Chapter 6: Frequency Domain
Anaysis
Instructor: Dr. Cailian Chen
Outline
1.
2.
3.
4.
5.
6.
Introduction of Frequency Response
Frequency Response of the Typical Elements
Bode Diagram of Open-loop System
Nyquist-criterion
Frequency Response Based System Analysis
Frequency Response of Closed-loop Systems


Concept
Graphical
method
Analysis
6-1 Introduction
Key problem of control: stability and system performance
Question: Why Frequency Response
Analysis:
1. Weakness of root locus method relies on the existence of
open-loop transfer function
2. Weakness of time-domain analysis method is that time
response is very difficult to obtain
 Computational complex
 Difficult for higher order system
 Difficult to partition into main parts
 Not easy to show the effects by graphical method
Frequency Response Analysis
Three advantages:
* Frequency response(mathematical modeling) can be
obtained directly by experimental approaches.
* Easy to analyze effects of the system with sinusoidal signals
* Convenient to measure system sensitivity to noise and
parameter variations
However, NEVER be limited to sinusoidal input
* Frequency-domain performances  time-domain
performances
Frequency domain analysis is a kind of (indirect method)
engineering method. It studies the system based on frequency
response which is also a kind of mathematical model.
Frequency Response
Example 5.1: RC circuit
R
xi (t)
xi (t)
C
xo(t)
t
ui  A sin(t  0 )
U 1m
U i (s)  2
s 2
U 1m
1
U 0 ( s) 
 2
Ts  1 s   2
Transfer function
G s 
1
1

RCs  1 Ts  1
By using inverse Laplase transform
U 1mT  Tt
U 1m
u 0 (t ) 
e 
sin(t   )
2 2
2
2
1 T 
1 T 
  -arctg T 
Steady state response
Transient response
Steady state response of u0
lim u0 
t 
U1m
1 T 
 U 1m
2
2
sin(t   )
1
1
sin(t  
)
1  jT
1  jT
When the input to a linear time-invariant (LTI) system is
sinusoidal, the steady-state output is a sinusoid with the same
frequency but possibly with different amplitude and phase.
Definition：Frequency response (or characteristic) is the ratio
of the complex vector of the steady-state output versus
sinusoidal input for a linear system.
1
G( j ) 
 A( )e j ( )
1  jT
1
1
A() 

;
2
2
1  jT
1  T
1
 ( )  (
)  arctgT
1  jT
0

Magnitude response
Phase response
The output has same magnitude and phase with input
Magnitude will be attenuated and phase lag is increased.
Magnitude of output versus input
Magnitude characteristic
Phase error of output and input
Phase characteristic
Generalized to linear time-invariant system
Transfer function of closed-loop system
G( s) 
C ( s) N ( s)
N (s)


R( s) D( s) ( s  p1 )(s  p2 )
(s  pn )
where p1 ,, pn are different closed-loop poles.
Given the sinusoidal input
r (t )  Ar sin t
Ar
R( s)  2
s  2
Ar
Ar
N (s)
C (s)  G(s)  2

 2
2
s 
D( s ) s   2
bn
b1
b2
a
a




 ... 
s  j s  j s  p1 s  p2
s  pn
c(t )  ae jt  ae jt  b1e  p1t  b2e  p2t  ...  bn e  pnt
n
  bi e  pit  (ae jt  ae jt )
i 1
(t  0)
 ct (t )  cs (t )
Transient response
Steady state response
For a stable closed-loop system, we have
- pi <0
Ar
Ar G( j )
a  G( s)  2
 ( s  j ) |s  j  
2
s 
2j
Ar
Ar G ( j )
a  G(s)  2
 ( s  j ) |s  j 
2
s 
2j
G( j) | G( j) | e jG( j )
G( j) | G( j) | e jG( j ) | G( j) | e jG( j )
Furthermore, we have
cs (t )  ae  jt  ae jt
e j (t G ( j ))  e j (t G ( j ))
 Ar | G ( j ) |
2j
 Ar | G ( j ) | sin(t  G ( j ))
 Ac sin(t   )
The magnitude and phase of steady state are as follows
Ac  Ar | G( j) |;
  G( j)
By knowing the transfer function G(s) of a linear system, the
magnitude and phase characteristic completely describe the steadystate performance when the input is sinusoid.
Frequency-domain analysis can be used to predict both timedomain transient and steady-state system performance.
Relation of transfer function and frequency characteristic of
LTI system (only for LTI system)
G ( s ) s  j  G ( j ) | G ( j ) | e jG ( j )
 Substitute s=jω into the transfer function
F s   L f t    f t e  st dt
Laplace transform
f t   L F s   
Inverse Laplace transform


