7-4 Control System Design by
Frequency Response Approach
1. Introduction: Basic requirements for closedloop systems in frequency domain
1) Performance specifications: A desired control
system should be stable with sufficiently fast transient
response, sufficiently small steady-state error, and
strong ability to counteract external disturbance.
In frequency domain, the performance specifica-tions
are usually given by
, Kg, c, K
or closed-loop performance indices Mr, b and so on.
Remark:
w  w
4
2
(4
z

1)

2
z
n
 c

1
w  w [(1  2z 2 )  (1  2z 2 ) 2  1] 2
n
 b
2)Frequency bands: In general, a control system
operates in low and medium frequency bands or
ranges while disturbance operates in high
frequency band.
b=(5~10)m
b
m
L
M
1
2
H

Low frequency range represents the steady-state
performance and is expected to be unity, that is, with
higher open-loop gain or some integral factors in
open-loop transfer function.
(j)
G (s )
1
m
b
G
F (s ) =
1 +G

Medium frequency region represents the transient
response and is expected to be sufficiently fast and well
damped, that is, with higher b and 0Mr1.4.
(j)
1
m
b

High frequency region represents the disturbance
attenuation ability and is expected to be sufficiently
attenuated, that is, the frequency response in high
frequency region be decreased quickly.
(j)
1
m
b

3) Information Obtainable from Open-Loop
Frequency Response
L(ω)
20dB/dec
-40dB/dec
M
20dB/dec
ωc
ω2 ωm
ω3
40dB/dec
The low frequency range (the range far below c) of the
locus indicates the steady-state behavior of the closed-loop
system.
L(ω)
20dB/dec
-40dB/dec
M
20dB/dec
ωc
ω2 ωm
ω3
40dB/dec
The medium-frequency region (the region near c) of the
locus indicates transient response and relative stability.
The high frequency region (the region far above c)
indicates disturbance attenuation.
Requirements on Open-Loop Frequency Response
To have a high value of the velocity error constant and
yet satisfactory relative stability, it is necessary to
reshape the open-loop frequency-response curve through
a compensator.
L(ω)
20dB/dec
-40dB/dec
M
20dB/dec
ωc
ω2 ωm
ω3
40dB/dec
The requirement on open-loop frequency response for
low frequency region: The gain in low-frequency region
should be large enough and should have an integral factor.
L(ω)
20dB/dec
-40dB/dec
M
20dB/dec
ωc
ω2 ωm
ω3
40dB/dec
G (s )
G
F (s ) =
1 +G
The requirement on open-loop frequency response for
medium frequency region: Near the gain crossover
frequency, the slope of the magnitude curve in the Bode
diagram should be -20 dB/decade.
MFR
-20dB/dec
L(ω)
20dB/dec
-40dB/dec
h
20dB/dec
ωc
ω2 ωm
ω3
40dB/dec
1/T s
K
The requirement on open-loop frequency response for
high frequency region: For the high-frequency region,
the gain should be attenuated as rapidly as possible to
minimize the effects of noise.
L(ω)
HFR
20dB/dec
-40dB/dec
h
20dB/dec
ωc
ω2 ωm
ω3
40dB/dec
1/T s
G
F (s ) =
1 +G
2. Lead Compensation
• Lead compensation essentially yields an appreciable
improvement in transient response and a small change in
steady-state accuracy.
• It may accentuate high-frequency noise effects.
R(s)
G c(s )
G(s)
-
Series compensation
C(s)
1). Mathematical model:
1
s+
T s +1
T , 0 < a <1
Gc (s ) = K c a
=Kc
1
aT s +1
s+
aT
f (w) = Р Gc ( j w) = tan - 1 T w - tan - 1 aT w
L(dB)
20dB / dec
Bode diagram for
10lg 
0
1
T
Gc
T s +1
=
K c a aT s +1
20lg 
m

