7-5 Relative Stability
1. Concept of Relative Stability
In designing a control system, it is required that the
system be stable. Furthermore, it is necessary that the
system have adequate relative stability.
Consider the two systems shown below:
j
j


(a)

Re

 indicates a closed-loop pole.
Re
(b)
System (a) is obviously more stable than system (b)
because the closed-loop poles of system (a) are
located farther left than those of system (b). The
Nyquist curves of the two systems show that the
closer the closed-loop poles are located to the j axis,
the closer the Nyquist curve is to the 1 point.
Im
Im
1
1
Re
Re
2. Phase and Gain margins
We can use the closeness of
approach of the G(j) locus
to the 1 point to measure
the margin of stability.
jv
1
Example. For large value of K, the system is unstable. As
the gain is decreased to a certain value, the G(j) locus
passes through the 1 point, which implies that with this
gain value, the system is on the verge of instability and
will exhibit sustained oscillations.
u
The step responses of a third order open-loop transfer
function with four different values of K:
j
(a)
h(t)
r(t)
-1
0
0
j
(b)
t
h(t)
-1
(a)
0
r(t)
0
j
(c)
(b)
t
h(t)
-1
r(t)
0
0
j
(c)
t
h(t)
(d)
r(t)
-1
0
0
t
Definitions of Phase and gain margins
1)The phase margin is defined as
g  1800  G ( j wc )H ( j wc )
Im
K g(dB)  0
where c is the gain
crossover frequency at which   0
GH(jc) =1.
[GH ]
1
Kg
1
0
c
Re
G( j )H( j )= 
2)The gain margin is defined as
1
Kg 
| G ( j w1 )H ( j w1 ) |
where 1 is the phase crossover frequency at which
GH(j1)=1800 .
In terms of decibels 20lg K g  20lg G ( j w1 )H ( j w1 ) (dB)
L( ) (dB)
60
50
40
30
20
10
0
p

2
p
3p

2
f (w)  G ( j w)H ( j w)
wc
1
2
4
6
20
8 10
g
w1
20 lg K g 40
Lgw
60
(rad / s )
Example Given the open-loop transfer function as
2
G (s ) =
1
s (s +1)( s +1)
5
Determine its phase and gain margins.
Solution: From
Gain margin:
G ( jwc )  1
K g  20lg G ( jw1 )
wc  1.2247
wc
j (wc )  90  tan wc  tan
=  154.530
5
1
1
g  180  j (wc )  1800 -154.530  25.470
w1 
5
K g  9.54
The gain and
phase margins
can also be
(dB) L (
30
20
10
)
[-20]
K=2
1/T
c
1
measured from
Bode diagram:
10
Kg
[-40]
[-60]
( )
0
-90
-180
-270
1
3. A few Comments on Phase and Gain margins
1) Phase and gain margins, as a measure of the closeness
to the 1 point, can be used as design criteria.
2) Both margins should be used in the determination of
relative stability (that is, gain or phase margin alone is
not enough ).
3) For minimum phase systems, both  and Kg must be
positive for the system to be stable. Negative margins
indicate instability.
1/ K g
j
-1
(dB)

1
c
0
Kg
0
0
c
-180
1
>0, Kg>1 (dB)>0, stable.
1/ K g
(dB)(度)
j
Kg
0
c

1
-1
0
0
-180
<0, Kg<1 (dB)<0, unstable.
c
1
4) Proper  and Kg ensure us against variations in the
system components and bound the behavior of the
closed-loop system near resonant frequency. In general,
300600 and Kg 6 dB. With these values, a
minimum-phase system has guaranteed stability, even if
the open-loop gain and time constants of the
components vary to a certain extent.
5)Conditionally stable systems will have two or
more phase crossover frequencies; some higher-order
systems with complicated numerator dynamics may
also have two or more gain crossover frequencies.

