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11/30/2015
1
ET 438a Automatic Control Systems Technology
Lesson 17: Combined Mode Control
1 lesson17et438a.pptx
Learning Objectives
lesson17et438a.pptx 2
After this presentation you will be able to:
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.
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2
Proportional-Integral Control
lesson17et438a.pptx 3
Control Mode Characteristics:
1. Proportional action produces fast response to large load changes.
2. Integral action drives output to zero steady-state error.
3. Adds one pole and one zero to system transfer function.
4. Used on systems that have large load changes and where proportional only
action fails to reduce steady-state error to acceptable limits.
Proportional-Integral Control
lesson17et438a.pptx 4
Mathematical Representations
Time Function: 0
0
Ipp vdt )t(eKK)t(eK)t(v
Laplace Function: )s(Es
KK)s(EK)s(V
Ip
p
s
sKK
)s(E
)s(V IpTransfer Function:
Proportional-Integral (PI) Controllers add a pole at s=0 and zero at s=0 to system
transfer function
zero
pole
All initial conditions set
to zero in transfer function
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3
Proportional-Integral Control Response
lesson17et438a.pptx 5
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
OP AMP Realization of PI Controller
lesson17et438a.pptx 6
PI Controller Circuit Derive the transfer function from the
generalized gain formula for inverting OP
AMP circuits
sCR
1sCR
sCR
sC
sCsCR
)s(A
RsC
sC
1RsC
R
sC
1R
)s(A
sC
1R)s(Z R)s(Z
)s(Z
)s(Z)s(A
i
f
i
f
V
i
f
i
f
V
ffii
i
fV
Simplify sCR
1
R
R
sCR
1
sCR
sCR)s(A
ii
f
ii
fV
CR
1K
R
RK
i
I
i
fp
Controller introduces a pole at s=0
and zero at s=-1/RfC
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4
Bode Plot of PI Controller
lesson17et438a.pptx 7
Let Ri = 10 kW, Rf = 100 kW and C=0.01 mF then produce Bode plot
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);
MatLAB Code
Bode Plot of PI Controller
lesson17et438a.pptx 8
Proportional
Action
Integral
Action
Integral
action is below
1000 rad/sec.
1/RfC set break
point
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5
PI Controller Design
lesson17et438a.pptx 9
Example 17-1: Design a PI OP AMP controller with Kp = 100 and an integral break
frequency of 100 rad/sec. Ri = 10kW
W
WW
W
M 0.1R
000,000,1100 000,10K RR R
RK
000,10R 100K CR
1
R
RK
f
pif
i
fp
ip
fi
fp
Answer
F 01.0C
F 101rad/sec 100M 0.1
1C
R
1C
CR
1
rad/sec 100
8
ff
m
W
Answer
Proportional-Derivative Control Mode
lesson17et438a.pptx 10
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:
Laplace Function:
Transfer Function:
sfrequenciehigh limit to Rate dt
)t(v d
vdt
)t(v dK
dt
)t(e dKK)t(eK)t(v 0ddpp
)s(EsK)s(EsKK)s(EK)s(V ddpp
sK1
sK1K
)s(E
)s(V
d
dp
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6
OP AMP Realization of PD Controller
lesson17et438a.pptx 11
OP AMP PD circuit
sK1
sK1K
)s(E
)s(V
d
dp
dI
fp
dI
Idd
RR
RK
RR
R CRK
Note: 1/Kd = the derivative action break point frequency
1/Kd = limiter action break point frequency
0≤≤1
f0 RR
PD Controller Design
lesson17et438a.pptx 12
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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7
Example 17-2 Solution (2)
lesson17et438a.pptx 13
Now compute the alpha value from the rate limit break frequency
Answer
Example 17-2 Solution (3)
lesson17et438a.pptx 14
