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Margins on Bode plot
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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
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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
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• 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
Mag
nitu
de (d
B)
10-1 100 101 102 103-40
-30
-20
-10
0
Phas
e (de
g)
Bode Diagram
Frequency (rad/s)
unstable
G(s)
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• 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
Mag
nitu
de (d
B)
10-1 100 101 102 1030
10
20
30
40
Phas
e (de
g)
Bode Diagram
Frequency (rad/s)
stable
G(s)
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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
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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
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-200
-150
-100
-50
0
50M
agni
tude
(dB
)
10-2 10-1 100 101 102 103 104-270
-225
-180
-135
-90
Phas
e (d
eg)
Bode DiagramGm = 9.92 dB (at 1.36 rad/s) , Pm = 25.1 deg (at 0.765 rad/s)
Frequency (rad/s)
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• 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
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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
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-150
-100
-50
0
50
100
150M
agni
tude
(dB
)
10-2 10-1 100 101 102 103 104-270
-225
-180
-135
-90
Phas
e (d
eg)
Bode DiagramGm = -12.1 dB (at 8.67 rad/s) , Pm = 11.4 deg (at 19.4 rad/s)
Frequency (rad/s)
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• 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
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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
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-200
-150
-100
-50
0
50
100M
agni
tude
(dB
)
10-2 10-1 100 101 102 103 104-270
-225
-180
-135
-90
Phas
e (d
eg)
Bode DiagramGm = 18.3 dB (at 8.67 rad/s) , Pm = -16.6 deg (at 3.42 rad/s)
Frequency (rad/s)
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• ans = 1.0e+002 * -1.7082 -0.2888 -0.0310 0.0040 + 0.0341i 0.0040 - 0.0341i
Two right half plane poles, unstable
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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
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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
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-40
-30
-20
-10
0
10M
agni
tude
(dB
)
10-3 10-2 10-1 100 101 102 103-90
-45
0
45
90
Phas
e (d
eg)
Bode DiagramGm = Inf , Pm = 118 deg (at 21.9 rad/s)
Frequency (rad/s)
poles =
-25.3788 -4.4559 -0.2653
stable
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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
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-40
-30
-20
-10
0
10M
agni
tude
(dB
)
10-3 10-2 10-1 100 101 102 10390
180
270
360
Phas
e (d
eg)
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
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-40
-30
-20
-10
0
10M
agni
tude
(dB
)
10-3 10-2 10-1 100 101 102 1030
45
90
135
180
225
270
Phas
e (d
eg)
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
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-40
-30
-20
-10
0
10M
agni
tude
(dB
)
10-3 10-2 10-1 100 101 102 103-90
-45
0
45
90
135
Phas
e (d
eg)
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
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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
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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
![Page 25: Margins on Bode plot. Margins on Nyquist plot Suppose: Draw Nyquist plot G(jω) & unit circle They intersect…](https://reader035.vdocuments.site/reader035/viewer/2022062311/5a4d1bed7f8b9ab0599e4d75/html5/thumbnails/25.jpg)
22 2 e.g.
nnssksG
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• 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
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# (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
![Page 31: Margins on Bode plot. Margins on Nyquist plot Suppose: Draw Nyquist plot G(jω) & unit circle They intersect…](https://reader035.vdocuments.site/reader035/viewer/2022062311/5a4d1bed7f8b9ab0599e4d75/html5/thumbnails/31.jpg)
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 = 03. Z = P + N = 0, no unstable poles in
closed-loop, stable
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Example:
1
4,1
4
j
jGs
sG
4:0at jG
0:at jG
2112
112
142 :Note
jj
jj
jjG
2radiuswith 2at centered circle a is jG
![Page 33: Margins on Bode plot. Margins on Nyquist plot Suppose: Draw Nyquist plot G(jω) & unit circle They intersect…](https://reader035.vdocuments.site/reader035/viewer/2022062311/5a4d1bed7f8b9ab0599e4d75/html5/thumbnails/33.jpg)
As you move aroundfrom ω = –∞ to 0–,to 0+, to +∞, you goaround “–1” c.c.w.once.
# encirclement N = – 1.
# unstable pole P = 1
1
4
s
sG
011 PNZ
![Page 34: Margins on Bode plot. Margins on Nyquist plot Suppose: Draw Nyquist plot G(jω) & unit circle They intersect…](https://reader035.vdocuments.site/reader035/viewer/2022062311/5a4d1bed7f8b9ab0599e4d75/html5/thumbnails/34.jpg)
i.e. # unstable poles of closed-loop = 0closed-loop system is stable.
Check:
c.l. pole at s = –3, stable.
sG
sGsGc
1..
