Control chap10

CONTROL SYSTEMS
THEORY
Sinusoidal Tools
CHAPTER 10
STB 35103

Objectives
To learn the definition of frequency
response
 To plot frequency response


Introduction


In previous chapters we learn to analyze
and design a control system using root
locus method.



Another method that can be used is
frequency response.

Introduction


What is frequency response?


The frequency response is a representation of
the system's open loop response to sinusoidal
inputs at varying frequencies.



The output of a linear system to a sinusoidal
input is a sinusoid of the same frequency but
with a different amplitude and phase.



The frequency response is defined as the
amplitude and phase differences between the
input and output sinusoids.

Introduction


Why do we use frequency response?


The open-loop frequency response of a system
can be used to predict the behaviour of the
closed-loop system .



we directly model a transfer function using
physical data.

Introduction


Frequency response is consists of:



Magnitude frequency response, M(ω)
Phase frequency response, ø(ω)

M o (ω )
M (ω ) =
M i (ω )

φ (ω ) = φo (ω ) − φi (ω )

Introduction


A transfer function Laplace form can be
change into frequency response using the
following expression:
G ( jω ) = G ( s ) s → jω



We can plot the frequency response in two
ways:




Function of frequency with separate magnitude
and phase plot.
As a polar plot.

Introduction


Magnitude and phase plot






Magnitude curve can be plotted in decibels
(dB) vs. log ω, where dB = 20 log M.
The phase curve is plotted as phase angle vs.
log ω.
Data for the plots can be obtained using
vectors on the s-plane drawn from the poles
and zeros of G(s) to the imaginary axis.

Introduction


Magnitude response at a particular frequency
is the product of the vector length from the
zeros of G(s) divided by the product of the
vector lengths from the poles of G(s) drawn to
points on the imaginary axis.
jω
jω1
A
O

O

B

X

C

X

D

A•B
M ( jω1 ) =
C•D
σ

Introduction


The phase response is the sum of the
angles from the zeros of G(s) minus the
sum of angles from the poles of G(s)
drawn to points on imaginary axis.
jω
jω1

θ1
O θ1
O

θ2

X

θ3

X

θ3

σ

φ ( jω1 ) = [θ1 + θ 2 ] − [θ3 + θ 4 ]

Introduction


Example 10.1


Find the analytical expression for the
magnitude frequency response and the phase
frequency response for a system G(s) = 1/
(s+2). Also, plot both the separate magnitude
and phase diagrams and the polar plot.

Introduction


Solution:


First substitute s=jω in the system function and
obtain
G ( jω ) =



1
jω + 2

We always put complex number as numerator so we
will multiply the above transfer function with the
complex conjugate.
1
jω + 2
1
− jω + 2
=
×
jω + 2 − jω + 2
(2 − jω )
= 2
(ω + 4)

G ( jω ) =

Introduction


In order for us to plot the magnitude frequency
response we must find the magnitude of the transfer
function.



Magnitude G(jω), M(ω)

G ( jω ) = G ( jω )g ( jω )∗
G


Where G(jω)* is the conjugate of G(jω), so the
magnitude for transfer function in the question is
G ( jω ) =
=

(2 − jω ) (2 + jω )
g 2
2
(ω + 4) (ω + 4)
1
(ω 2 + 4)

Introduction


The phase angle of G(jω), ø(ω)
B
φ (ω ) = − tan −1  ÷
 A

(2 − jω )
A
B
(ω 2 + 4)
1
= 2
× (2 − jω )
(ω + 4)

G ( jω ) =

ω 
φ (ω ) = − tan −1  ÷
2

Introduction


We can plot the magnitude frequency response and
phase frequency response

20 log M (ω ) = 1

ω 2 + 4 vs. log ω

φ (ω ) = − tan −1 ω 2 vs. log ω

Introduction


We can also plot the polar plot

M (ω ) < φ (ω ) = 1

ω 2 + 4 < − tan −1 (ω 2)

Introduction


Exercise 10.1


Convert the following transfer function to
frequency response. Find the magnitude
frequency response and phase frequency
response.
1
G(s) =
( s + 2)( s + 4)



Solution

1
( jω + 2)( jω + 4)
1
= 2 2
j ω + j 4ω + j 2ω + 8
1
=
8 − ω 2 + j 6ω

G ( jω ) =

Introduction
1
=
8 − ω 2 + j 6ω
1
8 − ω 2 − j 6ω
=
×
2
8 − ω + j 6ω 8 − ω 2 − j 6ω
8 − ω 2 − j 6ω
=
64 − 8ω 2 − j 48ω − 8ω 2 + ω 4 + j 6ω 3 + j 48ω − j 6ω 3 − j 2 36ω 2
8 − ω 2 − j 6ω
= 4
ω − 20ω 2 + 64

Introduction


Nyquist criterion


Nyquist criterion relates the stability of a
closed-loop system to the open-loop frequency
response and open-loop pole location.



This concept is similar to the root locus.



The most important concept that we need to
understand when learning Nyquist criterion is
mapping contours.

Introduction


Mapping contours


Mapping contours means we take a point on
one contours and put it into a function, F(s),
thus creating a new contours.

Introduction


When checking the stability of a system,
the shape of contour that we will use is a
counter that encircles the entire right halfplane.

