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
T hen: PM  angle indicated
GM  1 in value
k
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
Bode Diagram
30
G(s)
Magnitude (dB)
25
20
unstable
15
10
0
Phase (deg)
-10
-20
-30
-40
-1
10
0
10
1
10
Frequency (rad/s)
2
10
3
10
• If no wgc, gain never crosses 0dB or 1:
– Gain > 1: Closed loop system is unstable.
– Gain < 1: Closed loop system is stable
Bode Diagram
-10
stable
Magnitude (dB)
-15
G(s)
-20
-25
-30
40
Phase (deg)
30
20
10
0
-1
10
0
10
1
10
Frequency (rad/s)
2
10
3
10
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: closedloop system is stable
– PM<0, and at wpc right to wgc GM<0: closedloop system is unstable
Bode Diagram
Gm = 9.92 dB (at 1.36 rad/s) , Pm = 25.1 deg (at 0.765 rad/s)
50
Magnitude (dB)
0
-50
-100
-150
-200
-90
Phase (deg)
-135
-180
-225
-270
-2
10
-1
10
0
10
1
10
Frequency (rad/s)
2
10
3
10
4
10
• 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: closedloop system is stable
– PM<0, and at wpc right to wgc GM<0: closedloop system is unstable
Bode Diagram
Gm = -12.1 dB (at 8.67 rad/s) , Pm = 11.4 deg (at 19.4 rad/s)
150
Magnitude (dB)
100
50
0
-50
-100
-150
-90
Phase (deg)
-135
-180
-225
-270
-2
10
-1
10
0
10
1
10
Frequency (rad/s)
2
10
3
10
4
10
• 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: closedloop system is stable
– PM<0, and at wpc right to wgc GM<0: closedloop system is unstable
Bode Diagram
Gm = 18.3 dB (at 8.67 rad/s) , Pm = -16.6 deg (at 3.42 rad/s)
100
Magnitude (dB)
50
0
-50
-100
-150
-200
-90
Phase (deg)
-135
-180
-225
-270
-2
10
-1
10
0
10
1
10
Frequency (rad/s)
2
10
3
10
4
10
• 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
Bode Diagram
Gm = Inf , Pm = 118 deg (at 21.9 rad/s)
10
Magnitude (dB)
0
-10
poles =
-20
-30
-25.3788
-4.4559
-0.2653
-40
90
stable
Phase (deg)
45
0
-45
-90
-3
10
-2
10
-1
10
0
10
Frequency (rad/s)
1
10
2
10
3
10
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
Bode Diagram
Gm = -5.67 dB (at 11.5 rad/s) , Pm = -51.9 deg (at 21.9 rad/s)
10
Magnitude (dB)
0
-10
-20
-30
Phase (deg)
-40
360
270
poles =
4.7095 +11.5300i
4.7095 -11.5300i
-1.1956
-0.3235
Unstable
180
90
-3
10
-2
10
-1
10
0
10
Frequency (rad/s)
1
10
2
10
3
10
Bode Diagram
Gm = -5.91 dB (at 9.42 rad/s) , Pm = 45.3 deg (at 4.66 rad/s)
10
Magnitude (dB)
0
-10
-20
-30
-40
270
Phase (deg)
225
poles =
4.8503 + 7.1833i
4.8503 - 7.1833i
0.3993
-0.1000
Unstable
180
135
90
45
0
-3
10
-2
10
-1
10
0
10
Frequency (rad/s)
1
10
2
10
3
10
Bode Diagram
Gm = Inf , Pm = -82.2 deg (at 4.66 rad/s)
10
Magnitude (dB)
0
-10
-20
-30
-40
135
Phase (deg)
90
Poles =
28.9627
-4.4026 + 4.5640i
-4.4026 - 4.5640i
-0.2576
Unstable
45
0
-45
-90
-3
10
-2
10
-1
10
0
10
Frequency (rad/s)
1
10
2
10
3
10
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 complete
Nyquist 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
e.g.
Gs  
k
s  2 n s  
2
2
n
Encirclement of the -1 point
• 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
Nyquist Criterion Theorem
# (unstable poles of closed-loop) Z
= # (unstable poles of open-loop)
+ # encirclement
or: Z = P + N
To have closed-loop stable:
need Z = 0,
i.e. N = –P
P
N
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:
4
4
Gs  
, G j  
s 1
j  1
at   0 :
at    :

 Note:

G j   4
G j   0
4
2 j  1 2 j  1
G j   2 


2
j  1
j  1
j  1
 G j  is a circlecenteredat  2with radius  2
As you move around
from ω = –∞ to 0–,
to 0+, to +∞, you go
around “–1” c.c.w.
once.
# encirclement N = – 1.
4
G s  
s 1
# unstable pole P = 1
 Z  N  P  1  1  0
i.e. # unstable poles of closed-loop = 0
 closed-loop system is stable.
G s 
Check: Gc.. s  
1  G s 

4
s 1
4
s 1
1
4
4


s 1  4 s  3
c.l. pole at s = –3, stable.
Example:
1. Get G(jω) for
ω = 0+ to +∞
2. Use conjugate to
get G(jω) for
ω = –∞ to 0–
3. How to go from
ω = 0– to ω = 0+?
At ω ≈ 0 :
1
G s  
s
let : s  e j
  90,
s  j   


 0
  90,
s  j    0
 as   0 to 0 ,
1 1  j
 G s    e
s 


G j   90  90
as   0 to 0



  90  90
# 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 margins
2. 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 with
K = 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
 P = _______
o.l. stable,
Z = _______
c.l. stability: _______
Incorrect
Correct
Example:
G(s) stable, P = 0
G(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) :
Incorrect
Where’s “–1” ?
 # encirclement N = _______
 Z = P + N = _______
Make sense? _______
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+,
1
s2
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.
Incorrect
Example:
G(s) stable,
1. Get conjugate
for ω < 0
2. Connect ω = 0–
to ω = 0+.
Needs to go
one full circle
with radius ∞.
Two choices.
P=0
Choice a) :
N=0
Z=P+N=0
Incorrect!
closed-loop
stable
Choice b) :
N=2
Z=P+N
=2
 Closed
loop has
two
unstable
poles
Correct!
Which way is correct?
near s  0 ,
G s  
K0
2
s
K0
in this case
 s N in general
For stable & non-minimum phase systems,
K0  0
when s circles in c.c.w.
1
s
circlesin c.w.
Gs  circlesin c.w.
Example: G(s) has one unstable pole
 P = 1, no unstable zeros
1. Get conjugate
2. Connect
ω = 0–
to ω = 0+.
How?
One unstable
pole/zero
If 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 > –1
N = ______
Z=P+N
= ______
For A < –1
N = ______
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+,
 K0  0
 c.w. direction
If A < –1 < 0 :
N = ______
Z = P + N = ______
stability of c.l. : ______
If B < –1 < A : A=-0.2, B=-4, C=-20
N = ______
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: ______