Lecture 2 Basic ofmechanical and electrical controlsyste m
•
By
•
Oladokun Sulaiman
•
Note Revise : • Differential equation • Partial fraction
2.0 Objective • • • • • •
Free body and block diagram Block diagram Obtain the differential equation Obtain Laplace transform of the differential equations Solve the resulting algebraic transform Mathematical modeling of physical system
RECAP implified block diagram of closed-loop control system Process or plant
Compensator or controller Input U(s)
Output
Gc(s)
Gp(s) Y(s) Sensor
H(s)
•
Process or plant: – System to be controlled including actuators and power amplifiers
•
Sensor: – Instrumentation that measures output and converts it to t o a signal
•
Compensator or controller: – System added to enhance performance of control loop – Output or controlled variable is y (t ) is the variable we wish to control – Input is a measure of (but not always equal to) the desired system syst em output u (t ) – Error is desired output minus actual output: e(t )= )= y (t ) - u (t )
test waveform
•Refer to control engineering system page 19
Types of input 100
100 N N , , e e c c 50 r r o o F F
N
N , , e e c c 50 r r o o F F
0
N N , , e e c c r r o o F F
t1
t2 t3 Time, second
0
t4
t1
t2
t4
t3
Time, second
100
100 N N , , e e c c 50 r r o o F F
50
0
t1
t2 t3 Time, second
t4
0
t1
t2 Time, second
t3
Two step input response
Ramp Input Response
Time Constant • It is defined as time taken by a control plant to achieve output response equal to 63% of its desired des ired value.
Control System Response • Time response • Frequency response • Steady state response • Transient response • Undershoot • Overshoot • Settling time
Control system stability • Damping factor • Damping ratio • Routh’s stability criteria • Nyquist stability criteria
Control System Design Process
fig_01_11
•Control engineering system page 16
fig_01_11
2.1 Introduction : Mathematical modeling • In order to analyze and design a control system knowledge of its behavior through mathematical terms is essential • The mathematical equations equations are derived from law of physics i.e Newton • Analysis can be done in two operating condition: Steady and Transient • Change as a result of of input input or disturbance • Output will depend on the system variables and how they interact • Description of dynamic system is obtain from differential equation’ • models of the physical system • Solutions of these mathematical equations simulate the response of the physical system which they represent
2.2 Differential equation
• Input and output relationship of a linear measurement system - ordinary differential equation equation (ODE):
an
d n y dt
n
an− 1
+
1 d n −
dt
n− 1
+a1
+
dy dt
a0 y
+
• u = u = input, y = y = output; u and u and y varies y varies with t • n>m and a, b = constant coefficients • DE contain variable and rate of change of or derivative of the variable in control system • Ordinary differential equation (ODE) are main concern concern in control system , they contain singe dependent and independent variable which is usually time • The order DE relate to the index of the highest derivative
