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A Freq. Resp. Example (PID)
Wednesday 25 Oct 2013 EE 401: Control Systems Analysis and Design
• A Radar Tracking System Design a PID controller Specs:
o PM = 50o Ts < 4 sec
4( )
( 1)( 2)pG ss s s
( ) 1H s
Slide 1 of 8
A Freq. Resp. Example (PID)
Wednesday 25 Oct 2013 EE 401: Control Systems Analysis and Design
• A proportional Derivative (PD) controller: From Ts
Now,
Also,
Now, KI = 0, hence
( )c D PG s K s K
1
8
tan( )s mT
81.68 rad/s
4 tan(50 )
1( )cG j 1
1 1 1.7180 ( ) ( ) 59.9m pG j H j
1 1
( )
( ) ( )P
p
CosK
G j H j
(59.9 )1.1
0.454
Cos
0IK
1 1 1 1
1 ( )
( ) ( )I
D
p
K SinK
G j H j
1 0 (59.9 )1.1
1.7 1.7 0.454
Sin
11 1 1.68
( ) ( ) 189.2pG j H j
1 1( ) ( ) 0.467pG j H j
A Freq. Resp. Example (PID)
Wednesday 25 Oct 2013 EE 401: Control Systems Analysis and Design Slide 3 of 8
• A proportional Derivative (PD) controller: The open loop frequency response
( ) 1.1 1.1cG s s
-80
-60
-40
-20
0
20
40
Mag
nitu
de (
dB)
10-1
100
101
102
-180
-135
-90
Pha
se (
deg)
Bode Diagram: PD Controller
Frequency (rad/s)
PM = 50 (@1.68rad/s)
0 1 2 3 4 5 60
0.2
0.4
0.6
0.8
1
1.2
1.4CL Unit Step Response: PD Compensator
Time (seconds)
Am
plitu
de Closed-Loop Step Response
The closed-loop step response is very similar to the Lead Compensator
A Freq. Resp. Example (PID)
Wednesday 25 Oct 2013 EE 401: Control Systems Analysis and Design Slide 4 of 8
• A proportional, Integral, Derivative (PID) controller:• Effectively a Lead-Lag controller
KP =1.1 remains unchanged Lets try a range of integrator gains
o Start “small” KI = 0.005, 0.05, and 0.5
( ) Ic P D
KG s K K s
s
0.005
1 0.005 (59.9 )1.097
1.7 1.7 0.454ID K
SinK
0.05
1 0.05 (59.9 )1.113
1.7 1.7 0.454ID K
SinK
0.5
1 0.5 (59.9 )1.273
1.7 1.7 0.454ID K
SinK
A Freq. Resp. Example (PID)
Wednesday 25 Oct 2013 EE 401: Control Systems Analysis and Design Slide 5 of 8
• A PID controller:
Closed-Loop Step Response Open-Loop Frequency Response
( ) Ic P D
KG s K K s
s
0 1 2 3 4 5 6 7 8 9 100
0.2
0.4
0.6
0.8
1
1.2
1.4
Gc(s) = 1.095 + 0.005/s + 1.097 s
CL Unit Step Response: PID Compensator
Time (seconds)
Am
plitu
de
-100
-50
0
50
100
150
Mag
nitu
de (
dB)
10-3
10-2
10-1
100
101
102
-180
-150
-120
-90
Pha
se (
deg)
OL Bode Diagram: PID Controller (KI=0.005)
Frequency (rad/s)
0.005IK
A Freq. Resp. Example (PID)
Wednesday 25 Oct 2013 EE 401: Control Systems Analysis and Design Slide 6 of 8
• A PID controller:
Closed-Loop Step Response Open-Loop Frequency Response
( ) Ic P D
KG s K K s
s
0 1 2 3 4 5 6 7 8 9 100
0.2
0.4
0.6
0.8
1
1.2
1.4
Gc(s) = 1.095 + 0.050/s + 1.113 s
CL Unit Step Response: PID Compensator
Time (seconds)
Am
plitu
de
-100
-50
0
50
100
150
Mag
nitu
de (
dB)
10-3
10-2
10-1
100
101
102
-180
-150
-120
-90
Pha
se (
deg)
OL Bode Diagram: PID Controller (KI=0.050)
Frequency (rad/s)
0.05IK
A Freq. Resp. Example (PID)
Wednesday 25 Oct 2013 EE 401: Control Systems Analysis and Design Slide 7 of 8
• A PID controller:
Closed-Loop Step Response Open-Loop Frequency Response
( ) Ic P D
KG s K K s
s
0 1 2 3 4 5 6 7 8 9 100
0.2
0.4
0.6
0.8
1
1.2
1.4
Gc(s) = 1.095 + 0.500/s + 1.273 s
CL Unit Step Response: PID Compensator
Time (seconds)
Am
plitu
de
-100
-50
0
50
100
150
Mag
nitu
de (
dB)
10-3
10-2
10-1
100
101
102
-180
-150
-120
-90
Pha
se (
deg)
OL Bode Diagram: PID Controller (KI=0.500)
Frequency (rad/s)
0.5IK
A Freq. Resp. Example (PID)
Wednesday 25 Oct 2013 EE 401: Control Systems Analysis and Design Slide 8 of 8
• A PID controller: Closed-Loop pole locations
-1.4 -1.2 -1 -0.8 -0.6 -0.4 -0.2 0-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
CL Pole Locations: PID Compensator (KI=0.005)
Real Axis (seconds-1)
Imag
inar
y A
xis
(sec
onds
-1)
-1.4 -1.2 -1 -0.8 -0.6 -0.4 -0.2 0-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
CL Pole Locations: PID Compensator (KI=0.050)
Real Axis (seconds-1)
Imag
inar
y A
xis
(sec
onds
-1)
-1.4 -1.2 -1 -0.8 -0.6 -0.4 -0.2 0-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
CL Pole Locations: PID Compensator (KI=0.500)
Real Axis (seconds-1)
Imag
inar
y A
xis
(sec
onds
-1)
