106

= 0.2828 µF

(d-ii) For (c) L = 0.2828 H and C = 0.1414 µF.

SOLUTION TO 21.21. (a) From earlier developments 1 R1

Y H(s) = − in = − 1 Yout Cs + R2 (b) Let C = 1 F, and R1 = R2 = 1 Ω. (c) »Kf = 2*pi*3500

Kf = 2.1991e+04 »Km = 1/(Kf*1e-9) Km = 4.5473e+04 In the final design, R1 = R2 = 45.5 kΩ.

SOLUTION TO 21.22. (a) From problem 21 and voltage division, 1 R1

1 C2s

1 1 C1R1 R3C2 H(s) = − 1 × 1 =− 1 × 1 Cs + R3 + s+ s+ R2 C2s C1 R2 R3C2 (b) By inspection, let C1 = 0.1 F, R1 = 1 Ω, R2 = 10 Ω, R3 = 10 Ω, and C2 = 0.1 F, in which case H(s) = −

10 1 −10 × = s + 1 s + 1 (s + 1)2

(c) »Kf = 1e5; »Km = 0.1/(Kf*1e-9) Km = 1000 Hence, in the final design C1 = 1 nF, R1 = 1 kΩ, R2 = 10 kΩ, R3 = 10 kΩ, and C2 = 1 nF. (d) Cascade the circuit of figure P21.22 with another op amp section. For the first part of the design, again set ωp = 1 and use the same values as in part (c). The extra op amp section has the same values as the first section. As such, final values are the same as in part (c).

1 SOLUTION TO 21.23. The 2nd order NLP Butterworth transfer function is: HNLP (s) = 2 . The s + 2s + 1 design parameters and steps are detailed in the excel spread sheet below. An additional design called design C is also listed. For input attenuation, the resistor R1 is replaced by the voltage divider R3-R4 combination.

w0^2 1

w0/Q 1.41

Num 1.00

w0 1.0000

Q 0.7071

KNLP 1.00

KMA 22507.86

Kf=wp 6283.20

KmR 10000.00

KMB 22507.86

KMC 15915.46

Design A Design B Design C Saraga

RA ∞ 1 1 RA

RB 0 1 1-1/Q RA/3

K 1 2 3-1/Q 3-Apr

C1 2Q 1 1 rt(3)Q

C2 1/(2Q) 1/Q 1 1

R1 1 1 1 1/Q

R2 1 Q 1 1/rt(3)

alpha KNLP/K KNLP/K KNLP/K KNLP/K

R3 1/alpha 1/alpha 1/alpha 1/alpha

R4 1/(1-alpha) 1/(1-alpha) 1/(1-alpha) 1/(1-alpha)


RA ∞ 1.00 1.00 3.00

RB 0.0000 1.0000 -0.4142 1.0000

K 1.0000 2.0000 1.5858 1.3333

C1 1.4142 1.0000 1.0000 1.2247

C2 0.7071 1.4142 1.0000 1.0000

R1 1.0000 1.0000 1.0000 1.4142

R2 1.0000 0.7071 1.0000 0.5774

alpha 1.0000 0.5000 0.6306 0.7500

R3 1.0000 2.0000 1.5858 1.3333

R4 #DIV/0! 2.0000 2.7071 5.6569

w0 scale Design A Design B Design C Saraga

∞ 1.00 1.00 3.00

0.0000 1.0000 0.1140 1.0000

1.0000 2.0000 2.1140 1.3333

1.4142 1.0000 1.0000 1.2247

0.7071 1.4142 1.0000 1.0000

1.0000 1.0000 1.0000 1.4142

1.0000 0.7071 1.0000 0.5774

1.0000 0.5000 0.6306 0.7500

1.0000 2.0000 1.5858 1.3333

#DIV/0! 2.0000 2.7071 5.6569

wp scale Design A Design B Design C Saraga

∞ 1.00 1.00 3.00

0.000E+00 1.000E+00 1.140E-01 1.000E+00

1.000E+00 2.000E+00 2.114E+00 1.333E+00

2.251E-04 1.592E-04 1.592E-04 1.949E-04

1.125E-04 2.251E-04 1.592E-04 1.592E-04

1.000E+00 1.000E+00 1.000E+00 1.414E+00

1.000E+00 1.000E+00 1.000E+00 #DIV/0! 7.071E-01 5.000E-01 2.000E+00 2.000E+00 1.000E+00 6.306E-01 1.586E+00 2.707E+00 5.774E-01 7.500E-01 1.333E+00 5.657E+00

Km scale Design A Design B Design C Saraga

∞ 10000 10000 30000

0.000E+00 1.000E+04 1.140E+03 1.000E+04

1.000E+00 2.000E+00 2.114E+00 1.333E+00

1.000E-08 7.0711E-09 1.000E-08 1.000E-08

5.000E-09 1.000E-08 1.000E-08 8.165E-09

2.251E+04 2.251E+04 1.592E+04 2.757E+04

2.251E+04 1.000E+00 2.251E+04 #DIV/0! 1.592E+04 5.000E-01 4.502E+04 4.502E+04 1.592E+04 6.306E-01 2.524E+04 4.308E+04 1.125E+04 7.500E-01 2.599E+04 1.103E+05

KMS 19492.37

SOLUTION TO 21.24 AND 21.25. In problem 21.8, the transfer function information was computed in MATLAB as: % Numerators are each 1. Denominators are the polynomials d1 = 1.0000e+00 1.0000e+00 1.0000e+00 d2 = 1 1 Further we know from MATLAB that fcmin = 1.5521e+02 The Saraga design and Design A for d1, the second order section of each filter, are given by the excel spread sheet below, as well as two alternate designs labeled B and C.

w0^2 1

w0/Q 1.00

Num 1.00

w0 1.0000

Q 1.0000

KNLP 1.000

KMA 41016.58

Kf=wp 975.22

KmR 10000.00

KMB 20508.29

KMC 20508.29


RA ∞ 1 1 RA

RB 0 1 1-1/Q RA/3

K 1 2 3-1/Q 3-Apr

C1 2Q 1 1 rt(3)Q

C2 1/(2Q) 1/Q 1 1

R1 1 1 1 1/Q

R2 1 Q 1 1/rt(3)





RA ∞ 1.00 1.00 3.00

RB 0.0000 1.0000 0.0000 1.0000

K 1.0000 2.0000 2.0000 1.3333

C1 2.0000 1.0000 1.0000 1.7321

C2 0.5000 1.0000 1.0000 1.0000

R1 1.0000 1.0000 1.0000 1.0000

R2 1.0000 1.0000 1.0000 0.5774

alpha 1.0000 0.5000 0.5000 0.7500

R3 1.0000 2.0000 2.0000 1.3333

R4 #DIV/0! 2.0000 2.0000 4.0000


∞ 1.00 1.00 3.00

0.0000 1.0000 0.1140 1.0000

1.0000 2.0000 2.1140 1.3333

2.0000 1.0000 1.0000 1.7321

0.5000 1.0000 1.0000 1.0000

1.0000 1.0000 1.0000 1.0000

1.0000 1.0000 1.0000 0.5774

1.0000 0.5000 0.5000 0.7500

1.0000 2.0000 2.0000 1.3333

#DIV/0! 2.0000 2.0000 4.0000


∞ 1.00 1.00 3.00

0.000E+00 1.000E+00 1.140E-01 1.000E+00

1.000E+00 2.000E+00 2.114E+00 1.333E+00

2.051E-03 1.025E-03 1.025E-03 1.776E-03

5.127E-04 1.025E-03 1.025E-03 1.025E-03

1.000E+00 1.000E+00 1.000E+00 1.000E+00

1.000E+00 1.000E+00 1.000E+00 #DIV/0! 1.000E+00 5.000E-01 2.000E+00 2.000E+00 1.000E+00 5.000E-01 2.000E+00 2.000E+00 5.774E-01 7.500E-01 1.333E+00 4.000E+00


∞ 10000 10000 30000

0.000E+00 1.000E+04 1.140E+03 1.000E+04

1.000E+00 2.000E+00 2.114E+00 1.333E+00

5.000E-08 5.000E-08 5.000E-08 5.000E-08

1.250E-08 5.000E-08 5.000E-08 2.887E-08

4.102E+04 2.051E+04 2.051E+04 3.552E+04


KMS 35521.40

The first order (leaky integrator) section is common to both problems. This section consists of an input resistor (conductance) R1 (G1) connected to the inverting terminal with a parallel R2-C combination G1 feeding back from the output. The transfer function is: H(s) = . For the normalized design we Cs + G2 set G1 = G2 = 1 S (R1 = R2 = 1 Ω) and C = 1 F. This design can be scaled independently of the S&K 2nd order section. Hence we set Cnew = 50 nF. Thus Km = 20,508.29. Hence R1 = R2 = 20,508.29 Ω.