Example 2.1 : Spring mass balance damper x i
x o
k
F m
F s =
m
F D
c
k ( xi − xo ) + c( 2
d xo 2
d t
+
c d xo m d t
dxi
k +
m
dt xo
−
=
dxo dt
2
)=m
c d xi m d t
k +
m
d xo 2
dt
xi
2.3 Physical system modeling • To obtain linear approximation of physical system Time response solution is obtain: • Obtain the differential equation • Obtain Laplace transform of the differential equations • Solve the resulting algebraic transform
2.3a. Laplace transform Definition Definition of Laplace transformation of f (t ): ): ∞ ∞ L[ f (t )] = F ( s ) = ∫ e − st dt [ f (t )] = ∫ f (t ) e −st dt 0
0
where s = σ + jω = a complex variable – Inverse Laplace transformation – f (t ) = L-1[F -1[F (s)] – L Af [ Af (t )] )] = AL = AL[[f (t )] )] – L[f 1(t 1(t ) + f 2(t 2(t )] )] = L[f 1(t 1(t )] )] + L[f 2(t 2(t )] )] f (t ) A
For step input f (t ) = 0 t < t < 0 ∞ – = A t > t > 0 F ( s ) = L[ f (t ) ] = L[ A] =∫ Ae – Laplace transform: 0 t
A
=−
e
−∞ s t
0
−st
A
=
dt
•
Example 2
Find the time response xo response xo((t ) for this system if step input xi input xi (t )=1 )=1 and initial condition xo condition xo(0)=0 (0)=0 c k = k = 1 c=1
c
d x
0
d t
k ( xi- x0)
k
xi
xo
•
Differential equation:
•
For k For k =1, =1,c c = = 1;
•
Laplace transform
k ( xi − x0 ) − c c
•
d x
o
k x
d x
o
d t
•
d t
+ xo
1
=
s
+
( s
i
= xi
xo(t )
1 sX 0 ( s ) + X 0 ( s ) = s
( s )
k x
o
Inverse Laplace transform:
X o
=0
+ =
d t
Partial fraction
d x0
1
1
1)
+
() 1
t
− t
Transient response
Steady-state response
•Control engineering system page 33
table_02_01
t a b el _ 0 2 _ 0 1
table_02_02
2.3b. Transfer Functions • Defined as the ratio of the Laplace Transform of the output to the Laplace Transform of the input to the system • G(s) = Y(s)/X(s) •
X(s)
G(s)
Y(s)
Transfer Function • An assembly of linked components within a boundary. • The motor car is a good example; mechanical, electrical, control and suspension sub-systems within a body-chassis boundary. • A system may have one input and a related output dependent on the effect of that system (transfer (transfer function G ). • θ 0 = G θ I
•The boundary, represented as a "black box", may include a complex c omplex system which need not be analysed if G is provided. •More complex systems have interconnecting links to related systems. •A system must have input, process, output, and in most systems a source of power and a means of control.
Transfer Function Expression
Characteristics Equation
G ( s ) =
b1 S +b0 a 2 S
2
+a1 S +a 0
Characteristics Equation •Denominator of the transfer function equated to zero is the characteristics equation of the system •Characteristics equation of the system determines the response of the control system
2
a 2 S a1 S + a 0
=
0
Order of control systems • Zero Order System b1 s +b0 0
s ( s
2
a 1 s +a 2 )
+
b1 s + b0
• First Order System • 2nd Order System
s 1 ( s
2
b1 s s
2
( s
2
a 1 s +a 2 )
+
b0 +
a 1 s + a2 +
(Source: Instrumentation and Control Systems by Leslie Jackson)
)
Poles and Zeros • Roots of the Characteristics equation are called poles of the system Roots of numerator of the TF are called zeros of the system Example
1)
Ans
2 s
s
2
+
4
2)
3 s + 2
+
(1) zeros
=−
