K. W. Whites
EE 481 Course Syllabus
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EE 481 Microwave Engineering Fall 2008 Instructor:
Dr. Keith W. Whites Office: 317 Electrical Engineering/Physics (EEP) Building Email:
[email protected] Web: http://whites.sdsmt.edu Office hours: MWF 3:00-4:00 PM
To contact the instructor, please use e-mail rather than the telephone. All e-mail will be answered. The instructor will be available for assistance during the hours listed above, as well as other times when the office door is open. Catalog Description: (3-1) 4 credits. Prerequisite: 382 completed or concurrent. Presentation of basic principles, characteristics, and applications of microwave devices and systems. Development of techniques for analysis and design of microwave circuits. Time and Location: The lectures for this course will meet Monday, Wednesday, and Friday from 10:00-10:50 AM in room 251B EEP. Laboratory work will be performed in 230 EEP. There is no common laboratory time for this course. Course Reference Materials: The required materials for this course are • D. M. Pozar, Microwave Engineering, New York: John Wiley & Sons, third ed., 2004, which is available at the SDSMT Bookstore. • Additionally, the lecture notes K. W. Whites, EE 481 Microwave Engineering Lecture Notes, 2008, are available from the course web page. Grading:
30 % 30 % 20 % 20 %
– – – –
Two exams Laboratory Homework Final exam
Homework Policy: One homework set will generally be assigned each week. These homework assignments are to be turned in at the beginning of the class period on the due date. Late homework will be assessed a 10% per calendar day reduction in points. Labwork Policy: Near the middle of the semester, we will begin the first of approximately four to five labs for the course. These will involve the design, construction, and measurement of passive and active microwave circuits. The labs will also require the simulation of your circuits using Advanced Design System (ADS) from Agilent Technologies. Measurements will be performed in the Laboratory for Applied Electromagnetics and Communications (LAEC) located in room 230 EEP using Agilent 8753ES vector network analyzers. Laboratory work will be performed in pairs of students and open lab hours will be posted. Late lab reports will be assessed a 10% per calendar day reduction in points. Exam Policy: The exams will be closed book and closed notes with no formula sheets. Using or referring to equations stored in a calculator is not allowed, even if these equations come preprogrammed into the calculator. If you feel an exam problem was graded incorrectly, it must be South Dakota School of Mines and Technology
Revised 9/2/08
EE 481 Microwave Engineering
Lecture Notes
Keith W. Whites Fall 2008
Laboratory for Applied Electromagnetics and Communications Department of Electrical and Computer Engineering South Dakota School of Mines and Technology © 2008 Keith W. Whites
Whites, EE 481
Lecture 1
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Lecture 1: Introduction. Overview of Pertinent Electromagnetics.
Millimeter Wave Region
Microwave Region (λ = 30 cm to 8 mm)
RF Region
In this microwave engineering course, we will focus primarily on electrical circuits operating at frequencies of 1 GHz and higher. In terms of band designations, we will be working with circuits above UHF: Band
Frequency
HF
3 MHz-30 MHz
VHF
30 MHz-300 MHz
UHF
300 MHz-1 GHz
L
1-2 GHz
S
2-4 GHz
C
4-8 GHz
X
8-12 GHz
Ku
12-18 GHz
K
18-27 GHz
Ka
27-40 GHz
V
40-75 GHz
W
75-110 GHz
mm
110-300 GHz
RF, microwave and millimeter wave circuit design and construction is far more complicated than low frequency work. So why do it? © 2008 Keith W. Whites
Whites, EE 481
Lecture 1
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Advantages of microwave circuits: 1. The gain of certain antennas increases (with reference to an isotropic radiator) with its electrical size. Therefore, one can construct high gain antennas at microwave frequencies that are physically small. (DBS, for example.) 2. More bandwidth. A 1% bandwidth, for example, provides more frequency range at microwave frequencies that at HF. 3. Microwave signals travel predominately by line of sight. Plus, they don’t reflect off the ionosphere like HF signals do. Consequently, communication links between (and among) satellites and terrestrial stations are possible. 4. At microwave frequencies, the electromagnetic properties of many materials are changing with frequency. This is due to molecular, atomic and nuclear resonances. This behavior is useful for remote sensing and other applications. 5. There is much less background noise at microwave frequencies than at RF. Examples of commercial products involving microwave circuits include wireless data networks [Bluetooth, WiFi (IEEE Standard 802.11), WiMax (IEEE Standard 802.16), ZigBee], GPS, cellular telephones, etc. Can you think of some others?
Whites, EE 481
Lecture 2
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Lecture 2: Telegrapher Equations For Transmission Lines. Power Flow. Microstrip is one method for making electrical connections in a microwave circuit. It is constructed with a ground plane on one side of a PCB and lands on the other:
ε
Microstrip is an example of a transmission line, though technically it is only an approximate model for microstrip, as we will see later in this course. Why TLs? Imagine two ICs are connected together as shown:
A B
When the voltage at A changes state, does that new voltage appear at B instantaneously? No, of course not. If these two points are separated by a large electrical distance, there will be a propagation delay as the change in state (electrical signal) travels to B. Not an instantaneous effect. © 2008 Keith W. Whites
Whites, EE 481
Lecture 2
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In microwave circuits, even distances as small as a few inches may be “far” and the propagation delay for a voltage signal to appear at another IC may be significant. This propagation of voltage signals is modeled as a “transmission line” (TL). We will see that voltage and current can propagate along a TL as waves! Fantastic. The transmission line model can be used to solve many, many types of high frequency problems, either exactly or approximately: • Coaxial cable. • Two-wire. • Microstrip, stripline, coplanar waveguide, etc. All true TLs share one common characteristic: the E and H fields are all perpendicular to the direction of propagation, which is the long axis of the geometry. These are called TEM fields for transverse electric and magnetic fields. An excellent example of a TL is a coaxial cable. On a TL, the voltage and current vary along the structure in time t and spatially in the z direction, as indicated in the figure below. There are no instantaneous effects.
Whites, EE 481
Lecture 2
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E H E Δz
Fig. 1
A common circuit symbol for a TL is the two-wire (parallel) symbol to indicate any transmission line. For example, the equivalent circuit for the coaxial structure shown above is:
Analysis of Transmission Lines On a TL, the voltage and current vary along the structure in time (t) and in distance (z), as indicated in the figure above. There are no instantaneous effects. i ( z, t ) +
v ( z, t ) -
i ( z, t )
z
Whites, EE 481
Lecture 2
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How do we solve for v(z,t) and i(z,t)? We first need to develop the governing equations for the voltage and current, and then solve these equations. Notice in Fig. 1 above that there is conduction current in the center conductor and outer shield of the coaxial cable, and a displacement current between these two conductors where the electric field E is varying with time. Each of these currents has an associated impedance: • Conduction current impedance effects: o Resistance, R, due to losses in the conductors, o Inductance, L, due to the current in the conductors and the magnetic flux linking the current path. • Displacement current impedance effects: o Conductance, G, due to losses in the dielectric between the conductors, o Capacitance, C, due to the time varying electric field between the two conductors. To develop the governing equations for V ( z , t ) and I ( z , t ) , we will consider only a small section Δz of the TL. This Δz is so small that the electrical effects are occurring instantaneously and we can simply use circuit theory to draw the relationships between the conduction and displacement currents. This equivalent circuit is shown below:
Whites, EE 481
Lecture 2
i ( z , t ) R Δz
i ( z + Δz , t )
LΔz
+
v ( z, t )
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+
GΔz v ( z + Δz , t )
C Δz
-
-
Δz
Fig. 2
The variables R, L, C, and G are distributed (or per-unit length, PUL) parameters with units of Ω/m, H/m, F/m, and S/m, respectively. We will generally ignore losses in this course. In the case of a lossless TL where R = G = 0, a finite length of TL can be constructed by cascading many, many of these subsections along the total length of the TL: Rs + vs(t) -
L Δz C Δz
Δz
L Δz C Δz
Δz
LΔz C Δz
Δz
L Δz C Δz
RL
Δz
This is a general model: it applies to any TL regardless of its cross sectional shape provided the actual electromagnetic field is TEM. However, the PUL-parameter values change depending on the specific geometry (whether it is a microstrip, stripline, two-wire, coax, or other geometry) and the construction materials.
Whites, EE 481
Lecture 2
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Transmission Line Equations To develop the governing equation for v ( z , t ) , apply KVL in Fig. 2 above (ignoring losses) ∂i ( z , t ) (2.1a),(1) v ( z , t ) = LΔz + v ( z + Δz , t ) ∂t Similarly, for the current i ( z , t ) apply KCL at the node ∂v ( z + Δz , t ) (2.1b),(2) i ( z , t ) = C Δz + i ( z + Δz , t ) ∂t Then: 1. Divide (1) by Δz : v ( z + Δz , t ) − v ( z , t ) ∂i ( z , t ) (3) = −L Δz ∂t In the limit as Δz → 0 , the term on the LHS in (3) is the forward difference definition of derivative. Hence, ∂v ( z , t ) ∂i ( z , t ) = −L (2.2a),(4) ∂z ∂t 2. Divide (2) by Δz : i ( z + Δz , t ) − i ( z , t )
∂v ( z + Δz 0 , t )
= −C (5) Δz ∂t Again, in the limit as Δz → 0 the term on the LHS is the forward difference definition of derivative. Hence,
Whites, EE 481
Lecture 3
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Lecture 3: Phasor Wave Solutions to the Telegrapher Equations. Termination of TLs We will continue our TL review by considering the steady state response of TLs to sinusoidal excitation. Consider the following TL in the sinusoidal steady state: Zs
l
+ Z0, vp
Vs
ZL
z
We previously derived the wave equations for the voltage and current as ∂ 2v ( z , t ) 1 ∂ 2v ( z , t ) = 2 (1) 2 2 ∂z v p ∂t and
∂ 2i ( z , t ) 1 ∂ 2i ( z , t ) = 2 2 ∂z v p ∂t 2
(2)
For sinusoidal steady state, we will employ the phasor representation of the voltage and current as (3) v ( z , t ) = ℜe ⎡⎣V ( z ) e jωt ⎤⎦
i ( z , t ) = ℜe ⎡⎣ I ( z ) e jωt ⎤⎦
where V ( z ) and I ( z ) are spatial phasor functions.
© 2008 Keith W. Whites
(4)
Whites, EE 481
Lecture 3
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Substituting (3) into (1) gives d 2V ( z ) 1 ω2 2 = 2 ( jω ) V ( z ) = − 2 V ( z ) 2 dz vp vp
(5)
We define
β = ω LC
[rad/m] (2.12a),(6) as the phase constant for reasons that will be apparent shortly. (L and C are the usual TL per-unit-length parameters.) From (6)
β = ω LC = 2
2
ω2 v 2p
Substituting this into (5) gives d 2V ( z ) + β 2V ( z ) = 0 (7) 2 dz Similarly, from (4) and (2) we can derive d 2I ( z ) 2 + β I (z) = 0 (8) 2 dz Equations (7) and (8) are the wave equations for V and I in the frequency domain (i.e., the phasor domain). The solutions to these two second-order ordinary differential equations are (2.14a),(9) V ( z ) = Vo+ e − j β z + Vo− e + j β z and
I ( z ) = I o+ e − j β z + I o− e + j β z
Vo+ ,Vo− , I o+ , I o− are complex constants.
(10)
Whites, EE 481
Lecture 3
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We can confirm the correctness of these two solutions by direct substitution into (7) and (8). For example, substituting Vo+ e − j β z from (9) into (7) gives 2 Vo+ ( − j β ) e − jβ z + β 2V ( z ) =? 0 or − β 2V + e − j β z + β 2V + e − j β z =? 0 o
o + − jβ z o
which is indeed true. Therefore, V e
in (9) is a valid solution
to (7).
The constants I o+ and I o− in (10) can be expressed in terms of Vo+ and Vo− . In particular, it can be shown that Vo+ + (11) Io = Z0 Vo− − and (12) Io = − Z0 If we substitute (11) and (12) into (10) we find that Vo+ − j β z Vo− + j β z I (z) = e − e (2.14b),(13) Z0 Z0 and
V ( z ) = Vo+ e − j β z + Vo− e + j β z
(2.14a),(14)
Both of these equations should be committed to memory. They are the general form of phasor voltages and currents on transmission lines.
Whites, EE 481
Lecture 3
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The first terms in (13) and (14) are the phasor representation of waves propagating in the +z direction along the TL. The second terms in both equations represent waves propagating in the -z direction.
Discussion • As stated above, the first terms in (13) and (14) are the phasor representation of waves traveling in the +z direction. To see this, convert the first term in (14) to the time domain: + j ωt − β z ) ⎤ v ( z , t ) = ℜe ⎡⎣Vo+ e − j β z e jωt ⎤⎦ = ℜe ⎡ Vo+ e jφ e ( ⎣ ⎦ ⎡ ⎛ β ⎞ ⎤ = Vo+ cos (ωt − β z + φ + ) = Vo+ cos ⎢ω ⎜ t − z ⎟ + φ + ⎥ ⎣ ⎝ ω ⎠ ⎦ ⎡ ⎛ z ⎞ +⎤ + = Vo cos ⎢ω ⎜ t − ⎟ + φ ⎥ ⎜ ⎟ ⎢⎣ ⎝ v p ⎠ ⎥⎦ We can clearly see in this last result that we have a function of time with argument t − z / v p . From our previous discussions with TLs we recognize that this is a wave that is propagating in the +z direction with speed vp.
• Similarly, we can show that Vo− e + j β z (and I o− e + j β z ) are waves propagating in the -z direction.
Whites, EE 481
Lecture 4
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Lecture 4: TL Input Impedance, Time Average Power, Return and Insertion Losses. VSWR. Example N4.1: Determine an expression for the voltage at the input to the TL assuming Rs = Z0: l
Rs + Vs
+ Vg -
-
Z0 , β z
Zin
z=0
To calculate the input voltage Vg , we’ll first determine the effective impedance seen at the TL input terminals seen looking towards the load at z = 0. This is called the input impedance Z in . Forming the ratio of (19) and (20) from the previous lecture gives V ( −l ) 2Vo+ cos ( − β l ) = = − jZ 0 cot ( β l ) [Ω] Z in = + j 2Vo I ( −l ) − sin ( − β l ) Z0 In other words, the input impedance is purely reactive (2.46c) Z in = jX in where X in = − Z 0 cot ( β l )
A plot of this reactance is shown in Fig. 2.8c of the text.
© 2008 Keith W. Whites
Whites, EE 481
Lecture 4
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An equivalent circuit can now be constructed at the input to the TL by using Rs and Z in as
Using voltage division, − jZ 0 cot ( β l ) Z in Vg = Vs = Vs − jZ 0 cot ( β l ) + Z 0 Z in + Z 0
This circuit voltage Vg is also the voltage on the TL at z = −l . That is, from (19) in the previous lecture V ( z = −l ) = 2Vo+ cos ( − β l )
Since Vg = V ( z = −l ) , we can equate these two voltages giving − jZ 0 cot ( β l ) 2Vo+ cos ( β l ) = Vs − jZ 0 cot ( β l ) + Z 0
More often than not, expressions of this type are used to determine Vo+ in terms of Vs and Rs . We’ll see more on this topic in Lecture 5.
Input Impedance of a Transmission Line In problems like the one in the last example, it is helpful to have an analytical expression for the input impedance of an arbitrarily terminated TL.
Whites, EE 481
Lecture 4
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As we saw in the last lecture, the voltage and current everywhere on a homogeneous TL are V ( z ) = Vo+ e − j β z + Vo− e + jβ z (2.34a),(1) Vo+ − j β z Vo− + j β z and I ( z) = e − e (2.34b),(2) Z0 Z0 We can readily construct an input impedance expression for a TL of length l by dividing (1) and (2) for some arbitrary load reflection coefficient ΓL at z = 0: Vo− − j β l ⎞ + ⎛ + jβ l + + e ⎟ = Vo+ ( e + j β l + Γ L e − j β l ) V ( −l ) = Vo ⎜ e (3) Vo ⎝ ⎠ Vo+ ⎛ + j β l Vo− − j β l ⎞ Vo+ + j β l − + e ⎟= − Γ Le− jβ l ) I ( −l ) = e ( ⎜e Z0 ⎝ Vo ⎠ Z0
(4)
such that
+ + jβ l + Γ Le− jβ l ) V ( −l ) Vo ( e 1 + Γ Le− j 2 β l (2.43) Z in ≡ = = Z0 − j2βl 1 I ( −l ) Vo+ + j β l − Γ e L e − Γ Le− jβ l ) ( Z0 Substituting for Γ L and simplifying gives Z + jZ 0 tan ( β l ) Z in = Z 0 L [Ω] (2.44),(5) Z 0 + jZ L tan ( β l )
This is the input impedance for a lossless TL of length l and characteristic impedance Z0 with an arbitrary load ZL. Three special cases are: 1. With an open circuit load ( Z L = ∞ ), (5) yields
Whites, EE 481
Lecture 4
Z in = − jZ 0 cot ( β l ) [Ω]
Page 4 of 12
(2.46c),(6)
as we derived in the last lecture. 2. With a short circuit load ( Z L = 0 ), (5) yields Z in = jZ 0 tan ( β l ) [Ω]
(2.45c),(7)
A plot of this input reactance is shown in Fig. 2.6c. 3. With the resistive load Z L = Z 0 , (5) yields Z in = Z 0 [Ω] The input impedance is Z0 regardless of the length of the TL. All of these last three expressions should be committed to memory. You will use them often in microwave circuits. Note that both input impedances (6) and (7) are purely reactive, which is expected since neither type can dissipate energy, assuming lossless TLs.
Time Average Power Flow on TLs A hugely important part of microwave engineering is delivering signal power to a load. Examples include efficiently delivering power from a source to an antenna, or maximizing the power delivered from a filter to an amplifier.
Whites, EE 481
Lecture 4
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Often, the “power” we are ultimately concerned with is the time average power Pav, expressed as 1 ∗ Pav ( z ) = ℜe ⎡V ( z ) I ( z ) ⎤ (8) ⎣ ⎦ 2 This expression is similar to that used in circuit analysis. Substituting V(z) and I(z) from (1) and (2) into (8) gives + 2 o
1V 2 (9) Pav ( z ) = ℜe ⎡1 − Γ*L e − j 2 β l + Γ L e + j 2 β l − Γ L ⎤ ⎣ ⎦ 2 Z0 Notice that the second and third terms are conjugates so that − ( Γ Le
+ j 2βl
)
∗
+ Γ L e + j 2 β l = j 2ℑm ⎡⎣Γ L e + j 2 β l ⎤⎦
The real part of this sum is zero. Consequently, (9) simplifies to + 2 o
(
)
1V 2 1 − ΓL [W] (2.37),(10) 2 Z0 Since this power is not a function of z (true for a lossless and homogeneous TL), a z-dependence is no longer indicated for Pav. Pav =
It is important to reiterate that we’re assuming a lossless TL throughout this analysis. These results are not valid for lossy TLs. Equation (10) is very illuminating. It shows that the total time average power delivered to a load is equal to the incident time
Whites, EE 481
Lecture 4
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average power Vo+ ( 2 Z 0 ) minus the reflected time average 2 2 power Vo+ Γ ( 2 Z 0 ) . 2
The relative reflected time average power from an arbitrary load 2 on a lossless TL is the ratio of the two terms in (10) = Γ L . From (10) we see that if the load is entirely reactive so that Γ L = 1, then Pav = 0 and no time average power is delivered to the load, as expected. For all other passive loads, Pav > 0 . The time average power that is not delivered to the load can be considered a “loss” since the signal from the generator was intended to be completely transported – not returned to the generator. This return loss (RL) is defined as
(
RL = −10log10 Γ L
2
) = −20log
10
(Γ ) L
dB
(2.38),(11)
The two extremes for return loss with a passive load are: 1. A matched load where Γ L = 0 and RL = ∞ (no reflected power), and 2. A reactive load where Γ L = 1 and RL = 0 (all power reflected).