SOLUTION TO 21.26 AND 21.27. The relevent data from the solution of problem 21.10 is: k= 1 d1 = 1.0000e+00 7.6537e-01 1.0000e+00 d2 = 1.0000e+00 1.8478e+00 1.0000e+00 fcmin = 8.8800e+01 wcmin = 5.5795e+02

In providing the designs, we set forth all the possible S&K designs using two excel spreadsheets, one for each second order section. The designs for denominator d1 with numerator equal to 1 are: w0^2 1

0.7654

w0/Q

Num 1.00000

w0 1.0000

Q 1.3066

KNLP 1.0000

KMA 46834.37

Kf=wp 557.95

KmR 10000.00

KMB 17922.81

KMC 17922.81


RA ∞ 1 1 RA

RB 0 1 1-1/Q RA/3

K 1 2 3-1/Q 3-Apr

C1 2Q 1 1 rt(3)Q

C2 1/(2Q) 1/Q 1 1

R1 1 1 1 1/Q

R2 1 Q 1 1/rt(3)





RA ∞ 1.00 1.00 3.00

RB 0.0000 1.0000 0.2346 1.0000

K 1.0000 2.0000 2.2346 1.3333

C1 2.6131 1.0000 1.0000 2.2630

C2 0.3827 0.7654 1.0000 1.0000

R1 1.0000 1.0000 1.0000 0.7654

R2 1.0000 1.3066 1.0000 0.5774

alpha 1.0000 0.5000 0.4475 0.7500

R3 1.0000 2.0000 2.2346 1.3333

R4 #DIV/0! 2.0000 1.8100 3.0615


∞ 1.00 1.00 3.00

0.0000 1.0000 0.1140 1.0000

1.0000 2.0000 2.1140 1.3333

2.6131 1.0000 1.0000 2.2630

0.3827 0.7654 1.0000 1.0000

1.0000 1.0000 1.0000 0.7654

1.0000 1.3066 1.0000 0.5774

1.0000 0.5000 0.4475 0.7500

1.0000 2.0000 2.2346 1.3333

#DIV/0! 2.0000 1.8100 3.0615


∞ 1.00 1.00 3.00

0.000E+00 1.000E+00 1.140E-01 1.000E+00

1.000E+00 2.000E+00 2.114E+00 1.333E+00

4.683E-03 1.792E-03 1.792E-03 4.056E-03

6.859E-04 1.372E-03 1.792E-03 1.792E-03

1.000E+00 1.000E+00 1.000E+00 7.654E-01

1.000E+00 1.000E+00 1.000E+00 #DIV/0! 1.307E+00 5.000E-01 2.000E+00 2.000E+00 1.000E+00 4.475E-01 2.235E+00 1.810E+00 5.774E-01 7.500E-01 1.333E+00 3.061E+00


∞ 10000 10000 30000

0.000E+00 1.000E+04 1.140E+03 1.000E+04

1.000E+00 2.000E+00 2.114E+00 1.333E+00

1.000E-07 1.000E-07 1.000E-07 1.000E-07

1.464E-08 7.654E-08 1.000E-07 4.419E-08

4.683E+04 1.792E+04 1.792E+04 3.104E+04


KMS 40559.76

The designs for denominator d2 with numerator equal to 1 are:

w0^2 1

1.8478

w0/Q

Num 1.00

w0 1.00

Q 0.54

KNLP 1.00

KMA 19399.08

Kf=wp 557.95

KmR 10000.00

KMB 33117.77

KMC 17922.81


RA ∞ 1 1 RA

RB 0 1 1-1/Q RA/3

K 1 2 3-1/Q 3-Apr

C1 2Q 1 1 rt(3)Q

C2 1/(2Q) 1/Q 1 1

R1 1 1 1 1/Q

R2 1 Q 1 1/rt(3)





RA ∞ 1.00 1.00 3.00

RB 0.0000 1.0000 -0.8478 1.0000

K 1.0000 2.0000 1.1522 1.3333

C1 1.0824 1.0000 1.0000 0.9374

C2 0.9239 1.8478 1.0000 1.0000

R1 1.0000 1.0000 1.0000 1.8478

R2 1.0000 0.5412 1.0000 0.5774

alpha 1.0000 0.5000 0.8679 0.7500

R3 1.0000 2.0000 1.1522 1.3333

R4 #DIV/0! 2.0000 7.5703 7.3912


∞ 1.00 1.00 3.00

0.0000 1.0000 0.1140 1.0000

1.0000 2.0000 2.1140 1.3333

1.0824 1.0000 1.0000 0.9374

0.9239 1.8478 1.0000 1.0000

1.0000 1.0000 1.0000 1.8478

1.0000 0.5412 1.0000 0.5774

1.0000 0.5000 0.8679 0.7500

1.0000 2.0000 1.1522 1.3333

#DIV/0! 2.0000 7.5703 7.3912


∞ 1.00 1.00 3.00

0.000E+00 1.000E+00 1.140E-01 1.000E+00

1.000E+00 2.000E+00 2.114E+00 1.333E+00

1.940E-03 1.792E-03 1.792E-03 1.680E-03

1.656E-03 3.312E-03 1.792E-03 1.792E-03

1.000E+00 1.000E+00 1.000E+00 1.848E+00

1.000E+00 1.000E+00 1.000E+00 #DIV/0! 5.412E-01 5.000E-01 2.000E+00 2.000E+00 1.000E+00 8.679E-01 1.152E+00 7.570E+00 5.774E-01 7.500E-01 1.333E+00 7.391E+00


∞ 10000 10000 30000

0.000E+00 1.000E+04 1.140E+03 1.000E+04

1.000E+00 2.000E+00 2.114E+00 1.333E+00

1.000E-07 5.412E-08 1.000E-07 9.374E-08

8.536E-08 1.000E-07 1.000E-07 1.000E-07

1.940E+04 3.312E+04 1.792E+04 3.312E+04


KMS 17922.81

SOLUTION TO 21.28. For this problem we use the excel spread sheet given below. First we observe that H(s) =

Kˆ Kˆ 02 (= K NLP ) → H ( 0 s) = = 1 2 0 2 s2 + 0 s + 20 s2 + s + 1 ( ( 0 s) + 0 s) + 0 Q Q Q Kˆ

Thus after this type of frequency scaling, the new transfer function is: Hnew (s) =

0.7943 1 s2 + s +1 1.1286

The dc gain is of course 0.7943 and the modification of the circuit to achieve the correct dc gain is given in the spread sheet below via R3 and R4 which constitute a voltage divider that replaces R1.

w0^2 0.82306

0.8038

w0/Q

Num 0.65378

w0 0.9072

Q 1.1286

KNLP 0.7943

KMA 1131.42

Kf=wp 43982.40

KmR 10000.00

KMB 501.23

KMC 501.23


RA ∞ 1 1 RA

RB 0 1 1-1/Q RA/3

K 1 2 3-1/Q 3-Apr

C1 2Q 1 1 rt(3)Q

C2 1/(2Q) 1/Q 1 1

R1 1 1 1 1/Q

R2 1 Q 1 1/rt(3)