2, poles
s 2 s 3
6 s 2 +11 s + 6
+
1, − 2 ;
=−
6 s + 5
+
2.4 System modeling Step for drawing block diagram: • Step 1: Free body diagram • Step 2: Mathematical equations • Step 3: Block diagram
2.4a Mechanical system: Spring • Spring • Where k = stiffness, x= displacemnet • Fs= Fx
or F s
x x
F s
k
• Transfer function = k or k or 1/k 1/k • Output variable = transfer function × input variable k • Spring with free at both ends x
x2
1
x1(t )
• Block diagram
+
x1- x2
Fs = k ( x x 1 -– x – x x2) 2(t ))
k
F s(t )
4. Mechanical system: Mass • Fm = ma
F m(t )
x(t )
m
• Force F acting F acting on mass m 2
F m
=
m
d x dx
2
• Use D-operator where: D = d /dt and dt and D2 =d =d 2/dt 2/dt 2 x(t )
F m(t ) 1 mD 2
• Fm = mD2 mD2 x
Spring mass system xo(t )
xi(t ) m
• • • •
Equations: Fs = k ( xi – xi – x 0) Since Fm = ma and Fs = Fm mD2 mD2 x 0(t ) = Fs = k ( xi – xi – x 0)kxi(t)xi(t)-xo(t)-+ 0)kxi(t)xi(t)-xo(t)-+ xi(t )
xi(t ))- xo(t )
k
+ -
•
Block diagram for spring-mass system
Damper • c = c = damping coeff. • x = x = displacement • dx(t)/dt
= velocity
x( x(t )
F D(t ) cD
• FD( FD(t ) = cdx (t )/dt )/dt • FD( FD(t ) = cDx
Spring-damper system xo(t ) xi(t )
• Force on spring: Fs = k ( xi xi – xo – xo)) • Force on damper: FD = cDxo F s=F D
xi(t ) k
+
1/cD
-
Block diagram for spring-damper system
Spring-mass-damper system
F s1 s1
∀ •
F m
F D
F = ma Fs1 Fs1 - FD = Fm xi(t ) +
k
F s
F m +
-
-
F D
1
m D
cD
2
xo(t )
Spring Mass System Accelerati
ng Force
Dampin ing Spring
m
2
(m s
TF
=
+ f 2
+
dy dt
fs
= f
dt 2
dy dt
=ky
Force
d 2 y dt
Force
=m
d 2 y
+
G ( s)
+ky = F ( Distu sturbing
k ) Y ( s)
=
Y ( s) F ( s)
=
=
Force )
F ( s) 1
ms
2
+
f s + k
Example 2
fig_02_15 fig_02_15
•Draw free body diagram •Determine forces and direction i.e applied force to the right and impeding forces to the left spring, viscous damper, acceleration
•Write the differential equation
Contd •Take la place transform
) F (s ) MS 2 (s ) +f svX (s )+ KX (s = ( MS
•Or
2
f sv+K) X( s= ) F( s)
+
•Solve for transfer function •G(s)=X(s)/F(s) = 1/
( MS 2 + f vs + K )
fig_02_17
•Refer to control engineering system page 63
fig_02_18
fig_02_20
•Refer to control engineering system page 63
fig_02_20
fig_02_21
fig_02_21
fig_02_22
table_02_05
2.4b. Electrical system modeling modeling Electrical components: resistance R , capacitance C and C and inductance L Variables: voltage V and V and current i •
V R
+
Resistor -> VR = VR = iR
-
V L +
• •
Inductance
•
Capacitance->i Capacitance->i = = CdV /dt /dt = = CDV
-> VL = Ldi /dt /dt = = LDi
-
i
V C
+
Circuit theory • Series V = V 1 + V 2 • Parallel • V = V 1 = V 2 –
i = i 1 + i 2
V 1
+
-
V 2
V
+ V 1
V 2
V -
xamp e •
Vi = Vi = VL + VC + VC + VR ;
•
VL = Ldi /dt /dt = = LDi
•
CdVC /dt /dt = = i ;
•
Vo = iR
Equations: i 1 = i 2 + i 3 Vi – Vi – VA = L1Di 1 VA = L2Di 2 i 3 = (VA (VA – Vo) Vo)CD Vo = i 3R
VR = VR = Vo +
V L
V C
V i
VC = VC = /CD / i CD
V o
L1 +
C
V A
+
i1 i2
V i
i3 L2
R
-
V o
-
i3 -
V i
Block diagram
+ -
1 L1 D
V A
Use D-operator where D
d /dt and dt and D2
i1 +
i2
L2 D
V A
CD + -
V o
i3
R
V o
Example Summing the voltage around the loop -> assuming zero initial condition
•
Subtitute i(t)=dq(t)
