MESB374 System Modeling and Analysis
PID Controller Design
PID Controller
• Structure of Controller
• Effects of Proportional, Integral and Derivative Actions
• Design of PID Controllers
GC (s)
H(s)
+
Reference Input
R(s)
Error
E(s)++ Output
Y(s)
Disturbance
D(s)
Plant
G(s)Control Input
U(s)Gf (s)
Proportional plus Integral plus Derivative (PID) Control
0
( ) ( )t
CP CI CDu t K e t K e d K e t
21 2( ) CDCD CP CI
c
U s K s z s zK s K s KG s
E s s s
PID Controller
In time domain
In s-domain
1( )CP CI CDU s K K K s E s
s
1 2
1 2
CP CD
CI CD
K K z z
K K z z
current information passed information (prediction of) future information
• Derivative Action (KCD s): Provides added damping to the closed-loop system; reduces overshoot and oscillation in step response; tends to slow down the closed-loop system response. Dominant during initial transient, due to the effect of differentiation.
Effect of P-I-D Actions
• Integral Action (KCI /s): Eliminates steady-state error to step inputs; tends to destabilize the closed-loop system; has averaging effect. Dominant during steady-state by producing an accumulation of steady-state error to increase control effort.
• Proportional Action (KCP): Introduces immediate action due to error; improves system response time. Has similar control authority for both transient and steady-state.
Effect of PID Actions•Ex: For a given process
4 1
2 1 0.5 1R s
Y ss s s
4
(2 1)(0.5 1)s s Reference input
R(s) +
+ Output
Y(s)
Dis
turb
ance
D(s
)
•Unit step reference response •Unit step disturbance response
4 1
2 1 0.5 1D s
Y ss s s
Step Disturbance Response
Time (sec)
Am
plit
ud
e
0 2 4 6 8 10 120
0.5
1
1.5
2
2.5
3
3.5
4Step Reference Response
Time (sec)
Am
plit
ud
e
0 2 4 6 8 10 120
0.5
1
1.5
2
2.5
3
3.5
4
Effect of P Action•Objective: Design a system that has zero steady state error for step inputs with %OS<10% and Ts (2%)<6 [sec]
4
(2 1)(0.5 1)s s GC (s)+
Reference Input
R(s)
Error
E(s)++ Output
Y(s)
Disturbance
D(s)
Plant
Control Input
U(s)
1
0.05 1s
(A.) Let’s try Proportional control: GC(s)=KCP
CLTF:
42 1 0.5 1
4 11 12 1 0.5 1 0.05 1
CPC P
YRC P
CP
KG s G s s s
G sG s G s H s K
s s s
42 1 0.5 1
4 11 12 1 0.5 1 0.05 1
4 0.05 1
2 1 0.5 1 0.05 1 4
PYD
C PCP
CP
G s s sG s
G s G s H s Ks s s
s
s s s K
Effect of P Action(A.1) Design P Control using Root Locus:
1 2
2 2
1 2
4 1 1.125 2.55 0.05 0
7.1414 0,
14.1 0.25
CP
CP CP
K j
K K
CLCE:
4 11
2 1 0.5 1 0.05 1
14
10.05 0.5 2 20
CP
N s
CP
D s
Ks s s
Ks s s
-15
-10
Real Axis
-25 -20 -15 -10 -5 0 5
-5
0
5
10
15
Img. Axis
-7.5
(2 1) [rad]kP Z
kN N
3
5
3
0 7.5i i
P Z
p z
N N
2
1 2
( )0, 3 45 51 0
( )
1.235, 13.765
d D ss s
ds N s
s s
Not valid
Stability: 0<K CP <14.1
-0.667
53.76
Effect of P Action(A.2) Check for Steady State Error:
4
1 11 4
CPss ss
rCP
Ke r y
K
Stability: 0<K CP <14.1
Unit Step Reference Response
Unit Step Disturbance Response
0
1 4lim 0
1 4ss YD YDs
CP
y sG s Gs K
Larger gain results in smaller steady-state error.
Larger gain results in stronger attenuation of disturbance.
Do NOT forget
Think about the change of overshoot when gain increases
Effect of PI Actions
4
(2 1)(0.5 1)s s GC (s)+
Reference Input
R(s) Error
E(s)++ Output
Y(s)
Disturbance
D(s)
Plant
Control Input
U(s)
1
0.05 1s
(B.) Add integral action (PI control):
CLTF:
42 1 0.5 1
4 11 12 1 0.5 1 0.05 1
42 1 0.5 1
4 11 12 1 0.5 1 0.05 1
4 0.05 1
2 1 0.5 1 0.05 1 4
PICP
C PYR
PIC PCP
PYD
PIC PCP
CP PI
s zK
G s G s s s sG s
s zG s G s H s Ks s s s
G s s sG s
s zG s G s H s Ks s s s
s s
s s s s K s z
Effect of PI Actions(B.1) Design PI Control using Root Locus:
CLCE:
41 0
0.05 0.5 2 20
PI
N s
CP
D s
s z
Ks s s s
Real Axis
-25 -20 -15 -2 -1 0 1
5
Img. Axis
-15
-10
-5
0
10
15
(2 1) [rad]kP Z
kN N
3
5
3
0 7.3i i
P Z
p z
N N
2
1 2
( )0, 3 44 40 0
( )
0.9737, 13.6929
d D ss s
ds N s
s s
1 2
2 2
1 2
2 0.55 1 0.025 0
6.3246 0,
11 0
CP
CP CP
K j
K K
Not valid
Stability: 0<K CP <11
Choose zPI = -0.5:
-7.3
Effect of PI Actions(B.2) Check for Steady State Error:
1 0 1 0ss ss YRr
e r y G
Unit Step Reference Response
Unit Step Disturbance Response
0
1lim 0 0ss YD YDs
y sG s Gs
By using Integral action. steady state error is eliminated.
