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

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Mag

nitu

de (d

B)

10-1 100 101 102 103-40

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0

Phas

e (de

g)

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

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e (de

g)

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

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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)

• 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

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tude

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eg)

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

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tude

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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)

• 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

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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

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agni

tude

(dB

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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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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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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.

nnssksG

• 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 = 03. Z = P + N = 0, no unstable poles in

closed-loop, stable

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

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

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

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

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 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

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 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: _______

Incorrect Correct

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?

Choice a) :

Where’s “–1” ?# encirclement N = _______Z = P + N = _______

Make sense? _______

Incorrect

Choice b) :Where is

“–1” ?# encir.

N = _____Z = P + N

= _______closed-loop

stability _______

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

1s

Incorrect

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.

Choice a) :

N = 0

Z = P + N = 0

closed-loopstable

Incorrect!

Choice b) :N = 2Z = P + N

= 2Closedloop has two unstable poles

Correct!

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

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.

# 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

If connect c.w.:For A > –1

N = ______Z = P + N

= ______For A < –1

N = ______Z = ______

No contradiction. This is the correct way.

Example: G(s) stable, minimum phaseP = 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: ______