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Feedback Control Systems. Chapter 6. The Frequency-Response Design Method. Chapter 6. The Frequency-Response Design Method. Frequency Response. A linear system’s response to sinusoidal inputs is called the system’s frequency response. - PowerPoint PPT Presentation
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Lecture 8
Feedback Control Systems
President University Erwin Sitompul FCS 8/1
Dr.-Ing. Erwin SitompulPresident University
http://zitompul.wordpress.com
2 0 1 3
President University Erwin Sitompul FCS 8/2
Chapter 6
The Frequency-Response Design Method
Feedback Control Systems
President University Erwin Sitompul FCS 8/3
Frequency Response A linear system’s response to sinusoidal inputs is called the
system’s frequency response. Frequency response can be obtained from knowledge of the
system’s pole and zero locations.
( )( ),
( )
Y sG s
U s
0( ) sin( ) 1( ),u t A t t
02 2
0
( ) .( )A
U s u ts
L
02 2
0
( ) ( ) .A
Y s G ss
Consider a system described by
where
With zero initial conditions, the output is given by
Chapter 6 The Frequency-Response Design Method
President University Erwin Sitompul FCS 8/4
Frequency Response
*0 01 2
1 2 0 0
( ) .... ,n
n
Y ss p s p s p s j s j
Assuming all the poles of G(s) are distinct, a partial fraction expansion of the previous equation will result in
1 21 2 0 0( ) .... 2 sin( ), 0,np tp t p t
ny t e e e t t
01
0
Im( )tan .
Re( )
where
Transient response Steady-state response
1( ) ,
1G s
s
( ) sin(10 )u t t
Chapter 6 The Frequency-Response Design Method
President University Erwin Sitompul FCS 8/5
( )( )
( )
Y sG s
U s
Frequency ResponseExamine
00( ) ,( )
s jG j G s
0 0 0( ) ( ) ( ),G j G j G j
2 2
0 0 0( ) Re ( ) Im ( )G j G j G j
010
0
Im ( )( ) tan
Re ( )
G jG j
G j
0( ) 0.U j A
0 0 0( ) ( ) ( ),Y j G j U j
,M
.
But the input is
Therefore
0( ) ,Y j AM 0( ) sin( ).y t AM t • Meaning?
Magnitude
Phase
Chapter 6 The Frequency-Response Design Method
President University Erwin Sitompul FCS 8/6
Frequency Response A stable linear time-invariant system with transfer function
G(s), excited by a sinusoid with unit amplitude and frequency ω0, will, after the response has reached steady-state, exhibit a sinusoidal output with a magnitude M(ω0) and a phase Φ(ω0) at the frequency ω0.
The magnitude M is given by |G(jω)| and the phase Φ is given by G(jω), which are the magnitude and the angle of the complex quantity G(s) evaluated with s taking the values along the imaginary axis (s = jω).
The frequency response of a system consists of the frequency functions |G(jω)| and G(jω), which describe how a system will respond to a sinusoidal input of any frequency.
Chapter 6 The Frequency-Response Design Method
President University Erwin Sitompul FCS 8/7
Frequency Response Frequency response analysis is interesting not only because it
will help us to understand how a system responds to a sinusoidal input, but also because evaluating G(s) with s taking on values along jω axis will prove to be very useful in determining the stability of a closed-loop system.
As we know, jω axis is the boundary between stability and instability. Therefore, evaluating G(jω) along the frequency band will provide information that allows us to determine closed-loop stability from the open-loop G(s).
Chapter 6 The Frequency-Response Design Method
President University Erwin Sitompul FCS 8/8
Frequency Response
1( ) , 1.
1
TsD s K
Ts
Given the transfer function of a lead compensation
(a) Analytically determine its frequency response characteristics and discuss what you would expect from the result.
1( ) ,
1
TjD j K
Tj
2
1 1
2
1 ( )tan ( ) tan ( ).
1 ( )
TT TK
T
M = DΦ
• At low frequency, ω→0, M→K, Φ→0.• At high frequency, ω→∞, M→K/α, Φ→0.• At intermediate frequency, Φ > 0.
Chapter 6 The Frequency-Response Design Method
President University Erwin Sitompul FCS 8/9
Frequency Response(b) Use MATLAB to plot D(jω) with K = 1, T = 1, and α = 0.1 for
0.1 ≤ ω ≤ 100 and verify the prediction from (a).
1( ) .
0.1 1
sD s
s
Using bode(num,den), in this case bode([1 1],[0.1 1]), MATLAB produces the frequency response of the lead compensation.