Whites, EE 481
Lecture 5
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Lecture 5: Generator and Load Mismatches on TLs. Up to this point, we have focused primarily on terminated transmission lines that lacked a specific excitation. That is, the TL was semi-infinite and terminated by a load impedance. In this lecture, we’ll complete our review of TLs by adding a voltage source together with an arbitrary load (Fig. 2.19):
This TL model is very useful and applicable to a wide range of practical engineering situations. Quantities of interest in such problems include the input impedance (for matching purposes) and signal power delivered to the load. We will first consider the computation of the latter quantity assuming the TL is lossless. Proceeding, the voltage on the TL is expressed by
© 2008 Keith W. Whites
Whites, EE 481
Lecture 5
Page 2 of 9
V ( z ) = Vo+ e − j β z + Vo− e j β z
or where
V ( z ) = Vo+ ( e − j β z + Γ L e j β z ) ΓL =
Z L − Z0 Z L + Z0
(2.69),(1) (2.68),(2)
We’ll assume that the physical properties of the TL, the source and the load are known. This leaves the complex constant Vo+ as the only unknown quantity in (1). Generally speaking, we compute Vo+ by applying the boundary condition at the TL input. (Recall that we have already applied boundary conditions at the load.) This is accomplished by applying voltage division at the input: Z in Vin = Vg (3) Z in + Z g Observe that Vin is an electrical circuit quantity. However, at the input to the TL, voltage must be continuous from the generator to the TL. This implies that Vin must also equal V ( z = −l ) on the TL. Proceeding, then from (1) at the input V ( z = −l ) = Vo+ ( e + j β l + Γ L e − j β l )
(4)
Equating (3) and (4) to enforce the boundary condition at the TL input we find
Whites, EE 481
Lecture 5
Page 3 of 9
Vo+ ( e + j β l + Γ L e − j β l ) = or
Vo+ = Vg
Z in Vg Z in + Z g
−1 Z in jβ l − jβ l + Γ [V] e e ( ) L Z in + Z g
(2.70),(5)
Maximum Power Because the TL is lossless, the time average power Pav delivered to the input of the TL must equal the time average power delivered to the load. Therefore, ⎡ Vin* ⎤ 1 1 * Pav = ℜe[Vin I in ] = ℜe ⎢Vin * ⎥ 2 2 ⎣ Z in ⎦ 2
V ⎡ 1 ⎤ Pav = in ℜe ⎢ * ⎥ [W] 2 ⎣ Z in ⎦
or
(2.74),(6)
Now, substituting (3) into (6) gives 2
2
⎡ 1 ⎤ Z in (2.74),(7) ℜe ⎢ * ⎥ 2 Z in + Z g ⎣ Z in ⎦ If we define Z in = Rin + jX in and Z g = Rg + jX g , (7) becomes Pav =
Pav =
Vg 2
Vg
2
(R
in
Rin
+ Rg ) + ( X in + X g ) 2
2
(2.75),(8)
Whites, EE 481
Lecture 5
Page 4 of 9
Employing this last result, we’ll consider three special cases for Pav in an effort to maximize this quantity. We will assume that Z g is both nonzero and fixed: (1.) Load is matched to the TL: Z L = Z 0 . From (2), Γ L = 0 in this situation, which also implies that Z in = Z 0 . [This should be intuitive. If not, see (2.43).] Consequently, from (8) with Rin = Z 0 and X in = 0 : Pav,1 =
Vg
2
(Z
2
Z0
+ Rg ) + ( X g ) 2
0
(2.76),(9)
2
(2.) Generator is matched to an arbitrarily loaded TL: Z in = Z g and Γ L ≠ 0 . Specific values for β l , Z 0 , and Z L would need to be chosen so that Z in = Z g . Then from (8) and with Rin = Rg and X in = X g : Pav,2 =
or
Vg 2
2
(R
Pav,2 =
g
Rg
+ Rg ) + ( X g + X g ) 2
Vg
2
Rg
8 Rg2 + X g2
2
(2.78),(10)
Whites, EE 481
Lecture 6
Page 1 of 14
Lecture 6: The Smith Chart The Smith chart began its existence as a very useful graphical calculator for the analysis and design of TLs. It was developed by Phillip H. Smith in the 1930s. The Smith chart remains a useful tool today to visualize the results of TL analysis, oftentimes combined with computer analysis and visualization as an aid in design. The development of the Smith chart is based on the normalized TL impedance z ( z ) defined as Z ( z) 1+ Γ( z) z ( z ) ≡ = (1) Z0 1− Γ( z) where Z ( z ) = V ( z ) / I ( z ) is the total TL impedance at z and Γ ( z ) = Γ Le
− j2β z
(2)
is the generalized reflection coefficient at z. The real and imaginary parts of the generalized reflection coefficient Γ ( z ) will be defined as Γ ( z ) ≡ Γ r ( z ) + jΓ i ( z ) . Substituting this definition into (1) gives 1 + ( Γ r + jΓ i ) z ( z ) = (3) 1 − ( Γ r + jΓ i ) Now, we will define z ( z ) ≡ r + jx and separate (3) into its real and imaginary parts © 2008 Keith W. Whites
Whites, EE 481
Lecture 6
Page 2 of 14
1 + ( Γ r + jΓ i ) 1 − ( Γ r + jΓ i ) z ( z ) ≡ r + jx = ⋅ 1 − ( Γ r + jΓi ) 1 − ( Γ r + jΓ i )* *
=
1 + j 2Γi − ( Γ 2r + Γi2 ) 1 − 2Γ r + Γ r2 + Γi2
Equating the real and imaginary parts of this last equation gives 1 − ( Γ 2r + Γi2 ) 2Γ i r= and x = (2.55) 2 2 2 2 ( Γ r − 1) + Γi ( Γ r − 1) + Γi Rearranging both of these leads us to the final two equations 2 2 r ⎞ ⎛ ⎛ 1 ⎞ 2 (2.56a),(4) ⎜ Γr − ⎟ + Γi = ⎜ ⎟ 1+ r ⎠ ⎝ ⎝1+ r ⎠ 2 2 1⎞ ⎛1⎞ 2 ⎛ (2.56b),(5) and ( Γ r − 1) + ⎜ Γi − ⎟ = ⎜ ⎟ x x ⎝ ⎠ ⎝ ⎠ We will use (4) and (5) to construct the Smith chart. Definition: The Smith chart is a plot of normalized TL resistance and reactance functions drawn in the complex, generalized reflection coefficient [ Γ ( z ) ] plane.
To understand this, first notice that in the Γr-Γi plane: 1. Equation (4) has only r as a parameter and (5) has only x as a parameter. 2. Both (4) and (5) are families of circles. Consequently, we can plot (4) and (5) in the Γr-Γi plane while keeping either r or x constant, as appropriate.
Whites, EE 481
Lecture 6
Page 3 of 14
Plot (4) in the Γr-Γi plane: • For r = 0 : Γ 2r + Γi2 = 12 2
2
1⎞ ⎛ ⎛1⎞ • For r = 1: ⎜ Γ r − ⎟ + Γi2 = ⎜ ⎟ 2⎠ ⎝ ⎝2⎠ 2 2 1⎞ 1 ⎛ ⎛2⎞ 2 • For r = : ⎜ Γ r − ⎟ + Γi = ⎜ ⎟ 3⎠ 2 ⎝ ⎝3⎠ Plot these curves in the Γr-Γi plane: Γi = Im ⎡⎣Γ ( z ) ⎤⎦
Complex Γ(z) plane
1
r=0
-1
1/2
r=1/2
-1
1
Γ r = Re ⎡⎣Γ ( z ) ⎤⎦
r=1
Plot (5) in the Γr-Γi plane: 2 2 • For x = 1: ( Γ r − 1) + ( Γi − 1) = 12
• For x = −1: ( Γ r − 1) + ( Γ i + 1) = ( −1) 2
2
2
2
1 ⎞ ⎛ 1 ⎞ ⎛ • For x = 100 : ( Γ r − 1) + ⎜ Γi − ⎟ =⎜ ⎟ 100 ⎝ ⎠ ⎝ 100 ⎠ 1 2 2 : ( Γ r − 1) + ( Γ i − 100 ) = 1002 • For x = 100 2
2
Whites, EE 481
Lecture 6
Page 4 of 14
Plot these curves in the Γr-Γi plane: Γi = Im ⎡⎣Γ ( z ) ⎤⎦
Complex Γ(z) plane x=1
x=0.01
1
r=0
x=100 1
-1
Γ r = Re ⎡⎣Γ ( z ) ⎤⎦
x=-100 x=-0.01
-1 x=-1
Combining both of these curves (or “mappings”), as shown on the next page, gives what is called the Smith chart. As quoted from the text (p. 65): “The real utility of the Smith chart, however, lies in the fact that it can be used to convert from reflection coefficients to normalized impedances (or admittances), and vice versa, using the impedance (or admittance) circles printed on the chart.”
Additionally, it is very easy to compute the generalized reflection coefficient and normalized impedance anywhere on a homogeneous section of TL.
Whites, EE 481
Lecture 6
Page 5 of 14
Notice that the Γr and Γi axes are missing from the “combined” plot. This is also the case for the Smith chart.
Whites, EE 481
Lecture 6
Page 6 of 14
Important Features of the Smith Chart 1. By definition Γ ( z ) =
z ( z ) − 1 ( r + jx ) − 1 . Therefore = z ( z ) + 1 ( r + jx ) + 1
2 r + jx − 1 r − jx + 1 ( r − 1) + x Γ( z) = ⋅ = r + jx + 1 r − jx + 1 ( r + 1)2 + x 2 2
From this result, we can show that if r ≥ 0 then Γ ( z ) ≤ 1. This condition is met for passive networks (i.e., no amplifiers) and lossless TLs (real Z 0 ). Consequently, the standard Smith chart only shows the inside of the unit circle in the Γr-Γi plane. That is, Γ ( z ) ≤ 1 which is bounded by the r = 0 circle described by Γ 2r + Γ i2 = 1. 2. If z ( z ) is purely real (i.e., x = 0 ), then since 2Γ i x= 2 ( Γ r − 1) + Γi2 we deduce that Γ i = 0 (except possibly at Γ r = 1 ). Consequently, purely real z ( z ) values are mapped to Γ ( z ) values on the Γ r = ℜe ⎡⎣Γ ( z ) ⎤⎦ axis. 3. If z ( z ) is purely imaginary (i.e., r = 0 ) then from (4) Γ 2r + Γi2 = 12 which is the unit circle in the Γr-Γi plane.
Whites, EE 481
Lecture 6
Page 7 of 14
Consequently, purely imaginary z ( z ) values are mapped to Γ ( z ) values on the unit circle in the Γr-Γi plane.
Example N6.1: Using the Smith chart, determine the voltage reflection coefficient at the load and the TL input impedance.
Whites, EE 481
Lecture 7
Page 1 of 10
Lecture 7: Transmission Line Matching Using Lumped L Networks Impedance matching (or simply “matching”) one portion of a circuit to another is an immensely important part of microwave engineering. Additional circuitry between the two parts of the original circuit may be needed to achieve this matching. Why is impedance matching so important? Because: 1. Maximum power is delivered to a load when the TL is matched at both the load and source ends. This configuration satisfies the conjugate match condition. 2. With a properly matched TL, more signal power is transferred to the load, which increases the sensitivity of the device. 3. Some equipment (such as certain amplifiers) can be damaged when too much power is reflected back to the source. Factors that influence the choice of a matching network include: 1. The desire for a simple design, if possible. 2. Providing an impedance match at a single frequency is often not difficult. Conversely, achieving wide bandwidth matching is usually difficult.
© 2008 Keith W. Whites
Whites, EE 481
Lecture 7
Page 2 of 10
3. Even though the load may change, the matching network may need to perform satisfactorily in spite of this, or be adjustable. We will discuss three methods for impedance matching in this course: 1. L networks, 2. Single stub tuners (using shunt stubs), 3. Quarter wave transformers. You’ve most likely seen all three of these before in other courses, or in engineering practice.
Matching Using L Networks Consider the case of an arbitrary load that terminates a TL: Z0,
L
ZL
To match the load to the TL, we require Γ L = 0 . However, if Z L ≠ Z 0 additional circuitry must be placed between ZL and Z0 to bring the VSWR = 1, or least approximately so:
Whites, EE 481
Lecture 7
Page 3 of 10
For Γ L = 0 , this implies Z in = Z 0 . In other words, Rin = ℜe[ Z 0 ] and X in = 0 , if the TL is lossless. Note that we need at least two degrees of freedom in the matching network in order to transform Z L at the load to Z 0 seen at the input to the matching network. This describes impedance matching in general. For an L network specifically, the matching network is either (Fig. 5.2): RL > Z 0 :
RL < Z 0 : jX (jB)-1
Z0,
ZL
Zin
where Z L = RL + jX L . This network topology gets its name from the fact that the series and shunt elements of the matching network form an “L” shape. There are eight possible combinations of inductors and capacitors in the L network:
Whites, EE 481
Lecture 7
RL > Z 0 :
Page 4 of 10
RL < Z 0 :
ZL
ZL
ZL
ZL
ZL
ZL
ZL
ZL
Notice that this type of matching network is lossless; or at least the loss can potentially be made extremely small with proper component choices. As in the text, we’ll solve this problem two ways: first analytically, then using the Smith chart.
Analytical Solution for L Network Matching y Assume RL > Z 0 . Using Fig. 5.2(a):
−1
⎛ ⎞ 1 Z in = jX + ⎜ jB + (5.1),(1) ⎟ RL + jX L ⎠ ⎝ Through the proper choice of X and B we wish to force Z in = Z 0 . Solving (1) for the B and X that produce this outcome (by equating real and imaginary parts, as shown in the text) we find that
Whites, EE 481
Lecture 7
B=
Page 5 of 10
X L ± RL Z 0 RL ( RL − Z 0 ) + X L2 RL2 + X L2
X=
1 X L Z0 Z + − 0 B RL BRL
(5.3a),(2)
(5.3b),(3)
Comments: 1. Since RL > Z 0 , the argument is positive in the second square root of (2). (B must be a real number.) 2. Note that there are two possible solutions for B in (2). 3. X in (3) also has two possible solutions, depending on which B from (2) is used. y Assume RL < Z 0 . Using Fig. 5.2(b) with Z in = Z 0 , we obtain −1
⎛ ⎞ 1 Z in = Z 0 = ⎜ jB + (5.4),(4) ⎟ Z L + jX ⎠ ⎝ Solving this equation by equating real and imaginary parts as shown in the text gives X = ± RL ( Z 0 − RL ) − X L
B=±
1 Z0
Z 0 − RL RL
(5.6a),(5) (5.6b),(6)
Comments: 1. Since RL < Z 0 , the argument is positive for the square root in (5). 2. There are two solutions for both X and B. Use the top signs in both (5) and (6) for one solution and the bottom signs for the other.
Whites, EE 481
Lecture 7
Page 6 of 10
Smith Chart Solution for L Network Matching L-network matching can also be computed graphically using the Smith chart. This approach is less accurate than the analytical approach. However, more insight into the matching process is often obtained using the Smith chart. For example, the contribution each element makes to the matching is quite clear. The process of using the Smith chart to design the matching network is probably best illustrated by example. Example 5.1 in the text illustrates the design of an L network when RL > Z 0 (Fig. 5.2a). Here, we’ll give an example when RL < Z 0 .
Example N7.1 Design an L network to match the load 25 + j 30 Ω to a TL with Z 0 = 50 Ω at the frequency f = 1 GHz.
Since RL < Z 0 , we’ll use the circuit topology in Fig. 5.2b:
Whites, EE 481
Lecture 7
Page 7 of 10
We’ll solve this problem two ways: first with the Smith chart and then analytically. Steps for a Smith chart solution: Z 1 3 1. z L = L = + j p.u.Ω. Mark this point on the chart. Z0 2 5
2. The overall concept behind this type of L-network matching is to add a reactance x to z L such that the sum of −1 admittances b and ( z L + jx ) yield yin = 1 + j 0 = zin (the center of the Smith chart). In such a case, the TL sees a matched load. So, in this case we need to add a normalized impedance jx = − j 0.1 p.u.Ω in order to move to 1 + jx on an admittance chart. 3. Convert this impedance to an admittance value by reflecting through the origin to the diametrically opposed point on the constant VSWR circle. 4. Add the normalized susceptance b = 1.0 p.u.S to reach the center of the Smith chart. Here Γ = 0 and yin = 1 + j 0 , which means the TL now sees a matched load.
Whites, EE 481
Lecture 7
Page 8 of 10
50
0.7
1.4
1.2
0.8
0. 9
1.0
40
60 1. 8
0. 6
1.6
30
0.4
70 3. 0
0.
4
20
5 0.
2.0
0. 2
0.6
Start zL
0.8
4.0
0.3
0 1.
5.0
0.2
1+jx impedance circle (on Z chart)
8 0.
1+jx admittance circle (on Z chart)
80
0.6
10
1.0
6.0
7.0 8.0 9.0 10
0.1
0.4 20
50
20
10
5.0
4.0
3.0
2.0
1.8
1.6
1.4
1.2
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0
0
0.1
0.2 50
90
End
50
0.2 20
0.4 8.0 9.0 10
0.1
0. 6
6.0
1. 0
5.0
1.0
4. 0
0.8
Constant VSWR circle
0.3
0 3.
0.4
2.0
0. 5
0
0 16
0.6
11
4 0.
1.0
0. 9
1. 2
130
15 0
0.8
1.4
0.7
0 12
1.6
0.6
1. 8
0.2
K. W. Whites
140
Un-normalizing, we find that jX = jx ⋅ Z 0 = − j 0.1 ⋅ 50 = − j 5.0 Ω 1 jB = jb ⋅ Y0 = j1.0 ⋅ = j 0.02 S 50
100
170
7.0
8 0. 0.2
Whites, EE 481
Lecture 7
Page 9 of 10
What are the L and C values of these elements? We can identify the type of element by the sign of the reactance or susceptance: Inductor Z L = jω L
Impedance Admittance
YL =
1 jω L
=
−j ωL
Capacitor 1 −j ZC = = jω C ω C YC = jω C
y Since X < 0 , we identify this as a capacitor. Therefore, −j jX = = − j 5.0 Ω ω C2 For operation at 1 GHz, we need 1 C2 = = 31.8 pF 5 ⋅ 2π f y Since B > 0 , we also identify this as a capacitor. Therefore, jB = jω C1 = j 0.02 S For operation at 1 GHz, we need 0.02 C1 = = 3.18 pF 2π f
The final circuit is: C2=31.8 pF 50 Ω f = 1 GHz
ZL =25+j30 Ω
Zin
C1=3.18 pF
Whites, EE 481
Lecture 8
Page 1 of 8
Lecture 8: Single-Stub Tuning The second matching network we’ll discuss is the single-stub tuner (SST). The single-stub tuner uses a shorted or open section of TL attached at some position along another TL:
This is an example of a parallel SST, which is the only type we’ll study. (A series SST is shown in Fig. 5.4b of the text.) The shunt-connected section is called the stub. Although not necessary, all sections of TL will be assumed to have the same Z 0 and β . Why an open or shorted section? Because these are easy to fabricate, the length can easily be made adjustable and little to no power is dissipated in the stub. (An open stub is sometimes easier to fabricate than a short.)