14
14
1
s
s
34
414
ss
![Page 35: Margins on Bode plot. Margins on Nyquist plot Suppose: Draw Nyquist plot G(jω) & unit circle They intersect…](https://reader035.vdocuments.site/reader035/viewer/2022062311/5a4d1bed7f8b9ab0599e4d75/html5/thumbnails/35.jpg)
Example:1. Get G(jω) for
ω = 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
sG 1
![Page 36: Margins on Bode plot. Margins on Nyquist plot Suppose: Draw Nyquist plot G(jω) & unit circle They intersect…](https://reader035.vdocuments.site/reader035/viewer/2022062311/5a4d1bed7f8b9ab0599e4d75/html5/thumbnails/36.jpg)
jes :let
0
,90
js
0,90 js
9090,0 to0 as
je
ssG
11
9090jG 0 to0 as
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# encirclement N = _____
# open-loop unstable poles P = _____
Z = P + N = ________= # closed-loop unstable poles.
closed-loop stability: _______
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Example:Given:
1. G(s) is stable2. 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
![Page 40: Margins on Bode plot. Margins on Nyquist plot Suppose: Draw Nyquist plot G(jω) & unit circle They intersect…](https://reader035.vdocuments.site/reader035/viewer/2022062311/5a4d1bed7f8b9ab0599e4d75/html5/thumbnails/40.jpg)
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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? ________
![Page 42: Margins on Bode plot. Margins on Nyquist plot Suppose: Draw Nyquist plot G(jω) & unit circle They intersect…](https://reader035.vdocuments.site/reader035/viewer/2022062311/5a4d1bed7f8b9ab0599e4d75/html5/thumbnails/42.jpg)
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? ______
![Page 43: Margins on Bode plot. Margins on Nyquist plot Suppose: Draw Nyquist plot G(jω) & unit circle They intersect…](https://reader035.vdocuments.site/reader035/viewer/2022062311/5a4d1bed7f8b9ab0599e4d75/html5/thumbnails/43.jpg)
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 ∞?
![Page 44: Margins on Bode plot. Margins on Nyquist plot Suppose: Draw Nyquist plot G(jω) & unit circle They intersect…](https://reader035.vdocuments.site/reader035/viewer/2022062311/5a4d1bed7f8b9ab0599e4d75/html5/thumbnails/44.jpg)
2. To use Nyquist criterion, need complete Nyquist plot.
a) Get complex conjugateb) Connect ω = 0– to ω = 0+ through an
infinite circlec) Count # encirclement Nd) Apply: Z = P + N
o.l. stable, P = _______Z = _______c.l. stability: _______
![Page 45: Margins on Bode plot. Margins on Nyquist plot Suppose: Draw Nyquist plot G(jω) & unit circle They intersect…](https://reader035.vdocuments.site/reader035/viewer/2022062311/5a4d1bed7f8b9ab0599e4d75/html5/thumbnails/45.jpg)
Incorrect Correct
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Example:G(s) stable, P = 0G(jω) for ω > 0 as
given.1. Get G(jω) for
ω < 0 by conjugate
2. Connect ω = 0– to ω = 0+.But how?
![Page 47: Margins on Bode plot. Margins on Nyquist plot Suppose: Draw Nyquist plot G(jω) & unit circle They intersect…](https://reader035.vdocuments.site/reader035/viewer/2022062311/5a4d1bed7f8b9ab0599e4d75/html5/thumbnails/47.jpg)
Choice a) :
Where’s “–1” ?# encirclement N = _______Z = P + N = _______
Make sense? _______
Incorrect
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Choice b) :Where is
“–1” ?# encir.
N = _____Z = P + N
= _______closed-loop
stability _______
Correct
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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
1s
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Incorrect
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Example: G(s) stable, P = 01. Get conjugate
for ω < 02. Connect ω = 0–
to ω = 0+.
Needs to goone full circlewith radius ∞.Two choices.
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Choice a) :
N = 0
Z = P + N = 0
closed-loopstable
Incorrect!
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Choice b) :N = 2Z = P + N
= 2Closedloop has two unstable poles
Correct!
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Which way is correct?
For stable & non-minimum phase systems,
case in this ,0near 20
sKsGs
generalin 0Ns
K
00 K
c.c.w.in circles when s
c.w.in circles 1s
c.w.in circles sG
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Example: G(s) has one unstable poleP = 1, no unstable zeros
1. Get conjugate2. Connect
ω = 0–
to ω = 0+.How?One unstablepole/zeroIf connect in c.c.w.
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# encirclement N = ?If “–1” is to the left of A
i.e. A > –1then N = 0
Z = P + N = 1 + 0 = 1but if a gain is increased, “–1” could be
inside, N = –2Z = P + N = –1
c.c.w. is impossible
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If connect c.w.:For A > –1
N = ______Z = P + N
= ______For A < –1
N = ______Z = ______
No contradiction. This is the correct way.
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Example: G(s) stable, minimum phaseP = 0
G(jω) as given:get conjugate.Connect ω = 0–
to ω = 0+,00 Kdirection c.w.
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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)
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If C < –1 < B :N = ______Z = P + N = ______closed-loop stability: ______
If –1 < C :N = ______Z = P + N = ______closed-loop stability: ______