lecture 8 compensation-bode plot - aast
TRANSCRIPT
Control Systems I
Emam Fathyemail: [email protected]
http://www.aast.edu/cv.php?disp_unit=346&ser=68525
Lecture 8Compensation-Bode Plot
1
Relative Stability
11/29/2015 3
Gain cross-over point
ωg
Phase cross-over point
ωp
11/29/2015 4
ωp
Gain Margin
Unstable Stable
Stable
Unstable Stable
Stable
ωg
Phase Margin
Compensation design using the bode diagram
5
The compensator Gc(s) can be written
)(
)()(
ps
zsKsGc
The phase-lead compensation when z<p.
The phase-lag compensation when z>p.
phase-lead compensation
6
7
Bode Diagram
Frequency (rad/sec)
100
101
102
103
-180
-135
-90
Phase (
deg)
-60
-40
-20
0
20
40
60M
agnitu
de (
dB
)
PM=25° PM=40°
Phase –lead compensation
8
iv 2R
i
0v
1R
C 1
1
1)(
a
Ts
aTssGc
2
21
R
RRa
CRR
RRT
21
21
9
11
1)(
a
Ts
aTssGc
2
1aT m 1T
alg10alg20
m
dB
0
20
0
Bode diagram of the phase-lead network
aTm
1
1
1sin 1
a
amajG mc lg10)(lg20
10
Phase –lead compensation Steps:
1. Evaluate the uncompensated system phase margin when error
constants are satisfied.
2. Allowing for a small amount of safety , determine the necessary
additional phase lead, .
3. Evaluate a form Eq.
4. Evaluate -10loga and determine the frequency .
5. Evaluate T from Eq.
6. Check the result, and repeat the steps if necessary.
m
1
1sin 1
a
am
m
aTm
1
Design a lead controller such that the phase margin be 45°and the ramp error constant be 100.
)25( ss
k)(sC
)(sGc
-
)(sR
Example
Step 1: Consider with as a phase-lead controller.
Step 2: find k.
Step 3: Sketch the Bode plot for G(s).
1
1)(
s
sasGc
1a
100)25(
)(lim0
ss
kssGk c
sv 2500k
Example
)25(
2500)(
sssG
)125/(
100
ss
Step 4: Find the system PM and if it is not sufficient choose the required phase by: PM Existed - PM Desiredm
Bode Diagram
Frequency (rad/sec)
100
101
102
103
-180
-135
-90
Pha
se (
deg)
-60
-40
-20
0
20
40
60
Mag
nitu
de (
dB)
PM=25°
25 5 25 - 54 m 46.2a1
125sin
a
a
Step 5: Put the center of the controller in the new gain crossover frequency:
60new
c 0106.0a
new
c
1
1
1)(
s
sasGc
10106.0
10261.0
s
s
Bode Diagram
Frequency (rad/sec)
100
101
102
103
-180
-135
-90
Phas
e (d
eg)
-60
-40
-20
0
20
40
60
Mag
nitu
de (d
B)
PM=25°
91.3
91.3)log(10)(log20
ajG newc
60
Step 6: Check the controller.
100
101
102
103
104
-180
-135
-90
Phase (
deg)
Bode Diagram
Frequency (rad/sec)
-100
-50
0
50M
agnitu
de (
dB
)
)( jG
)()( jGjGc
48PM
phase-lag compensation
17
Design a lag controller for the following system such thatthe phase margin be 45° and the ramp error constant be100.
)20)(10( sss
k)(sC
)(sGc
-
)(sR
Example 2
Step 1: Consider with as a phase-lag controller.
Step 2: Find k
Step 3: Sketch the Bode plot for G(s).
1
1)(
s
sasGc
1a
100)20)(10(
)(lim0
sss
kssGk c
sv 20000k
)20)(10(
20000)(
ssssG
)120/)(110/(
100
sss
-100
-50
0
50
100M
agnitu
de (
dB
)
10-1
100
101
102
103
-270
-225
-180
-135
-90
Phase (
deg)
Bode Diagram
Frequency (rad/sec)
PM= -22°
5.5
Step 4: Find the system PM and if it is not sufficient choose the new gain crossover
frequency such that the PM is ok (reduce it a little).