Chapter 6 The Frequency-Response Design Method
President University Erwin Sitompul FCS 8/10
y t
nt
2
2 2( ) ,
2n
n n
G ss s
Frequency Response For second order system having the transfer function
we already plotted the step response for various values of ζ.
• The damping and rise time of a system can be determined from the transient-response curve.
Chapter 6 The Frequency-Response Design Method
President University Erwin Sitompul FCS 8/11
2
1( ) ,
( ) 2 ( ) 1n n
G ss s
Frequency Response The corresponding frequency response of the system can be
found by replacing s = jω
This G(jω) can be plotted along the frequency axis, for various values of ζ.
2
1( ) .
( ) 2 ( ) 1n n
G jj j
Chapter 6 The Frequency-Response Design Method
President University Erwin Sitompul FCS 8/12
Frequency Response
• The damping of a system can be determined from the peak in the magnitude of the frequency response curve.
• The rise time can be estimated from the bandwidth, which is approximately equal to ωn.
The transient-response curve and the frequency-response curve contain the same information.
?
Chapter 6 The Frequency-Response Design Method
President University Erwin Sitompul FCS 8/13
Frequency ResponseBandwidth is defined as the maximum frequency at which the
output of system will track an input sinusoid in a satisfactory manner.
By convention, the bandwidth is the frequency at which the output is attenuated to a factor of 0.707 times the input.
The maximum value of the frequency-response magnitude is referred to as the resonant peak Mr.
10.707
2
Chapter 6 The Frequency-Response Design Method
President University Erwin Sitompul FCS 8/14
Bode Plot TechniquesAdvantages of working with frequency response in terms of
Bode plots:1. Dynamic compensator design can be based entirely on
Bode plots.2. Bode plots can be determined experimentally.3. Bode plots of systems in series can be simply added,
which is quite convenient.4. The use of a logarithmic scale permits a much wider
range of frequencies to be displayed on a single plot compared with the use of linear scales.
Bode plot of a system is made of two curves, 1. The logarithm of magnitude vs. the logarithm of
frequency, log M vs. log ω, or also Mdb vs. log ω.2. The phase versus the logarithm of frequency
Φ vs. log ω.
Chapter 6 The Frequency-Response Design Method
President University Erwin Sitompul FCS 8/15
Bode Plot Techniques
1 2
1 2
( )( ) ( )( ) .
( )( ) ( )m
n
s z s z s zKG s K
s p s p s p
For root locus design method, the open-loop transfer function is written in the form
For frequency-response design method, s is replaced with jω to write the transfer function in the Bode form
1 20
( 1)( 1) ( 1)( ) .
( 1)( 1) ( 1)m
a b n
j j jKG j K
j j j
Chapter 6 The Frequency-Response Design Method
President University Erwin Sitompul FCS 8/16
Bode Plot Techniques
12
1
( )( ) ,
( )
s zKG s K
s s p
To draw the Bode plot of a transfer function, it must be rewritten in magnitude equation and phase equation, i.e.,
10 2
( 1)( ) .
( ) ( 1)a
jKG j K
j j
0 12
( ) ( 1) ( ) ( 1),a
KG j K jj j
Then
0 1
2
( 1),( )
( 1)( ) a
K jKG j
jj
0 12
log log log ( 1)( )
log log ( 1) ,( ) a
K jKG j
jj
0 1db2
20log 20log ( 1)( )
20 log 20log ( 1) .( ) a
K jKG j
jj
Phase Equation
Magnitude Equation
(log)
(db)
Magnitude Equation
Magnitude Equation
Chapter 6 The Frequency-Response Design Method
President University Erwin Sitompul FCS 8/17
Bode Plot TechniquesExamining all the transfer functions we have dealt with so far,
all of them are the combinations of the following four terms:
( )nj
12( ) 2 ( ) 1n nj j
1( 1)j
0K • Gain
• Pole or zero at the origin
• Simple pole or zero
• Quadratic poles or zeros
Once we understand how to plot each term, it will be easy to draw the composite plot, since log M and Φ are the additive combination of the magnitude logarithms and the phases of all terms.