1
T
( )
90
m
0

where it can be calculated by letting d/d=0 that (see
Appendix)
1
wm =
1+
T a
Therefore, we have
Gc ( j wm )
K ca
sin f m
1
aT
1
2
= - 10lg a
T w(1 - a )
tan f (wm ) =
1 + aT 2w2 wm =
fm
1- a
=
2 a
1- a
=
1+ a
which relates the maximum phase-lead angle and the
value of .
The phase-lead compensator
V 2 (s )
R2
R1Cs +1
T s +1
G c (s ) =
=
Ч
=aЧ
R2
V 1 (s ) R1 + R 2
aT s +1
R1Cs +1
R1 + R 2
C
R2
a=
R1 + R 2
R1
V1
R2
V2
R1R 2
T =
C
R1 + R 2
2). Lead Compensation Design Procedure
The design procedure is introduced via the following
example.
Example. Consider the following system with openloop transfer function G(s):
4
s (s + 2)
It is desired to design a compensator for the system so
that the static velocity error constant Kv is 20 sec-1, the
phase margin is at least 500, and the gain margin is at
least 10 dB.
Step 1: Determine the gain K to meet the steady-state
performance specification.
In this example, it is required that Kv=20 sec-1. Hence from
4
2K
K*
=
s (s + 2) s (0.5s +1)
we obtain that K=10.
Step 2: Let
G1 (s ) = KG (s )
Using the gain K thus determined, draw a Bode diagram of
G1(jw), the gain adjusted but uncompensated system.
Evaluate the phase margin.
In this example,
20
G1 (s ) = KG (s ) =
s (0.5s +1)
Draw the Bode diagram of G1(j), from which it can be
evaluated that
20
L(wc )  20lg
 0  wc  6.32
2
0.5wc
g = 180 + f (wc ) = 180 - 90 - tan - 1 (0.5wc ) = 17
A phase margin of 170 implies that the system is quite
oscillatory. Thus, satisfying the specification on the steady
state yields a poor transient-response performance, and
therefore, a phase lead compensator should be designed.
20
s(0.5s  1)
Magnitude (dB)
40
Bode Diagram
[-20db]
20
0
[-40db]
-20
-40
-90
Phase (deg)
G( s) 
-135
170
-180
-1
10
10
0
10
Frequency (rad/sec)
1
10
2
Step 3: Determine the necessary phase-lead angle to be
added to the system. Add an additional 50 to 120 to the
phase-lead angle required, because the addition of the
lead compensator shifts the gain crossover frequency to
the right and decreases the phase margin.
In this example:
f m = 50 - 17 + 5 = 38
Let the phase lead compensator be of the form
T s +1
G c (s ) = K c a
aT s +1
Then the open loop transfer function becomes
T s +1
G c (s )G (s ) = K c a
G (s )
aT s +1
Let
K = K ca
G1 (s ) = KG (s )
where K has been obtained as 10. Then
T s +1
G c (s )G (s ) =
G 1 (s )
aT s +1
Step 4: 1)Determine the attenuation factor  by using the
formula (1)/(1+)=sinm (the main idea is to use m to
compensate for the lack of the phase margin.
In this example, let :
fm
1- a
= arcsin
= 38
1+a
Hence,
1- a
Ю
= sin 38 = 0.5 Ю a =0.24
1+a
- 10 lg a Ю 6.2 dB
which should correspond to m, where
wm =
1
T a
2) Select new gain crossover frequency such that
wc' = wm
which can be found from the point  such that
20lgG1(j)=10lg=6.2 dB, that is, 9 rad/sec:
wm =
1
T a
= wc' = 9
20 lg G
[ 20]
6.3
wc
10lg
9
6.2 dB = 10lga
00
wm = wc' = 9
-1800
=170
[ 40]

Step 5. Determine the corner frequencies of the lead
compensator.
In this example, with =0.24, we obtain
1
= 4.41
T
1
= 18.4
aT
Step 6. Using the value of K determined in step 1 and that
of  determined in step 4, calculate constant Kc from
Kc =K / a
In this example,
K = K c a = 10 Ю K c = 10 / 0.24 = 41.7
Therefore, the
compensator is
s + 4.41
G c (s ) = 41.7
s +18.4
0.227s +1
= 10
0.054s +1
where
K = K c a = 10
G c (s ) 0.227s +1
=
10
0.054s +1
Step 7. Check the gain margin to be sure it is satisfactory.
If not, repeat the design process by modifying the polezero locations of the compensator until a satisfactory
result is obtained.
In this example, since after the compensation, the phase
angle never crosses 1800 line, the gain margin always
satisfies the requirement. This completes the design.
Summary of the design procedure
1. Determine the necessary phase-lead angle to be
added to the system:
f m = g '- g + (50  120 )
where ’ is the required phase margin.
2. Let the maximum phase angle of the compensator
be equal to m:
fm
and let
1- a
= arcsin
Ю a
1+a
Ю - 10 lg a
wc' = wm
3. Determine c’ by
letting
20lg G1 ( jw = 10lg a
20 lg G
[ 20]
6.3
wc
wc'
10lga
20lg G1 ( jw) = 10lga
[ 40]
5. Determine the corner frequencies by
'
c
w = wm =
1
T a