© 2008 Keith W. Whites
Whites, EE 481
Lecture 8
Page 2 of 8
We will study the SST from two perspectives. First, we will develop an analytical solution, followed by a Smith chart graphical solution. Referring to the figure above, the transformed load impedance at the stub position z = -d is Z + jZ 0 tan ( β d ) Z + jZ 0t = Z0 L (5.7),(1) Z ( z = −d ) = Z 0 L Z 0 + jZ L tan ( β d ) Z 0 + jZ Lt
where t ≡ tan ( β d ) . With a shunt connection, it is much simpler to work with admittances than impedances. So, we’ll define the transformed load admittance as Y = 1/ Z = G + jB . The distance d is chosen so that G = Y0 ( = 1/ Z 0 ) . As shown in the text, this condition leads to the solutions ⎧ 1 −1 tan t, t≥0 d ⎪⎪ 2π =⎨ (5.10),(2) λ ⎪1 π + tan −1 t ) , t < 0 ( ⎪⎩ 2π where 2 ⎧ 2⎤ ⎡ X ± R Z Z − R + X ( ) ( ) L L L L 0 0 ⎪ ⎣ ⎦ , RL ≠ Z 0 ⎪ − R Z t=⎨ L 0 (5.9),(3) ⎪ XL , RL = Z 0 ⎪− ⎩ 2Z 0 and Z L = RL + jX L .
Whites, EE 481
Lecture 8
Page 3 of 8
With this location of the stub, the transformed load admittance has a real part = Y0, which is almost a matched state. In general, however, this transformed YL will also have an imaginary part B. The length of the stub, ls, is chosen so that its input susceptance Bs = − B . Consequently, the parallel combination of the stub input susceptance and the transformed load admittance yield an input admittance Yin = Y0 , as seen from the source end of the TL. As shown in the text, this second condition provides the solutions ls 1 ⎛Y ⎞ = (5.11b),(4) tan −1 ⎜ 0 ⎟ short-circuit stub λ 2π B ⎝ ⎠ or ⎛B⎞ lo 1 =− tan −1 ⎜ ⎟ open-circuit stub (5.11a),(5) λ 2π ⎝ Y0 ⎠
where B is the transformed load susceptance at z = -d. Lengths of TL that are integer multiples of λ/2 can be added or subtracted from (2), (4), and (5) without altering the tuning.
Example N8.1: Match the load Z L = 35 − j 47.5 Ω to a TL with Z 0 = 50 Ω using a shunt, short-circuited single-stub tuner.
Whites, EE 481
Lecture 8
Page 4 of 8
Single Stub Tuning Using the Smith Chart We will now solve the single stub tuner problem using the Smith chart. In terms of quantities normalized to the characteristic impedance or admittance, the geometry is
Whites, EE 481
Lecture 9
Page 1 of 5
Lecture 9: Quarter-Wave-Transformer Matching For a TL in the sinusoidal steady state with an arbitrary resistive load (Fig. 2.16)
the input impedance of the right-hand TL is given as R + jZ1 tan β1l Z in = Z1 L (2.61),(1) Z1 + jRL tan β1l Now imagine that we have a special length l = λ1 / 4 of TL, as indicated in the figure above. At this frequency and physical length, the electrical length of the TL is 2π λ1 π = β1l = rad (2) λ1 4 2 Consequently, for a λ / 4 -length TL, tan β1l → ∞ . Using this
result in (1) gives Z12 Z in = RL
(2.62),(3)
This result is an interesting characteristic of TLs that are exactly λ/4 long. We can harness this characteristic to design a matching network using a λ/4-length section of TL. © 2008 Keith W. Whites
Whites, EE 481
Lecture 9
Page 2 of 5
Note that we can adjust Z1 in (3) so that Z in = Z 0 . In particular, from (3) with Z in = Z 0 we find Z1 = Z 0 RL (2.63),(4) In other words, a λ/4 section of TL with this particular characteristic impedance will present a perfect match ( Γ = 0 ) to the feedline (the left-hand TL) in the figure above. This type of matching network is called a quarter-wave transformer (QWT). Through the impedance transforming properties of TLs, the QWT presents a matched impedance at its input by appropriately transforming the load impedance. This is accomplished only because we have used a very special characteristic impedance Z1 , as specified in (4). Three disadvantages of QWTs are that: 1. A TL must be placed between the load and the feedline. 2. A special characteristic impedance for the QWT is required, which depends both on the load resistance and the characteristic impedance of the feedline. 3. QWTs work perfectly only for one load at one frequency. (Actually, it produces some bandwidth of “acceptable” VSWR on the TL, as do all real-life matching networks.)
Whites, EE 481
Lecture 9
Page 3 of 5
Real Loads for QWTs Ideally, a matching network should not consume (much) power. In (4) we can deduce that if instead of RL we had a complex load, then the QWT would need to be a lossy TL in order to provide a match. So, QWTs work better with resistive loads. However, if the load were complex, we could insert a section of TL to transform this impedance to a real quantity (is this possible?), and then attach the QWT. But, again, this would work perfectly for only one load at one frequency.
Adjusting TL Characteristic Impedance We see in (4) that the QWT requires a very specific characteristic impedance in order to provide a match Z1 = Z 0 RL With coaxial cable, twin lead and other similar TLs this is often not a practical solution for a matching problem. However, for stripline and microstrip adjusting the characteristic impedance is as simple as varying the width of the trace. Consequently, QWTs find wide use in these applications.
Whites, EE 481
Lecture 9
Page 4 of 5
As we’ll see in Lecture 12, the characteristic impedance of a microstrip
d
εr
W
as a function of W/d is Z0 @ΩD 175 150 125 100 75 50 25 2
4
6
8
10
Wêd
To construct this curve, it was assumed that ε r = 3.38 , which is the quoted specification for Rogers Corporation RO4003C laminate that we’ll be using in the lab.
Example N9.1: Design a microstrip QWT to match a load of 100 Ω to a 50-Ω line on Rogers RO4003C laminate. Estimate the fractional bandwidth under the constraint that no more than 1% of the incident power is reflected.
Whites, EE 481
Lecture 9
Page 5 of 5
From April 2006 High Frequency Electronics Copyright © 2006 Summit Technical Media
High Frequency Design
MATCHING NETWORKS
The Yin-Yang of Matching: Part 2—Practical Matching Techniques By Randy Rhea Consultant to Agilent Technologies
The conclusion of this article covers transmission line matching networks, plus a discussion of how characteristics of the load affects matching bandwidth and the choice of network topologies
The Standard QuarterWavelength Transmission Line Transformer
A
well-known distributed matching network is the quarter-wavelength long transmission line transformer. I will refer to this network as a type 11. The characteristic impedance of this line is given by Z0 =
RS RL
(41)
For example, a 100 ohm load is matched to a 50 ohm source using a 90° line with characteristic impedance 70.71 ohms. The matchable space of the quarter-wavelength transformer is small, essentially only the real axis on the Smith chart. Nevertheless, it enjoys widespread use. A quarter-wavelength line is also used in filter design as an impedance inverter to convert series resonant circuits to parallel resonance, and vice versa [4].
where ρload =
ZL′ − 1 ZL′ + 1
(43)
(Im [ρ ])
2
ρmax =
load
Re [ρload ]
+ Re [ρload ]
(44)
and the electrical length of the line is given by (see text)
θ12 = tan −1
4 a2 + b2 − b + 90° 2a
(45)
where a = Z12 X L
(46)
The General Transmission Line Transformer Perhaps less well-known is that a single series transmission line can match impedances not on the axis of reals. The matchable space of this type 12 network is plotted in Figure 9. The characteristic impedance of the series line is given by
Z12 = Z0
28
High Frequency Electronics
1 + ρmax 1 − ρmax
(42)
Figure 9 · By allowing line lengths other than 90°, the matchable impedance space for a single, series transmission line extends beyond the real axis.
Whites, EE 481
Lecture 10
Page 1 of 10
Lecture 10: TEM, TE, and TM Modes for Waveguides. Rectangular Waveguide. We will now generalize our discussion of transmission lines by considering EM waveguides. These are “pipes” that guide EM waves. Coaxial cables, hollow metal pipes, and fiber optical cables are all examples of waveguides. We will assume that the waveguide is invariant in the zdirection: y Metal walls b
μ, ε x z
a
and that the wave is propagating in z as e − j β z . (We could also have assumed propagation in –z.)
Types of EM Waves We will first develop an extremely interesting property of EM waves that propagate in homogeneous waveguides. This will lead to the concept of “modes” and their classification as • Transverse Electric and Magnetic (TEM), © 2008 Keith W. Whites
Whites, EE 481
• •
Lecture 10
Page 2 of 10
Transverse Electric (TE), or Transverse Magnetic (TM).
Proceeding from the Maxwell curl equations: xˆ yˆ zˆ ∇ × E = − jωμ H ⇒
or
∂ ∂x Ex
∂ ∂y Ey
∂ = − jωμ H ∂z Ez
∂Ez ∂E y − = − jωμ H x ∂y ∂z ⎛ ∂E ∂E ⎞ yˆ : − ⎜ z − x ⎟ = − jωμ H y ∂z ⎠ ⎝ ∂x ∂E y ∂Ex − = − jωμ H z zˆ : ∂x ∂y
xˆ :
However, the spatial variation in z is known so that ∂ ( e− jβ z ) = − j β ( e− jβ z ) ∂z Consequently, these curl equations simplify to ∂Ez + j β E y = − jωμ H x ∂y ∂E − z − j β Ex = − jωμ H y ∂x ∂E y ∂Ex − = − jωμ H z ∂x ∂y
(3.3a),(1) (3.3b),(2) (3.3c),(3)
Whites, EE 481
Lecture 10
Page 3 of 10
We can perform a similar expansion of Ampère’s equation ∇ × H = jωε E to obtain ∂H z + j β H y = jωε Ex (3.4a),(4) ∂y ∂H z − jβ H x − = jωε E y (3.4b),(5) ∂x ∂H y ∂H x − = jωε Ez (3.5c),(6) ∂x ∂y Now, (1)-(6) can be manipulated to produce simple algebraic equations for the transverse (x and y) components of E and H . For example, from (1): ⎞ j ⎛ ∂Ez + Hx = j β E y⎟ ωμ ⎜⎝ ∂y ⎠ Substituting for Ey from (5) we find j ⎡ ∂Ez 1 ⎛ ∂H z ⎞ ⎤ Hx = j j H β β + − − x ⎜ ⎟ jωε ⎝ ωμ ⎣⎢ ∂y ∂x ⎠ ⎦⎥ j ∂Ez j β ∂H z β2 = + 2 Hx − 2 ωμ ∂y ω με ω με ∂x
or,
Hx =
∂H z ⎞ j ⎛ ∂Ez ωε β − ∂y ∂x ⎠⎟ kc2 ⎝⎜
where kc2 ≡ k 2 − β 2 and k 2 = ω 2 με . Similarly, we can show that
(3.5a),(7) (3.6)
Whites, EE 481
Lecture 10
Hy = −
∂H z ⎞ j ⎛ ∂Ez + ωε β ∂x ∂y ⎟⎠ kc2 ⎜⎝
Page 4 of 10
(3.5b),(8)
Ex =
− j ⎛ ∂Ez ∂H z ⎞ + β ωμ ∂y ⎠⎟ kc2 ⎝⎜ ∂x
(3.5c),(9)
Ey =
∂Ez ∂H z ⎞ j ⎛ − + β ωμ ∂y ∂x ⎟⎠ kc2 ⎜⎝
(3.5d),(10)
Most important point: From (7)-(10), we can see that all transverse components of E and H can be determined from only the axial components Ez and H z . It is this fact that allows the mode designations TEM, TE, and TM. Furthermore, we can use superposition to reduce the complexity of the solution by considering each of these mode types separately, then adding the fields together at the end.
TE Modes and Rectangular Waveguides A transverse electric (TE) wave has Ez = 0 and H z ≠ 0 . Consequently, all E components are transverse to the direction of propagation. Hence, in (7)-(10) with Ez = 0 , then all transverse components of E and H are known once we find a solution for only H z . Neat!
Whites, EE 481
Lecture 10
Page 5 of 10
For a rectangular waveguide, the solutions for Ex , E y , H x , H y , and H z are obtained in Section 3.3 of the text. The solution and the solution process are interesting, but not needed in this course. What is found in that section is that
⎛ mπ ⎞ ⎛ nπ ⎞ = ⎜ ⎟ +⎜ ⎟ a ⎝ ⎠ ⎝ b ⎠ 2
kc ,mn Therefore,
2
m, n = 0,1,… ( m = n ≠ 0)
β = β mn = k 2 − kc2,mn
(11) (12)
These m and n indices indicate that only discrete solutions for the transverse wavenumber (kc) are allowed. Physically, this occurs because we’ve bounded the system in the x and y directions. (A vaguely similar situation occurs in atoms, leading to shell orbitals.) Notice something important. From (11), we find that m = n = 0 means that kc ,00 = 0 . In (7)-(10), this implies infinite field amplitudes, which is not a physical result. Consequently, the m = n = 0 TE or TM modes are not allowed. One exception might occur if Ez = H z = 0 since this leads to indeterminate forms in (7)-(10). However, it can be shown that inside hollow metallic waveguides when both m = n = 0 and
Whites, EE 481
Lecture 11
Page 1 of 7
Lecture 11: Dispersion. Stripline and Other Planar Waveguides. Perhaps the biggest reason the TEM mode is preferred over TE or TM modes for propagating communication signals is that ideally it is not dispersive. That is, the phase velocity of a TEM wave is not a function of frequency [ v p ≠ g (ω ) ] if the material properties of the waveguide are not functions of frequency. To see this, recall for a TEM wave that β = ω LC . Therefore, ω 1 vp = = β LC which is not a function of frequency, as conjectured, provided neither L nor C are functions of frequency. However, for either TE or TM modes, v p is a function of frequency regardless of the material properties of the waveguide. Take the rectangular waveguide as an example. In the last lecture, we found that 2 2 m π n π ⎛ ⎞ ⎛ ⎞ β = β mn = k 2 − kc2,mn and kc2,mn = ⎜ ⎟ +⎜ ⎟ a ⎝ ⎠ ⎝ b ⎠ where m, n = 0,1,2,… ( m = n ≠ 0 ) for TE modes, while m, n = 1,2,… for TM modes.
© 2008 Keith W. Whites
Whites, EE 481
Lecture 11
Page 2 of 7
For a CW signal carried by one of these modes, the phase velocity is
ω
v p ,mn =
ω 2 με − kc2,mn
which is clearly a function of frequency. Consequently, we have confirmed that TE and TM modes in a rectangular waveguide are dispersive. One special case is m = n = 0 . Since kc ,00 = 0 , then v p ≠ g ( f ) which means this is not a dispersive mode. However, the m = n = 0 mode is the TEM mode, which cannot exist in a hollow conductor waveguide.
The problem with (temporally) dispersive modes is that they can severely distort signals that have been modulated onto them as the carrier. As the signal propagates down the waveguide:
t
t
t
t
In communications, such distortion is often unacceptable. Therefore, the TEM mode is the one commonly used in microwave engineering. (For high power applications, hollow waveguides made be required; hence, one would need to somehow work around the distortion.)
Whites, EE 481
Lecture 11
Page 3 of 7
Since we prefer to work with the TEM mode of wave propagation, it is important that we use waveguides in our microwave circuits that will support TEM or “quasi-TEM” modes. Examples of such structures are: • Microstrip and covered microstrip, • Stripline, • Slotline, • Coplanar waveguide. In this course, we will work primarily with microstrip. Actually, in the lab we will exclusively use microstrip. Before delving into microstrip, however, let’s quickly overview some of these other TEM waveguides, beginning with stripline.
Stripline Stripline is a popular, planar geometry for microwave circuits. As shown in Fig. 3.22: Metal planes
b
εr W
Whites, EE 481
Lecture 12
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Lecture 12: Microstrip. ADS and LineCalc. One of the most widely used planar microwave circuit interconnections is microstrip. These are commonly formed by a strip conductor (land) on a dielectric substrate, which is backed by a ground plane (Fig. 3.25a): t d
εr
W
We will often assume the land has zero thickness, t. In practical circuits there will be metallic walls and cover to protect the circuit. We will ignore these effects, as does the text. Unlike the stripline, there is more than one dielectric in which the EM fields are located (Fig 3.25b): E
εr
H
This presents a difficulty. Notice that if the field propagates as a TEM wave, then c vp = 0
εr
© 2008 Keith W. Whites
Whites, EE 481
Lecture 12
Page 2 of 18
But which ε r do we use? The answer is neither because there is actually no purely TEM wave on the microstrip, but something that closely approximates it called a “quasi-TEM” mode. At low frequency, this mode is almost exactly TEM. Conversely, when the frequency becomes too high, there are appreciable axial components of E and/or H making the mode no longer quasi-TEM. This property leads to dispersive behavior. Numerical and other analysis have been performed on microstrip since approximately 1965. Some techniques, such as the method of moments, produce very accurate numerical solutions to equations derived directly from Maxwell’s equations and incorporate the exact cross-sectional geometry and materials of the microstrip. From these solutions, simple and quite accurate analytical expressions for Z 0 , v p , etc. have been developed primarily by curve fitting. The result is that at relatively “low” frequency, the wave propagates as a quasi-TEM mode with an effective relative permittivity, ε r ,e :
ε r ,e =
εr +1 εr −1 2
+
2
1 1 + 12 d W
(3.195),(1)
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Lecture 12
Page 3 of 18
The phase velocity and phase constant, respectively, are: c (3.193),(2) vp = 0
ε r ,e
β = k0 ε r ,e
(3.194),(3)
as for a typical TEM mode. In general,
1 ≤ ε r ,e ≤ ε r
(4)
The upper bound occurs if the entire space above the microstrip has the same permittivity as the substrate, while the lower bound occurs if in this situation the material is chosen to be free space. The characteristic impedance of the quasi-TEM mode on the microstrip can be approximated as ⎧ 60 ⎛ 8d W ⎞ + ln ⎜ ⎟ ⎪ W d 4 ε ⎝ ⎠ ⎪⎪ r ,e Z0 = ⎨ 120π ⎪ ⎪ ε r ,e ⎡⎢W + 1.393 + 0.667 ln ⎛⎜ W + 1.444 ⎞⎟ ⎤⎥ ⎪⎩ ⎝d ⎠⎦ ⎣d
W ≤1 d W >1 d
(3.196),(5) Alternatively, given a desired Z 0 and ε r , the necessary W d can be computed from (3.197).
Whites, EE 481
Lecture 12
Page 4 of 18
Again, (1) and (5) were obtained by curve fitting to numerically rigorous solutions. Equation (5) can be accurate to better than 1%.