•
Subtitute capacitor voltage charge relationship Q(T)=CvC(t)
•
Take la place transform, rearrange terms and simplify
•
Solve TF Vc(s)/V(s)
( vC s( ) vC s( )
2
L C s + R C s+ 1)C v 1/
=
s
2
= V( s)
L C
R
S + + L
1 L C
R-L-C Circuit L
1
di
R i + +
dt
1
C
C
∫ idt
T aking
∫ idt
e0 =
L aplace
transform
L s I ( s ) + R I ( s )
1 1
C s
=ei
1 1
C s
gives
I ( s ) = E i ( s )
I ( s ) = E 0 ( s )
•Combining two equations from previous slide gives T F
G ( s )
1
= = + + L C s
2
R C
s
1
TF of spring mass system and RLC circuit are mathematically similar and will give identical response
•Solution for electrical can also be done through KVL, KCL, • voltage divider, current divider
Steps for electrical modeling • Replace passive element value wit their impedance • Replace all source and time value with their Laplace transform • Assume transform current and current direction is each loop • Write KVL around each loop • Solve simultaneous equation for the output • Form the transfer function
fig_02_05
fig_02_05
fig_02_11
fig_02_11
2.3 Block diagram manipulation •
Block diagram – can be simplified to fewer blocks
•
Block diagram transformation and reduction –refer to table
•
Output input relationship – transfer function
Block diagram manipulation Series block diagram reduction
∀ ∀
θ
•
Transfer function =
θ
2 = F 1(D 1(D)θ 1 ;
θ
3 = F 2(D 2(D)θ 2
3 = F 1(D 1(D) F 2(D 2(D)θ 1 θ
3 / θ 1 = F 1(D 1(D) F 2(D 2(D)
θ F 1( D) D) 1
θ F ( D) D) 2 2
θ
θ 3
θ
F 1( D) D) F 2( D) D)
1
3
Parallel block diagram reduction
4 = θ 2 + θ 3 θ 4 = F 1(D 1(D)θ 1 + F 2(D 2(D)θ 1 θ 4 = [F [F 1(D 1(D) + F 2(D 2(D)]θ 1 Transfer function = θ 4 / θ 1 = F 1(D 1(D) + F 2(D 2(D) θ
θ
θ
1
F 1( s) s)
F 2( s) s)
θ
2
+
+ θ
θ 1
4
3
θ F 1( D)+ D)+ F 2( D) D)
4
Closed loop block diagram •
The negative feedback:
•
E(s) = U(s) – B(s) = U(s) – H(s)Y(s)
•
Y(s) = G(s)E(s) = G(s)[U(s) - H(s)Y(s)]
•
Y(s)[1 + G(s)H(s)] = G(s)U(s)
Y ( s ) G ( s ) = U ( s ) 1 + G ( s ) H ( s )
E ( s) s)
Inpu t +
G( s) s)
Output
B( B( s) s) H ( s) s)
U ( s )
G ( s )
1 + G ( s ) H ( s )
Close-loop transfer function = (Forward transfer function)/(1+ Open-loop transfer function)
Y ( s )
Example • Derive transfer function for x spring-mass k + mD system -
xo
i
2
k 2 mD
x0 = xi − x0
x
x xo xi
=
k mD 2 + k
i
k mD 2 + k
o
Multiple-loop feedback control system To eliminate G3(s)G4(s)H(s), move H2 behind G4(s) Rule 4: H 2
H 2 /G4 G4
G4
G3G4H1 = positive feedback control system Rule 6 – eliminate feedback loops
•Eliminate inner loop containing H 2/G 2/G4 and G2 and G3G4/(1-G 4/(1-G3G4H 1) 1)
Reduce the loop containing H3
Revise your digital last semester digital electronics
Tutorial Exercises •
Define transfer function of a control system.
•
Define what is time constant.
•
Describe the advantages of control and automation
Home work Chapter 2 : NISE Control -> • Answer all short questions • Problem 7,9,10,16,17,20,28 7,9,10,16,17,20,28
• Due date January 7
Summary • Free body and block diagram • Block diagram • Obtain the differential equation • Obtain Laplace transform of the differential equations • Solve the resulting algebraic transform • Mathematical modeling of physical system