Remark: If the required ’c is given, the design
procedure can be simplified as follows.
Example. Consider the following system with openloop transfer function G(s):
K
s (0.1s +1)
It is desired to design a compensator for the system so
that the static velocity error ess0.01, 450, and c 40.
Step 1: Determine the gain K to meet the steady-state
performance specification.
In this example, it is required that ess0.01. Hence K100.
Step 2: Using the gain K thus determined, draw a Bode
diagram of G (jw), the gain adjusted but uncompensated
system. Evaluate the phase margin.
In this example, with K=100,
100
G( s) 
s(0.1s  1)
From its Bode diagram (or by calculation),
20 lg G
 20dB / dec
 40dB / dec
44
0
1800
10
22 31
88
 6dB
g  17.90
wc  31rad / s
g  17.90
Obviously, a lead compensator is necessary.

Step 3: Determine the new crossover frequency. Since
wc we choose
it is required that =m40,
wc  44 rad / s
Step 4: Evaluate
20lg | G( jwc ) |
of the uncompensated system. In this example,
20lg | G( jwc ) | 6 dB
Step 5: Determine the corner frequencies of the
compensator. From
10 lg a  6 dB
we obtain
a  1/ 4
Let
wm  wc' 
1
 44
aT
we obtain
T  0.04544
Hence,
G c (s ) T s  1 0.04544s  1


K ca
aT s  1 0.01136s  1
This ends the design.
Bode diagrams for the system before and after the lead
compensation by using Matlab.
3. Lag Compensation
The primary function of a lag compensator is to provide
attenuation in high frequency range to give a system
sufficient phase margin.
1). Mathematical model:
1
s+
T s +1
T , b >1
Gc (s ) = K c b
=Kc
1
bT s +1
s+
bT
f (w) = Р Gc ( j w) = tan - 1 T w - tan - 1 bT w
G c (s ) T s +1
=
, b >1
K c b bT s +1
Bode diagram for
L(dB)
From which we
can obtain that
wm 
1
0
1
T
20dB / dec
m
10 lg 
1
T

20 lg 
(  )
T b

0
b 1
f (wm )  arcsin
b 1
f
90
Gc ( j w)
20lg
 20lg b
K c b w 1
T
m
The phase-lag compensator
The phase-lag compensation transfer function can be
obtained with the network shown in the following
Figure:
Gc ( s) 
R1
vi
R2
C
v0
Vo ( s )
R2Cs  1

