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, Aˆ , 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 ( R / I )
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)
6
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


2
2 2
 ω τ 1
1  ω2 τ 2
AR 
Also,
φ  tan
1
ω2 τ 2  1
2
(13-20a)
1 
I
1
1

tan

ωτ


tan


 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
9
Chapter 13
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  0 and ω
1) :
AR N  1/ ωτ and   
10
Chapter 13
Figure 13.2 Bode diagram for a first-order process.
11
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)
12