5.5new
c 4new
c
new
c
-100
-50
0
50
100
Magnitu
de (
dB
)
10-1
100
101
102
103
-270
-225
-180
-135
-90
Phase (
deg)
Bode Diagram
Frequency (rad/sec)
PM= -22°
5.5
-100
-50
0
50
100
Magnitu
de (
dB
)
10-1
100
101
102
103
-270
-225
-180
-135
-90
Phase (
deg)
Bode Diagram
Frequency (rad/sec)
Step 5: Find the gain of the system at the new gain crossover frequency and let:
0)log(20)(log20
ajG newc
0)log(2030 a
0316.010 20
30
a
Step 6: Put the right corner of the controller sufficiently far from
794.010
1
new
c
a 1
1)(
s
sasGc
179
15.2
s
s
30
new
c
Step 7: Check the designed controller.
-150
-100
-50
0
50
100
150
200
Magnitu
de (
dB
)
10-4
10-3
10-2
10-1
100
101
102
103
-270
-225
-180
-135
-90
Phase (
deg)
Bode Diagram
Frequency (rad/sec)
)( jG
)()( jGjGc
50PM
End of Lec
Design a lead controller for the followingsystem such that the phase margin be 45°and the ramp error constant be 100.
)20)(10( sss
k)(sC
)(sGc
+
-
)(sR
Step 1: Consider with as a phase-lead controller. 1
1)(
s
saksGc
1a
Note: If the plant has another gain k, let 11
1)(
a
s
sasGc
Step 2: Try to fix k according to performance request, otherwise let k=1
100)20)(10(
)(lim0
sss
kssGk c
sv 20000k
Example 6:
-100
-50
0
50
100
Magnitu
de (
dB
)
10-1
100
101
102
103
-270
-225
-180
-135
-90
Phase (
deg)
Bode Diagram
Frequency (rad/sec)
Step 3: Sketch the Bode plot of system (with the fixed k ) without controller.
)20)(10(
20000)(
ssssG
)120/)(110/(
100
sss
Step 4: Find the system PM and if it is not sufficient choose the required phase by:
PM Existed - PM Desiredm
705 (-20) - 54 m 32a1
170sin
a
a
PM= -20°
0)log(10)(log20
ajG newc
45new
c
15)log(10)(log20
ajG newc
Step 5: Put the center of controller in the new gain crossover frequency:
0039.0
a
new
c
1
1
1)(
s
sasGc
10039.0
11248.0
s
s
-100
-50
0
50
100
Magnitu
de (
dB
)
10-1
100
101
102
103
-270
-225
-180
-135
-90
Phase (
deg)
Bode Diagram
Frequency (rad/sec)
PM= -20°
15
45
Step 6: Check the controller.
-200
-150
-100
-50
0
50
100
Magnitu
de (
dB
)
Bode Diagram
Frequency (rad/sec)
10-1
100
101
102
103
104
-270
-225
-180
-135
-90
Phase (
deg)
)( jG
)()( jGjGc
18PM
Note: Design is not possible!
Why?
Design a lead controller for the following system such that the phase margin be 45° and the open
loop bandwidth be 10 rad/sec
2
1
s
)(sC)(sGc
+
-
)(sR
Step 1: Consider with as a phase-lag controller. 1
1)(
s
saksGc
1a
Note: If the plant has another gain k, let 11
1)(
a
s
saksGc
Step 2: Try to fix k according to performance request, otherwise let k=1
3log205
2
k7.17k
Why?
Example 7:
-60
-40
-20
0
20
40
Magnitu
de (
dB
)
100
101
102
103
-181
-180.5
-180
-179.5
-179
Phase (
deg)
Bode Diagram
Frequency (rad/sec)
2
7.17)(
ssG
Step 4: Find the system PM and if it is not sufficient choose the required phase by:
PM Existed - PM Desiredm
540 (0) - 54 m 8.5a1
145sin
a
a
Step 3: Sketch the Bode plot of system (with the fixed k ) without controller.