Example N12.1. Design a 50- Ω microstrip on Rogers RO4003C laminate with 1/2-oz copper and a standard thickness slightly less than 1 mm. Referring to the attached RO4003C data sheet from Rogers Corporation, we find that ε r = 3.38 ± 0.05 and d = 0.032". We will ignore all losses (dielectric and metallic). What does “1/2-oz copper” mean? Referring to the attached technical bulletin from the Rogers Corporation, copper foil thickness is more accurately measured through an areal mass. The term “1/2-oz copper” actually means “1/2 oz of copper distributed over a 1-ft2 area.” For 1-oz copper, t = 34 μm. For 2-oz copper, double this number and for ½-oz copper divide by 2. We will use (3.197) to compute the required W d to achieve a 50-Ω characteristic impedance:
Whites, EE 481
Lecture 12
Page 5 of 18
⎧ 8e A W ⎪ e2 A − 2 d ≤ 2 W ⎪ =⎨ d ⎪ 2 ⎧⎪ εr −1 ⎡ 0.61 ⎤ ⎫⎪ W >2 ⎨ B − 1 − ln ( 2 B − 1) + ⎬ ⎢ln ( B − 1) + 0.39 − ⎪⎩π ⎪⎩ ε r ⎦⎥ ⎪⎭ d 2ε r ⎣ (3.197),(6) To apply this equation, we first need to compute the constants A and B: Z εr +1 εr −1⎛ 0.11 ⎞ + + A= 0 0.23 (7) ⎜ ⎟ = 1.376 60 2 εr +1⎝ εr ⎠
B=
377π = 6.442 2Z 0 ε r
(8)
Next, we will arbitrarily assume that W d < 2 and use the simpler equation in (6). We find that W 8e1.376 = 2⋅1.376 = 2.317 . d e −2 Is this result less than 2? The answer is no. So, we need to recompute W d using the bottom equation in (6). We find here that W d = 2.316 , which is greater than 2 as assumed. So, with this result and d = 0.032", then W = 2.316 ⋅ 0.032" = 0.0741".
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Lecture 12
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A more common unit for width and thickness dimensions in microwave circuits is “mil” where 1 1 mil = " = 25.4 μm 1000 Therefore, 74.1 W = 0.0741" = " = 74.1 mils ( = 1.88 mm). 1000 This completes the design of the 50-Ω microstrip.
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Lecture 12
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Lecture 12
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Lecture 12
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Whites, EE 481
Lecture 13
Page 1 of 14
Lecture 13: Simple Quasi-Static Moment Method Analysis of a Microstrip. Computational electromagnetics (CEM) can provide accurate solutions for Z0 and other microstrip properties of interest including plots of E and H everywhere in space and J s and ρ s on the land or the ground plane. This can be accomplished regardless of the cross-sectional geometry of the microstrip, the thickness of the land or its conductivity. The method of moments (MM) is a very popular CEM technique. It is particularly useful for planar geometries such as microstrip, stripline, conformal antennas, etc. The MM was popularized by R. F. Harrington in 1965 with his book “Field Computations by Moment Methods.” Today, it is one of the most widely used CEM techniques. We’ll illustrate the MM technique with a solution to a quasistatic microstrip immersed in an infinite dielectric as shown: y Ground plane
ε
Land w
d x
That is, there is no substrate, per se. © 2008 Keith W. Whites
Whites, EE 481
Lecture 13
Page 2 of 14
Integral Equation We’ll imagine that a time harmonic voltage source has been applied across the two conductors: Line charge density [C/m]
y + + +++ + ++ +
ε -
-
-
- - - - - - ---- - - - - - -
+ V --
-
x
This causes a charge accumulation as shown. Next, the image method will be employed to create an equivalent problem for the fields in the upper half space ( y ≥ 0 ): O.b. point y r
ε
+ + +++ + ++ +
d
ε
r′
+ V -
d --- - -- -
Φe=0 naturally satisfied x
-V
+
In a previous EM course, you’ve likely learned that the electric potential Φe at a point r in a homogeneous space produced by a line charge density ρl ( r ′ ) is given by
Whites, EE 481
Lecture 13
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⎛ 1 ⎞ ρl ( r ′ ) ln ⎜ Φe ( r ) = ⎟ dl ′ ′ 2πε C∫′ r r − ⎝ ⎠ 1 2 2 ′ ′ ′ ρ r ln x x y y dx′ =− − + − ( ) ( ) ( ) l ∫ 2πε C ′ 1
(
)
(1)
(For example, see J. Van Bladel, Electromagnetic Fields. New York: Hemisphere Publishing, 1985.) It is very important to realize that this contour C ′ must include all charge densities in the space, which means we must include both conductors in this integral. To develop an equation from which we can solve for the charge density, we’ll apply the boundary condition Φe ( r ) = V ∀r ∈ {upper strip} (2) Now, using (2) in (1) and accounting for both the +ρl and –ρl strips yields 0 ⎡ 1 ⎧⎪ 2 2 ⎤ V =− ⎨ ∫ ρl ( r ′ ) ln ⎢ ( x − x′ ) + ( d − d ) ⎥ dx′ − 2πε ⎩⎪ top ⎣ ⎦
⎫ 2 2⎤ ′ ′ x x d d dx − + + ) ( ) ⎥ ⎬ ∫ ⎢⎣ ( ⎦ ⎭ bottom w 1 2 ⎡ln x − x′ − ln 2 ⎤ ′ ′ − + 4 ρ r x x d dx′ (3) or V = − ( ) ( ) ( ) l ∫ ⎢ ⎥ 2πε 0 ⎣ ⎦
ρl ( r ′ ) ln ⎡
(
)
Whites, EE 481
Lecture 13
Page 4 of 14
Recall that the unknown in (3) is the line charge density ρl. But how do we solve for this function? It varies along the strip so we can’t simply “pull” it out of the integral. Actually, (3) is called an integral equation because the unknown function is located in an integrand. You most likely haven’t encountered such equations before. Integral equations are very difficult to solve analytically. We’ll use a numerical solution method instead.
Basis Function Expansion In the moment method, we first expand ρl in a set of basis functions. For a simple MM solution, here we’ll use pulse basis functions and divide the strips into N uniform sections:
x1 2
xN − 1
2
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Lecture 13
Page 5 of 14
N
ρl ( r ′ ) ≈ ∑ α n Pn ( x′; xn −1 , xn )
where
(4)
n =1
What is not known in (4) are the amplitudes αn of the line charge density expansion. These are just numbers. So, instead of directly solving for the spatial variation of ρl in (3), now we’ll just be computing these N numbers, αn. Much simpler! However, we need to allow enough “degrees of freedom” in this basis function expansion (4) so that an accurate solution can be found. This is accomplished by choosing the proper type of expansion functions, a large enough N, etc. The next step in the MM solution is to substitute (4) into (3) w 1 ⎡N ⎤ ′ ′ ′ (5) ; , V =− α P x x x ( ) ∑ n n n −1 n ⎥ G ( x − x ) dx ⎢ ∫ 2πε 0 ⎣ n =1 ⎦ where
G ( x − x′ ) = ln ( x − x′ ) − ln
(
( x − x′ )
2
+ 4d 2
)
(6)
and is called the Green’s function. We can interchange the order of integration and summation in (5) since these are linear operators, except perhaps when x = x′ . In this case, the integrand becomes singular. We’ll consider this situation later in this lecture. Then, (5) becomes
Whites, EE 481
Lecture 13
V =−
1
w
N
α ∫ P ( x′; x ∑ 2πε n
n =1
or
V =−
Page 6 of 14
n −1
n
, xn ) G ( x − x′ ) dx′
0
xn
N
α ∫ G ( x − x′ ) dx′ ∑ 2πε 1
n =1
n
(7)
xn −1
Testing the Integral Equation In (7), we have N unknown coefficients αn to solve for, but only a single equation. We will generate a total of N equations by evaluating (7) at N points along the (top) strip. This process is called “testing” the integral equation. We’ll test (7) at the centers of each of the N segments, xm, giving xn 1 N V =− α n ∫ G ( xm − x′ ) dx′ m = 1,…, N (8) ∑ 2πε n =1 xn−1 This is the final system of equations that we will use to solve for all the coefficients αn.
Matrix Equation It is helpful to cast (8) into the form of a matrix equation [V ] = [ Z ] ⋅ [α ] where
Vm = V αn = αn
N ×1
N×N
(9)
N ×1
(10a) (10b)
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Lecture 13
Z mn = −
Page 7 of 14
xn
∫ G ( x − x′ ) dx′
1 2πε
(10c)
xn −1
The numerical solution to (9) is accomplished by “filling” or “populating” [V] and [Z], then solving a system of linear, constant coefficient equations. In particular, for • [V] – choose V = 1 V in (10a), for example. • [Z] – compute (10c) analytically, if possible, or by numerical integration. The “filling” of [V] is very simple, while filling [Z] is a bit more difficult. In this quasi-static microstrip example, though, it is possible to evaluate all of the terms analytically since a simple anti-derivative is available. In particular, with the center of the strip located at the origin as shown: y
ρl
(x,y) O.b. point r x Δ/2
-Δ/2
then the electrostatic potential at point r produced by a strip of width Δ supporting a constant line charge density ρl is given by Δ 2
ρ Φ e ( r ) = − l ∫ ln ⎡ 2πε −Δ 2 ⎢⎣
( x − x′ )
2
+ y 2 ⎤ dx′ ⎥⎦
(11)
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Lecture 14
Page 1 of 8
Lecture 14: Impedance and Admittance Matrices. As in low frequency electrical circuits, a matrix description for portions of microwave circuits can prove useful in simulations and for understanding the behavior of the subcircuit, among other reasons. Matrix descriptions are a very convenient way to integrate the effects of the subcircuit into a circuit without having to concern yourself with the specific details of the subcircuit. We will primarily be interested in ABCD and S matrices in this course, though Z and Y matrices will also prove useful. The ABCD and S parameters are probably new to you. As we’ll see, using these matrix descriptions is very similar to other two-port models for circuits you’ve seen before, such as Z and Y matrices.
Z Matrices As an example of Z matrices, consider this two-port network: I1 + V1 -
I2
[Z ]
+ V2 -
The Z-matrix description of this two-port is defined as © 2008 Keith W. Whites
Whites, EE 481
Lecture 14
⎡V1 ⎤ ⎡ Z11 ⎢V ⎥ = ⎢ Z ⎣ 2 ⎦ ⎣ 21
Page 2 of 8
Z12 ⎤ ⎡ I1 ⎤ ⋅⎢ ⎥ ⎥ Z 22 ⎦ ⎣ I 2 ⎦
(1)
≡[ Z ]
Z ij =
where
Vi Ij
(4.28) I k = 0, ∀k ≠ j
As an example, let’s determine the Z matrix for this T-network (Fig. 4.6): I1
ZA
ZB
+ V1 -
I2 +
ZC
V2 -
Applying (1) repeatedly to all four Z parameters, we find: V Z11 = 1 = Z A + ZC ( Z in at port 1 w/ port 2 o.c.) I1 I =0 2
Z12 = Z 21 = Z 22 =
V1 I2 V2 I1 V2 I2
⇒ V1 = I 2 Z C (think of I 2 as source) ∴ Z12 = Z C I1 = 0
⇒ V2 = I1Z C (think of I1 as source) ∴ Z 21 = Z C I 2 =0
= Z B + ZC
( Z in at port 2 w/ port 1 o.c.)
I1 = 0
Collecting these calculations, then for this T-network:
Whites, EE 481
Lecture 14
⎡ Z A + ZC = Z [ ] ⎢ Z C ⎣
Page 3 of 8
ZC ⎤ Z B + Z C ⎥⎦
Notice that this matrix is symmetric. That is, Z ij = Z ji for i ≠ j . It can be shown that [ Z ] will be symmetric for all “reciprocal” networks. What’s the usefulness of an impedance matrix description? For one thing, if a complicated circuit exists between the ports, one can conveniently amalgamate the electrical characteristics into this one matrix. Second, if one has networks connected in series, it’s very easy to combine the Z matrices. For example: I1 + V1 -
I1′ + V1′ I1′′ + V1′′ -
[ Z ′]
I 2′ + ′ V - 2
[ Z ′′]
I 2′′ + ′′ V - 2
I2 + V2 -
[Z ] By definition ⎡V ′′ ⎤ ⎡ I ′′ ⎤ ⎡V ′ ⎤ ⎡I ′⎤ 1 1 1 ⎢ ⎥ = [ Z ′] ⋅ ⎢ ⎥ and ⎢ ⎥ = [ Z ′′] ⋅ ⎢ 1 ⎥ ⎢⎣V2′′ ⎥⎦ ⎢⎣ I 2′′ ⎥⎦ ⎢⎣V2′ ⎥⎦ ⎢⎣ I 2′ ⎥⎦
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Lecture 14
Page 4 of 8
From the figure we see that I1′ = I1′′ , I 2′ = I 2′′ , and that V1 = V1′ + V1′′ , V2 = V2′ + V2′′ . So, summing the above two matrix equations gives ⎡ V ′ + V ′′ ⎤ ⎡I ′⎤ ⎡I ′⎤ 1 1 1 ⎢ ⎥ = [ Z ′] ⋅ ⎢ ⎥ + [ Z ′′] ⋅ ⎢ 1 ⎥ ⎢⎣V2′ + V2′′ ⎥⎦ ⎢⎣ I 2′ ⎥⎦ ⎢⎣ I 2′ ⎥⎦ Also from the figure, note that I1 = I1′ and I 2 = I 2′ . Therefore, ⎡V1 ⎤ ⎡ I1 ⎤ ′ ′′ = Z + Z ⋅ (2) [ ] [ ] } ⎢I ⎥ ⎢V ⎥ { ⎣ 2⎦ ⎣ 2⎦ [Z ]
From this result, we see that for a series connection of two-port networks, we can simply add the Z matrices to form a single “super” Z matrix (3) [ Z ] = [ Z ′] + [ Z ′′] that incorporates the electrical characteristics of both networks and their mutual interaction.
Y Matrices A closely related characterization is the Y-matrix description of a network: I1 + V1 -
By definition:
I2
[Y ]
+ V2 -
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Lecture 15
Page 1 of 6
Lecture 15: S Parameters and the Scattering Matrix. While Z and Y parameters can be useful descriptions for networks, S and ABCD parameters are even more widely used in microwave circuit work. We’ll begin with the scattering (or S) parameters. Consider again the multi-port network from the last lecture, which is connected to N transmission lines as: V1+ , I1+
t1
t3
Z0 − 1
Z0
− 1
V ,I
V2+ , I 2+
V3+ , I 3+
t2
[S ]
Z0
V3− , I 3−
tN
VN+ , I N+
Z0
V2− , I 2−
VN− , I N−
Rather than focusing on the total voltages and currents (i.e., the sum of “+” and “-” waves) at the terminal planes t1 ,…., tn , the S parameters are formed from ratios of reflected and incident voltage wave amplitudes. When the characteristic impedances of all TLs connected to the network are the same (as is the case for the network shown above), then the S parameters are defined as
© 2008 Keith W. Whites
Whites, EE 481
Lecture 15
⎡V1− ⎤ ⎡ S11 ⎢ ⎥ ⎢ ⎢ ⎥=⎢ ⎢VN− ⎥ ⎢⎣ S N 1 ⎣ ⎦
Page 2 of 6
S1N ⎤ ⎡V1+ ⎤ ⎥⋅⎢ ⎥ ⎥ ⎢ ⎥ S NN ⎥⎦ ⎢⎣VN+ ⎥⎦
⎡⎣V − ⎤⎦ = [ S ] ⋅ ⎡⎣V + ⎤⎦ where [ S ] is called the scattering matrix. or
(4.40),(1)
As we defined in the last lecture, the terminal planes are the “phase = 0” planes at each port. That is, with − j β z −t j β z −t Vn ( zn ) = Vn+ e n ( n n ) + Vn− e n ( n n ) n = 1,…, N then at the terminal plane tn Vn ( zn = tn ) ≡ Vn = Vn+ + Vn− Each S parameter in (1) can be computed as
Vi − Sij = + Vj
(4.41),(2) Vk+ = 0, ∀k ≠ j
Notice in this expression that the wave amplitude ratio is defined “from” port j “to” port i: Si j Let’s take a close look at this definition (2). Imagine we have a two-port network:
Whites, EE 481
Lecture 15
V1+
Z0
t1
t2
[S ]
V2+
Z0
V1−
Then, for example,
Page 3 of 6
V2−
V1− S11 = + V1
V2+ = 0
Simple enough, but how do we make V2+ = 0 ? This requires that: 1. There is no source on the port-2 side of the network, and 2. Port 2 is matched so there are no reflections from this port. Consequently, with V2+ = 0 ⇒ S11 = Γ11 , which is the reflection coefficient at port 1. Next, consider V2− S2 1 = + V1
V2+ = 0
Again, with a matched load at port 2 so that V2+ = 0 , then S 21 = T21 which is the transmission coefficient from port 1 to port 2. It is very important to realize it is a mistake to say S11 is the reflection coefficient at port 1. Actually, S11 is this reflection coefficient only when V2+ = 0 .
Whites, EE 481
Lecture 16
Page 1 of 10
Lecture 16: Properties of S Matrices. Shifting Reference Planes. In Lecture 14, we saw that for reciprocal networks the Z and Y matrices are: 1. Purely imaginary for lossless networks, and 2. Symmetric about the main diagonal for reciprocal networks. In these two special instances, there are also special properties of the S matrix which we will discuss in this lecture.
Reciprocal Networks and S Matrices In the case of reciprocal networks, it can be shown that t (4.48),(1) [S ] = [S ]
where [ S ] indicates the transpose of [ S ] . In other words, (1) is a statement that [ S ] is symmetric about the main diagonal, which is what we also observed for the Z and Y matrices. t
Lossless Networks and S Matrices The condition for a lossless network is a bit more obtuse for S matrices. As derived in your text, if a network is lossless then
© 2008 Keith W. Whites
Whites, EE 481
Lecture 16
Page 2 of 10
[ S ] = {[ S ] } which is a statement that [ S ] is a unitary matrix. t −1
*
(4.51),(2)
This result can be put into a different, and possibly more useful, t form by pre-multiplying (2) by [ S ]
[ S ] ⋅ [ S ] = [ S ] ⋅ {[ S ] } t
*
t
t −1
= [I ]
(3)
[ I ] is the unit matrix defined as
0⎤ ⎡1 [ I ] = ⎢⎢ % ⎥⎥ ⎢⎣0 1 ⎥⎦ Expanding (3) we obtain i→
k↓
⎡ S11 S21 " S N 1 ⎤ ⎡ S11* ⎢ * ⎢S # ⎥ ⎢ S21 S22 12 ⎢ ⎥⋅ ⎢ # ⎥ ⎢ # % ⎢ ⎥ ⎢ * " S S NN ⎦ ⎣ S N 1 ⎣ 1N
j→
S12* " S1*N ⎤ 0⎤ ⎥ ⎡1 * # ⎥ ⎢ S22 % ⎥ (4) = ⎥ ⎥ ⎢ % ⎢⎣0 1 ⎥⎦ * ⎥ " S NN ⎦
=[ S ]
t
Three special cases – y Take row 1 times column 1: * S11S11* + S 21S 21 + " + S N 1S N* 1 = 1 Generalizing this result gives
(5)
Whites, EE 481
Lecture 16
N
∑S k =1
ki
Ski* = 1
Page 3 of 10
(4.53a),(6)
In words, this result states that the dot product of any column of [ S ] with the conjugate of that same column equals 1 (for a lossless network). y Take row 1 times column 2: * S11S12* + S 21S 22 + " + S N 1S N* 2 = 0 Generalizing this result gives N
∑S k =1
ki
S kj* = 0 ∀ ( i, j ) , i ≠ j
(4.53b),(7)
In words, this result states that the dot product of any column of [ S ] with the conjugate of another column equals 0 (for a lossless network). y Applying (1) to (7): If the network is also reciprocal, then [ S ] is symmetric and we can make a similar statement concerning the rows of [ S ] .
That is, the dot product of any row of [ S ] with the conjugate of another row equals 0 (for a lossless network).