Now find the value of RI
From previous calculations Rd = 100 kW and 0.1
Answer
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8
Example 17-2 Solution (4)
lesson17et438a.pptx 15
Find Rf and Ro use proportional gain and previously computed values
Answer
Example 17-2 Solution (5)
lesson17et438a.pptx 16
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
11/30/2015
9
PD Bode Plot
lesson17et438a.pptx 17
Use MatLAB to generate the Bode plot of the PD controller designed in Example 17-2
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);
MatLAB Code
Define
variables
Compute
parameters
Generate transfer
function and plot
response
PD Bode Plot
lesson17et438a.pptx 18
Computed transfer function
-0.1 s - 10
---------------
0.001 s + 1
Derivative
Break f Rate limit
Break f
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10
Proportional + Integral + Derivative
Controllers
lesson17et438a.pptx 19
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
Mathematical Relationships for PID
Controllers
lesson17et438a.pptx 20
Time Function:
Laplace Function:
Transfer Function:
0ddp
0
Ipp vdt
)t(v dK
dt
)t(e dKKdt )t(eKK)t(eK)t(v
)s(EsK)s(EsK)s(Es
K)s(EK)s(V
)s(EsK)s(EsKK)s(Es
KK)s(EK)s(V
ddI
p
ddp
Ip
p
2
d
2
dIp
sKs
sKsKK
)s(E
)s(V
11/30/2015
11
OP AMP Realization of PID Controller
lesson17et438a.pptx 21
sK1
sK1
s
sKK
)sE
)s(V
d
dIp
ii
I
ddd
d1
1
1i
ip
CR
1K
CRK
RR
R
RR
RK
Single OP AMP realization of PID control
Notice that the parameters are dependent on
each other. Changing alpha will also change
Kp.
2
d
2
ddIIp
sKs
sKs1KKKK
)sE
)s(V
Modify system response by changing the values of
Kp, Kd, KI and .
OP AMP Realization of PID Controller
lesson17et438a.pptx 22
Multiple OP AMPs allow independent adjustment of PID parameters.
sCR
1
)s(E
)s(V
II
I
RI
Ci
+
-Ei(s) VI(s)
RidCd +
- Vd(s)
Rd
R1 +
- V(s)
Rf
R2
sCR1
sCR
)s(E
)s(V
did
ddd
21
did
dd
II1
f
RR Where
sCR1
sCR
CR
1
R
R
)s(E
)s(V
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12
OP AMP Realization of PID Controller
lesson17et438a.pptx 23
Multiple OP AMP Realization
Compute common denominator
2
IdIidII
2
IIdd
2
IdIidII
did
1
f
IIdid
IIdd
didII
did
1
f
did
dd
II1
f
sCCRRsCR
sCRCR
sCCRRsCR
sCR1
R
R
)s(E
)s(V
sCRsCR1
sCRsCR
sCR1sCR
sCR1
R
R
)s(E
)s(V
sCR1
sCR
sCR
1
R
R
)s(E
)s(V
II
I
CRK
1
1
fp
R
RK ddd CRK
didd CRK
Define the following relationships between components and parameters
Multiple OP AMP PID Realization
lesson17et438a.pptx 24
Divide numerator and denominator by RICI and make defined substitutions
2
IdIid
II
II
II
2
IIdd
II
did
IIII
1
f
sCCRRCR
1sCR
CR
1
sCRCRCR
1sCR
CR
1
CR
1
R
R
)s(E
)s(V
2
d
2
ddIIp
sKs
sKsKKKK
)s(E
)s(V
II
I
CRK
1
1
fp
R
RK ddd CRK
didd CRK
Where
Controller adds two zeros and two poles to
System. Change system response by changing
Parameters Kp KI, Kd .
11/30/2015
13
PID Controller Design
lesson17et438a.pptx 25
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
Example 17-3 Solution (2)
lesson17et438a.pptx 26
Find the feedback
resistor value for the
differentiator
Answer
Now find the input resistor values for the summing amplifier
Answer
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14
Example 17-3 Solution (3)
lesson17et438a.pptx 27
Find the input resistor
value for the
differentiator
Solve for Rid and
substitute in all
known values
Answer
End Lesson 17: Combined Mode
Control
ET 438a Automatic Control Systems Technology
lesson17et438a.pptx 28