Post on 21-Apr-2017
درنقنصدق اننالما،،، نـقـدر
LECTURE (5)
Analysis and Design of Control
Systems using Frequency Response
Assist. Prof. Amr E. Mohamed
Agenda
Introduction
Frequency Response
Bode Plots
Gain and Phase Margins
Stability Analysis
Bandwidth and Cutoff Frequency
Compensation and Controller Design in the Frequency Domain
2
Frequency Response
For a stable, linear, time-invariant (LTI) system, the steady state
response to a sinusoidal input is a sinusoid of the same frequency but
possibly different magnitude and different phase.
Sinusoids are the Eigenfunctions of convolution.
If input is 𝐴 𝑐𝑜𝑠(𝜔𝑜𝑡 + 𝜃) and steady-state output is 𝐵 cos(𝜔𝑜𝑡 + 𝜑),
then the complex number𝐵
𝐴𝑒𝑗(𝜑−𝜃) is called the frequency response of
the system at frequency 𝜔𝑜.
3
)(
)(
jT
sT 𝐶(𝑠)
𝐶(𝑗𝜔)
𝑅(𝑠)
𝑅(𝑗𝜔)
4
L.T.I systemtAtr sin)( )sin()( tBty
Magnitude: Phase:A
B
G(s)
H(s)
＋ -
)(ty)(tr
)()(1
)(
)(
)(
sHsG
sG
sR
sY
jsjs
Magnitude: Phase:)()(1
)(
jHjG
jG
)]()(1[
)(
jHjG
jG
Steady state response
Bode Plots
The Bode form is a method for the frequency domain analysis.
The Bode plot of the function G(jω) is composed Bode of two plots:
One with the magnitude of G(jω) plotted in decibels (dB) versus log10(jω).
The other with the phase of G(jω) plotted in degree versus log10(ω).
Feature of the Bode Plots
Since the magnitude of G(jω) in the Bode plot is expressed in dB, products and
division factors in G() become additions and subtraction, respectively
The phase relations are also added and subtracted from each other algebraically.
The magnitude plot of Bode of G(jω) can be approximated by straight-lines
segments which allow the simple sketching of the bode plot without detailed
computation.
5
6
1
210log
decDecade :
1
22log
octOctave :
1 10 100
Logarithmic coordinate
2 3 4 20
dB
Mag(dB)
Phase (deg)
1 1 1 1 1 1
This is a sheet of 5 cycle, semi-log paper.This is the type of paper usually used forpreparing Bode plots.
(rad/sec)
(rad/sec)
7
A number to decibel conversion
)number(log20 10
decade
8
Bode Plots In order to simplify the Bode plot, it is convenient to use the so-called Bode form
Given: the open loop T.F 𝐺 𝑠 𝐻(𝑠) =𝐴 𝑠+𝑧1 𝑠+𝑧2 … 𝑠+𝑧𝑚
𝑠𝑞 𝑠+𝑝1 𝑠+𝑝2 …(𝑠+𝑝𝑛)(𝑠2+𝑎𝑠+𝜔𝑜
2)
Where A, m, q and n are real constants
The bode form can be expressed as:
𝐺 𝑠 𝐻(𝑠) =𝐾 1 +
𝑠𝑧1
1 +𝑠𝑧2
… 1 +𝑠𝑧𝑚
𝑠𝑞 1 +𝑠𝑝1
1 +𝑠𝑝2
… 1 +𝑠𝑝𝑛
1 +𝑎𝜔𝑜
𝑠 +𝑠𝜔𝑜
2
𝐺 𝑗𝜔 𝐻 𝑗𝜔 =𝐾 1 +
𝑗𝜔𝑧1
1 +𝑗𝜔𝑧2
… 1 +𝑗𝜔𝑧𝑚
𝑗𝜔𝑞 1 +𝑗𝜔𝑝1
1 +𝑗𝜔𝑝2
… 1 +𝑗𝜔𝑝𝑛
1 +𝑎𝜔𝑜
𝑗𝜔 +𝑗𝜔𝜔𝑜
2
9
Basic Terms
Constant gain G s H s = K or G jω 𝐻 jω = K
Integral factors G s H s =1
𝑠or G jω 𝐻 jω =
1
jω
Derivative factors G s H s = 𝑠 or G jω 𝐻 jω = jω
First order factors G s H s = 1 +𝑠
𝑎
±1or G jω 𝐻 jω = 1 +
𝑗𝜔
𝑎
±1
Quadratic factors G s H(𝑠) = 1 +2z𝑤
𝑛
𝑠 +𝑠
𝑤𝑛
2 ±1
or
G jω 𝐻 jω = 1 +2z𝑤
𝑛
𝑗𝜔 +𝑗𝜔
𝑤𝑛
2 ±1
10
The Bode Plot for a constant gain K
)( jT
4K
4/1K
4K
4/1K
|G(jω)H(jω)| = 20𝑙𝑜𝑔10𝐾
zerojj ))H(G(
11
The Bode Plot for a derivative factor 3j
3j
2j
2j
j
j
dB/decade20 nslope
90n
1
12
The Bode Plot for an integral factor
3j
1
2j
1j
3j
2j
1j
dB/decade20 nslope
90n
13
The Bode Plot for a first order factor 𝟏 +𝑠
𝑎Magnitude:
Phase:
])(1log[10
)(1log20)1(
2
2
a
aaj
dB
aaj
1tan)1(
01log100 dBa
a
adB
adB
aaja
log20log20
log201
oGHa
a 00tan0 1
oGHa
a 90tan 1
01.32log1011 dBja
14
The Bode Plot for a Second order factor
2
21)()(
nn
jjjHjG
z
1,)log(40
1,)2log(20
1,0
)()(
nn
n
n
jHjG
1,180
1,90
1,0
)()( 0
0
n
o
n
n
jHjG
2
12
22
2
10
2
)(1
2
tan)()(2))(1()()(
2)(1log20)()(2))(1()()(
n
n
nn
nndB
nn
jHjGjjHjG
jHjGjjHjG
2
21)()(
nn
sssHsG
z
15
The Bode Plot for a Second order factor
16
2
21)()(
nn
jjjHjG
z
Bode plots for
𝐺 𝑠 𝐻(𝑠) = 𝑠;
𝐺(𝑠)𝐻(𝑠) =1
𝑠;
𝐺(𝑠)𝐻(𝑠) = 1 +𝑠
𝑎
𝐺(𝑠)𝐻(𝑠) =1
1 +𝑠
𝑎
17
Example 5.1: Draw the Bode plot for a system transfer function T(s)