RA ∞ 1.00 1.00 3.00

RB 0.0000 1.0000 0.1140 1.0000

K 1.0000 2.0000 2.1140 1.3333

C1 2.2573 1.0000 1.0000 1.9549

C2 0.4430 0.8860 1.0000 1.0000

R1 1.0000 1.0000 1.0000 0.8860

R2 1.0000 1.1286 1.0000 0.5774

alpha 0.7943 0.3972 0.3758 0.5957

R3 1.2589 2.5179 2.6613 1.6786

R4 4.8621 1.6588 1.6019 2.1917


∞ 1.00 1.00 3.00

0.0000 1.0000 0.1140 1.0000

1.0000 2.0000 2.1140 1.3333

2.4881 1.1023 1.1023 2.1548

0.4883 0.9766 1.1023 1.1023

1.0000 1.0000 1.0000 0.8860

1.0000 1.1286 1.0000 0.5774

0.7943 0.3972 0.3758 0.5957

1.2589 2.5179 2.6613 1.6786

4.8621 1.6588 1.6019 2.1917


∞ 1.00 1.00 3.00

0.000E+00 1.000E+00 1.140E-01 1.000E+00

1.000E+00 2.000E+00 2.114E+00 1.333E+00

5.657E-05 2.506E-05 2.506E-05 4.899E-05

1.110E-05 2.220E-05 2.506E-05 2.506E-05

1.000E+00 1.000E+00 1.000E+00 8.860E-01

1.000E+00 1.129E+00 1.000E+00 5.774E-01

7.943E-01 3.972E-01 3.758E-01 5.957E-01

1.259E+00 2.518E+00 2.661E+00 1.679E+00

4.862E+00 1.659E+00 1.602E+00 2.192E+00


∞ 10000 10000 30000

0.000E+00 1.000E+04 1.140E+03 1.000E+04

1.000E+00 2.000E+00 2.114E+00 1.333E+00

5.000E-08 5.000E-08 5.000E-08 5.000E-08

9.813E-09 4.430E-08 5.000E-08 2.558E-08

1.131E+03 5.012E+02 5.012E+02 8.682E+02

1.131E+03 5.657E+02 5.012E+02 5.657E+02

7.943E-01 3.972E-01 3.758E-01 5.957E-01

1.424E+03 1.262E+03 1.334E+03 1.645E+03

5.501E+03 8.315E+02 8.029E+02 2.148E+03

KMS 979.83

1   SOLUTION TO 21.29. (a) HHP (s) = H NLP c = . At s = j p ,  s   2  c c + 2 +1  s   s 

  1 1 HHP ( j p ) = HNLP  c  = = 2  j p  5.5 2  5.5  c  + 2  c  + 1 − −j 2 +1 j  j   7  7   p  p Thus in MATLAB, »Magfp = 1/abs(1 - (5.5/7)^2 -j*sqrt(2)*(5.5/7)) Magfp = 8.5091e-01 »Attenfp = -20*log10(Magfp)

Attenfp = 1.4023e+00 »Magfs = 1/abs(1 - (5.5/1)^2 -j*sqrt(2)*(5.5/1)) Magfs = 3.3040e-02 »Attenfs = -20*log10(Magfs) Attenfs = 2.9619e+01 Thus the attenuation at fp is 1.4023 dB and that at fs is 29.619 dB. (b) From problem 21.15, the transfer function is 1 LC 2  1 1 s2 + + s+  C L LC and the values of L and C realizing the 2nd order Butterworth NLP transfer function can be computed according to Hcir (s) =

1 1 2 1 C + = 2, =1 ⇒ + = 2 ⇒ C 2 − 2 2C + 2 = 0 C L LC C 2 (C − 2)2 = 0 ⇒ C = 2 F ⇒ L = 2 H (c) Here Km = 1000. Ls →

L C → s

1 1 LKm C

and s

1 s Km → → s. Thus in Cs C C C C

MATLAB, »wc = 2*pi*5.5e3 wc = 3.4558e+04 »Km = 1000; »C = sqrt(2); L = sqrt(2); »Lhp = Km/(C*wc) Lhp = 2.0462e-02 »Chp = 1/(Km*wc*L) Chp = 2.0462e-08 Therefore, the resistors take on values of 1 kΩ, the inductor is changed to a capacitor of value of Chp = 20.46 nF and the capacitor is changed to an inductor of value Lhp = 20.46 mH.

SOLUTION TO 21.30. (a) Ω=

p

, Amax = 2 dB,

1

= max = 10 0.1×2 − 1 = 0.76478, Ωc = 6 10

0.1× 2

»Wc=1/(10^0.2-1)^(1/6) Wc = 1.0935e+00

= 1.0935 −1

(b) »wchp = 2*pi*5e3/Wc wchp = 2.8730e+04 »fchp = wchp/(2*pi) fchp = 4.5725e+03 Thus

chp

=

p

Ωc

= 28.73 krad/s.

( c ) (i) NLP → HP transformation: scale by wchp. Chp =

1

1 = 17.404 F, Lhp = = 34.807 H Llp chp Clp chp

(ii) Magnitude scale to obtain proper value of capacitors. »Clp = 1; »Llp = 2; »Chp = 1/(Llp*wchp) Chp = 1.7404e-05 »Lhp = 1/(Clp*wchp) Lhp = 3.4807e-05 »Cnewhp = 100e-9; »Km = Chp/Cnewhp Km = 1.7404e+02 »Lnewhp = Km*Lhp Lnewhp = 6.0578e-03 K m = 174.02, Chp,new =

Chp Km

= 100 nF , Lhp,new = K m Lhp = 6.06 mH

SOLUTION TO 21.31. 1 The 2nd order NLP Butterworth transfer function is: HNLP (s) = 2 . Using the transformation s + 2s + 1 s to 1/s, we obtain the NHP Butterworth transfer function: s2 s2  1 HNHP (s) = HNLP   = 2 = K∞ 2 s s + 2s + 1 s + d(1) s + d(2)

INPUT:

d(1) 1.414213562

d(2) 1

K 1.33334

NHP Params

w0 1

Q 0.70710678

alpha 0.74999625

NHP Crt Pars

HP Crt Params

C1 = Q C2 = rt(3) 0.707106781 1.73205081 C1new=C1/(Km*Kf) 3.06293E-09

C2new 7.5026E-09

R1 = 1 1 R1new=R1*Km 12247.44871

K 1

Kf 18849.6

R2 =1/(rt(3)Q) 0.816496581

R

R2new=R2*Km 10000

Rnew=R*KmR 30000

Km 12247.44871

1

KmR 30000

R/3 0.333333333 Rnew/3 10000

A plot of the design without input attenuation is shown below. Notice that as predicted the gain is 4/3.

R 10k

Vin

C 3.063n

C0 7.503n

15 Vpls

Vminus 15

XOpAmp R2 10k R0 12.247k

IVm R1 30k

S&K HP-Small Signal AC-3 +100.000

(V)

+316.228

+1.000k

Frequency +3.162k

(Hz)

+10.000k

+1.000

+500.000m

+0.000e+000

MAG(V(IVM))

Input attenuation requires that we replace C1 with a series combination of capacitors in which C1 = C3 + C4 and (1/C3)/(1/C4 + 1/C3) = alpha. Here then, C1 = C3 + C4 and alpha = C4/(C3 + C4) = C4/C1. Thus C4 = alpha*C1 and C3 = (1 – alpha)*C1. Thus Input Attenuation

C3 = (1 - alpha)C1 7.65744E-10

C4 = alpha*C1 2.29719E-09

SOLUTION TO 21.32. The fourth order Butterworth NLP transfer function can be obtained from tables or from MATLAB as follows: »[z,p,k] = buttap(4) z= [] p= -3.8268e-01 + 9.2388e-01i -3.8268e-01 - 9.2388e-01i -9.2388e-01 + 3.8268e-01i -9.2388e-01 - 3.8268e-01i k= 1 »% Second Order Sections

»n1 = 1; »d1 = poly([p(1),conj(p(1))]) d1 = 1.0000e+00 7.6537e-01 1.0000e+00 »n2 = 1; »d2 = poly([p(3),conj(p(3))]) d2 = 1.0000e+00 1.8478e+00 1.0000e+00 » Thus, 1 1 HNLP (s) = 2 × 2 s + 0.76537s + 1 s + 1.8478s + 1 having frequency response »w = 0:0.01:3.5; »h = freqs(k*poly(z),poly(p),w); »plot(w,abs(h)) »grid »xlabel('Normalized Frequency') »ylabel('Magnitude 4th Order Butterworth') 1 0.9

Magnitude 4th Order Butterworth

0.8 0.7 0.6 0.5 0.4 0.3 TextEnd

0.2 0.1 0

0

0.5

1

1.5 2 Normalized Frequency

2.5

The Saraga design parameters are given in the following Excel tables:

3

3.5

INPUT: NLP H(s) NHP H(s)

d(1) 0.76537 0.76537

d(2) 1 1

K 1.33334

Kf 18849.6

Km 22630.24168

KmR 60000

alpha 0.74999625

NHP Crt Pars

C1 = Q 1.306557613

C2 = rt(3) 1.732050808

R1 = 1 1

R2 =1/(rt(3)Q) 0.441886576

R 1

R/3 0.333333333

Scale by w0HP

1.306557613

1.732050808

1

0.441886576

1

0.333333333

HP Crt Params

C1new=C1/(Km*Kf) 3.06293E-09

C2new 4.0604E-09

R1new=R1*Km 22630.24168

R2new=R2*Km 10000

Rnew=R*KmR 60000

Rnew/3 20000

w0LP/HP 1 1

Q 0.541184111 0.541184111

Input Attenuation

C3 = (1 - alpha)C1 7.65744E-10

KNLP/NHP 1 1

w0LP/HP 1 1

Q 1.306557613 1.306557613

C4 = alpha*C1 2.29719E-09


d(1) 1.8478 1.8478

d(2) 1 1

K 1.33334

Kf 18849.6

Km 10000

KmR 30000

alpha 0.74999625

NHP Crt Pars

C1 = Q 0.541184111

C2 = rt(3) 1.732050808

R1 = 1 1

R2 =1/(rt(3)Q) 1.066827827

R 1

R/3 0.333333333

Scale by w0HP

0.541184111

1.732050808

1

1.066827827

1

0.333333333

HP Crt Params

C1new=C1/(Km*Kf) 2.87106E-09

C2new 9.18879E-09

R1new=R1*Km 10000

R2new=R2*Km 10668.27827

Rnew=R*KmR 30000

Rnew/3 10000

Input Attenuation

C3 = (1 - alpha)C1 7.17777E-10

C4 = alpha*C1 2.15329E-09

SOLUTION TO 21.33. Using MATLAB, »fp = 5e3; fs = 1.5e3; »wp = 2*pi*fp; ws = 2*pi*fs; »Amax = 3; Amin = 40; »n = buttord(wp,ws,Amax,Amin,'s') n =4 »[z,p,k] = buttap(n) z =[] p= -3.8268e-01 + 9.2388e-01i -3.8268e-01 - 9.2388e-01i -9.2388e-01 + 3.8268e-01i

KNLP/NHP 1 1

-9.2388e-01 - 3.8268e-01i k= 1 »d1 = real(poly([p(1),p(2)])) d1 = 1.0000e+00 7.6537e-01 1.0000e+00 »d2 = real(poly([p(3),p(4)])) d2 = 1.0000e+00 1.8478e+00 1.0000e+00 In general, 2 1 1 s  1 HNHP (s) = HNLP   = = × 2 2 s 2 2 (1 0LP )  1  1 2 ( 0LP ) 0LP s + s + (1 0LP) + + ( 0LP ) Q  s Q  s

=

( 0HP )2 s2 s2 + 0HP s + ( 0HP ) 2 Q

The S&K Saraga design for d1 is given by the following excel spreadsheet: INPUT: NLP H(s) NHP H(s)

d(1) 0.76537 0.76537

d(2) 1 1

K 1.33334

Kf 31416

Km 45260.48336

KmR 60000

alpha 0.74999625

NHP Crt Pars

C1 = Q 1.306557613

C2 = rt(3) 1.732050808

R1 = 1 1

R2 =1/(rt(3)Q) 0.441886576

R 1

R/3 0.333333333

Scale by w0HP

1.306557613

1.732050808

1

0.441886576

1

0.333333333

HP Crt Params

C1new=C1/(Km*Kf) 9.18879E-10

C2new 1.21812E-09

R1new=R1*Km 45260.48336

R2new=R2*Km 20000

Rnew=R*KmR 60000

Rnew/3 20000

Input Attenuation

C3 = (1 - alpha)C1 2.29723E-10

KNLP/NHP 1 1

w0LP/HP 1 1

C4 = alpha*C1 6.89156E-10

The S&K Saraga design for d2 is given by the following excel spreadsheet:

Q 1.306557613 1.306557613


d(1) 1.8478 1.8478

d(2) 1 1

K 1.33334

Kf 31416

Km 20000

KmR 60000

alpha 0.74999625

NHP Crt Pars

C1 = Q 0.541184111

C2 = rt(3) 1.732050808

R1 = 1 1

R2 =1/(rt(3)Q) 1.066827827

R 1

R/3 0.333333333

Scale by w0HP

0.541184111

1.732050808

1

1.066827827

1

0.333333333

HP Crt Params

C1new=C1/(Km*Kf) 8.61319E-10

C2new 2.75664E-09

R1new=R1*Km 20000

R2new=R2*Km 21336.55655

Rnew=R*KmR 60000

Rnew/3 20000

Input Attenuation

C3 = (1 - alpha)C1 2.15333E-10

KNLP/NHP 1 1

w0LP/HP 1 1

C4 = alpha*C1 6.45986E-10

This completes the design.

SOLUTION TO 21.34. For the woofer, 8 8 H(s) = = L Ls + 8 s + 8 L Thus,

8 = 2000 × 2π ⇒ L = 636 µH. L For the tweeter, 8 8Cs s H(s) = 1 = = 1 + 8 8Cs + 1 s + Cs 8C

Thus,

1 = 2000 × 2π ⇒ C = 9.95 µF. 8C

SOLUTION TO 21.35. For the woofer, 1 HNLP (s) = 2 s + 2s + 1 The transfer function of the following circuit

Q 0.541184111 0.541184111

is 1 LC H(s) = 1 1 2 s + s+ LC C 1 1 ⇒ C = 0.70711 F and since = 1, L = 2 = 1.4142 H. Frequency scaling the C LC 0.70711 element values by Kf = 4000π and magnitude scaling by Km = 8 yields C = = 7.0337 µF and Km K f 1.4142K m L= = 0.90032 mH: Kf Thus

2=

For the tweeter we first realize the NLP Butterworth transfer function as above to obtain as above

with Thus C = 0.70711 F and L = 2 = 1.4142 H. We now apply the frequency transformation s →

C to each element (capacitors become inductors and inductors become capacitors according to s figure 21.24) and we obtain the HP circuit topology

where Cnew = 7.0337 µF and Lnew = 0.90032 mH.

SOLUTION TO 21.36. Consider figure (a). Let the current entering the RC network from Z1 be dentoted by Ifa. Let the voltage from this point to ground be denoted Vfa. Then Vout,a = V fa + I faZ1 = H a (s)Vin For figure (b) with a similar denotation of voltage and current, we have   V I fb  Z k −1 1  fb  Z Vout,b = V fb +  I fb + Z  1 = V fb + I fb 1 + V fb =  (2k − 1)V fb +  k k k Z1  1  k   k − 1 If

(2k − 1)V fb ≅ V fa and I fb ≅ I fa

(**)

then Vout,b ≅

Vout ,a 1 H a (s)Vin = k k

Z1 be large relative k −1 to what it sees in the RC network. Hence, in general, k must be close to 1. Thus only small gain enhancements are possible. For such a potentially sensitive approach to gain enhancement, it might be better simply to add another op amp stage as op amps are comparatively inexpensive. For gain enhancement, k < 1. However, for the (**) to be valid, we require that

CHAPTER 22 PROBLEM SOLUTIONS SOLUTION TO PROBLEM 22.1 (a) For figure P22.1a, T0 = 2 and ω0 = π . Let t0 = -1 in equation 22.5b. Then f(t) = δ(t) and 1

c n = 0.5 ∫ (t)e − jnπt dt = 0.5

for all n

−1

From equation 22.6, an = 1 and bn = 0 for all n. Finally from equation 22.2 f (t) = 0.5 +

∞

∑ cos(nπt)

n=1

(b) For figure P22.1b, T0 = 2 and ω0 = π . Let t0 = 0 in equation 22.5b. Then f(t) = - δ(t -1) and 1

c n = −0.5 ∫ (t − 1)e − jnπt dt = −0.5e − jnπ −1

From equation 22.6, bn = 0 for all n, and an = -1 for n even an = 1 for n odd Finally from equation 22.2 f (t) = −0.5 + cos(πt) − cos(2πt) + cos(3πt) − cos(4πt) + K