For stable systems having two or more gain crossover
frequencies, the phase margin is measured at the highest
gain crossover frequency.
Example Given the open-loop transfer function as
K
K
G (s ) 
 G ( j w) 
3
(s  1)
( j w  1)3
Determine its phase and gain margins when K is chosen
as 4 and 10, respectively.
Solution: When K=4,
G ( j w) 
4
(1  w2 )3
, G ( j w)  3 tan 1 w
Letting G(j)=1, it can be solved that
1
3
Therefore,
wc  16  1  1.233
g  1800  G ( j wc )  1800  152.9  27.1
On the other hand, from
G ( jw1 )  3 tan 1 w1  1800
we have
w1  3
G ( jw1 ) 
Therefore,
4
(1  3)3

1
2
Kg  2
When K=10,
G ( j w) 
10
(1  w2 )3
, G ( j w)  3 tan 1 w
Letting G(j)=1, it can be solved that
1
3
wc  100  1  1.91,
Therefore,
g  1800  G ( j wc )  1800  187 0  7 0  00
Similarly, from
G ( jw1 )  3tan 1 w1  1800  w1  3
we have
G ( j w1 ) 
Therefore,
10
(1  3)3

4
Kg  1
5
The system is unstable!
5
1
4
4. Correlation between Step Transient Response
and Frequency Response in the Standard
Second-Order System
r (t )
R(s)
wn2
s ( s  2zwn )
1) For the unit-step input, the output is
- zwn t
e
c (t ) = 1 sin ( wd t + b )
2
1- z
with
2
wd  wn 1  z
c (t )
C (s )
The maximum percent overshoot is
M p =e
- pV/ 1- V2
´ 100%
Note that the overshoot becomes excessive for values of
<0.4.
2) For the open-loop transfer function, the phase margin
is
g = 180 +ÐG ( j wc ) = tan
0
-1
2z
(4z 4 +1) - 2z 2
See the appendix for the deduction of the above
equation.
90
g
2z
70
(4z 4 +1) - 2z 2
/ deg
= tan
-1
80
60
50
40
30
20
10
0
0
0.2
0.4
0.6
0.8
1

1.2
1.4
1.6
1.8
2
Curve  versus 
Thus a phase margin of 600 corresponds to a
ratio of 0.6.
damping
We shall summarize the correlation between the step
transient response and frequency response of the
standard second-order system.
1)  and  are related approximately by a straight line for
0   0.6, as follows :
=  /100.
For higher order systems having a dominant pair of
closed-loop poles, this relationship may be used as a
rule of thumb in estimating the damping ratio from the
frequency response.
2) From
wd = wn 1 - z 2
wr = wn 1 - 2z 2
it is clear that r and d are almost the same for small
values of . Therefore, r is indicative of the speed of
the transient response of the system [ts=3.5/( n)].
3) From
M p =e
Mr =
- pV/ 1- V2
´ 100%
1
2z 1- z
2
, 0 < z £ 0.707
The smaller the value of  is, the larger the values of Mr
and Mp are.
This figure clearly
shows the correlation
between Mp and Mr.
Note that if  >0.707,
there is no resonant peak;
however, oscillations
always exist in time
domain for 0<<1.
Mr
Mp
Mp=c(tp)1
5. Correlation between Step Transient
Response and Frequency Response in general
systems (time domain and frequency domain)
For higher-order systems having a dominant pair of
complex conjugate closed-loop poles, the following
relationships generally exist between the step
transient response and frequency response:
1) The value of Mr is indicative of the relative stability.
Satisfactory transient performance should be 1.0<Mr
<1.4 (0 dB<Mr<3dB, 0.4 << 0.7);
For values of Mr>1.5, the step transient response may
exhibit excessive overshoot.
2) The magnitude of the resonant frequency r is indicative
of the speed of the transient response. The larger the value
of r, the faster the time response is.
3) The resonant peak frequency r and the damped natural
frequency d for the step transient response are very close
to each other for lightly damped systems.
5. Cutoff Frequency and Bandwidth (Performance
index for closed-loop systems)
1) Definition
dB
20 lg ( j 0)
20 lg 0.707( j 0)
wb
Lgw
The bandwidth is the frequency range 0b, where b
(called cutoff frequency) is the frequency at which the
magnitude of the closed-loop frequency response is 3 dB
below its zero-frequency value (j0).
Example. Given a first-order system
r (t )
R(s)
1
Ts
Determine its cutoff frequency b.
b=1/T ! ts=3T, Tb
wb  1/ T
c (t )
C (s )
3dB
Example. Given a second-order system
wn2
s (s  2zwn )
Determine its cutoff frequency b.
|  ( jw) |
Since
1
(1  w2 / wn2 ) 2  4z 2 w2 / wn2
| ( j 0) | 1
we obtain that
(1  wb2 / wn2 )2  4z 2wb2 / wn2  2
wb  wn [(1  2z 2 )  (1  2z 2 ) 2  1]
1
2
The bandwidth b can be approximately related to the
natural frequency n of the system: ts=3/n,
b=Kn; hence, n b.
2) General case
• A large bandwidth corresponds to a fast response;
• The bandwidth b can be approximately related to the
natural frequency n of a system;
•For the system to follow arbitrary inputs accurately, it
must have a large bandwidth. From the viewpoint of noise,
however, the bandwidth should not be too 1arge. Thus
there are conflicting requirements on the bandwidth, and a
compromise is usually necessary for good design.
Summary of Frequency Response Analysis
1. Definition of frequency response
The frequency response of a system is defined as the
steady-state response of the system to a sinusoidal input
signal, which, as we have proved, can be fully
characterized by its sinusoidal transfer function:
G ( j w) = G ( j w) ÐG ( j w)
Y
Amplitude
ratio
of
the
output
G ( j w)  
sinusoid to the input sinusoid
X
Phase shift of the output sinusoid
ÐG ( jw) =
with respect to the input sinusoid
2. Bode and Nyquist plots
(dB)
L( )
[-20]
[-40]
20lgk
0
( )
0
-90
-180
-270
1
3
[-20]
Im
Type 2