Example N16.1 In a homework assignment, the S matrix of a two port network was given to be
Whites, EE 481
Lecture 16
Page 4 of 10
⎡0.2 + j 0.4 0.8 − j 0.4 ⎤ S = [ ] ⎢ 0.8 − j 0.4 0.2 + j 0.4⎥ ⎣ ⎦ t Is the network reciprocal? Yes, because [ S ] = [ S ] .
Is the network lossless? This question often cannot be answered simply by quick inspection of the S matrix. Rather, we will systematically apply the conditions stated above to the columns of the S matrix: y C1 ⋅ C1* : ( 0.2 + j 0.4 )( 0.2 − j 0.4 ) + ( 0.8 − j 0.4 )( 0.8 + j 0.4 ) = 1 y C2 ⋅ C2* : Same = 1 y C1 ⋅ C2* : ( 0.2 + j 0.4 )( 0.8 + j 0.4 ) + ( 0.8 − j 0.4 )( 0.2 − j 0.4 ) = 0 y C2 ⋅ C1* : Same = 0 Therefore, the network is lossless. As an aside, in Example N15.1 of the text, which we saw in the last lecture, ⎡ 0.1 j 0.8⎤ S [ ]= ⎢ ⎥ ⎣ j 0.8 0.2 ⎦ This network is obviously reciprocal, and it can be shown that it’s also lossy.
Example N16.2 (Text Example 4.4). Determine the S parameters for this T-network assuming a 50- Ω system impedance, as shown.
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Lecture 16
Page 5 of 10
V1+
V2+
Z 0 = 50 Ω
V1
VA
Z 0 = 50 Ω
V2
V1−
V2− Z in
[S ]
First, take a general look at the circuit: y It’s linear, so it must also be reciprocal. Consequently, [ S ] must be symmetric (about the main diagonal). y The circuit appears unchanged when “viewed” from either port 1 or port 2. Consequently, S11 = S 22 . Based on these observations, we only need to determine S11 and S 21 since S 22 = S11 and S12 = S 21 . Proceeding, recall that S11 is the reflection coefficient at port 1 with port 2 matched: V1− S11 = Γ11 V + =0 = + 2 V1 V + =0 2
The input impedance with port 2 matched is Z in = 8.56 + 141.8 ( 8.56 + 50 ) Ω = 50.00 Ω which, not coincidentally, equals Z 0 ! With this Zin: Z − Z0 S11 = in =0 Z in + Z 0 which also implies S 22 = 0 .
Next, for S 21 we apply V1+ with port 2 matched and measure V2− :
Whites, EE 481
Lecture 17
Page 1 of 7
Lecture 17: S Parameters and Time Average Power. Generalized S Parameters. There are two remaining topics concerning S parameters we will cover in this lecture. The first is an important relationship between S parameters and relative time average power flow. The second topic is generalized scattering parameters, which are required if the port characteristic impedances are unequal.
S Parameters and Time Average Power There is a simple and very important relationship between S parameters and relative time average power flow. To see this, consider a generic two-port connected to a TL circuit: V2+
V1+ Γ1 V1
V2 Γ 2
V1−
V2−
By definition, V1− = S11V1+ + S12V2+ V2− = S 21V1+ + S 22V2+ At port 1, the total voltage is V1 = V1+ + V1− © 2008 Keith W. Whites
(1) (2) (3)
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Lecture 17
Page 2 of 7
and the total time average power at that port is comprised of the two terms (see 2.37): Pinc =
V1+
2
and Pref =
V1−
2
2Z 0 2Z 0 Further, since port 2 is matched the total voltage there is V2 V + =0 = V2−
(4),(5)
(6)
2
Consequently, for this circuit the transmitted power is
Ptrans V + =0 = 2
V2−
2
(7)
2Z 0
Using the results from (4), (5), and (7), we will consider ratios of these time average power quantities at each port and relate these ratios to the S parameters of the network. y At Port 1. Using (4) and (5), the ratio of reflected and incident time average power is:
V1−
2
Pref V1− = = + 2 + Pinc V V1 1
2
(8)
From (1) and noticing port 2 is matched so that V1− S11 = + V1 V + =0 2
then in (8):
Pref Pinc
= S11 V2+ = 0
2
(9)
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Lecture 17
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This result teaches us that the relative reflected time average 2 power at port 1 equals S11 when port 2 is matched. y At Port 2. Using (7) and (4), the ratio of transmitted and incident time average power is:
Ptrans V + =0 2
Pinc
=
V2−
2
+ 2 1
(10)
V
However, from (2) and with V2+ = 0 , then
Ptrans Pinc
= S21
2
(11)
V2+ = 0
This result states that the relative transmitted power to port 2 2 equals S21 when port 2 is matched. Equations (9) and (11) provide an extremely useful physical interpretation of the S parameters as ratios of time average power. Note that this interpretation is valid regardless of the loss (or even gain) of the network. However, if the network is lossless we can use (9) and (11) to develop other very useful relationships. Recall that for a lossless network, ⎡ S11 S12 ⎤ [S ] = ⎢S S ⎥ 22 ⎦ ⎣ 21 must be unitary. As a direct result of this
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Lecture 18
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Lecture 18: Vector Network Analyzer. A network analyzer is a device that can measure S parameters over a range of frequencies. There are two types: 1. Scalar network analyzer. Measures only the magnitude of the S parameters. 2. Vector network analyzer (VNA). Measures both the magnitude and phase of the S parameters. The latter is generally a much more expensive piece of equipment. The VNA is basically a sophisticated transmitter and receiver pair with vast signal processing capabilities. Here is a block diagram of a typical vector network analyzer (text p. 183):
© 2008 Keith W. Whites
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Lecture 18
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We will examine the basic subsystems of a VNA in this lecture and some important topics concerning the sources of error and calibration of the VNA. The following pages are from “Network Analyzer Basics,” Agilent Product Note E206. (Agilent Technologies is the company that was formed when Hewlett Packard Corporation spun-off its test and measurement business.)
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Lecture 19
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Lecture 19: Proper Microwave Laboratory Practices. Microwave circuit measurements are very different than electrical measurements at lower frequencies. Here are four key differences: 1. We often use a network analyzer to make S parameter measurements rather than traditional instruments for voltage or current measurements. 2. Precision connections are necessary for accurate and repeatable measurements. 3. Special tools are used to tighten coaxial connectors. 4. Electrostatic protection is absolutely necessary. You will be using an Agilent 8753ES Vector Network Analyzer (VNA) to make all of your measurements this semester. This VNA operates from 30 MHz to 6 GHz.
© 2008 Keith W. Whites
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Lecture 19
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A VNA measures both the magnitude and phase of S parameters. Note that a VNA is intrinsically making frequency domain measurements, i.e., sinusoidal steady state.
Types of Coaxial Connectors There are nine types of coaxial connectors that you may encounter in most RF and microwave engineering laboratories: 5. APC-7 6. APC-3.5 7. 2.92 mm 8. 2.4 mm 9. 1.85 mm The right-hand column lists metrology-grade connectors. 1. BNC 2. Type F 3. Type N 4. SMA
The photograph on p. 130 of your text shows Type N, TNC, SMA, APC-7 and 2.4 mm connectors:
In this course, you will primarily be using the SMA connector.
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Proper Microwave Coaxial Connections We will use only SMA connectors in the EE 481 laboratory. Do NOT finger tighten these, or any other, precision microwave connectors. Instead, use a black-handled torque wrench, which provides the proper 5 in-lbs of torque required for SMA.
If you over-tighten a connection, it is possible you will damage the VNA connector, the cable or your microwave circuit. There may be different types of torque wrenches lying about the lab. Be certain you are using the BLACK-handled torque wrench. Standard operating procedures for making microwave coax connections include: 1) When initially making coaxial connections: y Make sure you are electrostatically grounded. y Be certain there are no metal filings or other debris inside the connectors.
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Lecture 20
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Lecture 20: Transmission (ABCD) Matrix. Concerning the equivalent port representations of networks we’ve seen in this course: 1. Z parameters are useful for series connected networks, 2. Y parameters are useful for parallel connected networks, 3. S parameters are useful for describing interactions of voltage and current waves with a network. There is another set of network parameters particularly suited for cascading two-port networks. This set is called the ABCD matrix or, equivalently, the transmission matrix. Consider this two-port network (Fig. 4.11a): I1 + V1 -
I2 ⎡A B⎤ ⎢C D ⎥ ⎣ ⎦
+ V2 -
Unlike in the definition used for Z and Y parameters, notice that I 2 is directed away from the port. This is an important point and we’ll discover the reason for it shortly. The ABCD matrix is defined as
⎡V1 ⎤ ⎡ A B ⎤ ⎡V2 ⎤ ⎢ I ⎥ = ⎢C D ⎥ ⋅ ⎢ I ⎥ ⎦ ⎣ 2⎦ ⎣ 1⎦ ⎣
© 2008 Keith W. Whites
(4.63),(1)
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It is easy to show that
A=
C=
V1 V2
,
B=
I 2 =0
I1 V2
V1 I 2 V =0 2
, D= I2 =0
I1 I2
V2 = 0
Note that not all of these parameters have the same units. The usefulness of the ABCD matrix is that cascaded two-port networks can be characterized by simply multiplying their ABCD matrices. Nice! To see this, consider the following two-port networks: I1 + V1 -
I 2′
I2 ⎡ A1 ⎢C ⎣ 1
B1 ⎤ D1 ⎥⎦
+ V2 -
+ V2′ -
I3 ⎡ A2 ⎢C ⎣ 2
B2 ⎤ D2 ⎥⎦
+ V3 -
In matrix form
and
⎡V1 ⎤ ⎡ A1 ⎢ I ⎥ = ⎢C ⎣ 1⎦ ⎣ 1
B1 ⎤ ⎡V2 ⎤ ⋅ D1 ⎥⎦ ⎢⎣ I 2 ⎥⎦
(4.64a),(2)
⎡V ′ ⎤ ⎡ A ⎢ 2 ⎥=⎢ 2 ⎢⎣ I 2′ ⎥⎦ ⎣C2
B2 ⎤ ⎡V3 ⎤ ⋅ D2 ⎥⎦ ⎢⎣ I 3 ⎥⎦
(3)
When these two-ports are cascaded,
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Lecture 20
I1 + V1 -
I 2′ + + V2 V2′ - -
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I3
I2
⎡ A1 ⎢C ⎣ 1
B1 ⎤ D1 ⎥⎦
⎡ A2 ⎢C ⎣ 2
B2 ⎤ D2 ⎥⎦
+ V3 -
it is apparent that V2′ = V2 and I 2′ = I 2 . (The latter is the reason for assuming I 2 out of the port.) Consequently, substituting (3) into (2) yields ⎡V1 ⎤ ⎡ A1 B1 ⎤ ⎡ A2 B2 ⎤ ⎡V3 ⎤ (4.65),(4) ⎢ I ⎥ = ⎢C D ⎥ ⋅ ⎢C D ⎥ ⋅ ⎢ I ⎥ 1⎦ ⎣ 2 2⎦ ⎣ 3⎦ ⎣ 1⎦ ⎣ 1 We can consider the matrix-matrix product in this equation as describing the cascade of the two networks. That is, let ⎡ A3 B3 ⎤ ⎡ A1 B1 ⎤ ⎡ A2 B2 ⎤ (5) ⎢ C D ⎥ = ⎢C D ⎥ ⋅ ⎢ C D ⎥ 3⎦ 1⎦ ⎣ 2 2⎦ ⎣ 1 ⎣ 3 ⎡V1 ⎤ ⎡ A3 B3 ⎤ ⎡V3 ⎤ (6) so that ⎢ I ⎥ = ⎢C D ⎥ ⋅ ⎢ I ⎥ 3⎦ ⎣ 3⎦ ⎣ 1⎦ ⎣ 3 I1
where
+ V1 -
I3 ⎡ A3 ⎢C ⎣ 3
B3 ⎤ D3 ⎥⎦
+ V3 -
In other words, a cascade connection of two-port networks is equivalent to a single two-port network containing a product of the ABCD matrices. It is important to note that the order of matrix multiplication must be the same as the order in which the two ports are
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Lecture 21
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Lecture 21: Signal Flow Graphs. Consider the following two-port network (Fig 4.14a): a1
t1
t2 a2
[S ]
Z0
Z0
b1
b2
A signal flow graph is a diagram depicting the relationships between signals in a network. It can also be used to solve for ratios of these signals. Signal flow graphs are used in control systems, power systems and other fields besides microwave engineering. Key elements of a signal flow graph are: 1. The network must be linear, 2. Nodes represent the system variables, 3. Branches represent paths for signal flow. For example, referring to the two-port above, the nodes and branches are (Fig 4.14b): a1
S 21
b2
S11
S 22 b1
S12
a2
© 2008 Keith W. Whites
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Lecture 21
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4. A signal yk traveling along a branch between nodes ak and b j is multiplied by the gain of that branch: ak
That is,
S jk
bj
b j = S jk ak
5. Signals travel along branches only in the direction of the arrows. This restriction exists so that a branch from ak to b j denotes a proportional dependence of b j on ak , but not the reverse.
Solving Signal Flow Graphs Signal flow graphs (SFGs) can form an intuitive picture of the signal flow in a network. As an application, we will develop SFGs in the next lecture to help us calibrate out systematic errors present when we make measurements with a VNA. Another useful characteristic is that we can solve for ratios of signals directly from a SFG using a simple algebra. There are four rules that form the algebra of SFGs: 1. Series Rule. Given the two proportional relations
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Lecture 21
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V2 = S 21V1 and V3 = S32V2 V3 = ( S32 S 21 )V1
then
(4.69),(1)
In a SFG, this is represented as (Fig. 4.16a): S21 V1
S21S32
S32 V2
V3
V1
V3
In other words, two series paths are equivalent to a single path with a transmission factor equal to a product of the two original transmission factors. 2. Parallel Rule. Consider the relation: V2 = S aV1 + SbV1 = ( Sa + Sb )V1
(4.70),(2)
In a SFG, this is represented as (Fig 4.16b): Sa V1
V2
Sa + Sb V1
V2
Sb
In other words, two parallel paths are equivalent to a single path with a transmission factor equal to the sum of the original transmission coefficients. 3. Self-Loop Rule. Consider the relations V2 = S 21V1 + S 22V2
(4.71a),(3)
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V3 = S32V2
and
(4.71b),(4)
We will choose to eliminate V2 . From (3)
V2 (1 − S22 ) = S 21V1
⇒ V2 =
S21 V1 1 − S22
Substituting this into (4) gives S S V3 = 32 21 V1 1 − S 22
(4.72),(5)
In a SFG, this is represented as (Fig 4.16c): S22 S21 V1
S 21 1 − S 22
S32 V2
V3
V1
S32 V2
V3
S32 S 21 1 − S 22 V1
V3
In other words, a feedback loop may be eliminated by dividing the input transmission factor by one minus the transmission factor around the loop. 4. Splitting Rule. Consider the relationships: V4 = S 42V2 V2 = S 21V1 and V3 = S32V2
(6) (7) (8)
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The SFG is S42
S21 V1
V4
S32 V2
V3
From (6) and (7) we find that V4 = S 42 S 21V1 . Hence, if we use the Series Rule “in reverse” we can define: V4 = S 42V4′ and V4′ = S21V1 In a SFG, this is represented as (Fig 4.16d): S42
S21 V1
S21
V4
S32 V2
S42
V4'
S32
S21 V3
V4
V1
V2
V3
In other words, a node can be split such that the product of transmission factors from input to output is unchanged.
Example N21.1. Construct a signal flow graph for the network shown below. Determine Γin and VL using only SFG algebra. a1 Γ S Γin b1
t1
t2
a2 ΓL b2
VL
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Lecture 22
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Lecture 22: Measurement Errors. TRL Calibration of a VNA. As we discussed in Lecture 18, a VNA measures both the magnitude and phase of S parameters. However, there will invariably be significant errors in these microwave measurements that must be “removed” somehow if we are to obtain accurate results. There are three general types of errors: 1. Systematic: repeatable errors due to imperfections in components, connectors, test fixture, etc. 2. Random: vary unpredictability with time and cannot be removed. From noise, connector repeatability, etc. 3. Drift: caused by changes in systems characteristics after a calibration has been performed due to temperature, humidity and other environmental variables. Using well-designed and maintained equipment in an unchanging environment is about all we can do to minimize random errors. A similar environment helps minimize drift errors, or the network analyzer can be recalibrated. The effects of systematic errors can be largely removed from the S parameters using “calibration.” (In the context of microwave
© 2008 Keith W. Whites
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Lecture 22
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measurements, “calibration” has a much different meaning than “calibrating” low-frequency equipment.) To do this calibration, we need to assume a general model for the effects of systematic errors, such as that shown in Fig 4.20:
We’ll define ⎡⎣ S m ⎤⎦ as the S parameters that are actually measured by the VNA. These include all of the errors we mentioned earlier. The error boxes (with parameters [ S ] ) and how these are specifically connected to the DUT form the model of the systematic errors. The parameters [ S ′] are those we desire to know. These are the S parameters of the DUT, which also, unfortunately, contain random errors. The purpose of network analyzer calibration is to determine the numerical values of all the S (or ABCD) parameters in the error model at each frequency of interest.
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For coaxial measurements, we often use precision Short, Open, Load and Thru (SOLT) standards as loads connected to the test ports. With these known standards as loads, we make several Sparameter measurements to construct enough equations from which we can numerically determine the error parameters.
Thru-Reflect-Line (TRL) Calibration SOLT standards are difficult to implement for VNA measurements of microstrip and similar circuits. Instead, the Thru-Reflect-Line (TRL) method is more commonly used. The TRL calibration method is very cleverly designed. It doesn’t rely on precisely known standards and it uses only three simple connections to completely characterize the error model. The three connections for TRL calibration of microstrip are: 1. Thru. Directly connect port 1 to 2, at the desired reference planes, using matched microstrip. 2. Reflect. Terminate a microstrip connected to each port with a load that produces a large reflection, say an open or short. These can be imperfect loads.
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3. Line. Connect the two ports together through a microstrip approximately λ/4 longer than the Thru (at the center frequency). We will step through each of these connections and outline the solutions for the S parameters using signal flow diagrams. 1. Thru Standard. The configuration for this measurement is shown in Fig 4.21a. The measured S matrix is defined as [T ] :
Notice that: a. The [ S ] matrices for the two error boxes are assumed to be identical. This simplifies things for us right now, though this is not assumed in actual VNA TRL “cal kits.” b. The reflection planes for the DUT are coincident. Consequently, this is called a “zero length Thru.” You’ll use this in the lab. c. S 21 = S12 for a reciprocal error box.
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Lecture 23
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Lecture 23: Basic Properties of Dividers and Couplers. For the remainder of this course we’re going to investigate a plethora of microwave devices and circuits – both passive and active. To begin, during the next six lectures we will focus on different types of power combiners, power dividers and directional couplers. Such circuits are ubiquitous and highly useful. Applications include: • Dividing (combining) a transmitter (receiver) signal to many antennas. • Separating forward and reverse propagating waves (can also use for a sort of matching). • Signal combining for a mixer. As a simple example, a two-way power splitter would have the form (Fig 7.1a): P1
Divider or Coupler
P2 = α P1 P3 = (1 − α ) P1
where α ∈ \ and 0 ≤ α ≤ 1. The same device can often be used as a power combiner: © 2008 Keith W. Whites
Whites, EE 481
Lecture 23
P1 = P2 + P3
Divider or Coupler
Page 2 of 9
P2 P3
We see that even the simplest divider and combiner circuits are three-port networks. It is common to see dividers and couplers with even more than that. So, before we consider specific examples, it will be beneficial for us to consider some general properties of three- and four-port networks.