Vi ( s) ( R1  R2 )Cs  1
R1  R 2
b
R2
T  (R1  R 2 )C
T s 1
Gc (s ) 
bT s  1
2). Lag Compensation Design Procedure
The design procedure is introduced via the following
example.
Example. Consider the following system with openloop transfer function G(s):
G (s )
G (s ) =
1
s (s +1)(0.5s +1)
It is desired to compensate the system so that the static
velocity error constant Kv is 5 sec-1, the
phase margin is at least 40°and the gain margin is at
least 10 dB.
Step 1: Determine gain K to satisfy the requirement on
the given static velocity error constant.
In this example, from
G1 (s ) = K
1
s (s +1)(0.5s +1)
we obtain that
Kv =K =5
Draw the Bode diagram of the gain-adjusted but
uncompensated system G1(j):= KG(j).
[40]
[20]
14 dB
[60]
=200
=200 implies that the gain-adjusted but uncompensated system is unstable.
Since in this example, there is no requirement about c, it
is possible to use a lag compensator to satisfy the required
performance specifications.
[20]
'
c
w
00
900
1800
[20]
[40]
[60]
Step 2: Find the new crossover frequency point c’ such
that
0
0
g   g * d  g * (5  12 )
where * is the required phase margin with addition of 50
to 120 to compensate for the phase lag of the lag
compensator.
In this example,
it can be found
that *=400
corresponds to
0.7. We choose  00
900
=120. Hence,
’=520.
1800
[20]
[40]
[60]
400
0.7
With ’=520, it can be found from the Bode diagram that
’ corresponds to 0.5 rad/s. Therefore, we choose c’=0.5
rad/s.
[20]
[40]
00
900
[60]
520
400
1800
0.5 0.7
Step 3: Choose the corner frequency  = 1/T
(corresponding to the zero of the lag compensator) 1
octave to 1 decade below the new gain crossover frequency.
In this example, we choose
1
w = = 0.1
T
1/T
20lg
Step 4: Determine  such that
20 lg b  20 lg G1 ( j wc' )  0
which will bring the magnitude curve down to 0 dB at
this new gain crossover frequency. Then determine the
corner frequency =1/T.
In this example, it can be measured that
20 lg G1 ( j wc' )  20 dB
Therefore,
1
20 lg + 20 lg10 = 0 Ю b = 10
b
[20]
[40]
20lg
00
[60]
520
900
400
1800
0.1
0.5 0.7
Since we have obtained that =10 and
1
w = = 0.1
T
1
= 0.01
the corner frequency w =
bT
Step 5: Finally, let
K
K = K cb Ю K c =
b
In this example,
5
K = K c b = 5 Ю K c = = 0.5
10
Therefore, the compensator is
T s +1
0.1s +1
Gc (s ) = K c b
=5
bT s +1
0.01s +1
5(0.1s +1)
Gc (s )G (s ) =
s (s +1)(0.5s +1)(0.01s +1)
[40]
[20]
c’
14 dB
[60]
’=400
=200
Example. Consider a unity-feedback system with its
open-loop transfer function as
G (s ) 
1
s (0.5s  1)
It is required that the steady-state error for unit ramp
input be 0.05 and the phase margin of the system be at
least 450.
Step 1: According to the requirement of the steady-state
error, we need that
1
G1 (s )  K
s (0.5s  1)
with
K  K v  20
Then, plotting the Bode diagram of the gain-adjusted
but uncompensated transfer function yields
20 dB /dec
40 dB /dec
20
G1 (s ) 
s (0.5s  1)
from which it can be obtained that
wc  6.3 rad /s
f (wc )  900  tan 1 (0.5wc )  1620
Therefore, the phase margin is
0
g  18
.
20 dB / dec
40 dB / dec
20
G1 (s ) 
s (0.5s  1)

Step 2: With ’=450 and an additional 50 to compensate
for the phase-lag, we obtain
wcў = 1.5rad / s
f (wc )  1300
at which it can be found that 20lgG1(jc’)=20dB.
20 dB / dec
40 dB / dec
20
G1 (s ) 
s (0.5s  1)
wcў
To make c’ be the new crossover frequency, the following
equation should be satisfied
- 20 lg b = - 20 lg G1 ( j wc' ) = - 20 dB
Ю b = 10
[20]
c’
20lgG1(jc’)=20dB
20lg
00
900
1800
0.15
c’=1.5
[40]
Step 3: Determine corner frequencies of the
compensator:
G c (s ) T s +1
=
K c b bT s +1
b = 10
1
bT
20 dB / dec
1
Choose
= 0.1Чwcў = 0.15
T
1
Then
= 0.015
bT
Therefore, the compensator is
G c (s ) = 20
6.67s +1
6.67s +1
= K cb
66.7s +1
66.7s +1
1
T

20 lg b

Step 4: Finally, let
K
K = K cb Ю K c =
b
In this example,
20
K = K c b = 20 Ю K c =
=2
10
The open-loop transfer function of the compensated
system is then obtained as
20
6.67s  1
G (s )Gc (s ) 

s (0.5s  1) 66.7s  1
1
s(0.5s  1)
Gc ( s )  20
Bode Diagram
6.67 s  1
66.7 s  1
100
Magnitude (dB)
[-20db]
50
[-40db]
[-20db]
0
[-40db]
-50
-90
Phase (deg)
G( s) 
-135
c '  1.5
-180
-2
10
10
-1
10
Frequency (rad/sec)
0
10
1
The time response of the uncompensated and
compensated system are show in the following figure:
uncompensated
compensated
A Few Comments on Lag Compensation
1. Lag compensators are essentially low-pass filters.
Therefore, lag compensation permits a high gain at low
frequencies.
2. The closed-loop pole located near the origin gives a very
slowly decaying transient response, although its
magnitude will become very small because the zero of
the lag compensator will almost cancel the effect of this
pole. However, the transient response (decay) due to this
pole is so slow that the settling time will be adversely
affected.
3. c’<c  ts.
3. The phase lag-lead compensator