0
0
 0




Type 1
0
Type 0
Re
3. Nyquist stability criterion
A system is stable if and only if from
N=(PZ)/2
we can obtain that Z=0, where P is the number of the
open-loop poles in the right-hand half s-plane and N is the
number of the encirclements of the point (1, j0) of
G(j)H(j) in a counterclockwise direction as  varies
from 0 to + .
Its Bode diagram counterpart is:
The system is stable if and only if from
N+ N=(P – Z)/2
we can obtain that Z=0 (20lgG(j)>0).
4. Extension to the cases when G(s)H(s) involves poles
and zeros at the origin
j
jw
s-plane
j
1

T
j0+
Radius 
Radius 
Re
=0

Re
1
T
=0
Radius
r 
Example:
Ks
G (s ) 
(T 1s  1)(T 2s  1)
j0+
5. Gain and phase margins
Im
g  180  G ( j wc )H ( j wc )
0
with c : GH(jc) =1.
1
Kg 
| G ( j w1 )H ( j w1 ) |
with 1: GH(j1)=1800 .
K g(dB)  0
1
Kg
1
 0
[GH ]
0
c
Re
G( j )H( j )= 
L( ) (dB)
60
50
40
30
20
10
0
p

2
p

3p
2
f (w)  G ( j w)H ( j w)
wc
1
2
4
6
20
8 10
g
w1
20 lg K g 40
Lgw
60
(rad / s )
Appendix
The deduction of phase margin for the standard
second-order system:
r (t )
R(s)
n2
s( s  2n )
c (t )
C (s )
Letting
| G ( j wc ) | 1 
wn2
wc wc2  4z 2wn2
1
Hence,
wc  wn
(4z 4  1)  2z 2
At this frequency,
G ( j wc )  j w  ( j wc  2zwn )
Therefore,
 900  tan 1
(4z 4  1)  2z 2
g  1800  G ( j wc )  900  tan 1
 tan 1
2z
(4z 4  1)  2z 2
2z
(4z 4  1)  2z 2
2z