12
1
13
5.1
)()(
12
12
13
3
)()(
21
3)()(
sss
s
sHsG
sss
s
sHsG
sss
ssHsG
Bode Mag. plot for Example 5.1:a. components;b. composite 18
Example 5.1: Draw the Bode plot for a system transfer function T(s)
Bode phase plot for Example 5.1:a. components;b. composite 19
12
1
13
5.1
)()(
12
12
13
3
)()(
21
3)()(
sss
s
sHsG
sss
s
sHsG
sss
ssHsG
Example 5.2: Draw the Bode plot for a system transfer function T(s)
Bode Mag. plot for Example 5.2:a. components;b. composite
108.05
12
13
06.0
)()(
2522
3)()(
2
2
sss
s
sHsG
sss
ssHsG
20
Example 5.2: Draw the Bode plot for a system transfer function T(s)
Bode phase plot for Example 5.2:a. components;b. composite
21
108.05
12
13
06.0
)()(
2522
3)()(
2
2
sss
s
sHsG
sss
ssHsG
Relative stability (Gain and Phase Margins)
A transfer function is called minimum phase when all the poles and zeros
are LHP and non-minimum-phase when there are RHP poles or zeros.
The phase margin (PM): is that amount of additional phase lag at the gain
crossover frequency required to bring the system to the verge of instability.
𝑃𝑀 = 180 + phase of GH measured at the gain crossover frequency 0 𝑑𝐵
The gain margin (GM): is the reciprocal of the magnitude |G(jw)| at the
frequency at which the phase angle is
𝐺𝑀 = 0 − 𝑑𝐵 𝑜𝑓 𝐺𝐻 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑎𝑡 𝑡ℎ𝑒 𝑝ℎ𝑎𝑠𝑒 𝑐𝑟𝑜𝑠𝑠𝑜𝑣𝑒𝑟 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 −180
Minimum phase system Stable
22
23
Open loop transfer function :
Closed-loop transfer function :
)()( sHsG
)()(1 sHsG
Open loop Stability poles of in LHP)()( sHsG
)0,0()0,1(Re
Im
RHP
Closed-loop Stability
poles of in left side of (-1,0))()( sHsG
Gain and Phase Margins (Cont.) The gain margin indicates the system gain can be increased by a factor of GM
before the stability boundary is reached.
The phase margin is the amount of phase shift of the system at unity
magnitude that will result instability
Fig. 5.17 Phase and gain margin definition
(a) Stable system (b) Unstable system
24
Example 5.3:
For the system shown in Fig.5.18, determine the gain margin, phase
margin, phase-crossover frequency, and gain-crossover frequency.
Solution:
The Bode plot for this system is shown in Fig. 5.19.
)(sR )(sC
1025
)1(202
ssss
s
25
Example 5.3:
4.0131
9.9293 dB
0.4426
103.6573º
The gain margin = 9.929 and the phase margin = 103.7 degree
26
Bandwidth and Cutoff Frequency
-3 dB
cutoff frequency
bandwidth
27
Using Matlab For Frequency Response
Instruction: We can use Matlab to run the frequency response for the
previous example. We place the transfer function in the form:
The Matlab Program
28
>> num = [5000 50000];>> den = [1 501 500];>> Bode (num,den)
]500501[
]500005000[
)500)(1(
)10(50002
ss
s
ss
s
Relationship between System Type and Log-Magnitude Curve
Consider the unity-feedback control system. The static position,
velocity, and acceleration error constants describe the low-frequency
behavior of type 0, type 1, and type 2 systems, respectively.
For a given system, only one of the static error constants is finite and
significant.
The type of the system determines the slope of the log-magnitude
curve at low frequencies.
Thus, information concerning the existence and magnitude of the
steady-state error of a control system to a given input can be
determined from the observation of the low-frequency region of the
log-magnitude curve.
29
Determination of Static Position Error Constants.
Consider the type-0
unity-feedback control
system,
30
Determination of Static Position Error Constants.
Consider the type-1
unity-feedback control
system,
31
Determination of Static Position Error Constants.
Consider the type-2
unity-feedback control
system,
32
33