SOLUTION TO PROBLEM 22.2 (a) T0 = 1 and ω0 = 2π . Let t0 = 0 in equation 22.5b. Then f (t) = e − ( ln(2)) t and from equation 22.5b 1

cn = ∫

e −( ln(2)) t e − j 2nπt dt =

0

1

∫e 0

=

0.5 (ln(2) + j2nπ)

(b) Using the above result for cn c0 =0.5 = 0.7213 , ln2 c1 = 0.5 = 0.7213e -j1.4609 ln2+ j2π c2 = 0.5 = 0.397e -j1.516 ln2+ j4π

−( ln(2)+ j 2nπ ) t

dt =

[

]

−1 e −( ln(2)+ j 2nπ ) − 1 (ln(2) + j2nπ)

From equation 22.6 d0 = c0 = 0.7213 d1 =2 c1 = 0.158 , θ 1 =-1.461x180/π =-83.7o d2 =2 c2 = 0.0795 θ 2 =-1.516x180/π =-86.84 o Thus, f(t) in the form of equation 22.3 is f(t) = 0.7213 + 0.158cos(2πt - 83.7o) + 0.0795os(2πt - 86.84o)

SOLUTION TO PROBLEM 22.3. (a)

T0 = 1 and ω0 = 2π . Let t0 = 0 in equation 22.5b. Then f (t) = e − ( ln(2)) t [u(t) − u(t − 0.5)]

and from equation 22.5b 0.5

cn =

∫

e −( ln(2)) t e − j 2nπt dt =

0

∫ e −( ln(2)+ j 2nπ) t dt = (ln(2) + j2nπ) [e −0.5( ln(2)+ j 2nπ) − 1]

0.5

−1

0

(−1)n 2 = (ln(2) + j2nπ) 1−

(b) Using the above result for cn, and MATLAB to evaluate the numerical result, »n= 0; »c0= (1- (-1)^n/sqrt(2))/(log(2) + j*2*n*pi c0 4.2256e-01 »n=1; »c1= (1- (-1)^n/sqrt(2))/(log(2) + j*2*n*pi) c1 = 2.9612e-02- 2.6843e-01i »abs(c1) ans = 2.7006e-01 »degreec1=angle(c1)*180/pi degreec1 = -8.3705e+01 »n=2; »c2= (1- (-1)^n/sqrt(2))/(log(2) + j*2*n*pi) c2 = 1.2817e-03- 2.3237e-02i

»abs(c2) ans = 2.3272e-02 degreec2= angle(c2)*180/pi degreec2 = -8.6843e+01 From equation 22.6 and equation 22.3 f(t) = 0.4226 + 0.54cos(2πt - 83.7o) + 0.04654os(2πt - 86.84o)

SOLUTION PROBLEM 22.4. (a) f(t) = cos(4t) sin(2t) = 0.5[ sin(6t) - sin (2t)] . The fundamental angular frequency of f(t) is ω0= 2 rad/s. The given f(t) can be expressed as f(t) = -0.5 sin( ω0t) + 0.5sin(3ω0t) . Observe that b1= -0.5, b3= 0.5 and all other ai and bi are zero. From equation 22.4 , d1= 0.5 /-90o and d3 = 0.5 /90o. From equations 22.6a and 22.6b. c1 = 0.25j and c3 = -0.25j. All other cn are zero for n positive (b) f(t) = sin2 (4t) cos2 (8t)= 0.5[1 - cos(8t)]x0.5 1+ cos(16t) = 0.25 [ 1 - cos(8t) + cos(16t) - cos(8t) cos(16t)] = 0.25 - 0.375 cos(8t) + 0.25cos(16t) - 0.125cos(24t) The fundamental angular frequency of f(t) is ω0= 8 rad/s. The given f(t) can be expressed as f(t) = 0.25 -0.375 cos( ω0t) + 0.25cos(2ω0t) - 0.125sicos(3ω0t). Observe that a0= 0.5, a1=-0.375, a2= 0.25 , a3= -0.125 and all other ai and bi are zero. Fom equation 22.4, d0= 0.25, d1= 0.375 /180o , d2 = 0.25 /0o, and d3 = 0.125 /180o, From equations 22.6a and 22.6b. c0 = 0.25, c1 = -0.375 and c2 = 0.25, c3 = -0.125. All other cn are zero for n positive. (c) f(t) = 2 + 1.5 sin(500t)- 2cos(2000t)]cos(10 6 t) =2cos(106 t) + 0.75sin(1000500t) + 0.75sin(999500t) - cos(1002000t) - cos(998000t) The fundamental angular frequency of f(t) is ω0= 500 rad/s. The given f(t) can be expressed as f(t) =2cos( 2000ω0t) + 0.75sin(2001ω0t) + 0.75sin(1999ω0t) - cos(2004ω0t) - cos(1996ω0t) Observe that: a1996 = -1, b1999= 0.75, a2000= 2, b2001= 0.75 , a2004= -1, and all other ai and bi are zero. From equation 22.4 , d1996 = 1 /180o , d1999 = 0.75 /-90o, d2000 = 2 /0o , d2001 = 0.75/-90o, d2004 = 1 /180o , and all other di are zero. From equations 22.6a and 22.6b. c1996 = -0.5 , c1999 = -j0.375, c2000 = 1 ,

c2001 = -j0.375o, dc2004 =-0.5 , and all other cn are zero for n positive..

SOLUTION PROBLEM 22.5 By inspesction, the derivative of f(t) is ∞ A A f '( t) = − ∑ A (t − nT ) = − f (t) T n=−∞ T where fδ(t) is shown in figure 22/7, with its Fourier series given by equation 22.20b, i.e.

A 2A ∞ f (t) = + ∑ cos(n 0t) T T n=1 Therefore f '( t) = −

2A ∞ ∑ cos(n 0 t) T n=1

The dc component is the average value of f(t) and is given by 0.5A. Other terms in the Fourier series of f(t) are obtained by integrating the cosine terms in the above expression. The result is

f (t) = 0.5A −

A ∞ ∑ sin(n 0t) nπ n=1

SOLUTION TO PROBLEM 22.6 Denote by f5(t) the waveform of figure P22.5, with A= 0.5 and T =1. Then by inspection f (t) = f prob5 (−t) + 0.5 Substituting the result of problem 22.5 into the above equation, we have 0.5 ∞ f (t) = 0.75 + ∑ sin(n 0 t) nπ n=1

SOLUTION PROBLEM 22.7 Consider the square wave g(t) shown in figure 22.4 with its Fourier series given by equation 22.13. By inspection, the derivative of f(t) is f '( t) = −

4 g(t) f'(t) = - 4 g(t) T T

Substituting equation 22.13 into the above expression, we have 8A ∞ sin(n 0t) f '( t) = − ∑ πT n=1,odd n The dc component is the average value of f(t) and is given by 0.5A. Other terms in the Fourier series of f(t) are obtained by integrating the sine terms in the above expression. The result is

f (t) = 0.5A +

4A ∞ cos(n 0t) ∑ π 2 n=1,odd n2

SOLUTION PROBLEM 22.8 The method used below is simpler than that suggested in the hint. We first sketch the waveform of f'(t) and observe that it may be expressed as the sum of two periodic rectangular pulse trains: f'(t)= 1 [ fp (t + 0.5αT) +fp (t - 0.5αT)] αT where fp(t) is sketched in figure 22.8. Using equation 22.23, we have ∞

2Asin(nαπ) f'(t)= 1 { [αA + ∑ cos(nω0 (t+0.5αT))] αT nπ n=1 ∞

-[αA + ∑

n=1

2Asin(nαπ) cos(nω0 (t-0.5αT))]} nπ ∞

2Asin(nαπ) = 1 {∑ [cos(nω0 (t+0.5αT))-cos(nω0 (t-0.5αT)] αT n=1 nπ \

∞

- 4Asin2 (nαπ) = 1 [∑ sin(nω0 t)] αT n=1 nπ The dc component is the average value of f(t) and is given by αA. Other terms in the Fourier series of f(t) are obtained by integrating the sine terms in the above expression. The result is ∞

sin(nαπ) 2 f(t) = αA + 2A { ∑ [ ] cos(nω0 t)} n 2 απ n=1

SOLUTION TO PROBLEM 22.9 We first sketch the waveform of f'(t) and observe that it may be expressed as the sum of a periodic rectangular pulse train and a periodic impulse train: f'(t)= 1 fp (t + 0.5αT) - fδ (t ) αT where fp(t) is sketched in figure 22.8, and fδ(t) in figure 22.7. Using equations 22.23 and 22.20b, we have

