Margins on Bode plot
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Margins on Bode plot
Margins on Nyquist plot
Suppose:• Draw Nyquist plot
G(jω) & unit circle• They intersect at point A• Nyquist plot cross neg.
real axis at –k
in value1kGM
indicated angle :Then PM
Relative stability from margins• One of the most widely used methods in
determine “how stable the system is”
• Margins on based on open-loop transfer function’s frequency response
• Basic rule:– PM>0 and GM>0: closed-loop system stable– PM + Mp 70– As PM or GM 0: oscillates more– PM=0 and GM=0: sustained oscillation– PM<0: unstable
• If no wgc, gain never crosses 0dB or 1:– Gain > 1: Closed loop system is unstable.– Gain < 1: Closed loop system is stable
10
15
20
25
30
Magnitude (dB
)
10-1
100
101
102
103
-40
-30
-20
-10
0
Phase (deg)
Bode Diagram
Frequency (rad/s)
unstable
G(s)
• If no wgc, gain never crosses 0dB or 1:– Gain > 1: Closed loop system is unstable.– Gain < 1: Closed loop system is stable
-30
-25
-20
-15
-10
Magnitude (dB
)
10-1
100
101
102
103
0
10
20
30
40
Phase (deg)
Bode Diagram
Frequency (rad/s)
stable
G(s)
Relative stability from margins• If there is one wgc and multiple wpc’s all >
wgc– PM>0, all GM>0, and closed-loop system is
stable
• If there is one wgc but > one wpc’s– Closed-loop system is stable if margins >0– PM and GM reduce simultaneously– PM and GM becomes 0 simultaneously, at
which case the closed loop system will have sustained oscillation at wgc=wpc
Relative stability from margins• If there is one wgc, and multiple wpc’s• And if system is minimum phase (all zeros in
left half plane)• And if gain plot is generally decreasing
– PM>0, all GM>0: closed-loop system is stable
– PM>0, and at wpc right to wgc GM>0: closed-loop system is stable
– PM<0, and at wpc right to wgc GM<0: closed-loop system is unstable
-200
-150
-100
-50
0
50M
agn
itu
de
(dB
)
10-2
10-1
100
101
102
103
104
-270
-225
-180
-135
-90
Ph
ase
(deg
)
Bode DiagramGm = 9.92 dB (at 1.36 rad/s) , Pm = 25.1 deg (at 0.765 rad/s)
Frequency (rad/s)
• ans =
1.0e+002 *
-1.7071
-0.2928
-0.0168
-0.0017 + 0.0083i
-0.0017 - 0.0083i
All poles negative (in left half plane)
Closed loop system is stable
Relative stability from margins• If there is one wgc, and multiple wpc’s• And if system is minimum phase (all zeros in
left half plane)• And if gain plot is generally decreasing
– PM>0, all GM>0: closed-loop system is stable
– PM>0, and at wpc right to wgc GM>0: closed-loop system is stable
– PM<0, and at wpc right to wgc GM<0: closed-loop system is unstable
-150
-100
-50
0
50
100
150M
agn
itu
de
(dB
)
10-2
10-1
100
101
102
103
104
-270
-225
-180
-135
-90
Ph
ase
(deg
)
Bode DiagramGm = -12.1 dB (at 8.67 rad/s) , Pm = 11.4 deg (at 19.4 rad/s)
Frequency (rad/s)
• ans =
1.0e+002 *
-1.7435
-0.0247 + 0.1925i
-0.0247 - 0.1925i
-0.1748
-0.0522
Closed loop system poles are all negative
System is stable
Relative stability from margins• If there is one wgc, and multiple wpc’s• And if system is minimum phase (all zeros in
left half plane)• And if gain plot is generally decreasing
– PM>0, all GM>0: closed-loop system is stable
– PM>0, and at wpc right to wgc GM>0: closed-loop system is stable
– PM<0, and at wpc right to wgc GM<0: closed-loop system is unstable
-200
-150
-100
-50
0
50
100M
agn
itu
de
(dB
)
10-2
10-1
100
101
102
103
104
-270
-225
-180
-135
-90
Ph
ase
(deg
)
Bode DiagramGm = 18.3 dB (at 8.67 rad/s) , Pm = -16.6 deg (at 3.42 rad/s)
Frequency (rad/s)
• ans =
1.0e+002 *
-1.7082
-0.2888
