Transcript RF amplifier

Lesson 17: Combined Mode Control
ET 438a Automatic Control Systems
Technology
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lesson17et438a.pptx
Learning Objectives
After this presentation you will be able to:



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Describe the common control mode combinations used in analog
control systems
List the characteristics of combined control modes
Write the time, Laplace and transfer functions of combined
control modes
Identify the Bode plots of combined control modes
Design OP AMP circuits that realize theoretical combined
control mode performance.
lesson17et438a.pptx
Proportional-Integral Control
Control Mode Characteristics:
1.
2.
3.
4.
3
Proportional action produces fast response to large load changes.
Integral action drives output to zero steady-state error.
Adds one pole and one zero to system transfer function.
Used on systems that have large load changes and where proportional only
action fails to reduce steady-state error to acceptable limits.
lesson17et438a.pptx
Proportional-Integral Control
Mathematical Representations

Time Function:
Laplace Function:
Transfer Function:
v( t )  K p  e( t )  K p  K I   e( t ) dt  v0
0
 Kp  KI 
V(s)  K p  E(s)  
  E(s)
s


V (s)
K  s
 Kp   I 
E (s)
 s 
All initial conditions set
to zero in transfer function
zero
pole
Proportional-Integral (PI) Controllers add a pole at s=0 and zero at s=0 to system
transfer function
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Proportional-Integral Control Response
Time plots of PI controller
output for piece-wise linear
error input
Proportional mode
gives instantaneous response
to error while integral mode
decrease error over time.
Constant error produces
linearly increasing controller
output
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OP AMP Realization of PI Controller
Derive the transfer function from the
generalized gain formula for inverting OP
AMP circuits
PI Controller Circuit
A V (s) 
 Zf (s)
Zi (s)
Zi (s)  R i
Kp 
Rf
Ri
KI 
1
Ri C
Simplify
lesson17et438a.pptx
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Cs
1 
1 




 R f 

C

s
R

 f C  s 
C  s 

A V (s) 

C  s R i
Ri
Controller introduces a pole at s=0
and zero at s=-1/RfC
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Zf (s)  R f 
Cs

 R f  C  s 
C  s   R f  C  s  1

A V (s) 

Ri C s
Ri  C s
R Cs
1   Rf
1
A V (s)   f




Ri
Ri  C s
 Ri  C s Ri  C s 
Bode Plot of PI Controller
Let Ri = 10 kW, Rf = 100 kW and C=0.01 mF then produce Bode plot
MatLAB Code
ri=input('Enter value of input resistance: ');
c=input('Enter value of capacitance: ');
rf=input('Enter value of feedback resistance: ');
% compute transfer function model parameters for
% PI controller
% compute numerator parameter
tau=rf*c;
% compute parameter for denominator
tau1 = ri*c;
% build transfer function
% denominator form a1*s^2+a2s+a3
Av=tf([tau 1],[tau1 0])
%plot graph
bode(Av);
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Bode Plot of PI Controller
Integral
action is below
1000 rad/sec.
Integral
Action
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Proportional
Action
1/RfC set break
point
PI Controller Design
Example 17-1: Design a PI OP AMP controller with Kp = 100 and an integral break
frequency of 100 rad/sec. Ri = 10kW
Kp 
Rf
Ri
Kp 
Rf
Ri

1
K p  100 R i  10,000 W
Rf C
 R f  K p  K p  10,000 W 100  1,000,000 W
R f  1.0 MW
Answer
  100 rad/sec
1
1

 C
Rf C
Rf 
1
 110 8 F
1.0 MW100 rad/sec 
C  0.01 mF
Answer
C
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Proportional-Derivative Control Mode
Proportional-Derivative (PD) Control combines proportional and
derivative actions. Used in processes that have sudden load changes
that proportional only cannot handle.
Mathematical Representations
Time Function:
v ( t )  K p  e( t )  K p  K d 

10
d e( t )
d v( t )
   Kd 
 v0
dt
dt
d v( t )
 Rate limit to high frequencie s
dt
Laplace Function:
V(s)  K p  E(s)  K p  K d  s  E(s)    K d  s  E(s)
Transfer Function:
 1 Kd  s 
V(s)
 Kp  

E(s)
1



K

s
d


lesson17et438a.pptx
OP AMP Realization of PD Controller
OP AMP PD circuit
 1 Kd  s 
V(s)
 Kp  

E(s)
1



K

s
d


Kd  R d  C
Note: 1/Kd = the derivative action break point frequency
1/Kd = limiter action break point frequency
0≤≤1
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lesson17et438a.pptx

