Transcript Slide 1

Frequency Response Analysis
Chapter 13
Sinusoidal Forcing of a First-Order Process
For a first-order transfer function with gain K and time constant τ ,
the response to a general sinusoidal input, x  t   A sin ωt is:
y t  

1
KA
2 2
ω τ
ωτet / τ  ωτ cos ωt  sin ωt

(5-25)
Note that y(t) and x(t) are in deviation form. The long-time
response, yl(t), can be written as:
KA
y t  
sin  ωt  φ  for t  
(13-1)
ω2 τ 2  1
where:
φ   tan 1  ωτ 
1
Chapter 13
Figure 13.1 Attenuation and time shift between input and output
sine waves (K= 1). The phase angle φ of the output signal is given
by φ  Time shift / P  360 , where t is the (period) shift and P
is the period of oscillation.
2
Frequency Response Characteristics of
a First-Order Process
Chapter 13
For x(t )  A sin wt , y  t   Aˆ sin  ωt  φ  as t   where :
Aˆ 
KA
ω2 τ 2  1
and φ   tan 1  ωτ 
1. The output signal is a sinusoid that has the same frequency, w,
as the input.signal, x(t) =Asinwt.
2. The amplitude of the output signal, Â , is a function of the
frequency w and the input amplitude, A:
Aˆ 
KA
ω2 τ 2  1
(13-2)
3. The output has a phase shift, φ, relative to the input. The
amount of phase shift depends on w.
3
Chapter 13
Dividing both sides of (13-2) by the input signal amplitude A
yields the amplitude ratio (AR)
Aˆ
K
AR  
(13-3a)
A
ω2 τ 2  1
which can, in turn, be divided by the process gain to yield the
normalized amplitude ratio (ARN)
AR N 
1
ω2 τ 2  1
(13-3b)
4
Shortcut Method for Finding
the Frequency Response
Chapter 13
The shortcut method consists of the following steps:
Step 1. Set s=jw in G(s) to obtain G  jω  .
Step 2. Rationalize G(jw); We want to express it in the form.
G(jw)=R + jI
where R and I are functions of w. Simplify G(jw) by
multiplying the numerator and denominator by the
complex conjugate of the denominator.
Step 3. The amplitude ratio and phase angle of G(s) are given
by:
AR  R 2  I 2
Memorize 
  tan 1 ( I / R)
5
Example 13.1
Find the frequency response of a first-order system, with
Chapter 13
1
G s 
τs  1
(13-16)
Solution
First, substitute s  jω in the transfer function
1
1
G  jω  

τjω  1 jωτ  1
(13-17)
Then multiply both numerator and denominator by the complex
conjugate of the denominator, that is,  jωτ  1
 jωτ  1
 jωτ  1
G  jω  
 2 2
 jωτ  1  jωτ  1 ω τ  1

1
ω τ 1
2 2
j
 ωτ 
ω τ 1
2 2
 R  jI
(13-18)
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R
where:
Chapter 13
I
1
(13-19a)
ω τ 1
ωτ
2 2
(13-19b)
ω τ 1
2 2
From Step 3 of the Shortcut Method,
2
1

  ωτ 
AR  R 2  I 2   2 2    2 2 
 ω τ 1   ω τ 1 
or
AR 
Also,


2
2 2
 ω τ 1
1  ω2 τ 2
1
ω2 τ 2  1
2
(13-20a)
1 
I
φ  tan    tan 1  ωτ    tan 1  ωτ  (13-20b)
R
7
Chapter 13
Complex Transfer Functions
Consider a complex transfer G(s),
Ga  s  Gb  s  Gc  s 
G s 
G1  s  G2  s  G3  s 
Substitute s=jw,
Ga  jω  Gb  jω  Gc  jω 
G  jω  
G1  jω  G2  jω  G3  jω 
(13-22)
(13-23)
From complex variable theory, we can express the magnitude and
angle of G  jω  as follows:
G  jω  
Ga  jω  Gb  jω  Gc  jω 
(13-24a)
G1  jω  G2  jω  G3  jω 
G  jω   Ga  jω   Gb  jω   Gc  jω  
 [G1  jω   G2  jω   G3  jω  
]
(13-24b)
8
Chapter 13
Bode Diagrams
• A special graph, called the Bode diagram or Bode plot,
provides a convenient display of the frequency response
characteristics of a transfer function model. It consists of
plots of AR and φ as a function of w.
• Ordinarily, w is expressed in units of radians/time.
Bode Plot of A First-order System
Recall:
AR N 
1
ω2 τ 2  1
and φ   tan 1  ωτ 
 At low frequencies ( ω  0 and ω
1) :
AR N  1 and   
 At high frequencies ( ω   and ω
1) :
AR N  1/ ωτ and   
9
Chapter 13
Figure 13.2 Bode diagram for a first-order process.
10
Chapter 13
• Note that the asymptotes intersect at ω  ωb  1/ τ, known as the
break frequency or corner frequency. Here the value of ARN
from (13-21) is:
AR N  ω  ωb  
1
 0.707
11
(13-30)
• Some books and software defined AR differently, in terms of
decibels. The amplitude ratio in decibels ARd is defined as
AR d  20 log AR
(13-33)
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Integrating Elements
Chapter 13
The transfer function for an integrating element was given in
Chapter 5:
Y s K
G s 