The lead compensator improves settling time
and phase margin, but increases the bandwidth.
However, phase-lag compensator when applied
properly improves phase margin but usually
results in a longer settling time.
It is natural, therefore, to consider using a
combination of the lead-lag compensator, so
that the advantages of both schemes can be
utilized.
The transfer function of a lag-lead compensator can be
written as
ж 1 +T s цж 1 +T s ц
1 чз
2
ч
Gc (s ) = K c зз
чз
ч
1
+
a
T
s
1
+
b
T
s
и
1 ши
2 ш
lead
1
(b > 1, a =
b
)
lag
It is usually assumed that the two corner frequencies of
the lag portion are lower than the two corner frequencies
of the lead portion.
L(dB)
10
0
1
T2
1
T2
1
T1
1
 T1

10
(  )
90
1
0
90

Example. Consider a unity-feedback system with its
open-loop transfer function as
K
G (s ) 
s (0.5s  1)(s  1)
It is desired that the static velocity error constant be 10/s,
the phase margin be 500, and the gain margin be 10 dB or
more.
According to the requirement of the steady-state error, we
need that
K  K v  10
Then, plotting the Bode diagram of the gain-adjusted but
uncompensated transfer function yields
[-40]
1.5
[-40]
=320 shows that the system is unstable.
A lag-lead compensator is utilized:
[-40]
After the compensation:
[-40]
For more detail, see p. 513.
4. Feedforward Compensation
We consider the following system
G c (s )
C
R
-
E
G(s)
where Gc(s) is called feedforward compensator and can
be used to improve the steady-state error while keeping
the system stability. The closed-loop transfer function
is
C (s )
G (s )
= (1 +G c (s ))
R (s )
1 +G (s )
It is required that
E (s ) = R (s ) - C (s ) = 0
Since
E (s ) = R (s ) - C (s ) =
1 - G (s ) G c (s )
1 +G (s )
R (s )
a natural choice seems
G c (s ) = G - 1 (s )
In that case
E (s ) = R (s ) - C (s ) = 0
However,
1
G c (s ) =
G (s )
sometimes is difficult to be physically realized,
especially when G(s) has zeros in right half s plane.
Therefore, approximation compensation is necessary.
Example. The original system is
R
G (s )
C
To improve the steady-state performance, an integrator can
be used. However, such a compensator could cause
instability:
R
K
s
G (s )
C
To solve such a conflicting requirement, feedforward
compensator can be utilized.
Example. The block diagram of system is
Gc (s )
R
C
0.855
s (0.1s +1)(0.05s +1)
Design a Gc(s) as simple as possible such that there is no
steady-state error when r(t)=t·1(t).
Solution: The original system is Type 1 system.
Therefore, without compensation, the static velocity error
is nonzero. Since
E (s ) = R (s ) - C (s ) =
1 - G (s ) G c (s )
1 +G (s )
R (s )
70
We can choose
Gc (s ) =G - 1 (s )
which, however, is not the simpler one. An alternative is
to write
1  G (s )G c (s )
E (s )  C (s )  R (s ) 
R (s )
1  G (s )
s (0.1s  1)(0.05s  1) G c (s )0.855 1

 2
s (0.1s  1)(0.05s  1)  1
s
Using Final value theorem,
ess  lims 0 sE (s )
s (0.1s  1)(0.05s  1) G c (s )0.855 1
 lims 0

s (0.1s  1)(0.05s  1)  1
s
0.005s 3  0.01s 2  s Gc (s )0.855 1
ess  lims 0

s (0.1s  1)(0.05s  1)  1
s
Let
s
G c (s )0.855  s  G c (s ) 
0.855
Substituting Gc(s) into E(s) yields
0.005s 3  0.01s 2
1
ess  lims 0

s (0.1s  1)(0.05s  1)  1 s
0.005s 2  0.01s
 lims 0
0
s (0.1s  1)(0.05s  1)  1
Gc(s)=s/0.885 is obviously simpler than Gc(s)=1/G(s).
This completes the design.
Appendix
T w(1  a )
f (w)  Gc ( j w)  tan T w  tan aT w  tan
1  aT 2 w2
1
1
1
d f (w)
d
0 
 tan f (w)  0
dw
dw
1
d T w(1  a )  T (1  a )[1  aT 2w2 ]
wm 


0

T a
d w 1  aT 2w2 
(1  aT 2w2 )2
Therefore,
T w(1  a )
tan f (wm ) 
1  aT 2w2
wm 
1
aT
1 a

2 a