Chapter 6 The Frequency-Response Design Method
President University Erwin Sitompul FCS 8/18
Bode Plot Techniques0K
Magnitude
Phase
Gain
Chapter 6 The Frequency-Response Design Method
President University Erwin Sitompul FCS 8/19
Bode Plot Techniques( )nj
Magnitude
Phase
90n
log log( )n n jj
Pole or zero at the origin
Chapter 6 The Frequency-Response Design Method
President University Erwin Sitompul FCS 8/20
Bode Plot TechniquesAs example, Bode plot of a zero (jω) at origin will be as
follows:
Chapter 6 The Frequency-Response Design Method
President University Erwin Sitompul FCS 8/21
Approximation Magnitude Phase
Bode Plot Techniques1( 1)j Simple pole or zero
1
1
1
1( 1)j
2( 1) 1j j
( 1)j
1 0
( 1) 45j
( ) 90j
( 1) 1j
( 1) 1j j
( 1)j j
• The point where ωτ = 1 or ω = 1/τ is called the break point.
( 1)j
Chapter 6 The Frequency-Response Design Method
President University Erwin Sitompul FCS 8/22
Bode Plot TechniquesAs example, Bode magnitude plot of a simple zero (jωτ+1) is
given below, with τ = 10.• The break point lies at ω = 1/τ = 0.1.
db
10( 1)j
db
1
20log( 2) 3 db( 1)j
db
120log( )( 1)j
• Correction of Asymptote
Chapter 6 The Frequency-Response Design Method
President University Erwin Sitompul FCS 8/23
Bode Plot Techniques The corresponding Bode phase plot of a simple zero (jωτ+1)
is given as:
1 1 0
1( 1) 45j
1( ) 90j
• Corrections of Asymptotes by 11°,at ω = 0.02 and ω = 0.5.
• Corresponds to 1/5ωbreak and 5ωbreak
Chapter 6 The Frequency-Response Design Method
President University Erwin Sitompul FCS 8/24
Bode Plot TechniquesQuadratic poles or zeros
12( ) 2 ( ) 1n nj j
• Asymptotes can be used for rough sketch.• Afterwards, correction must be made
according to the value of damping factor ζ.
Chapter 6 The Frequency-Response Design Method
President University Erwin Sitompul FCS 8/25
Bode Plot TechniquesChapter 6 The Frequency-Response Design Method
President University Erwin Sitompul FCS 8/26
5
Bode Plot TechniquesChapter 6 The Frequency-Response Design Method
President University Erwin Sitompul FCS 8/27
Bode Plot: ExamplePlot the Bode magnitude and phase for the system with the transfer function
2000( 0.5)( ) .
( 10)( 50)
sKG s
s s s
0.5
10 50
2000 0.5( 1)( )
10( 1) 50( 1)
j
j jKG jj
Convert the function to the Bode form,
0.5
10 50
2( 1).
( 1)( 1)
j
j jj
• ωb1 = 0.5, ωb2 = 10, and ωb3 = 50
• One pole at the origin
5 terms will be drawn separately and finally composited1 1 1
0.5 10 502, ( ) , ( 1), ( 1) , ( 1) .j j jj
Chapter 6 The Frequency-Response Design Method
President University Erwin Sitompul FCS 8/28
Bode Plot Techniques
40
60
20
0
–20
–40
db2
1( )j
0.5( 1)j
1
10( 1)j
1
50( 1)j
ωb1 = 0.5 ωb2 = 10 ωb3 = 50
: Rough composite
–20 db/dec
–20 db/dec
–40 db/dec
0 db/dec
Chapter 6 The Frequency-Response Design Method
President University Erwin Sitompul FCS 8/29
Bode Plot Techniques
40
60
20
0
–20
–40
db
ωb1 = 0.5 ωb2 = 10 ωb3 = 50
: Rough composite
: 3-db-corrected composite
+3 db–3 db
–3 db
Final Result
Chapter 6 The Frequency-Response Design Method
President University Erwin Sitompul FCS 8/30
Bode Plot Techniques
2
1( )j
0.5( 1)j
1
10( 1)j
1
50( 1)j
ωb1 = 0.5 ωb2 = 10 ωb3 = 50
0.1
2.52
5010
250
+11°
–11° –11°
+11° –11°
+11°
Chapter 6 The Frequency-Response Design Method
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Bode Plot Techniquesωb1 = 0.5 ωb2 = 10 ωb3 = 50
Final Result
Chapter 6 The Frequency-Response Design Method
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Homework 8No.1, FPE (6th Ed.), 6.4.
Hint: Draw the Bode plot in logarithmic and semi-logarithmic scale accordingly.
No.2.
(a) Derive the transfer function of the electrical system given above.
(b) If R1 = 10 kΩ , R2 = 5 kΩ and C = 0.1 μF, draw the Bode plot of the system.
Chapter 6 The Frequency-Response Design Method
Due: Thursday, 21 November 2013.