By using Integral action, the effect of a constant disturbance can also be eliminated.
Has transient performance been improved? Not much
Effect of PID Actions
(C.) Add derivative action (PID control):
CLTF:
1 2
1 2
1 2
1 2
42 1 0.5 1
4 111
2 1 0.5 1 0.05 1
42 1 0.5 1
4 111
2 1 0.5 1 0.05 1
4 0.05 1
2 1 0.5 1 0.05 1 4
CDC P
YRC P
CD
PYD
C PCD
CD
s z s zK
G s G s s s sG s
s z s zG s G s H sK
s s s s
G s s sG s
s z s zG s G s H sK
s s s s
s s
s s s s K s z s z
1 21C CP CI CD CD
s z s zG s K K K s K
s s
4
(2 1)(0.5 1)s s GC (s)+
Reference Input
R(s) Error
E(s)++ Output
Y(s)
Disturbance
D(s)
Plant
Control Input
U(s)
1
0.05 1s
Effect of PID Actions(C.1) Design PI Control using Root Locus:
CLCE:
1 2
41 0
0.05 0.5 20 20N s
CD
D s
s z s z
Ks s s s
Real Axis
-20 -10 -15 -2 -1 0 1
5
Img. Axis
-15
-10
-5
0
10
15
(2 1) [rad]kP Z
kN N
23
2
0 9.95i i
P Z
p z
N N
3 2
1 2 3
( )0, 4 56.6 184.8 168 0
( )
9.92, 2.60, 1.62
d D ss s s
ds N s
s s s
Stability: 0<K CD
Choose z1 = -0.5, z 2=-2.1
-9.95
-2.6
-2.1
-9.92 -1.62
Effect of PID Actions(C.2) Check for Steady State Error:
1 0 1 0ss ss YRr
e r y G
Unit Step Reference Response
Unit Step Disturbance Response
0
1lim 0 0ss YD YDs
y sG s Gs
By using Integral action. steady state error is eliminated.
By using Integral action, the effect of a constant disturbance can also be eliminated.
Has transient performance been improved? Yes
PID Controller Design via Root Locus
Design ProcedureStep1: Select the position of the two zeros such that the root locus will intersect with the desired performance region.Step 2: Pick the controller gain KC such that CL poles are in the performance regionStep 3: Find the corresponding PID gain using the above formula.
2
1 21 2
1 2
,CD C
CDCD CP CIC C CP C
CI C
K KN s K s z s zK s K s K
G s K K K z zD s s s
K K z z
Pole-Zero structure of PID Controller:
PID Controller adds one open-loop pole at origin and two open-loop zeros, z1 and z2. These two open-loop zeros could be either real or complex conjugate pair.
Design of PID ControllerEx: ( Motion Control of Hydraulic Cylinders )
M
qqININ
Recall the example of the flow control of a hydraulic cylinder that takes into account the capacitance effect of the pressure chamber. The plant transfer function is:
2
2 2
( )( )
( ) 2p p
IN p p p
CV sG s
Q s s s
VV CC
BB
AA
where 6 rad/sec, 0.1, 0.2.p p pC
Output
V(s)
2
2 22p p
p p p
C
s s
GC (s)+
Reference Velocity
R(s)
Error
E(s)Plant
Control Input
QIN (s)
Dis
turb
ance
D
(s)
++
Design of PID Controller
1 2
2 22
22 2 2
0.2 6 0.2 6( )
2 1.2 61.885 18.755 1.885 18.755
P p
IN p p p
p p
CV sG s
Q s s s s ss j s j
We would like to design a controller such that the closed loop system is better damped (smaller OS%)
CLTF:
( ) ,
with ( ) , and ( )
C P CCL
P C C P C
C PC
C P
V s K N s N sG s
R s D s D s K N s N s
N s N sG s G s
D s D s
Output
V(s)
2
2 22p p
p p p
C
s s
GC (s)+
Reference Velocity
R(s)
Error
E(s)Plant
Control Input
QIN (s)
Dis
turb
ance
D
(s)
++
Design of PID Controller
2
1 2
1 2
0.2 61 0C
s z s zK
s s p s p
2
1 21 2
1 2
,CD C
CDCD CP CIC C CP C
CI C
K KN s K s z s zK s K s K
G s K K K z zD s s s
K K z z
arctan
ln %X
Closed-loop Characteristic Equation
PID controller design
Real Axis
-30 -25 -20 -15 -10 -5 0
-5
0
5
10
15
Img. Axis
20
-15
-10
-20
Can a PID controller be designed to satisfy the transient design specifications (smaller overshoot and faster settling time) ?
t TTS S
S
(2%)4 4
Transient performance region