-60
-40
-20
0
20
40
Magnitu
de (
dB
)
100
101
102
103
-181
-180.5
-180
-179.5
-179
Phase (
deg)
Bode Diagram
Frequency (rad/sec)
0)log(10)(log20
ajG newc
7.6new
c
63.7)log(10)(log20
ajG newc
Step 5: Put the center of controller in the new gain crossover frequency:
63.7
062.0
a
new
c
1
1
1)(
s
sasGc
1062.0
13596.0
s
s
7.6
-80
-60
-40
-20
0
20
40
Magnitu
de (
dB
)
100
101
102
103
-180
-135
Phase (
deg)
Bode Diagram
Frequency (rad/sec)
Step 6: Check the controller.
)( jG
)()( jGjGc
45PM
sec/7radBW loopopen
3
Controller is not ok
Try again
Step 2: Try to fix k according to performance request, otherwise let k=1
3log205
2
k
Open loop bandwidth is near to gain crossover frequency so:
7.17k
3log207
2
k7.35k
Design a lead controller for the following system such that the phase margin be 45° and the open
loop bandwidth be 10 rad/sec
2
1
s
)(sC)(sGc
+
-
)(sR
Example 7(Continue):
-60
-40
-20
0
20
40
Magnitu
de (
dB
)
100
101
102
103
-181
-180.5
-180
-179.5
-179
Phase (
deg)
Bode Diagram
Frequency (rad/sec)
2
7.35)(
ssG
Step 3: Sketch the Bode plot of system (with the fixed k ) without controller.
Step 4: Find the system PM and if it is not sufficient choose the required phase by:
PM Existed - PM Desiredm
540 (0) - 54 m 8.5a1
145sin
a
a
0)log(10)(log20
ajG newc
9new
c
63.7)log(10)(log20
ajG newc
Step 5: Put the center of controller in the new gain crossover frequency:
0461.0
a
new
c
1
1
1)(
s
sasGc
10461.0
12674.0
s
s
-60
-40
-20
0
20
40
Magnitu
de (
dB
)
100
101
102
103
-181
-180.5
-180
-179.5
-179
Phase (
deg)
Bode Diagram
Frequency (rad/sec)
63.7
-100
-50
0
50
100
Magnitu
de (
dB
)
Bode Diagram
Frequency (rad/sec)
10-1
100
101
102
103
-180
-135
Phase (
deg)
Step 6: Check the controller.
)( jG
)()( jGjGc
45PM
sec/14radBW loopopen
3
Controller is not ok
Try again
Step 2: Try to fix k according to performance request, otherwise let k=1
3log207
2
k
Open loop bandwidth is near to gain crossover frequency so:
7.35k
3log206
2
k5.25k
Design a lead controller for the following system such that the phase margin be 45° and the open
loop bandwidth be 10 rad/sec
2
1
s
)(sC)(sGc
+
-
)(sR
Example 7(Continue):
2
5.25)(
ssG
Step 3: Sketch the Bode plot of system (with the fixed k ) without controller.
Step 4: Find the system PM and if it is not sufficient choose the required phase by:
PM Existed - PM Desiredm
540 (0) - 54 m 8.5a1
145sin
a
a
-60
-40
-20
0
20
40
Magnitu
de (
dB
)
100
101
102
103
-181
-180.5
-180
-179.5
-179
Phase (
deg)
Bode Diagram
Frequency (rad/sec)
0)log(10)(log20
ajG newc
8new
c
63.7)log(10)(log20
ajG newc
Step 5: Put the center of controller in the new gain crossover frequency:
0519.0
a
new
c
1
1
1)(
s
sasGc
10519.0
1301.0
s
s
-60
-40
-20
0
20
40
Magnitu
de (
dB
)
100
101
102
103
-181
-180.5
-180
-179.5
-179
Phase (
deg)
Bode Diagram
Frequency (rad/sec)
63.7
-80
-60
-40
-20
0
20
40
Magnitu
de (
dB
)
100
101
102
103
-180
-135
Phase (
deg)
Bode Diagram
Frequency (rad/sec)
Step 6: Check the controller.
)( jG
)()( jGjGc
45PM
sec/10radBW loopopen
3
Controller is ok