mesb374 system modeling and analysis pid controller design
DESCRIPTION
MESB374 System Modeling and Analysis PID Controller Design. PID Controller. Structure of Controller Effects of Proportional, Integral and Derivative Actions Design of PID Controllers. Disturbance D (s). Reference Input R ( s ). Control Input U ( s ). +. Output Y (s). Error E ( s ). - PowerPoint PPT PresentationTRANSCRIPT
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