∞

2Asin(nαπ) f'(t)= 1 [αA + ∑ cos(nω0 (t+0.5αT)) αT nπ n=1

∞

- [A + 2A T T

∑

cos(nω0 t)]

n=1 ∞

sin(nαπ) = 2A {∑[ cos(nω0 (t+0.5αT)) -cos(nω0 t)]} T n=1 nαπ The dc component is the average value of f(t) and is given by 0.5αA. Other terms in the Fourier series of f(t) are obtained by integrating the sine terms in the above expression. The result is ∞

sin(nαπ) f(t) =0.5αA + A { ∑ [ sin(nω0 (t+0.5αT)) -sin(nω0 t)]} nπ n=1 nαπ It remains to rewrite the expression in the form of equation 22.2. To this end, let b = sin(nαπ)/(nαπ) and re-write the terms within [ ] as follows:

bsin(nαπ)cos(nω0 t) + {bcos(nαπ) - 1}sin(nω0 t) Hence, f(t) in the form of equation 22.2 has the coefficients, for n=1,2... an =

A sin2 (nαπ) απ 2 n2

bn =

dn =

a2n + b2n =

A {sin(nαπ)cos(nαπ) -nαπ} απ 2 n2

A sin4 (nαπ) + {sin(nαπ)cos(nαπ) -nαπ}2 απ 2 n2

A sin2 (nαπ) +(nαπ)[nαπ - sin(2nαπ)] απ 2 n2 n θ n = tan-1 (-b a ) =

n

The result is item 6 of table 22.4

SOLUTION PROBLEM 22.10. Observe that the present f(t) can be derived from that of problem 22.9 by (a) replacing t by -t; and (b) replacing α by β. Thus the Fourier series for f(t) is: ∞

f(t) =0.5βA +∑[ an cos(nω0 t) + bnsin(nω0 t) ] n=1

where an =

A sin2 (nβπ) 2 βπ n2

bn =

dn =

a2n + b2n =

-A {sin(nβπ)cos(nβπ) -nβπ} βπ 2 n2

A sin4 (nβπ) + {sin(nβπ)cos(nβπ) -nβπ}2 2 2 βπ n

A sin2 (nβπ) +(nβπ)[nβπ - sin(2nβπ)] 2 2 βπ n n θ n = tan-1 (-b an ) =

SOLUTION PROBLEM 22.11. Following the hint, we have the second derivative of f(t) given by f ''( t) =

1 1 f (t + T) − f (t) (1− )T (1− )T

where fd(t) is given in example 22.5. Notice that we have focused on the part of the waveform over [–αT, (1–α)T]. By making use of equation 22.20b, we obtain ∞ 2A f ''( t) = ∑ [cos(n 0 t + 2n π) − cos(n 0 t)] (1− )T 2 n=1

=

∞ −4A ∑ [sin(n 0t + n π)sin(n π)] (1− )T 2 n=1

Therefore, ∞   4A sin(n π)   sin(n t + n π) ∑ 0  (1− )T 2 n=1  ( n 0) 2  A ∞  A sin(n π) = +∑ 2 2 cos(n 0t + (n − 0.5)π ) 2 n=1  n π (1− ) 

f (t) = [ f (t)]ave +

Letting T = 1 and α = 0.25 we obtain,    π 16A sin n    4 A  f (t) = + ∑  cos ( n2πt + (0.25n − 0.5)π ) 2 n=1  3n 2π 2    ∞

Therefore, d0 = 0.5A , and dn =

 π 16Asin n  4

. 3n 2π 2 It follows that d1 = 0.38211A and d2 = 0.13509A.

SOLUTION TO PROBLEM 22.12.

Denote by fp(t, α) the period rectangular waveform of figure 22.5, with A=1. Then we can express the present f(t), with T= 4, as the sum of 3 terms: f(t) = 3fp (t-0.125T, 0.25) + 4 fp (t - 0.5T, 0.5) -2 From equation 22.14b, and equation 22.12c, for n= 1,2,... cn = 3 sin(0.25nπ)e-jnω0x0.125T + 4 sin(0.5nπ)e-jnω0x0.5T πn πn = 3 sin(0.25nπ)e-j0.25n π + 4 sin(0.5nπ)e-jnπ πn πn The numerical values of the first few Fourier series coefficients are: c0 = averge value of f(t) = 1 (1 + 4 - 2) = 0.75 4 c1 = 3 sin(0.25π)e-j0.25 π + 4 sin(0.5π)e-j π = -0.7958 -j 0.4775 π π 3 4 -j0.5 π c2 = sin(0.5π)e + sin(π)e-j2π =-j0.4775 2π 2π

Solution Problem 22.13 (a) For sinusoidal steady analysis, the transfer function is H(jω) =

YL 1 1 1 = = = YL +Y C + Y R 1 +ZLYC + ZLYR (1 - ω2 LC) + jωL (1 - 4×10 -5 ω2 ) + j×10--3 ω R

The transfer function evaluated at various input frequencies are listed below. H(0) = 1 H(j377) = 0.2128 /-175.4o H(j3x377) = 0.0199/-178.7o H(j5x377) = 0.0071/-179.2o Using equation 15.7 and superposition, we obtain the steady state output voltage (in V): vout (t) = 200 + 200× 2×0.2128cos (377t -175.4o ) +60× 2×0.0199cos (3×377t +30o -178.7 o ) +80× 2×0.0071cos (5×377t +50o -17.2 o )

= 200 + 2×42.55cos (377t -175.4o ) + 2×1.196cos (3×377t -148.7 o ) + 2×0.5668cos (5×377t -129.2 o ) (b) From equation derived in P11.39, Vout,eff= 2002 +42.552 + 1.1962 +0.56682 = 204.48 V The average power absorbed by the 10 kΩ resistor is Pav =204.48 104

2

= 4.1812 W

Solution Problem 22.14 One correction in the problem statement: in the angle expression, 5000 should be 10000. (a) Using the identity cos(x)cos(y)= 0.5cos(x+y) + 0.5cos(x-y), we have vin (t) = 0.1cos (998,000t) +0.2cos (999,000t) + 2cos (1,000,000t) + 0.2cos (1,001,000t) + 0.1cos (1,002,000t)

V

(b) The transfer function has a constant magnitude of 10, and a phase shift proportional to the deviation from ωc. at ω = ωc + 2ωm, the phase shift is -9 degrees. From these facts, we can write directly vout (t) = cos (998,000t +9 o ) +2cos (999,000t + 4.5o ) + 20cos (1,000,000t) + 2cos (1,001,000 - 4. 5o ) + cos (1,002,000t - 9o )

V

(c) Using the identity x+y x+y cos(x)cos(y) = 2 cos( )cos( ) 2 2 we can group the terms in vout(t) and re-write it as

vout(t) = 2 cos( 2ω mt - 9o)cos( ω ct) +4cos( ω mt - 4.5o)cos( ω ct) + 20cos(ω ct) = [20 + 4cos(ω mt - 4.5o) +2 cos(2ω m - 9o)]cos(ω ct) = g(t)cos( ω ct) Thus

g(t) = 20 + 4cos( ω mt - 4.5 o) +2cos(2ω mt - 90)

With td = 78.54 µs, then ωmtd = 1000 x 78.54x10-6 = 0.07854 rad, or 4.5 degrees. We have

10 f(t - td) = 20[1 + 0.2cos( ω m(t - td) +0.1cos( 2ω m(t - td)] = 20[1 + 0.2cos(ω mt - 4.5o) + 0.1 cos( 2ω mt - 9o)] = g(t)

SOLUTION PROBLEM 22.15. (a) This proof is a special case of the general proof given in the solution to Problem 22.16. See the solution to problem 22.16, below.