Basic Properties of Three-Port Networks As we’ll show here, it’s not possible to construct a three-port network that is: 1. lossless, 2. reciprocal, and 3. matched at all ports. This basic property of three-ports limits our expectations for power splitters and combiners. We must design around it. To begin, a three-port network has an S matrix of the form: ⎡ S11 S12 S13 ⎤ (7.1),(1) [ S ] = ⎢⎢ S21 S22 S23 ⎥⎥ ⎢⎣ S31 S32 S33 ⎥⎦
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Lecture 23
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If the network is matched at every port, then S11 = S 22 = S33 = 0 . (It is important to understand that “matched” means Γ1 , Γ 2 and Γ 3 = 0 when all other ports are terminated in Z 0 .) If the network is reciprocal, then S 21 = S12 , S31 = S13 and S32 = S 23 . Consequently, for a matched and reciprocal three-port, its S matrix has the form: ⎡ 0 S12 S13 ⎤ (7.2),(2) [ S ] = ⎢⎢ S12 0 S23 ⎥⎥ ⎢⎣ S13 S 23 0 ⎥⎦ Note there are only three different S parameters in this matrix. Lastly, if the network is lossless, then [ S ] is unitary. Applying (4.53a) to (2), we find that 2 2 S12 + S13 = 1 (7.3a),(3) S12 + S 23 = 1
(7.3b),(4)
S13 + S 23 = 1
(7.3c),(5)
2
2
2
2
and applying (4.53b) that: S13* S 23 = 0 * S 23 S12 = 0 S12* S13 = 0
(7.3d),(6) (7.3e),(7) (7.3f),(8)
From (6)-(8), it can be surmised that at least two of the three S parameters must equal zero. If this is the case, then none of the equations (3), (4) or (5) can be satisfied. [For example, say
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Lecture 23
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S13 = 0 . Then (6) and (8) are satisfied. For (7) to be satisfied and S 23 ≠ 0 , we must have S12 = 0 . But with S12 and S13 both zero, then (3) cannot be satisfied.]
Our conclusion then is that a three-port network cannot be lossless, reciprocal and matched at all ports. Bummer. This finding has wide-ranging ramifications. However, one can realize such a network if any of these three constraints is loosened. Here are three possibilities: 1. Nonreciprocal three-port. In this case, a lossless three-port that is matched at all ports can be realized. It is called a circulator (Fig 7.2): Port 2
Port 1
Port 3
⎡0 0 1 ⎤ [ S ] = ⎢⎢1 0 0⎥⎥ ⎢⎣0 1 0 ⎥⎦
Notice that Sij ≠ S ji . 2. Match only two of the three ports. Assume ports 1 and 2 are matched. Then, ⎡ 0 S12 S13 ⎤ (7.7),(9) [ S ] = ⎢⎢ S12 0 S23 ⎥⎥ ⎢⎣ S13 S23 S33 ⎥⎦
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Lecture 24
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Lecture 24: T-Junction and Resistive Power Dividers. The first class of three-port network we’ll consider is the Tjunction power divider. We will look at lossless, nearly lossless and lossy dividers in this and the next lecture. A simple lossless T-junction network is shown in Fig. 7.6: V2+ V1+
Γ2
Z 0 Γ1
V1−
Z1
V2−
Z2
V3+
Γ3 Yin
V3−
There are two basic constraints we need to incorporate into this power splitter: 1. The feedline should be matched. 2. The input time average power Pin should be divided between ports 2 and 3 in a desired ratio. In the text, this ratio is defined as X:Y where: y X / ( X + Y ) ⋅100% of the incident power is delivered to one output port, and y Y / ( X + Y ) ⋅100% of the incident power is delivered to the other. © 2008 Keith W. Whites
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Lecture 24
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For example: y 1:1 means 50% of the incident time average power is delivered to each output port. y 2:1 means 67% of the incident time average power is delivered to one output port and the remaining to the other. Referring to the circuit above, in order to enforce the first constraint on the power splitter requires that 1 1 1 Yin = + = (7.25),(1) Z1 Z 2 Z 0 Consequently, to divide the incident power between the two output ports, we simply need to adjust the characteristic impedances of the two TLs.
Because port 1 is matched, the input time average power is simply: 2 1 V0 (2) Pin = 2 Z0 where V0 is the phasor voltage at the junction.
The output powers can be computed similarly as 2 2 1 V0 1 V0 and P2 = P1 = 2 Z1 2 Z2 Dividing (3) and (4) by (2) we find
(3),(4)
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Lecture 24
P1 1/ Z1 Z 0 = = Pin 1/ Z 0 Z1
and
Page 3 of 10
P2 1/ Z 2 Z 0 = = Pin 1/ Z 0 Z 2
(5),(6)
Because the network is lossless: P1 P2 + =1 Pin Pin Substituting (5) and (6) into this expression gives Z0 Z0 1 1 1 + = 1 so that + = Z1 Z 2 Z1 Z 2 Z 0 Consequently, not only have we split the power between the output ports, but in light of (1) we have also ensured that the feedline is matched.
So, once we have specified the desired ratios for the output port powers, we can use (5) and (6) to compute the required characteristic impedances of these TLs: Z0 Z0 Z1 = and Z 2 = (7),(8) P1 Pin P2 Pin That’s basically it for the design of a simple T-junction power divider. An example of this design process is given in Example 7.1 of the text, which we’ll cover later. From a practical standpoint, there are two important points that arise with T-junction power splitters:
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Lecture 24
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1. Junction effects. At the junction of the TLs, there is likely to be an accumulation of excess charge. Take a microstrip junction for example: + ++++ ++++ +
These charges attract oppositely-signed charges on the ground plane: +++ +
E
- - - -
This time-varying electric field is a displacement current, of course. We can model this effect as a lumped capacitor connected to ground, as shown in Fig. 7.6. 2. Characteristic impedance of the output lines. It is not too practical to have these Z1 and Z2 characteristic impedances in the system. We generally like to work with just one system impedance, Z0. To compensate for this, we can use QWTs for matching:
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Lecture 24
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V2+ V1+
Z in,1
Z0 V2−
Z 0 Yin V1−
Z in,2
V3+ Z0 V3−
Using QWTs makes this power splitter narrow-banded, unfortunately. Here, instead of Z1 and Z2, the impedances of interest in the power splitter design are Zin,1 and Zin,2. From (1), the match condition now becomes 1 1 1 + = (9) Z in,1 Z in,2 Z 0 and from (5) and (6), the power division constraints become Z Z P1 P2 = 0 and = 0 (10),(11) Pin Z in,1 Pin Z in,2
Example N24.1 (text example 7.1). Design a 1:2, T-junction power divider in a 50-Ω system impedance.
We’ll choose to use the network in Figure 7.6 with B = 0:
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Lecture 25
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Lecture 25: Wilkinson Power Divider. The next three port network we will consider is the Wilkinson power divider (Fig 7.8b): λ/4 Z0
Port 2
Z 0,Q
Port 1
R
Z0 Z 0,Q
λ/4
Z0
Port 3
This is a popular power divider because it is easy to construct and has some extremely useful properties: 1. Matched at all ports, 2. Large isolation between output ports, 3. Reciprocal, 4. Lossless when output ports are matched. There is much symmetry in this circuit which we can exploit to make the S parameter calculations easier. Specifically, we will excite this circuit in two very special configurations (symmetrically and anti-symmetrically), then add these two solutions for the total solution. This mathematical process is called an “even-odd mode analysis.” It is a technique used in many branches of science such as quantum mechanics, antenna analysis, etc. © 2008 Keith W. Whites
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Lecture 25
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We will now show that for a 1:1 Wilkinson power divider, Z 0,Q = 2 Z 0 and R = 2 Z 0 . To simplify matters, as in the text, we will: 1. Normalize all impedances to Z 0 , 2. Not draw the return line for the TL. For example, a TL with characteristic impedance 2Z 0 will be delineated as 2
Hence, the Wilkinson power divider shown in the first figure above and with matched terminations can be drawn as
z0,Q z0,Q
Even-Odd Mode Analysis of the Wilkinson Power Divider In the even-odd mode analysis for the S parameters, we will first excite this network symmetrically at the two output ports, followed by an anti-symmetrical excitation.
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Lecture 25
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• Symmetric excitation (even mode):
z0,Q z0,Q
Notice that I = 0 because we have symmetric excitation. Hence, V2 = V3 and we can bisect this circuit as shown to simplify the analysis (Fig. 7.10a): x =- /4
V2e 1
/4 2
x =0
V1e Γ
z0,Q
zine
1 Vg 2 = 2V
r/2
o.c. o.c.
We can recognize this circuit as a QWT. Consequently, z0,2 Q e zin = (7.33),(1) 2 or z0,Q = 2 zine (2)
We want the output ports to be matched. Therefore, zine = 1 ⇒ z0,Q = 2 Since zine = 1, then by voltage division at the output port zine 1 e V2 = e Vg 2 = Vg 2 = V zin + 1 2
(3)
(4)
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Lecture 25
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Next, to find V1e we’ll use the TL equation V ( x ) = V + ( e − j β x + Γe j β x ) so that
V ( 0 ) = V + (1 + Γ ) = V1e
(5)
Therefore, 2π λ −j ⋅ ⎞ ⎛ j 2λπ ⋅ λ4 V ( − λ 4) = V ⎜ e + Γe λ 4 ⎟ ⎝ ⎠ = jV + (1 − Γ ) = V2e = NV +
(7.34),(6)
(4)
Γ is the reflection coefficient at port 1 seen looking towards the normalized load of 2 Ω/Ω. Therefore, 2 − z0,Q 2 − 2 Γ= = (7) N 2 + z0,Q ( 3) 2 + 2
Substituting V + from (6) into (5) and using (7) we find that V V1e = ⋅ (1 + Γ ) = N − jV 2 j (1 − Γ ) (7) • Anti-symmetric excitation (odd mode): V2 1
/4 2
z0,Q
B
/4
2V
r/2 A
z0,Q
2
1
r/2 1 V3 1
-2V
(8)
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Lecture 25
Page 5 of 11
Since the circuit is fed anti-symmetrically, V3 = −V2 and the voltage = 0 at points A and B. Hence, to simplify the analysis, we can bisect the circuit with grounds as shown (Fig 7.10b): V2o
V1o Γ
2
zino
2V
For zino , notice that the load is a short circuit and the TL is λ 4 long (1/2 rotation around the Smith chart). This means zino = ∞ . Therefore, to match port 2 (and 3) for odd mode excitation, select r = 1 ⇒ r = 2 [Ω/Ω] (9) 2 Further, because zino = ∞ , then with r = 2 and port 2 matched: r 2 ⋅ 2V = (10) V2o = NV r 2 +1 (9) Even and odd solutions are eigenvectors. Any solution can be determined by summing appropriately weighted eigenvectors. With this information, we’ll be able to deduce most of the S parameters. But first, let’s determine zin,1 so we can compute S11 . Terminating ports 2 and 3 gives the circuit in Fig. 7.11a:
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Lecture 26
Page 1 of 11
Lecture 26: Quadrature (90º) Hybrid. Back in Lecture 23, we began our discussion of dividers and couplers by considering important general properties of threeand four-port networks. This was followed by an analysis of three types of three-port networks in Lectures 24 and 25. We will now move on to (reciprocal) directional couplers, which are four-port networks. As in the text, we will consider these specific types of directional couplers: 1. Quadrature (90º) Hybrid, 2. 180º Hybrid, 3. Coupled Line, and 4. Lange Coupler. We will begin with the quadrature (90º) hybrid. Fig 7.21 shows this coupler implemented with microstrip as a 1:1 power divider:
Because of symmetry, we can simplify the analysis of this circuit considerably using even-odd mode analysis. This process
© 2008 Keith W. Whites
Whites, EE 481
Lecture 26
Page 2 of 11
is similar to what we did in the last lecture with the Wilkinson power divider.
Even-Odd Mode Analysis of the Quadrature Hybrid The normalized (wrt Z 0 ) TL circuit is shown in Fig 7.22, minus the return lines:
A symmetric (even mode) excitation of this circuit is shown in Fig. 7.23a: A1e = 1
A4e =
λ /8
1
and an anti-symmetric (odd mode) excitation is shown in Fig. 7.23b:
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Lecture 26
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A1o = 1 − A4o =
1
1
1
λ /8
Observe that the circuit and its boundary conditions remain the same in both the even and odd mode configurations. It is only the excitation that changes. Because of this and the circuit being linear, the total solution is simply the sum of the even and odd mode solutions. Each solution (even and odd) is simpler to determine than the complete circuit, which is why we employ this technique. • Even mode. Because the voltages and currents must be the same above and below the line of symmetry (LOS) in Fig 7.23a, then I = 0 at the LOS ⇒ open circuit loads at the ends of λ/8 stubs, as shown. Referring to the definition of Bi ( i = 1,…, 4 ) in Fig 7.22, we can write from Fig 7.23a that for the even mode excitation: B1e = Γ e A1e , B2e = Te A1e (1a) B3e = B2e = Te A1e , B4e = B1e = Γ e A1e (1b) where A1e = 1 2 , and Γ e and Te are the reflection and transmission coefficients for the even mode configuration.
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Lecture 26
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• Odd mode. Because the voltages and currents must have opposite values above and below the LOS in Fig 7.23b, then V = 0 along the LOS ⇒ short circuit loads at the ends of λ/8 stubs, as shown. B1o = Γ o A1o , B2o = To A1o (2a) (2b) B3o = − B2o = −To A1o , B4o = − B1o = −Γ o A1o where A1o = 1 2 and Γ o and To are reflection and transmission coefficients for the odd mode configuration.
Then,
• Total solution. The total solution is the sum of the voltages in both circuits. From this fact, we can deduce that the total Bi coefficients will be the sum of (1) and (2): 1 1 B1 = B1e + B1o = Γ e + Γ o (7.62a),(3) 2 2 1 1 B2 = B2e + B2o = Te + To (7.62b),(4) 2 2 1 1 B3 = B3e + B3o = Te − To (7.62c),(5) 2 2 1 1 B4 = B4e + B4o = Γ e − Γ o (7.62d),(6) 2 2 Likewise, the incident wave coefficients are 1 1 A1 = A1e + A1o = + = 1 2 2 1 1 A4 = A4e + A4o = − = 0 2 2
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Lecture 26
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These match the assumed excitation in the original circuit on p. 2. To finish the calculation of the S parameters for the quadrature hybrid, we need to determine the reflection and transmission coefficients for the even- and odd-mode configurations. Your text shows that the solutions for Γ e and Te are −1 Γ e = 0 and Te = (7.64),(7),(8) (1 + j ) 2 Here we’ll derive solutions for Γ o and To . From Fig 7.23b: 12 1
Γo
To 1
1
2
1 1
We have three cascaded elements, so we’ll use ABCD parameters to solve for the overall S parameters of this circuit. y Elements 1 and 3. These are short circuit stubs of length λ 8 , which appear as the shunt impedance 2π λ π Z in = jZ 0 tan β l where β l = ⋅ = λ 8 4 Z in Therefore, = j , or YN = − j Z0 From the inside flap of your text:
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Lecture 27
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Lecture 27: The 180º Hybrid. The second reciprocal directional coupler we will discuss is the 180º hybrid. As the name implies, the outputs from such a device can be 180º out of phase. There are two primary objectives for this lecture. The first is to show that the S matrix of the 180º hybrid is ⎡0 1 1 0 ⎤ ⎢ ⎥ − j ⎢1 0 0 −1⎥ (7.101),(1) [ S ] = ⎢1 0 0 1 ⎥ 2 ⎢ ⎥ ⎣0 −1 1 0 ⎦ with reference to the port definitions in Fig. 7.41: (Σ) (Δ)
1
2 180º Hybrid
4
3
The second primary objective is to illustrate the three common ways to operate this device. These are: 1. In-phase power splitter: Input Isolation
1
2 180º Hybrid
4
3
Through Coupled
With input at port 1 and using column 1 of [S], we can deduce that port 1 is matched, the outputs are ports 2 and 3 (which are in phase) and port 4 is the isolation port. © 2008 Keith W. Whites
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Lecture 27
Page 2 of 8
2. Out-of-phase power splitter: Isolation Input
1
2
Through
180º Hybrid 4
Coupled
3
With input at port 4 and using column 4 of [S], we can deduce that port 4 is matched, the outputs are ports 2 and 3 (which are completely out of phase) and port 1 is the isolation port. 3. Power combiner: Sum (Σ) Difference (Δ)
1
2 180º Hybrid
4
3
Input A Input B
With inputs at ports 2 and 3 and using columns 2 and 3 of [S], we can deduce that both ports 2 and 3 are matched, port 1 will provide the sum of the two input signals and port 4 will provide the difference. Because of this, ports 1 and 4 are sometimes called the sum and difference ports, respectively. There are different ways to physically implement a 180º hybrid, as shown in Fig. 7.43. We’ll focus on the ring hybrid and specifically consider the first two applications described above.
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Lecture 27
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Ring Hybrid The ring hybrid (aka the rat race) is shown in Fig. 7.43a:
We’ll analyze this structure using the same even-odd mode approach we applied to the Wilkinson power divider and the branch line coupler in the previous two lectures. In the present case, the physical symmetry plane bisects ports 1 and 2 from 3 and 4 in the figure above. 1. In-phase power splitter. Assume a unit amplitude voltage wave incident on port 1: Port 2 Port 1
1
B2
B1
Port 3
B3
B4 Port 4
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Lecture 27
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As in Lecture 26, proper symmetric and anti-symmetric excitations of this device are required to produce the even and odd mode problems, as shown in Fig. 7.44:
1 1
1
1
1
1
1
1
As we derived in Lecture 26, 1 1 B1 = Γ e + Γ o 2 2 1 1 B2 = Te + To 2 2 1 1 B3 = Γ e − Γ o 2 2 1 1 B4 = Te − To 2 2
(7.102a),(2) (7.102b),(3) (7.102c),(4) (7.102d),(5)
Each of the even and odd solutions for Bi ( i = 1,…, 4 ) can be found by cascading ABCD matrices, then converting to S parameters. Since the ports are terminated by matched loads, we can directly determine Γ e and Te from these S parameters. o
o
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Lecture 28
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Lecture 28: Coupled Line and Lange Directional Couplers. These are the final two directional couplers we will consider. They are closely related and based on two TLs that interact with each other, but are not physically connected.
Coupled Line Directional Coupler When two TLs are brought near each other, as shown in the figure below (Fig. 7.26), it is possible for power to be coupled from one TL to the other.
This can be a serious problem on PCBs where lands are close together and carry signals changing rapidly with time. EMC engineers face this situation in high speed digital circuits and in multiconductor TLs. For coupled line directional couplers, this coupling between TLs is a useful phenomenon and is the physical principle upon which the couplers are based. © 2008 Keith W. Whites
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Lecture 28
Page 2 of 11
Consider the geometry shown in Fig 7.27: 1 +++
2 +++
+++
+++
εr - --
- --
- --
- --
When voltages are applied, charge distributions will be induced on all of the conductors. The voltages and total charges are related to each other through capacitance coefficients Cij: 1
C12
2
V1
V2 C11
By definition (Q = CV ):
C22
Q1 = C11V1 + C12V2 Q2 = C21V1 + C22V2
where • C11 = capacitance of conductor 1 with conductor 2 present but grounded. • C22 = capacitance of conductor 2 with conductor 1 present but grounded. • C12 = mutual capacitance between conductors 1 and 2. (If the construction materials are reciprocal, then C21 = C12 .) By computing only these capacitances and the quasi-TEM mode wave speed, we’ll be able to analyze these coupled line problems. Why? Assuming TEM modes, then
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Lecture 28
Z=
Page 3 of 11
L LC 1 = = C C v pC
(1)
Notice that L doesn’t appear here. Hence we only need v p and C , as conjectured. This is a widely used approach in all TL problems, not just microstrip or coupled lines.