(5-34)
U s s
AR  G  jω  
K
K

jω ω
φ  G  jω  K       90
(13-34)
(13-35)
Second-Order Process
A general transfer function that describes any underdamped,
critically damped, or overdamped second-order system is
K
G s  2 2
(13-40)
τ s  2ζτs  1
12
Substituting s  jω and rearranging yields:
AR 
K

Chapter 13
1 ω τ
2 2

2
(13-41a)
  2ωτζ 
 2ωτζ 
φ  tan 1 

1  ω 2 τ 2 
2
(13-41b)
Figure 13.3 Bode diagrams for second-order processes.
13
Time Delay
Chapter 13
Its frequency response characteristics can be obtained by
substituting s  jω ,
G  jω  e jωθ
(13-53)
which can be written in rational form by substitution of the
Euler identity,
G  jω  e jωθ  cos ωθ  j sin ωθ
(13-54)
From (13-54)
AR  G  jω   cos 2 ωθ  sin 2 ωθ  1
(13-55)
 sin ωθ 
φ  G  jω   tan 1  

cos
ωθ


or
φ  ωθ
(13-56)
14
Chapter 13
Figure 13.6 Bode diagram for a time delay, e θs.
15
Chapter 13
Figure 13.7 Phase angle plots for e θs and for the 1/1 and 2/2
Padé approximations (G1 is 1/1; G2 is 2/2).
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Process Zeros
Consider a process zero term,
Chapter 13
G  s   K ( sτ  1)
Substituting s=jw gives
G  jω   K ( jωτ  1)
Thus:
AR  G  jω   K ω2 τ 2  1
φ  G  jω    tan 1  ωτ 
Note: In general, a multiplicative constant (e.g., K) changes
the AR by a factor of K without affecting φ .
17
Chapter 13
Frequency Response Characteristics of
Feedback Controllers
Proportional Controller. Consider a proportional controller with
positive gain
Gc  s   K c
(13-57)
In this case Gc  jω  Kc , which is independent of w.
Therefore,
AR c  Kc
(13-58)
and
φc  0
(13-59)
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Chapter 13
Proportional-Integral Controller. A proportional-integral (PI)
controller has the transfer function (cf. Eq. 8-9),

 τI s 1 
1 
Gc  s   K c 1 
  Kc 

τ
s
τ
s
I 

 I 
Substitute s=jw:
(13-60)

 jwτ I  1 

1 
1
Gc  jw  K c 1 
  Kc 
  Kc 1 
 τ I jw 
 j wτ I 
 τI w

j

Thus, the amplitude ratio and phase angle are:
AR c  Gc  jω   K c 1 
1
 ωτ I 
2
 Kc
 ωτ I 
2
1
ωτ I
φc  Gc  jω  tan 1  1/ ωτ I   tan 1  ωτ I   90
(13-62)
(13-63)
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Chapter 13
 10 s  1 
G
s

2
Figure 13.9 Bode plot of a PI controller, c  


10
s


20
Ideal Proportional-Derivative Controller. For the ideal
proportional-derivative (PD) controller (cf. Eq. 8-11)
Chapter 13
Gc  s   K c 1  τ D s 
(13-64)
The frequency response characteristics are similar to those of a
LHP zero:
AR c  Kc
 ωτ D 
2
1
φ  tan 1  ωτ D 
(13-65)
(13-66)
Proportional-Derivative Controller with Filter. The PD controller
is most often realized by the transfer function
 τDs 1 
Gc  s   K c 

ατ
s

1
 D

(13-67)
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Chapter 13
Figure 13.10 Bode
plots of an ideal PD
controller and a PD
controller with
derivative filter.
Idea: Gc  s   2  4 s  1
With Derivative
Filter:
 4s  1 
Gc  s   2 

 0.4 s  1 
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PID Controller Forms
Chapter 13
Parallel PID Controller. The simplest form in Ch. 8 is


1
Gc  s   K c 1 
 τDs 
 τ1s

Series PID Controller. The simplest version of the series PID
controller is
 τ1s  1 
Gc  s   K c 
(13-73)
  τ D s  1
 τ1s 
Series PID Controller with a Derivative Filter.
 τ1s  1  τ D s  1 
Gc  s   K c 


 τ1s   τ D s  1 
23
Chapter 13
Figure 13.11 Bode
plots of ideal parallel
PID controller and
series PID controller
with derivative filter
(α = 1).
Idea parallel:
1


Gc  s   2 1 
 4s 
 10s

Series with
Derivative Filter:
 10s  1  4 s  1 
Gc  s   2 


10
s
0.4
s

1



24
Nyquist Diagrams
Chapter 13
Consider the transfer function
1
G s 
2s  1
(13-76)
with
AR  G  jω  
1
 2ω   1
2
(13-77a)
and
φ  G  jω   tan 1  2ω
(13-77b)
25
Chapter 13
Figure 13.12 The Nyquist diagram for G(s) = 1/(2s + 1)
plotting Re  G  jω   and Im  G  jω   .
26
Chapter 13
Figure 13.13 The Nyquist diagram for the transfer
function in Example 13.5:
5(8s  1)e6 s
G( s) 
(20s  1)(4s  1)
27