Introduction






The number of closed-loop poles in the right half
plane (also equals zeros of 1+ G(s)H(s)), Z
The number of open-loop poles in the right half
plane , P
The number of counterclockwise rotations about
(-1,0), N

N=P-Z


The above relationship is called the Nyquist
Criterion; and the mapping through G(s)H(s) is
called the Nyquist Diagram of G(s)H(s)

Introduction


Examples to determine the stability of a
system
P = 0, N = 0,

Z = P − N = 0, the system is stable

P = 0, N = −2, (clockwise '− ve ')
Z = 0 − (−2) = 2,system unstable

Sketching the Nyquist Diagram


The contour that encloses the right half-plane
can be mapped through the function G(s)H(s)
by substituting points along the contour into
G(s)H(s).



The points along the positive extension of the
imaginary axis yield the polar frequency
response of G(s)H(s).



Approximation can be made to G(s)H(s) for
points around the infinite semicircle by
assuming that the vectors originate at the
origin.



Example 10.4


Sketch a nyquist diagram based on the block
diagram below.



Solution


The open loop transfer function G(s),

500
G (s) =
( s + 1)( s + 10)( s + 3)


Replacing s with jω yields the frequency response of
G(s)H(s), i.e.

G ( jω ) =

500
( jω + 1)( jω + 10)( jω + 3)

(−14ω 2 + 30) − j (43ω − ω 3 )
= 500
(−14ω 2 + 30) 2 + (43ω − ω 3 ) 2



Magnitude frequency response

G ( jω ) =


500
(−14ω 2 + 30) 2 + (43ω − ω 3 ) 2

Phase frequency response
3
B
−1  −(43ω − ω ) 
∠G ( jω ) = tan  ÷ = tan 
÷
A
−14ω 2 + 30 


−1



Using the phase frequency response and magnitude
frequency response we can calculate the key points
on the Nyquist diagram. The key points that we will
calculate are:
 Frequency when it crosses the imaginary and real
axis.
 The magnitude and polar values during the
frequency that crosses the imaginary and real
axis.
 The magnitude and polar values when frequency
is 0 and ∞.



When a contour crosses the real axis, the imaginary
value is zero. So, the frequency during this is,

(−14ω 2 + 30) − j (43ω − ω 3 )
500
(−14ω 2 + 30) 2 + (43ω − ω 3 ) 2


(−14ω 2 + 30)
(43ω − ω 3 )
= 500 
−j

2
2
3 2
(−14ω + 30) + (43ω − ω )
( −14ω 2 + 30) 2 + (43ω − ω 3 ) 2 

real

imaginary

(43ω − ω 3 )
=0
2
2
3 2
(−14ω + 30) + (43ω − ω )

We need to find the frequency when imaginary is
zero by finding the value of ω that could produce
zero imaginary value.
 There are actually two conditions that could produce
zero imaginary.
 First


0
=0
2
2
3 2
(−14ω + 30) + (43ω − ω )

 Second

(43ω − ω 3 )
=0
∞



For the first condition, in order to get the numerator
equals to zero we must find the root value of the
numerator polynomial.

(43ω − ω 3 )
0
=
2
2
3 2
(−14ω + 30) + (43ω − ω )
( −14ω 2 + 30) 2 + (43ω − ω 3 ) 2
(43ω − ω 3 ) = 0

ω1 = 0
ω2 = 6.5574
ω3 = −6.5574

There are three frequencies
where the contour crosses the
real axis.



For the second condition, the frequency values in the
denominator that could produce zero imaginary value
is infinity, ∞.

(43ω − ω 3 )
(43ω − ω 3 )
=
2
2
3 2
(−14ω + 30) + (43ω − ω )
∞

(−14ω 2 + 30) 2 + (43ω − ω 3 ) 2 = ∞
ω =∞



When a contour crosses the imaginary axis, the real
value is zero.

(−14ω 2 + 30) − j (43ω − ω 3 )
500
(−14ω 2 + 30) 2 + (43ω − ω 3 ) 2


(−14ω 2 + 30)
(43ω − ω 3 )
= 500 
−j

2
2
3 2
(−14ω + 30) + (43ω − ω )
( −14ω 2 + 30) 2 + (43ω − ω 3 ) 2 

real

imaginary

(−14ω 2 + 30)
=0
2
2
3 2
(−14ω + 30) + (43ω − ω )



There are two conditions that could produce zero real
value.
 First

0
=0
2
2
3 2
(−14ω + 30) + (43ω − ω )
 Second

(−14ω 2 + 30)
=0
∞



Calculate the frequency values for the first condition.

(−14ω 2 + 30)
0
=
2
2
3 2
(−14ω + 30) + (43ω − ω )
(−14ω 2 + 30) 2 + (43ω − ω 3 ) 2
(−14ω 2 + 30) = 0

ω = ±1.4639


Calculate the frequency values for the second
condition

(−14ω 2 + 30)
( −14ω 2 + 30)
=
2
2
3 2
(−14ω + 30) + (43ω − ω )
∞
ω=∞



Now that we know the frequencies of the key points
in our polar plot we will now calculate the
magnitudes and phase for each key points frequency.

Cross real

Cross imaginary



The new contour can be plot based on the key points
in the previous table.

ω = 6.5574

ω =∞

C

ω = 1.4639

A

ω=0



Note that the semicircle with a infinite
radius, i.e., C-D, is mapped to the origin if
the order the denominator of G(s) is
greater than the order the numerator of
G(s).

Control chap10

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