FOR THE REMAINING PARTS , WE USE THE FOLLOWING MATLAB CODE : % chapter 22, problem 15. %part (b). Vmax= 30*pi; Vmin=0; T=4;R = 1; C = 1; voutmin= Vmin+(Vmax - Vmin)/(1+ exp(0.5*T/(R*C))) voutmax = Vmax - (Vmax -Vmin)/(1+ exp(0.5*T/(R*C))) t1= 0: 0.05: 2; vseg1= Vmax+(voutmin- Vmax)*exp((-t1/(R*C))); t2= 2:0.05: 4; vseg2= Vmin+(voutmax- Vmin)*exp(-(t2-T/2)/(R*C)); t= [ t1 t2]; v= [vseg1 vseg2]; plot(t,v) grid xlabel('t in seconds') ylabel('Steady State Output Voltage in V')

%part (c) error1= 100*(12.235- voutmin)/voutmin error2= 100*(82.013 - voutmax)/voutmax TO OBTAIN

(b) voutmin = 1.1235e+01 voutmax = 8.3013e+01 (c) error1 = 8.9045e+00 error2 = -1.2048e+00

SOLUTION PROBLEM 22.16. CORRECTION: In the problem statement,

s

should be s .

For simplicity, let us consider the case when the transfer function is a voltage ratio, i.e., H(s) = Vout./Vin. If a constant input vin(t) = Vcon is applied to the stable network, then the vout(∞) = KV con ,

independent of the initial conditions. This is because the zero-input response for a stable network approaches zero as t approaches infinite. To see this observe that the zero-state response is given by  K vout (t) = L−1 s + 1

(

KV V  KV  × con  = L−1  con − con  = KVcon 1− e −t s  s +1   s

)u(t)

from which vout(∞) = KV con as asserted. For the remainder of our proof we make use of the fact that in steady state, vout(t+T) = vout(t) with t = 0 in our case. Specifically, after the first order network has reached steady state, the vout(t) waveform will be periodic as shown in figure P22.15c, where the time reference has been chosen so that vout,min occurs at t = 0. Recall equation 8.19

[

]

x(t) = x(∞) + x(t +0 ) − x(∞) e −(t −t0 )

Applying this equation to the interval [0, T/2], we have b0 = KVmax + (a0 − KVmax ) e −0.5T / ≡ KVmax + (a0 − KVmax ) Note that

(1)

= e −0.5T / . Similarly, applying equation 8.19 to the interval [ T/2, T] leads to a0 = KVmin + (b0 − KVmn )

(2)

Equations (1) and (2) can be written as a single matrix equation − 1 

1   a0  (1− )KVmax  = −   b0   (1− )KVmin 

(3)

Solving equation (3) by matrix inverse (or Cramer's rule) results in a0  b  =  0

−1  2 − 1 1 

[v out (t)]min = a0 = K = KVmin +

1   (1− )KVmax  K    (1− )KV  = + 1  1  min  

1  Vmax  K  Vmax + Vmin    V  = + 1 V    min   max + Vmin 

Vmax + Vmin V + Vmin + Vmin − Vmin = K max +1 +1

Vmax − Vmin V − Vmin V −V = KVmin + K max = KVmin + K max 0.5T min / +1 1+ 1/ 1+ e

and

[v out (t)]max = b0 = K Vmax ++ 1Vmin = K Vmax +

Vmin + Vmax − Vmax +1

V − Vmin V −V = KVmax − K max = KVmax − K max 0.5Tmin / 1 + 1/ 1+ e This complete completes the derivation of the desired formulas.

SOLUTION TO PROBLEM 22.17 There is one correction in the problem statement. 1-kHz should be 5.1mHz. (a) The transfer function is, according to equation 4.3, and using the given element values R1 = 10 kΩ, Rf = 50 kΩ and C = 20 mF: Rf 1 V Z Y R R 1 =1 -5 H(s) = out = - f = - 1 = = Vin Z1 Yf 1 1 + R Cs 1 + 1000s f Cs + Rf (b) Here we have We have K= -5, t = 1000 s, Vmax = 1 V, Vmin = -1 V and T = 1/f = 1/0.0051= 196.15 s. In using the equations derived in problem 22.16, we note that the subscripts min and max should be switched in this case because K is negative. The following MATLAB codes perform the needed numerical calculations. f= 5.1e-3; Cf=20e-3; Rf=50e3; Rs= 10e3; T=1/f; K= -Rf/Rs; tau=Rf*Cf; ['part (b)'] Vmax=1; Vmin=-1; voutmax= K*( Vmin + (Vmax-Vmin)/(1+exp(0.5*T/tau))) voutmin= K*( Vmax - (Vmax-Vmin)/(1+exp(0.5*T/tau))) answers from MATLAB are: voutmax = 0.2449 voutmin = -0.2449 (c) ['part (c)']

t1= 0:0.005*T:0.5*T; v1= 5 + (-0.245 -5).*exp(-(t1/tau)) ; t2= 0.5*T:0.005*T:T; v2= -5 + (0.245 +5).*exp(-(t2-0.5*T)/tau) ; t=[t1,t2]; v=[v1,v2]; plot (t,v) xlabel('time in seconds') ylabel('vout in V') grid

0.3

vout in V

0.2 0.1 0 -0.1 -0.2 -0.3

0

50

100 time in seconds

150

200

SOLUTION PROBLEM 22.18. vi (t) = cos (ωt) = cos(6000t) vo (t) = 10v i + v2i + v3i = 10cos(ωt) + cos2 (ωt) + cos3 (ωt) = 10cos(ωt) +0.5 +0.5cos(2ωt) + 0.75cos(ωt) + 0.25cos(3ωt) = 0.5 + 10.75 cos(ωt) + 0.5cos(2ωt) 0.25cos(3ωt) The effective values of various components of the output are: dc 0.5

fundamental 10.75/√2

2nd harmonic 0.5/√2

The total harmonic distortion is 0.559 × 100% = 5.2% 10.75

3rd harmonic 0.25/√2

total harmonic 0.559/√2

and the average power at the fundamental frequency is 2 0.5× 10.75 = 0.5778 W 100

SOLUTION TO PROBLEM 22.19. vo (t) = V0 + V1 cos(ωt) + V2 cos(2ωt) + V3 cos(3ωt) We use equation 22.30 and the values of vo(t) read from the oscilloscope to compute Vk. The total harmonic distortion is given by

H.D. =

V22 + V23 ×100 % V1

MATLAb codes: vo0= 10; vo60= 5.2; vo120= -4.6; vo180= -9.6; V0= (vo0+2*vo60+2*vo120 +vo180)/6 V1= (vo0+ vo60 - vo120 -vo180)/3 V2= (vo0 - vo60 - vo120 +vo180)/3 V3= (vo0-2*vo60+2*vo120 -vo180)/6 % total harmonic distortion HD= 100*sqrt( V2^2 +V3^2)/V1 The following answers are obtained frm MATLAB output. V0 = 0.2667 V V1 = 9.8000 V V2 = -0.0667 V3 = 0 V HD = 0.6803 (percent)

SOLUTION PROBLEM 22.20. We shall follow the solution given in example 22.12, and only indicate the needed changes below. There are two corrections in example 22.12: (1) 20cos(ωαT) should be 20cos(0.5ωαT) and (2) 1.842cos(ωt) should be 1.842 cos(3ωt). The new input is vi (t) = 0.9 cos(6000t) The positive peak of the output sine wave is clipped for αT.

0.9 ×20cos(0.5ωαT) = 15 → απ = cos-1 (15/18) → α = 0.1864 Equation 22.35 becomes

Equations 2.36 becomes

Using A=1, and α = 0.1864in item #2 of table 22.4, equation 22.37 becomes

and equation 22.38 becomes

Equation 22.39 becomes

from which the harmonic distortions are: third order fifth order

1.076/16.57 = 6.595% 0.55/16.57 = 3.319%

–

SOLUTION PROBLEM 22.21. Corrections: (1) cos( 1 )e cos( 1

– −  )e

2 − 1 +π

RC

2 − 1 +π 

RC

 

= cos( 2 ) should be

= cos( 2 ) and (2) on page 943, equation 22.46 , θ2 should be –θ2, i.e., equation

22.46 should be cos( 1

– − )e 

2 − 1 +π 

RC

 

.

The proof is similar to that given on page 943 for the half-wave rectifier case, except for some minor changes described below. For the case of a full-wave rectifier, the output voltage waveform is a modification of figure 22.17 as shown below.