-0.0310
0.0040 + 0.0341i 0.0040 - 0.0341i
Two right half plane poles, unstable
Conditionally stable systems• Closed-loop stability depends on the overall
gain of the system
• For some gains, the system becomes unstable
• Be very careful in designing such systems
• Type 2, or sometimes even type 1, systems with lag control can lead to such
• Need to make sure for highest gains and lowest gains, the system is stable
Relative stability from margins• If there are multiple wgc’s
– Gain plot cannot be generally decreasing– There may be 0, or 1 or multiple wpc’s
– If all PM>0: closed-loop system is stable
– If one PM<0: closed-loop system is unstable
-40
-30
-20
-10
0
10M
agn
itu
de
(dB
)
10-3
10-2
10-1
100
101
102
103
-90
-45
0
45
90
Ph
ase
(deg
)
Bode DiagramGm = Inf , Pm = 118 deg (at 21.9 rad/s)
Frequency (rad/s)
poles =
-25.3788 -4.4559 -0.2653
stable
Relative stability from margins• If there are multiple wgc’s
– Gain plot cannot be generally decreasing– There may be 0, or 1 or multiple wpc’s
– If all PM>0: closed-loop system is stable
– If one PM<0: closed-loop system is unstable
-40
-30
-20
-10
0
10M
agn
itu
de
(dB
)
10-3
10-2
10-1
100
101
102
103
90
180
270
360
Ph
ase
(deg
)
Bode DiagramGm = -5.67 dB (at 11.5 rad/s) , Pm = -51.9 deg (at 21.9 rad/s)
Frequency (rad/s)
poles =
4.7095 +11.5300i 4.7095 -11.5300i -1.1956 -0.3235
Unstable
-40
-30
-20
-10
0
10M
agn
itu
de
(dB
)
10-3
10-2
10-1
100
101
102
103
0
45
90
135
180
225
270
Ph
ase
(deg
)
Bode DiagramGm = -5.91 dB (at 9.42 rad/s) , Pm = 45.3 deg (at 4.66 rad/s)
Frequency (rad/s)
poles =
4.8503 + 7.1833i 4.8503 - 7.1833i 0.3993 -0.1000
Unstable
-40
-30
-20
-10
0
10M
agn
itu
de
(dB
)
10-3
10-2
10-1
100
101
102
103
-90
-45
0
45
90
135
Ph
ase
(deg
)
Bode DiagramGm = Inf , Pm = -82.2 deg (at 4.66 rad/s)
Frequency (rad/s)
Poles =
28.9627 -4.4026 + 4.5640i -4.4026 - 4.5640i -0.2576
Unstable
Limitations of margins• Margins can be come very complicated
• For complicated situations, sign of margins is no longer a reliable indicator of stability
• In these cases, compute closed loop poles to determine stability
• If transfer function is not available, use Nyquist plot to determine stability
Stability from Nyquist plot
The completeNyquist plot:
– Plot G(jω) for ω = 0+ to +∞
– Get complex conjugate of plot,that’s G(jω) for ω = 0– to –∞
– If G(s) has pole on jω-axis, treat separately
– Mark direction of ω increasing
– Locate point: –1
22 2
e.g.nnss
ksG
• As you follow along the G(jω) curve for one complete cycle, you may “encircle” the –1 point
• Going around in clock wise direction once is +1 encirclement
• Counter clock wise direction once is –1 encirclement
Encirclement of the -1 point
# (unstable poles of closed-loop) Z= # (unstable poles of open-loop) P
+ # encirclementN
or: Z = P + N
To have closed-loop stable:need Z = 0, i.e. N = –P
Nyquist Criterion Theorem
That is: G(jω) needs to encircle the “–1” point counter clock wise P times.
If open loop is stable to begin with, G(jω) cannot encircle the “–1” point for closed-loop stability
In previous example:1. No encirclement, N = 0.
2. Open-loop stable, P = 0
3. Z = P + N = 0, no unstable poles in closed-loop, stable
Example:
1
4,
1
4
jjG
ssG
4:0at jG
0:at jG
21
12
1
12
1
42 :Note
j
j
j
j
jjG
2radiuswith 2at centered circle a is jG
As you move around
from ω = –∞ to 0–,
to 0+, to +∞, you go
around “–1” c.c.w.
once.
# encirclement N = – 1.
# unstable pole P = 1
1
4
ssG
011 PNZ
i.e. # unstable poles of closed-loop = 0
closed-loop system is stable.