Even-Odd Mode Characteristic Impedances To simplify the problem analysis, we’ll assume the two strips are identical and located on top of a dielectric. Because of the symmetry, we can use an even-odd mode solution approach. • Even mode. The voltages and currents are the same on both strips, as shown in Fig 7.28a:
Hence, the electric field has even symmetry about the plane of symmetry (POS). This plane is called an “H-wall.” Notice that no E field lines from conductor 1 (2) terminate on conductor 2 (1). Consequently, the two halves are decoupled, as we expect. This leads us to the equivalent circuit:
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Lecture 28
1
Page 4 of 11
2
V1
V2 C11
C22
We define the effective capacitance to ground for either conductor in this configuration as Ce = C11 = C22 (7.68),(2) Then using (1) 1 Z 0,e = (7.69),(3) v p ,eCe which is the characteristic impedance of either TL when both are operated in the even mode. This is called the even mode characteristic impedance. • Odd mode. The voltages and currents are opposite on each line as shown in Fig 7.28(b):
Notice here that the electric field lines are perpendicular to the POS. Therefore, similar to image theory, we can consider the POS as an equipotential surface (or an “E wall”). That is, as a ground plane where V = 0 .
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Lecture 28
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This approach leads to the equivalent capacitance circuit: Sum = C12 1
2C12
2C12 2
V1
V2 C11
C22
The effective capacitance to ground of either conductor in this configuration is then Co = 2C12 C11 = C11 + 2C12 (7.70),(4) so that, from (1) Z 0,o =
1 v p ,oCo
(7.71),(5)
This is the odd mode characteristic impedance. It is the characteristic impedance seen when a voltage wave is launched on the structure with odd symmetry, as shown in Fig. 7.28b. The computation of Ce and Co required in (3) and (5) is a bit involved. Alternatively, the text presents two examples of graphical design data for specific geometries. Fig 7.29 contains design data for striplines and Fig 7.30 for microstrips with ε r = 10 .
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Lecture 29
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Lecture 29: Microwave Filter Design by the Insertion Loss Method. The next major topic we’re going to cover in this course is microwave filter design. Its theoretical basis is exactly the same as low frequency analog filters, as you saw in your electronics courses. For example, you’ll see Butterworth and Chebyshev filters, but designed to operate at microwave frequencies. Implementation of these filters is different, however. For example, we won’t use discrete inductors and capacitors.
Insertion Loss Method We will begin this process with the design of analog filters, but perhaps with more detail than you’ve seen before. There are different methods for systematically designing filters, but the insertion loss method is probably the most prominent. In this technique, the relative power loss due to a lossless filter with reflection coefficient Γ (ω ) : Γ (ω )
[S ]
© 2008 Keith W. Whites
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Lecture 29
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is specified in the loss ratio PLR defined as: P Po PLR = inc = Pload Po ⎡1 − Γ (ω )2 ⎤ ⎣ ⎦ or
2 PLR = ⎡1 − Γ (ω ) ⎤ ⎣ ⎦
−1
(8.49),(1)
In Section 4.1, the text shows that Γ (ω ) is an even function of 2 ω. This implies that Γ (ω ) can be expanded in a polynomial series in ω 2 . 2
In particular, for a linear and time invariant system, Γ (ω ) is a rational function meaning It can be expressed as a quotient of real polynomials M (ω 2 ) and N (ω 2 ) . We’ll choose: M (ω 2 ) 2 (8.51),(2) Γ (ω ) = 2 2 M (ω ) + N (ω ) 2
Using (2) in (1)
PLR = 1 +
M (ω 2 ) N (ω 2 )
(8.52),(3)
This is valid for any linear, time invariant system that is an even function of ω.
Types of Low Pass Filters There are four types of low pass filters discussed in the text that are all based on (3):
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Lecture 29
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1. Maximally Flat, Butterworth, Binomial Filter. For this type of low pass filter: 2N
⎛ω ⎞ PLR = 1 + k 2 ⎜ ⎟ ⎝ ωc ⎠ where N =filter order and ωc = cutoff frequency.
(8.53),(4)
If k = 1, then PLR = 2 at ω = ωc , which is the 3-dB frequency: −1 PLR
Flat Po Po 2 Pass band
ωc
Stop band
ω
For large ω and with k = 1, then 2N
2N
⎛ω ⎞ ⎛ω ⎞ PLR ≈ k 2 ⋅ ⎜ ⎟ =12 ⋅ ⎜ ⎟ (5) ω ω ⎝ c⎠ ⎝ c⎠ From this result we learn that the insertion loss IL, defined as IL = 10log ( PLR ) , (8.50)
increases by 20N dB/decade in the stop band for the maximally flat low pass filter.
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Lecture 29
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2. Equal Ripple or Chebyshev Filter. For this type of low pass filter: 2 2⎛ ω ⎞ PLR = 1 + k TN ⎜ ⎟ (8.54),(6) ω ⎝ c⎠ where TN ( x ) is the Chebyshev polynomial. A typical plot of −1 PLR in (6) is −1 PLR
Po
Equal ripple
Po 1+ k 2
Pass band
ωc
ω
Stop band
Generally, N is chosen to be an odd integer when the source and load impedances are equal (two-sided filters). For large ω ωc and using the large argument form of TN , (6) becomes
k 2 ⎛ 2ω ⎞ PLR ≈ ⎜ ⎟ 4 ⎝ ωc ⎠
2N
(7)
As with the Butterworth filter, (7) also increases at 20N dB/decade, but with the extra factor 22 N (8) 4 compared to (5). Consequently, there is more roll off with the Chebyshev low pass filter. For example,
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Lecture 30
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Lecture 30: Scaling of Low Pass Prototype Filters. Stepped Impedance Low Pass Filters. In the last lecture, we discussed the design of prototype low pass filters where Rs = RL = 1 Ω and ωc = 1 rad/s. Of course, one generally is not going to implement the prototype filter. So what good is it? It is possible to scale and transform the low pass prototype filter to obtain a low pass, high pass, band pass and band stop filters for any impedance “level” ( Rs = RL ) and cutoff frequency. Nice! The process of filter design has three basic steps as discussed in the last lecture: (1) collect the filter specifications, (2) design the low pass prototype filter, (3) scale and convert the prototype. The first two steps were performed in the previous lecture. We’ll now consider the last step, beginning with scaling. There are two types of scaling for low pass prototype circuits, impedance scaling and frequency scaling: 1. Impedance Scaling. Since the filter is a linear circuit, we can multiply all the impedances (including the terminating resistances) by some factor without changing the transfer function of the filter. Of course, the input impedances will change. © 2008 Keith W. Whites
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Lecture 30
Page 2 of 13
If the desired source and load impedances equal R0 , then (8.64a),(1) • X L′ = R0 X L = ω ( R0 L ) . Therefore, L′ = R0 L . 1⎛R ⎞ C • X C′ = R0 X C = − ⎜ 0 ⎟ . Therefore, C ′ = . (8.64b),(2) R0 ω⎝ C ⎠ • Rs′ = R0 ⋅1 = R0 . (8.64c),(3)
• RL′ = R0 ⋅ RL = R0 RL .
(8.64d),(4)
2. Frequency Scaling. As defined for the prototype ωc = 1 rad/s. To scale for a different cutoff frequency, we substitute
ω→
ω ωc
(8.65),(5)
Applying this to the inductive and capacitive reactances in the prototype filter we find ⎛ L⎞ L • X L′ = ω L ω → ω = ω ⎜ ⎟ . Therefore, L′ = . (8.66a),(6) ωc ωc ⎝ ωc ⎠ 1 1 ⎛ω ⎞ C = ⎜ c ⎟ . Therefore, C ′ = . (8.66b),(7) • X C′ = ωC ω → ω ω ⎝ C ⎠ ωc ωc
For a one-step impedance and frequency scaling, we can combine (1)-(4), (6) and (7) to obtain RL • Lk ′ = 0 k (8.67a),(8)
ωc
• Ck ′ =
Ck ω c R0
(8.67b),(9)
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• Rs′ = R0
(10)
• RL′ = R0 RL (11) where k = 1,…, N as in Fig. 8.25. For example, in the circuit of Fig. 8.25a, C1 = g1 , L2 = g 2 , C3 = g 3 , etc.
Example N30.1. Design a 3-dB, equi-ripple low pass filter with a cutoff frequency of 2 GHz, 50-Ω impedance level and at least 15-dB insertion loss at 3 GHz.
The first step is to determine the order of the filter needed to achieve the required IL at the specified frequency. From equation (7) in the previous lecture for ω ωc k 2 ⎛ 2ω ⎞ PLR ≈ ⎜ ⎟ 4 ⎝ ωc ⎠
2N
(12)
(This is just an approximation here since ω / ωc = 1.5 .) What value do we use for k? From Fig. 8.21 −1 PLR
1
Equal ripple
1 1+ k 2
Pass band
ωc
Stop band
ω
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Lecture 30
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we see that the passband ripple equals 1 + k 2 . So, with A = ripple in dB, then 10log (1 + k 2 ) = A k = 10 A /10 − 1 (13) so that Consequently, for A = 3 dB then k = 0.998 ≈ 1. Therefore, equation (12) becomes PLR ≈ 32 N / 4 so that
N 1 3 5
10logPLR 3.5 dB 22.6 dB 41.7 dB
Fig. 8.27b w/ |ω/ωc|-1=0.5 6 dB 19 dB 35 dB
The third column is the more accurate number since it originates from the plot in Fig. 8.27b. The second column is less accurate because we used (12) with ω / ωc = 1.5 , which is not 1. For this filter, we’ll choose N = 3 to meet the IL specification. From Table 8.4 (3.0-dB ripple), we find the immitance values to be g1 = 3.3487 , g 2 = 0.7117 , g3 = g1 and g 4 = 1. Using (8)-(11) with R0 = 50 Ω, f c = 2 GHz and arbitrarily choosing the prototype circuit having the fewest inductors L2′
Rs′ + Vs
-
C1′
C3′
RL′
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Lecture 30
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then, • RS ′ = RL′ = R0 = 50 Ω C g 3.3487 • C1′ = C3′ = 1 = 1 = = 5.33 pF 9 ω c R0 ω c R0 2π ⋅ 2 ⋅10 ⋅ 50 RL Rg 50 ⋅ 0.7117 • L2′ = 0 2 = 0 2 = = 2.83 nH 9 2π ⋅ 2 ⋅10 ωc ωc The response of this filter was computed in ADS and is shown below. This filter was also designed in ADS using the Filter DesignGuide feature for automatic filter design. S-PARAMET ERS S_Param SP1 Start=0.5 GHz Stop=4 GHz Step=0.01 GHz
Port P1 Num=1
Term Term1 Num=1 Z=50 Ohm
C C1 C=5.33 pF
L L1 L=2.83 nH R=
C C2 C=5.33 pF
Term Term2 Num=2 Z=50 Ohm
Port P2 Num=2
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dB(S(2,1)) dB(S(1,1))
0
-10
-20
-30 0.5
1.5
2.5
3.5
4.0
3.5
4.0
freq, GHz
phase(S(2,1)) phase(S(1,1))
200
100
0
-100
-200 0.5
1.5
2.5
freq, GHz
Using SmartComponents in ADS can greatly speed up the filter design process. Here using the low pass filter from the “Filter DG – All Networks” palette. The filter is designed using the Filter DesignGuide, which is activated by pointing to DesignGuide -> Filter. DT DA_LCLowpassDT1_lpdesign1 DA_LCLowpassDT1 Fp=2 GHz Fs=3 GHz Ap=3 dB As=15 dB N=3 ResponseType=Chebyshev Rg=50 Ohm Rl=50 Ohm
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Lecture 31
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Lecture 31: Stub Synthesis. Kuroda’s Identities. Stub Low Pass Filters. Another method for synthesizing microwave filters without lumped elements is to use shorted and opened stubs to realize the inductances and capacitances of the filter. However, the separations between the stubs are generally not negligible and thus will degrade the filter performance if they are neglected. Kuroda’s identities are transformations that prove useful with this type of problem.
Stub Synthesis The text calls this stub synthesis process “Richard’s transformation.” Here we’ll show a simpler approach. In Section 2.3 of text, we derived the stub input impedances: • Short circuit TL: Z in = jZ 0′ tan β l (2.45c),(1) (2.46c),(2) • Open circuit TL: Yin = jY0′ tan β l To be consistent with Section 8.4, the prime indicates an impedance-scaled (i.e., normalized) value.
© 2008 Keith W. Whites
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Lecture 31
Page 2 of 13
From (1) with l < λ 4 (or β l < π 2 ), the input impedance for a short circuit stub is a positive reactance (i.e., an effective inductance) while from (2), the open circuit stub presents a negative reactance (i.e., an effective capacitance). We’ll use these two properties to construct effective inductances and capacitances for filters. From (1), it is apparent that we cannot express the input impedance in the form ω L since ω appears in the tangent function. We can conclude that this effective inductance varies with frequency. If this is the case, then which f would one choose? One choice is l = λ 8 ( β l = π 4 ), which is halfway between 0 and λ 4 (beyond this, the reactance changes sign). So, with l = λ 8 then tan β l = 1 and (1) becomes Z in = jZ 0′ Now, for an inductor at ω = ωc Z L = jωc L Equating (3) and (4) we see that by choosing Z 0′ = ωc L
(3) (4)
(5) then this λ/8-long TL has the same input impedance as an inductor with inductance L.
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We will employ (5) in the design of stub filters. In such an application, the filter coefficients gk will be associated with unscaled component values (i.e., the unprimed Lk and Ck values in Section 8.4). So, from (5) Z 0′ 1 Z0 ≡ = (6) N ωc L = L
ωc
(5)
ωc
where the unprimed quantity indicates an unscaled coefficient. This relationship in (6) is very useful. It shows us that we can realize an effective inductance with filter coefficient Lk by using a short circuit TL with an unscaled characteristic impedance Z 0 = Lk (7) that is λ/8-long at ω = ω c . Similarly, one can show that a filter coefficient Ck can be effectively realized by an open circuit stub with unscaled characteristic impedance 1 Z0 = (8) Ck that is λ 8 -long at ω = ω c .
These relationships are shown in Fig. 8.34:
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These effective L and C values for the stubs change with frequency. This affects, and generally degrades, the filter performance for f ≠ f c .
Kuroda’s Identities Now that we can construct stubs to perform as effective inductors and capacitors in a filter [but with L ( f ) and C ( f ) ], we must next address the effects that occur when the stubs are separated from each other. This is an effect we ignored in the low pass prototype filter. We assumed it contained lumped elements that were interconnected without any time delay between them. In other words, all the elements existed at a point in space.
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In microwave circuits, this restriction may be difficult to realize in the physical construction. Hence, the distances between stubs may not be electrically small. The four Kuroda identities allow us to add redundant TLs to the microwave filter circuit and transform it into a more practical form. By definition, a redundant TL is a matched TL of λ 8 length. Because it is matched at both ends, these “unit elements” have no effect on the filter at the center frequency. However, they will have an effect at other frequencies. The four Kuroda identities are shown in Table 8.7:
Each box represents a unit element with the indicated characteristic impedance and length λ 8 . The lumped elements
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Lecture 31
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represent short- or open-circuit stubs acting as normalized (i.e., unscaled) series or shunt TLs. For example, the first figure in entry (a) represents: λ 1 Z2
Z1
λ 8 Z2
8
Z1
The text shows a proof of the first Kuroda identity entry in Table 8.7. We’ll prove the second one in the following example.
Example N31.1. Prove the second Kuroda identity in Table 8.7.
The left hand circuit is λ
Z1
8 Z in
λ
8
Z2
From this circuit
Z in = jZ1 tan β l = jZ1Ω where Ω ≡ tan β l . Zin is an un-normalized value because of Z1.
Cascading ABCD matrices for this circuit:
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Lecture 32
Page 1 of 11
Lecture 32: High Pass and Bandpass Microwave Filters. Resonant Stub Filters. As has been stated in recent lectures, the low pass prototype can also be used to design high pass, bandpass, and bandstop filters, in addition to low pass filters. To achieve this, the prototype filter must be “converted,” in addition to impedance and frequency scaled. We will discuss the design process for high pass and bandpass filters in this lecture.
High Pass Filter Transformation The frequency substitution
ω→
−ωc
ω
(8.68),(1)
in a transfer function converts a low pass filter response to a high pass one. Referring to the low pass prototype filters in Fig. 8.25, we will use (1) to convert the impedances of the series inductances and the shunt capacitances. • Series Inductance. With Z L = jω L and substituting (1):
© 2008 Keith W. Whites
Whites, EE 481
Lecture 32
jω Lk → − j
Page 2 of 11
ωc 1 Lk = ω jω Ck ′
(2)
1 (8.69a),(3) ω c Lk We can deduce from this result that the series inductances of the low pass prototype filter are converted to series capacitances. This is indicative of a high pass filter. where
Ck ′ ≡
The purpose of the negative sign in (1) is also apparent from (3): it yields physically realizable capacitor values. • Shunt Capacitance. With YC = jωC and substituting (1): ω 1 jωCk → − j c Ck = ω jω Lk ′
(4)
1 (8.69b),(5) ω Ck Here we see that the substitution in (1) transforms the shunt capacitances in the low pass prototype to shunt inductances. This is also indicative of a high pass filter.
where
Lk ′ ≡
Including impedance scaling, the complete conversion of a low pass prototype circuit to a high pass filter is accomplished using 1 Ck ′ = (8.70a),(6) R0ω c Lk
Lk ′ =
R0 ω c Ck
(8.70b),(7)
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Lecture 32
Page 3 of 11
Bandpass Filter Transformation The conversion of the low pass prototype to bandpass or bandstop filters is only slightly more involved than for high pass filters. The conversion to a bandpass filter is accomplished with the substitution 1⎛ω ω ⎞ (8.71),(8) ω→ ⎜ − 0⎟ Δ ⎝ ω0 ω ⎠ ω − ω1 where Δ= 2 (8.72),(9)
ω0
is the fractional bandwidth of the passband: P
Po Po 2
ω1 ω 2 ω0 ωc
ω
Notice that ω0 is not the center frequency ωc. The text defines ω 0 = ω1ω 2 (8.73) which is the geometric mean of ω1 and ω2, rather than the more common arithmetic-mean definition. This was done to make the design equations simpler.