The exponential decay of vo(t) starts with the value Vmcos(θ1) at θ = θ1 – π (instead of θ1 – 2π, as in the half-wave rectifier case). Therefore, in equation 22.45, change T to T/2 and 2π to π. In other words

v 0 (t ) = (Vm cos( 1))

  1 T −   t −  2 −  RC     e

= (Vm cos( 1)) e

In equation 22.46, make correction (2) above, and change 2π to π, i.e.,

cos( 1

– −  )e

2 − 1 +π 

RC

 

= cos( 2 )

−

t − ( 1− RC

)

The desired proof is complete.

SOLUTION PROBLEM P22.22 Refer to figure 22.17. Assuming θ1 = 0, we have v0 (t) = Vmexp( -t ) = Vm 1 - ( t ) + ( t )2 - ( t )3 + ... RC RC RC RC Assuming θ2 = 0, we compute the average value of vo(t) over the time interval [ 0, T]. For the case RC >> T, we can approximate vo(t) over this interval by keeping only the first two terms of the infinite series. Thus v0 (t) ≅ Vm 1 - t RC which indicates that the plot of vo vs. t over the interval [ 0 T] is approximately a straight line. Therefore the average of the vo(t) over [ 0 T] is equal to vo(T/2). Thus, for the case RC >> T, or equivalently ωRC >> 2π,

Vdc ≅ v o T = Vm 1 - T = Vm 1 - π 2RC ωRC 2

SOLUTION PROBLEM 22.23 Given values are: C = 20e-6 F, R =100kΩ, Vm = 20 V; f = 60 Hz. From equation 22.48b Vdc ≅ (1 -

π )V = (1 - 1 )Vm = (1 1 )×20 = 19.916 V ωRC m 2fRC 5 2×60×10 ×20×10-6

To calculate the ripple factor, we first calculate θ2 from equation 22.47, and then use the result in equation 22.50. θ 2 = cos-1 ( 2π ) = 0.1289 rad RC

ripple factor ≅

1 - cos(θ 2 )

=0.2406%

3 [ 1 +cos(θ 2 )] For the diode average and peak currents, use equations 22.49a and 22.52 Idc = Vdc = 19.917 = 0.199 ×10-3 A R 100,000 id,peak ≅ VmωCsin(θ 2 ) = 19.4 ×10-3 A

SOLUTION PROBLEM 22.24 Equations 22.44 - 22.52 are derived for a half-wave rectifier. For a full-wave rectifier, some of these equations will be modified slightly as given below. The difference arises from changing T to T/2. The new equation for θ2 is derived in problem 22.21, and repeated below. The given values are: C = 20e-6 F, R =100kΩ, Vm = 20 V; f = 60 Hz. A modification of equation 22.48b gives Vdc = (1 -

π )V = (1 - T )Vm = (1 1 )×20 = 19.96 V 2ωRC m 4RC 5 4×60×10 ×20×10-6

To calculate the ripple factor, we first calculate θ2 from equation derived in problem 22.21, and then use the result in equation 22.50. π θ 2 ≅ cos-1 (e- ωRC ) =5.23 degrees ripple factor ≅

1 - cos(θ 2 )

=0.12%

3 [ 1 +cos(θ 2 )] For the diode average and peak currents, use equations 22.49a and 22.52 Idc = Vdc = 19.96 = 0.1996 ×10-3 A R 100,000 id,peak ≅ VmωCsin(θ 2 ) = 13.75 ×10-3 A _

SOLUTION PROBLEM 22.25. CORRECTIONS: (1) 195 Ω should be 1950 Ω. (2) 100 µF should be 10 µF. Since this problem only requires an estimate of the answer, we can use reasonable approximations to simply the solution. Let H(s) = Vo (s) I(s) be the transfer function for the linear circuit to the right of the diodes. Then, Z par (s) V (s) V (s) H(s) = in × o = Zin (s) I(s) Vin (s) R + Z par (s) where Z par (s) is the impedance of the parallel C-RL. The first step is to find the magnitude H ( j ) . As long as 1/RL << ωC and 1/(ωC) << R, the following approximations are valid: Z par ( j ) =

1

1 ≅ j C+ RL

1 C

which means that the parallel impedance is essentially that of the capacitor, and Z par ( j ) Z par ( j ) 1 ≅ ≅ R + Z par ( j ) R RC and again since 1/(ωC) << R, Zin ( j ) ≅

1 C

Using these approximations, Z par ( j ) 1 1 R H ( j ) = Zin ( j ) × ≅ × = R + Z par ( j ) C RC ( RC) 2 It is given that Idc = 0.01 A. From the short pulse property, the input current i(t) consists of very short pulses at 120 Hz and all ac components of i(t) have peak magnitudes approximately equal to twice the average dc value. Hence the peak magnitudes are 0.02 A. Therefore the magnitude of 120-Hz component of the output voltage is H ( j2π × 120) × 0.02 ≅

R

( RC )

2

× 0.02 =

(

1950 2π × 120 × 1950 × 10 × 10−6

)

2

× 0.02 = 0.18042 V

Hence, the effective value is 0.18042

2 = 0.12757 V and the ripple factor is 0.12757 × 100 = 0.42524 % 30

SOLUTION PROBLEM 22.26 Since this problem only requires an estimate of the answer, we can use reasonable approximations to simply the solution. Let H(s) = Vo (s) I(s) be the transfer function for the linear circuit to the right of the diodes. Then H(s) = Vo = Vin × Vo I I Vin Under the condition 1/(ωC) << R and 1/(ωC) << RL, we have Zin (s) ≅ ZC(s) = 1 Cs and

Vo ≅ 1 Vin ωRC

2

Using these approximations, H(jω) ≅ Zin Vout = 1 ( 1 )2 = R( 1 )3 Vin ωC ωRC ωRC For a full-wave rectifier, the fundamental frequency is 120 Hz, and ω = 240π rad/s. It is given that Idc = 0.01 A. From the short pulse property, the input current i(t) consists of very short pulses at 120 Hz and all ac components of i(t) have peak magnitudes approximately equal to twice the average dc value. Hence the peak magnitudes are 0.02 A. Therefore the peak magnitude of 120-Hz component of the output voltage is

(

1 H(j120π) x0.02 ≅ = R( 1 )3 ×0.02= 975 × ωRC 120π×975×16×10-6 Hence, the effective value is 0.012/ 2 =0.0848 V and the ripple factor is

) × 0.02 = 0.012 V 3

0.0848 ×100% = 0.2827 % 30

SOLUTION PROBLEM 22.27 Using H(jω) of figure P22.27 and the given formula for cn , we have 0.5ωs

0.25ωs

H(jω) e -jnT0 ω dω = 1 ωs

cn = 1 ωs -0.5 ωs

e-jnT0 ω dω = -0.25 ωs

s 1 [ e-jnT0 ω ]0.25ω -0.25 ωs -jnT 0ω s

= 1 sin (nπ ), n = 1,2,... nπ 2 For n = 0, we have c0= d0 = a0 /2 = [Average value of H(jω) ]= 0.5 . For n = 1,2,..., from equation 22.6

b n = 0, dn=an = 2 sin (n π ), nπ 2 The Fourier series representation of H(jω) is

H(jω) = 0.5 + 2 cos(T0ω) - - 1 cos(3T0ω) + - 1 cos(5T0ω) - - 1 cos(7T0ω + ... π 3 3 7 A plot of H(jω) vs. ω curve using the first 11 terms (n=0,1,...11) of the Fourier series is given below together with the MATLAB codes.

1.5

H(jw)

1 0.5 0 -0.5

0

20

40 60 w in radians

%chapter 2, problem 27. ws= 100; T0=2*pi/ws; w= 0: ws/200:ws; d0w = 0.5; d1w = (2/pi)*cos(T0*w); d3w= -(2/pi/3)*cos(3*T0*w); d5w= (2/pi/5)*cos(5*T0*w); d7w= -(2/pi/7)*cos(7*T0*w); d9w= +(2/pi/9)*cos(9*T0*w); d11w= -(2/pi/10)*cos(11*T0*w); H =d0w +d1w + d3w +d5w +d7w + d9w +d11w; plot(w,H) xlabel(' w in radians') ylabel('H(jw)') grid

80

100

R a DeCarlo and P Lin Linear Circuit Analysis s

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