Check:
c.l. pole at s = –3, stable.
sG
sGsGc
1..
14
14
1
s
s
3
4
41
4
ss
Example:
1. Get G(jω) forω = 0+ to +∞
2. Use conjugate to get G(jω) forω = –∞ to 0–
3. How to go from ω = 0– to ω = 0+? At ω ≈ 0 :
s
sG1
jes :let
0
,90
js
0,90 js
9090,0 to0 as
je
ssG
11
9090jG 0 to0 as
# encirclement N = _____
# open-loop unstable poles P = _____
Z = P + N = ________
= # closed-loop unstable poles.
closed-loop stability: _______
Example:
Given:1. G(s) is stable
2. With K = 1, performed open-loop sinusoidal tests, and G(jω) is on next page
Q: 1. Find stability margins2. Find Nyquist criterion to determine
closed-loop stability
Solution:1. Where does G(jω) cross the unit
circle? ________Phase margin ≈ ________
Where does G(jω) cross the negative real axis? ________
Gain margin ≈ ________
Is closed-loop system stable withK = 1? ________
Note that the total loop T.F. is KG(s).
If K is not = 1, Nyquist plot of KG(s) is a scaling of G(jω).
e.g. If K = 2, scale G(jω) by a factor of 2 in all directions.
Q: How much can K increase before GM becomes lost? ________
How much can K decrease? ______
Some people say the gain margin is 0 to 5 in this example
Q: As K is increased from 1 to 5, GM is lost, what happens to PM?
What’s the max PM as K is reduced to 0 and GM becomes ∞?
2. To use Nyquist criterion, need complete Nyquist plot.
a) Get complex conjugate
b) Connect ω = 0– to ω = 0+ through an infinite circle
c) Count # encirclement N
d) Apply: Z = P + N
o.l. stable, P = _______
Z = _______
c.l. stability: _______
Incorrect Correct
Example:
G(s) stable, P = 0
G(jω) for ω > 0 as given.
1. Get G(jω) forω < 0 by conjugate
2. Connect ω = 0– to ω = 0+.But how?
Choice a) :
Where’s “–1” ?
# encirclement N = _______
Z = P + N = _______
Make sense? _______
Incorrect
Choice b) :
Where is“–1” ?
# encir.N = _____
Z = P + N= _______
closed-loopstability _______
Correct
Note: If G(jω) is along –Re axis to ∞ as ω→0+, it means G(s) has in it.
when s makes a half circle near ω = 0, G(s) makes a full circle near ∞.
choice a) is impossible,but choice b) is possible.
2
1
s
Incorrect
Example: G(s) stable, P = 0
1. Get conjugatefor ω < 0
2. Connect ω = 0–
to ω = 0+.
Needs to goone full circlewith radius ∞.Two choices.
Choice a) :
N = 0
Z = P + N = 0
closed-loopstable
Incorrect!
Choice b) :
N = 2
Z = P + N= 2
Closedloop has two unstable poles
Correct!
Which way is correct?
For stable & non-minimum phase systems,
case in this ,0near 20
s
KsGs
generalin 0Ns
K
00 K
c.c.w.in circles when s
c.w.in circles 1s
c.w.in circles sG
Example: G(s) has one unstable pole
P = 1, no unstable zeros
1. Get conjugate
2. Connectω = 0–
to ω = 0+.How?One unstablepole/zeroIf connect in c.c.w.
# encirclement N = ?
If “–1” is to the left of A
i.e. A > –1
then N = 0
Z = P + N = 1 + 0 = 1
but if a gain is increased, “–1” could be inside, N = –2
Z = P + N = –1
c.c.w. is impossible
If connect c.w.:
For A > –1N = ______
Z = P + N
= ______
For A < –1N = ______
Z = ______
No contradiction. This is the correct way.
Example: G(s) stable, minimum phase
P = 0
G(jω) as given:
get conjugate.
Connect ω = 0–
to ω = 0+,00 Kdirection c.w.
If A < –1 < 0 :N = ______Z = P + N = ______stability of c.l. : ______
If B < –1 < A : A=-0.2, B=-4, C=-20N = ______Z = P + N = ______closed-loop stability:
______
Gain margin: gain can be varied between (-1)/(-0.2) and (-1)/(-4),
or can be less than (-1)/(-20)
If C < –1 < B :N = ______Z = P + N = ______closed-loop stability: ______
If –1 < C :N = ______Z = P + N = ______closed-loop stability: ______
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