1
c 
c 
F s e st ds

Fourier transform
F  j   F  f  t     f  t  e  jt dt

1 
1
jt
f  t   F  F  s   
F
j

e
d  Inverse Fourier transform




2 j
Vector of frequency characteristic
It is a complex vector, and has three forms:
jv
G(j)
V ()
Algebraic
form
G  j   U    jV  
A()
Polar form
()
0
U() u
Exponential form
G( j)  G( j) G( j)  A() ()
G( j)  G( j) e jG( j )  A()e j ( )
A    G  j   U 2    V 2  
 V   
    arctan 

U





U    A  cos  
V    A  sin   
We have learned following mathematical models：
differential equation, transfer function and frequency
response
Differential
equation
d
s
dt
d
j 
dt
Transfer
function
L{g (t )}
Frequency
response
s  j
L1{G(s)}
Impulsive
response
14
Example 5.2: Given the transfer function
C ( s)
1
G( s) 
 2
R( s) s  3s  4
Differential equation：
d 2c(t )
dc(t )

3
 4c(t )  r (t )
2
dt
dt
Frequency response:
G( j ) 
c( j )
1
1


s( j ) ( j )2  3( j )  4 4   2  3 j
15
6-2 Frequency Characteristic of Typical
Elements of system
How to get frequency characteristic?
Input a sinusoid signal to the control system
Measure the amplitude and phase of the steady-state output
Change frequency
Get the amplitude ratio of the output versus input
Get the phase difference between the output and input
Are the measured data enough？
N
y
Data processing
16
Nyquist Diagram
• Magnitude characteristic diagram
A-ω plot
• Phase characteristic diagram
-ω plot
• Gain-phase characteristic diagram
Bode
diagram
Nyquist diagram
Polar form or algebraic form: A and  define a vector for a particular
frequency ω.
G( j)  G( j) e j ( )  X ()  jY ()
ω
0
1/(2τ)
A(ω)
1
0.89
(ω)
0
-26.6
1/τ
2/τ
3/τ
4/τ
0.70 0.45 0.32 0.24 0.2
7
-45 -63.5 -71.5 -76 -78.7
jY ( )
 
0
5/τ
 ( )
A( )
1
  0 X ( )

Nyquist Diagram of RC Circuit
∞
0
-90
Bode Diagram
• Bode Diagram: Logarithmic plots of magnitude
response and phase response
• Horizontal axis: lgω (logarithmic scale to the base of
10)（unit: rad/s）
• Log Magnitude
In feedback-system, the unit commonly used for the logarithm of
the magnitude is the decibel (dB)
L()  20lg G( j)  20lg A()
Property 1: As the frequency doubles, the decibel value increases
by 6 dB.
As the frequency increases by a factor of 10, the decibel value
increases by 20 dB.
• Log scale and linear scale

0 1 2 3 4 5 6 7 8 9 10
one decade
（a)线性分度
0
L

1
一倍频程
2
3
一倍频程
4 5 6 7 8 9 10
octave octave
一倍频程
20
30 40 50 60 80 100
一倍频程
十倍频程
十倍频程
十倍频程
one decade
one decade
（b)对数分度
• Notation：
– Logarithmic scale use the nonlinear compression of horizontal
scale. It can reflect a large region of frequency variation.
Especially expand the low-frequency range.
– Logarithmic magnitude response simplify the plotting.
Multiplication and division are changed into addition and
subtraction.
– We cannot sketch ω =0 on the horizontal scale. The smallest ω
can be determined by the region of interest.
L( )(dB)
0.1
1
10
• Given T=1，plot the Bode
Diagram by using Matlab
bode([1],[1 1])

 ( )()
0.1
1
10

Frequency Characteristic of Typical Elements

Seven typical elements
L( )(dB)
1.Proportional element
Frequency characteristic G( j )  K
20lgK
0dB
 ( )()
It is independent on ω.
The corresponding magnitude and phase
characteristics are as follows:

0°

jY ( )
 A( )  K
 ( )  0

K
0
X ( )
 L( )  20 lg K
 ( )  0

2. Integration element
Frequency characteristic
1
1
G( j ) 
 e
j 
j

2
The corresponding magnitude and phase
characteristics are as follows:
 A( )  1 
 ( )  90

 L( )  20 lg 
 ( )  90

L( )(dB)
jY ( )
0dB
 
0
X ( )

°
°
-20dB/dec


 ( )()

3. Derivative Element
Frequency characteristic
G( j )  j  e
j

2
The corresponding magnitude and phase
characteristics are as follows:
 A( )  
 ( )  90

 L( )  20 lg 
 ( )  90

L( )(dB)
jY ( )
20dB/dec
0dB


 0
0
X ( )

 ( )()
°
°

4.Inertial Element
Frequency characteristic
1
G ( j ) 
1  jT
Magnitude and phase responses
1

 A( ) 

1   2T 2

 ( )  arctgT
Rewrite it into real and imaginary parts
1
T
G ( j ) 

j
 X ( )  jY ( )
2 2
2 2
1  T
1  T
jY ( )
[ X ()  0.5]  Y ()  0.5
2
Nyquist
2
2
diagram is half of
the circle with center at (0.5,0)
and radius 0.5.
 
T 0
1   2T 2
1
1   2T 2
0.5
 ( )
A( )
1
  0X ( )

Log magnitude and phase characteristics are as follows:
1

L
(

)

20lg
 20lg 1   2T 2


1   2T 2

 ( )  arctgT
Low-frequency region： ω<<1/T, L(ω)≈-20lg1=0
Asymptote
High-frequency region： ω>>1/T, L(ω)≈-20lgωT
L( )(dB)
The frequency where the
low- and high-frequency
asymptotes
meet
is
called
the
break
frequency (ω＝1/T).
渐近线
交接频率
渐近线

精确曲线
0.1/T
1/T
10/T
 ( )()

The true modulus has a
value of L()  10lg (1+1)  -3dB
0.1/T
1/T
10/T
• Remarks：
①
The error of true modulus and asymptote
ωT
ΔL(ω)
0.1
0.2
0.5
1
2
5
－0.04
－0.17
－0.97
－3.01
－0.97
10
－0.17 －0.04
It can be seen that the error at the break
frequency is biggest.
② (ω) is symmetrical at all rotations about the point
ω＝1/T,(ω)=-45°
ωT
0.1
0.2
0.25
0.5
1
2
3
4
5
8
10
(ω)
6
 4
27
45
63
72
76
79
83
84
5. First Derivative Element
Frequency characteristic
G( j )  1  jT
 A( )  1   2T 2

 ( )  arctgT
Nyquist Diagram
 L( )  20lg 1   2T 2

 ( )  arctgT
Inertial Element
First Derivative Element
Frequency characteristics are the inverse each other
• Log magnitude characteristic is symmetrical about the
line of 0dB
• Phase characteristic is symmetrical about 0 degree
6. Second order oscillation element (Important)
G ( j ) 
1
2

1  2  j 2
n
n
0
j ω=∞
ω=0
ζ=0.2—0.8
-0.5
  0, G( j 0)  10o
1
  n ,G( jn ) 
  90o
2
  , G( j)  0  180o
-1
-1.5
-0.5
0
0.5
1
Nyquist Diagram
1.5
L( )  20 lg (1   2 /  n ) 2  4 2 ( /  n ) 2
2
 ( )  arctg
2 / n
1  ( / n ) 2
• For ω<<ωn, L(ω)≈0
• For ω>>ωn ,
L(ω)≈-40lgω/ωn
=-40(lg ω-lgωn)
(dB)
40
20
0 0.1
-20
40dB/dec
1
10 ω/ωn
-40dB/dec
1
10 ω/ωn
(o)
180
0
0.1
-180
Bode Diagram
Bode Diagram
Remarks:
The low- and high-frequency asymptote intersect
at ω= ωn，i.e. the undamped natural frequency。
 Unlike a first-order element which has a singlevalued deviation between the approximation and
accurate moduli, the discrepancy depends upon the
damping ratio ξ.
The true magnitude may be below or above the
straight-line approximate magnitude.
 The resonant peak Mr is the maximum value of
L(ω)
7. Delay Element
G( j )  e
 L( )  0
 ( )  T

 j T
 A( )  1
 ( )  T

jY ( )
1
0
L( )(dB)
L( )  0

 ( )()
0

  0 X ( )
Nyquist Diagram is a circle
°