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Lecture 32
Page 4 of 11
As we did with the high pass filter, we’ll apply the transformation (8) to the series inductors and shunt capacitors in the low pass prototype circuit. • Series Inductance. With Z L = jω L and substituting (8): 1 ⎛ ω ω0 ⎞ Lk 1 ω0 Lk jω Lk → j L j ω − = + ⎜ ⎟ k N ω jω N Δ ⎝ ω0 ω ⎠ Δ (8) 0Δ N Lk ′
(10)
1/ Ck ′
From this result we see that the series inductors in the low pass prototype are transformed to a series LC combination with elements LR Lk ′ = k 0 (series element 1 of 2) (8.74a),(11) ω0Δ Δ (series element 2 of 2) (8.74b),(12) ω 0 Lk R0 where we’ve also included the impedance scaling. and
Ck ′ =
• Shunt Capacitance. With YC = jωC and substituting (8): 1 ⎛ ω ω0 ⎞ Ck 1 ω0 C k (13) jωCk → j C j ω − = + ⎜ ⎟ k N ω ω ω j ω Δ Δ Δ (8) N 0 ⎝ 0 ⎠ N Ck ′
1/ Lk ′
From this result we see that the shunt capacitors in the low pass prototype are transformed to a parallel LC combination with elements Ck Ck ′ = (shunt element 1 of 2) (8.74d),(14) ω 0 R0 Δ
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Lecture 32
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R0 Δ (shunt element 2 of 2) ω 0Ck where we’ve also included impedance scaling. Lk ′ =
Table 8.6 in transformations:
the
text
succinctly
(8.74c),(15)
summarizes
these
How do we implement these filters in microstrip? Consider a second order bandpass filter: L1′
1
C1′
L2′
C2′
1
We could use the first Kuroda identity to transform L1′ to a shunt capacitance, but what about the series capacitance C1′ ? Kuroda’s fourth identity transforms a series capacitance to a series capacitance. That’s no help here. We’re stuck!
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Lecture 33
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Lecture 33 – Active Microwave Circuits: Two-Port Power Gains. We are going to focus on active microwave circuits for the remainder of the semester. There are many types of active circuits such as amplifiers, oscillators, and mixers. We will concentrate only on amplifiers. It is often a much more involved process to design and construct active circuits that operate correctly than passive ones. Reasons for this include: • A bias network is required, • The devices are nonlinear, • Unintended oscillations produced by circuit instability. More care, patience, and experience are often required in the design of active RF and microwave circuits than purely passive ones. The analysis of such circuits is usually very difficult given the nonlinear behavior of the devices. For linear amplifiers, though, a linear analysis is applicable which helps simplify matters. For this reason, we will focus on linear, small signal amplifiers. Furthermore, we will use measured (or given) S parameters for the devices (transistors) rather than detailed device parameters (β, Cπ, rπ, etc.). Consequently, we can treat the transistor as a
© 2008 Keith W. Whites
Whites, EE 481
Lecture 33
Page 2 of 11
two port, but possibly one with gain. This approach works well for the steady state analysis of linear, small signal amplifiers. For other types of active circuits, such as oscillators, mixers, or power amplifiers, the nonlinear behavior of the circuit devices must be explicitly accounted for, which precludes the use of S parameters. Much more difficult. One big difference with active devices is that the magnitude of the S parameters may be greater than one. Often it is only S21 that has this characteristic, with port 1 serving as the input and port 2 the output. With passive devices, S parameters with magnitudes greater than unity are physically impossible.
Types of Power Gains Referring to a generic two-port network circuit such as Pin
PL Z0
Z0
there are three commonly used definitions for power gain. P G= L 1. Operating Power Gain: Pin
(1)
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Lecture 33
Page 3 of 11
This is the ratio of the time-average power dissipated in a load to the time-average power delivered to network. P GA = av ,n 2. Available Gain: (2) Pav , S This is the ratio of the maximally available time-average power from the network to the maximally available timeaverage power from the source. P GT = L 3. Transducer Gain: (3) Pav , S This is the ratio of the time-average power dissipated in the load to the maximally available time-average power from the source. It is this latter transducer gain that we used in EE 322 Electronics II – Wireless Communication Electronics to characterize the performance (i.e. gain) of the active devices in circuits. Among other applications, these three definitions of power gain are used to design different types of amplifiers: 1. Operating Power Gain, G. Maximum linear output power amplifiers. 2. Available Power Gain, GA. Low Noise Amplifiers (LNAs).
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Lecture 33
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3. Transducer Power Gain, GT. Simultaneously conjugate matched input and output ports (leads to maximum linear gain).
Power Gain Expressions We will now derive analytical expressions for these power gains in terms of the S parameters of the network, as well as the source and load impedances. These will prove central to the design of linear microwave amplifiers. Referring to this generic two-port circuit (Fig. 11.1): V1+
ZS VS
+ -
+ V1 Γ S Γin Z in
V1−
V2+ t1
[S] (wrt Z0)
t2 Γ out Γ L
V2−
TLs are infinitesimally short, with characteristic impedance Z0.
ZL
Z out
then by the definition of the S parameters we can write V1− = S11V1+ + S12Γ LV2− (11.2a),(4) and V2− = S 21V1+ + S 22Γ LV2− (11.2b),(5) In these equations we have used the relationship V2+ = Γ LV2− . As we showed in Lecture 21 using signal flow graphs
Whites, EE 481
Lecture 33
V1− Γ S S Γin = + = S11 + L 12 21 1 − Γ L S 22 V1 Similarly, it can be show that Γ S S V2− Γ out = + = S 22 + S 12 21 1 − Γ S S11 V2 Next, by voltage division at the source Z in V1 = VS = V1+ + V1− = V1+ (1 + Γin ) Z in + Z S Z in VS V1+ = so that Z in + Z S 1 + Γ in
Page 5 of 11
(11.3a),(6)
(11.3b),(7)
(8)
(9)
Now, using Γin = ( Z in − Z 0 ) ( Z in + Z 0 ) and after some algebra, (9) can be reduced to 1 − Γ S VS ⋅ (11.4),(10) V1+ = 1 − Γ S Γin 2 There are four different time-average power quantities we need to determine in order to compute (1)-(3): 1. Pin: Time-average power provided by the source | V1+ |2 Pin = 1− | Γin |2 ) ( 2Z 0 Substituting for V1+ from (10) gives | VS |2 |1 − Γ S |2 2 1 | | Pin = − Γ ( ) in 8Z 0 |1 − Γ S Γin |2
(11.5),(11)
(11.5),(12)
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Lecture 34
Page 1 of 12
Lecture 34 – Amplifier Stability. You’ve seen in EE 322 that a simple model for a feedback oscillator has an amplifier and a feedback network connected as:
Oscillation occurs at the output power P0 and frequency f0 where:
Anytime a portion of a circuit has gain, the circuit may oscillate, even though that is not the intended function of the circuit. For example, an amplifier circuit has gain and if it’s not properly designed, it may oscillate. This concept is referred to as circuit stability. A circuit that is not stable may oscillate. This “oscillation” may not be easily detected. That is, you will probably not measure nice sinusoids at the circuit nodes.
© 2008 Keith W. Whites
Whites, EE 481
Lecture 34
Page 2 of 12
Instead, the circuit may just not “work,” or you will see DC voltages that “are not possible,” or the waveforms are incredibly “noisy”, etc. A very frustrating experience, especially if you don’t understand circuit stability. Even a brief period of oscillation could permanently damage a circuit because of large voltages and power levels that might be generated. A thorough stability analysis requires a large signal analysis of the nonlinear circuit. Very difficult. Here, we will perform a much simpler two-port, S-parameter analysis. This is sufficient only for linear, small signal circuit applications. This analysis can be considered accurate as a test for startup instabilities. It can also provide a starting point for large signal analysis.
Negative Resistance Consider the generic linear, small-signal transistor amplifier circuit we saw in the last lecture:
Whites, EE 481
Lecture 34
Z S Z in
Γ S Γin
Page 3 of 12
Z out Z L
Γ out Γ L
We will imagine the transistor is characterized by the S matrix: ⎡ S11 S12 ⎤ (1) [ St ] = ⎢ S S ⎥ 22 ⎦ ⎣ 21 Further, we will assume the matching networks are passive so that Γ S < 1 and Γ L < 1 (2) Oscillation is possible in this circuit if a signal “incident” on the input or output port of the transistor is “reflected” with a gain > 1. That is, if Γ in > 1 or Γ out > 1 (3) the circuit may become unstable and oscillate. This could occur when noise in the circuit is incident on either port, is reflected with gain, is reflected again at the corresponding matching network, and so on. It is then possible that at some frequency, such noise could be amplified repeatedly to such a level that the device is forced into nonlinear operation and instability.
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Lecture 34
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In order for Γ > 1, the real part of the impedance seen looking into the port must be negative. That is, (4) Rin < 0 or Rout < 0 for instability. This region lies outside the unit circle on a Smith chart. As we derived in Lecture 21
Γ L S12 S21 (11.3a),(5) 1 − Γ L S22 Γ S S Γ out = S22 + S 12 21 (11.3b),(6) 1 − Γ S S11 Using these in (3), we find that for the circuit to be unconditionally stable: Γ S S Γ in = S11 + L 12 21 < 1 (11.19a),(7) 1 − Γ L S22 Γin = S11 +
and
Γ out = S22 +
Γ S S12 S21 <1 1 − Γ S S11
(11.19b),(8)
If Γ S > 1 or Γ L > 1, then (7) and (8) are requirements only for conditional stability.
Stability Circles It can be very helpful to generate a graphical depiction of the range of Γ S and Γ L values that may lead to instability. Stability circles are one way to do this. They are particularly helpful
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Lecture 34
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because of the extra information we gain because they are drawn on the Smith chart. Stability circles define the boundary between stable and potentially unstable Γ L or Γ S . To determine these boundaries, we will set Γin = 1 (or Γ out = 1) and draw these curves in the Γ L (or Γ S ) plane. For the load stability circle, from (5) we find Γ S S S11 + L 12 21 = 1 1 − Γ L S22
(11.20),(9)
After some manipulation, as shown in the text, (9) can be rearranged to Γ L − CL = RL (10) which is an equation for a circle in the complex Γ L plane.
( S22 − S11* Δ )
*
In (10)
CL =
(11.25a),(11) | S 22 |2 − | Δ |2 is the center of the circle in the complex Γ L plane, and S12 S21 (11.25b),(12) RL = 2 2 | S22 | − | Δ | is the radius, where
Δ = S11S 22 − S12 S 21
(11.21),(13)
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Lecture 34
Page 6 of 12
For a conditionally stable circuit, the load stability circle may look something like this on the extended Smith chart: Im ( Γ L )
RL CL
Re ( CL )
Im ( CL ) Re ( Γ L )
Again, this circle defines those Γ L values where Γin = 1. Inside the load stability circle (and with Γ L ≤ 1) are those Γ L that produce a stable or unstable circuit. We don’t know which yet. The converse can be said for those Γ L outside of the circle (and with Γ L ≤ 1). So how do we identify the stable region? Very easily, as it turns out. We will choose a special load ZL that will quickly uncover the region of stability. Specifically, we’ll choose Z L = Z 0 ⇒ Γ L = 0 so that from (5) Γin =| S11 | (14) Consequently, if S11 < 1, then the region containing the origin in the Γ L plane (and Γ L ≤ 1) is the stable region. Otherwise, if S11 > 1 , then the region inside the stability circle (and Γ L ≤ 1) is the stable region. These two situations are illustrated below.
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Lecture 35
Page 1 of 9
Lecture 35 – Single Stage Amplifier: Design for Maximum Gain. Amplifiers must be designed for different performance requirements that depend on the application. Examples of these different requirements are maximum gain, maximum output power, specific gain, circuit stability with varying load impedance, wide bandwidth, and low noise. Often, the first amplifier in a microwave-frequency receiver will be a low noise amplifier (LNA). After the signal level is raised well above the noise level, gain often becomes more important than noise. We will first consider the design of amplifiers that are needed for large gain rather than for their noise or bandwidth characteristics. To realize maximum gain, the input and output matching networks are simultaneously conjugate matched to the transistor. We also need the entire amplifier system (the transistor and matching networks) to be matched to the system impedance. We will use the transducer gain approach to design the amplifier of Figure 11.2 for maximum gain.
© 2008 Keith W. Whites
Whites, EE 481
Lecture 35
Γ S Γin
Page 2 of 9
Γ out Γ L
In this approach, we begin with the transducer gain we developed in Lecture 33: GT = GS G0GL (1) 1− | Γ S |2 GS = (11.16a),(2) where |1 − Γin Γ S |2 G0 = S 21 and
2
1− | Γ L |2 GL = |1 − S22Γ L |2
(11.16b),(3) (11.16c),(4)
Maximum power transfer from the input matching network (IMN) to the transistor occurs when the two are conjugate matched so that Z in = Z S* . For a real-valued system impedance, this leads to the requirement Γ in = Γ*S (11.36a),(5) Similarly, maximum power is transferred from the transistor to the output matching network (OMN) when (11.36b),(6) Γ out = Γ*L When (5) and (6) are realized in the amplifier circuit, maximum transducer gain will be obtained (for lossless matching
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Lecture 35
Page 3 of 9
networks). The value of this maximum transducer gain, GTmax , is found by substituting (5) into (1) 1 1− | Γ L |2 2 GTmax = (11.37),(7) | S 21 | 2 2 1− | Γ S | |1 − S 22Γ L | In order to achieve this gain, the IMN and OMN must be simultaneously designed so that (5) and (6) are satisfied. To develop these design equations, we substitute the expressions for Γin and Γ out derived in Lecture 33 into (5) and (6) giving S S Γ (11.38a),(8) Γ*MS = S11 + 12 21 ML 1 − S22Γ ML S S Γ (11.38b),(9) Γ*ML = S22 + 12 21 MS 1 − S11Γ MS where the ‘M’ subscript indicates values obtained with conjugate matching.
These are two equations from which we can solve for the two unknowns Γ MS and Γ ML . As shown in the text, these solutions are Γ MS
B1 − B12 − 4 | C1 |2 = 2C1
(11.40a),(10)
B2 − B22 − 4 | C2 |2 Γ ML = and (11.40b),(11) 2C2 (not ‘+’ as in the text – see Gilmore and Besser, vol. II, p. 79). In (10) and (11):
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and
Lecture 35
Page 4 of 9
B1 = 1+ | S11 |2 − | S 22 |2 − | Δ |2 B2 = 1+ | S 22 |2 − | S11 |2 − | Δ |2 * C1 = S11 − S 22 Δ C2 = S 22 − S11* Δ Δ = S11S 22 − S12 S 21
(11.41a),(12) (11.41b),(13) (11.41c),(14) (11.41d),(15) (11.21),(16)
These design equations will produce physically realizable, passive matching networks provided the transistor is unconditionally stable (or has been “stabilized”). The final step is to construct these lossless passive matching networks. This will be illustrated in the following example.
Example N35.1 (Text example 11.3). For the amplifier circuit shown earlier in this lecture, design input and output matching networks to achieve maximum transducer gain at 4 GHz. The S parameters for the transistor, referenced to 50 Ω, are:
f(GHz) 3.0 4.0 5.0
S11 0.80( − 89° 0.72( − 116° 0.66( − 142°
S21 2.86(99° 2.60( 76° 2.39(54°
S12 0.03(56° 0.03(57° 0.03( 62°
S22 0.76( − 41° 0.73( − 54° 0.72( − 68°
The first step in this design process is to ensure that the device is unconditionally stable at 4.0 GHz.
Whites, EE 481
Lecture 36
Page 1 of 10
Lecture 36 – Single Stage Amplifier: Design for Specific Gain. Maximum gain amplifiers discussed in the previous lecture are designed to provide the maximum gain possible from the circuit. If a specific gain value less than the maximum is required, then another design approach must be used, which is the topic of this lecture. Additional motivation for amplifiers with less than maximum gain is related to the gain-bandwidth product concept. Increased gain often occurs at the expense of bandwidth and vice versa. Generally speaking, more bandwidth from an amplifier may be obtained if the gain is reduced.
Unilateral Transistor Before introducing the method of design for specific gain, we will first discuss the concept of a unilateral transistor. This is an approximation in which S12 = 0 is assumed for a transistor. This is in contrast to a bilateral transistor where S12 ≠ 0 . The unilateral approximation can substantially simplify the design of an amplifier circuit while, in some cases, introducing acceptable error in the gain calculation.
© 2008 Keith W. Whites
Whites, EE 481
Lecture 36
Page 2 of 10
The assumption that S12 = 0 has some important consequences for the input and output reflection coefficients in a two-port network: Z S Z in
Γ S Γin
Z out Z L
Γ out Γ L
In particular, from (11.3a) for the input reflection coefficient and S12 = 0 : Γ S S Γin = S11 + L 12 21 = S11 (11.3a),(1) 1 − Γ L S22 while from (11.3b) for the output reflection coefficient Γ S S Γ out = S22 + S 12 21 = S 22 (11.3b),(2) 1 − Γ S S11
Using (1) and (2) in the expression for GT (from (11.13) in the text) provides what is called the unilateral transducer gain, GTU, given as 1− | Γ S |2 1− | Γ L |2 2 | S 21 | GTU = (11.15),(3) |1 − Γ S S11 |2 |1 − S 22Γ L |2 The error in gain calculated by the unilateral approximation is sometimes quantified through the ratio of GTU to the regular (or
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Lecture 36
Page 3 of 10
“bilateral”) transducer gain GT. As given in the text, this ratio is bounded as 1 GT 1 (11.45),(4) < < 2 2 (1 + U ) GTU (1 − U ) where U is the unilateral figure of merit defined as S11 S 21 S12 S 22 U= 2 2 1 − S11 1 − S 22
(
)(
)
(11.46),(5)
The unilateral assumption is often found acceptable when the error is less than approximately 0.5 dB or so. We have found that this expression in (4) is actually not correct. You will be investigating this in your homework.
Constant Gain Circles The unilateral transducer gain in (3) is the product of three terms GTU = GSU G0GLU (6) where and
GSU =
1 − ΓS
2
1 − S11Γ S
2
, GLU =
G0 = S 21
2
1 − ΓL
2
1 − S 22Γ L
2
(7),(8) (9)
The S parameters of the transistor are fixed by the device itself as well as the particular biasing.
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Lecture 36
Page 4 of 10
To achieve a specific gain, which is less than the maximum, from the transistor amplifier circuit, mismatch will be purposefully designed into the input and/or output “matching” networks to yield the gain we seek. In other words, the source and load gain factors in (7) and (8) will be adjusted downward from their maximum values GSU max and GLU max in order to achieve the specified gain. The maximum GS occurs when there is a conjugate match at the transistor input, so that for a unilateral device * (10) Γ S = Γ*in = N S11 (1)
Likewise, the maximum GL occurs when there is a conjugate match at the transistor output so that * (11) Γ L = Γ*out = N S 22 (2)
Applying (10) in (7) and (11) in (8) leads to an expression for the maximum unilateral transducer gain, GTU max , given as 1 1 2 (12) GTU max = S ≡ GSU max G0GLU max 21 2 2 1 − S11 1 − S 22 where the maximum source and load gain factors for the unilateral device are 1 (11.47a),(13) GSU max = 2 1 − S11
Whites, EE 481
Lecture 36
GLU max =
and
Page 5 of 10
1 1 − S 22
(11.47b),(14)
2
We will define normalized gain factors gS and gL to quantify the amount of reduction in the source and load gain factors needed to achieve the desired gain. These factors are given as 2 1 − ΓS GS 2 (11.48a),(15) = − gS ≡ S 1 11 2 GSU max 1 − S11Γ S
(
1 − ΓL GL gL ≡ = GLU max 1 − S 22Γ L
2
and
2
)
(
1 − S 22
2
)
(11.48b),(16)
so that 0 ≤ g S ≤ 1 and 0 ≤ g L ≤ 1. As shown in the text, (15) and (16) can be cast in the forms Γ S − CS = RS and Γ L − CL = RL , (17),(18) respectively, where CS = RS = and
CL = RL =
g S S11*
(11.51a),(19)
1 − (1 − g S ) S11
2
1 − g S 1 − S11
2
1 − (1 − g S ) S11
2
(11.51b),(20)
1 − (1 − g L ) S22
2
(11.52a),(21)
1 − g L 1 − S 22
2
1 − (1 − g L ) S22
2
(
)
* g L S 22
(
)
(11.52b),(22)