RI
RI  Rd
Kp 
Rf
RI  Rd
R0  Rf
PD Controller Design
Example 17-2: Design a PD control that has a proportional gain of 10
a derivative action break point of 100 rad/sec and a limiter frequency
break point of 1000 rad/sec. C = 0.1 mF. Draw the OP AMP circuit
and place all values on the schematic.
Example 17-2 Solution
Answer
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Example 17-2 Solution (2)
Now compute the alpha value from the rate limit break frequency
Answer
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Example 17-2 Solution (3)
Now find the value of RI
From previous calculations Rd = 100 kW and 0.1
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Answer
Example 17-2 Solution (4)
Find Rf and Ro use proportional gain and previously computed values
Answer
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Example 17-2 Solution (5)
Draw OP AMP Schematic and label components
Transfer function
Negative sign causes 180 degree phase shift due to inverting configuration. Add
Inverting OP AMP circuit with gain of -1 to remove this
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PD Bode Plot
Use MatLAB to generate the Bode plot of the PD controller designed in Example 17-2
MatLAB Code
Define
variables
Compute
parameters
Generate transfer
function and plot
response
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ri=11.11e3;
c=0.1e-6;
rf=1.111e6;
rd=1e5;
% compute transfer function model parameters for
% PD controller
% compute parameters
kd=rd*c;
alpha=ri./(ri+rd);
kp=-rf./(ri+rd);
% build transfer function
% denominator form a1*s^2+a2s+a3
% numerator for b1*s^2+b2s+b3
Av1=kp*tf([kd 1],[alpha*kd 1])
%plot graph
bode(Av1);
PD Bode Plot
Computed transfer function
-0.1 s - 10
--------------0.001 s + 1
Derivative
Break f
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Rate limit
Break f
Proportional + Integral + Derivative
Controllers
Proportional-Integral-Derivative (PID) Controller Characteristics:
1. Proportional mode provides fast response to large process
load changes
2. Integral mode removes the steady-state (offset) error from
the output
3. Derivative mode improves system stability and improves
response to rapid load changes.
4. Widely used in industry
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Mathematical Relationships for PID
Controllers

Time Function:
v( t )  K p  e( t )  K p  K I   e(t ) dt  K p  K d 
0
Laplace Function:
Transfer Function:
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lesson17et438a.pptx
d e( t )
d v( t )
   Kd 
 v0
dt
dt
 Kp  KI 
V(s)  K p  E(s)  
  E(s)  K p  K d  s  E(s)    K d  s  E(s)
s




K 
V(s)  K p  E(s)   I   E(s)  K d  s  E(s)    K d  s  E(s)
 s 


 K I  s  Kd  s2 
V(s)
 Kp  
2 
E(s)
 s    Kd  s 
OP AMP Realization of PID Controller
V(s)
 K  s   1  Kd  s 
 K p   I   

Es) 
 s  1    K d  s 
 K I  K I  K d  1  s  K d  s 2 
V(s)
 K p  

Es) 
s    Kd  s2


Kp 
Single OP AMP realization of PID control
Modify system response by changing the values of
Kp, Kd, KI and .

R1
R1  R d
K d  R d  Cd
KI 
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lesson17et438a.pptx
Ri
R i  R1
1
R i  Ci
Notice that the parameters are dependent on
each other. Changing alpha will also change
Kp.
OP AMP Realization of PID Controller
Multiple OP AMPs allow independent adjustment of PID parameters.
VI (s)
1

E(s) R I  C I  s
Ci
Ei(s)
VI(s)
-
RI
+
Vd (s)
 R d  Cd  s

E(s) 1  R id  C d  s
R2
Rd
-
Cd
22
Rid
+
lesson17et438a.pptx
Vd(s)
Rf
-
R1
V(s)
+
R d  Cd  s 
V(s) R f  1




E(s) R1  R I  CI 1  R id  Cd  s 
Where R1  R 2
OP AMP Realization of PID Controller
Multiple OP AMP Realization
Compute common denominator
R d Cds 
V(s) R f  1




E(s) R 1  R I C Is 1  R idCd s 
R d Cd sR I C Is  
V(s) R f  1  R idCd s




E(s) R 1  R I C Is1  R idCd s  1  R idCd sR I C Is 

1  R idCd s
R d C d R I C Is 2
V(s) R f 




2
E(s) R 1  R I C Is   R id R I Cd C Is
R I C Is   R id R I Cd C Is 2 
Define the follow relationships between components and parameters
1
 R ICI
KI
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lesson17et438a.pptx
K d  R d Cd
Rf
Kp 
R1
K d  R idCd
Multiple OP AMP PID Realization
Divide numerator and denominator by RICI and make defined substitutions
 1

 1 
 1 
2
R idCd s  
R d Cd R I C Is 
 

R
C
R
C
R
C
V(s) R f  I I  I I 
 I I



E(s) R 1 
 1 
 1 
2

R I C Is  
R id R I Cd C Is


R
C
R
C


 I I
 I I
 K I  K I K ds  K ds 2 
V(s)
 Kp 

2
E(s)
s


K
s
d


Where
1
 R ICI
KI
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lesson17et438a.pptx
K d  R d Cd
Controller adds two zeros and two poles to
System. Change system response by changing
Parameters Kp KI, Kd .
Kp 
Rf
R1
K d  R idCd
PID Controller Design
Example 17-3: Design a PID controller using the circuit above.
Proportional gain is 5. The derivative time constant is 0.5
seconds. The integral gain is 0.143 and =0.1. The capacitor for
the integrator Ci=10 mF and the differentiator, Cd= 1 mF
Rf = 1 MW. Find values for all other components.
Example 17-3 Solution
Define
parameters
Answer
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Example 17-3 Solution (2)
Find the feedback
resistor value for the
differentiator
Answer
Now find the input resistor values for the summing amplifier
Answer
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lesson17et438a.pptx
Example 17-3 Solution (3)
Find the input resistor
value for the
differentiator
Solve for Rid and
substitute in all
known values
Answer
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lesson17et438a.pptx
End Lesson 17: Combined Mode
Control
ET 438a Automatic Control Systems Technology
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lesson17et438a.pptx