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CONTROL SYSTEMS THEORY Sinusoidal Tools CHAPTER 10 STB 35103

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Page 1: Control chap10

CONTROL SYSTEMS THEORY

Sinusoidal Tools

CHAPTER 10STB 35103

Page 2: Control chap10

Objectives To learn the definition of frequency

response To plot frequency response

Page 3: Control chap10

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.

Page 4: Control chap10

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. 

Page 5: Control chap10

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.

Page 6: Control chap10

Introduction Frequency response is consists of:

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

( )( )

( )

( ) ( ) ( )

o

i

o i

MM

M

Page 7: Control chap10

Introduction A transfer function Laplace form can be

change into frequency response using the following expression:

We can plot the frequency response in two ways: Function of frequency with separate magnitude and

phase plot. As a polar plot.

( ) ( )s j

G j G s

Page 8: Control chap10

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.

Page 9: Control chap10

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.

XXO

jω1

σ

ACB

OD

1( )M j A •B

C•D

Page 10: Control chap10

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.

XXO

jω1

σO

1 1 2 3 4( ) [ ] [ ]j

13

1 32

Page 11: Control chap10

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.

Page 12: Control chap10

Introduction Solution:

First substitute s=jω in the system function and obtain

We always put complex number as numerator so we will multiply the above transfer function with the complex conjugate.

1( )

2G j

j

2

1( )

2

1 2

2 2

(2 )

( 4)

G jj

j

j j

j

Page 13: Control chap10

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)

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

( ) ( ) ( )G j G j G j

2 2

2

(2 ) (2 )( )

( 4) ( 4)

1

( 4)

j jG j

Page 14: Control chap10

Introduction The phase angle of G(jω), ø(ω)

2

2

(2 )( )

( 4)

1(2 )

( 4)

jG j

j

A B

1( ) tanB

A

1( ) tan2

Page 15: Control chap10

Introduction We can plot the magnitude frequency response and

phase frequency response

220log ( ) 1 4 vs. log M

1( ) tan 2 vs. log

Page 16: Control chap10

Introduction We can also plot the polar plot

2 1( ) ( ) 1 4 tan ( 2)M

Page 17: Control chap10

Introduction Exercise 10.1

Convert the following transfer function to frequency response. Find the magnitude frequency response and phase frequency response.

Solution

1( )

( 2)( 4)G s

s s

2 2

2

1( )

( 2)( 4)

1

4 2 8

1

8 6

G jj j

j j j

j

Page 18: Control chap10

Introduction

2

2

2 2

2

2 2 4 3 3 2 2

2

4 2

1

8 6

1 8 6

8 6 8 6

8 6

64 8 48 8 6 48 6 36

8 6

20 64

j

j

j j

j

j j j j j

j

Page 19: Control chap10

Introduction

Page 20: Control chap10
Page 21: Control chap10

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.

Page 22: Control chap10

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.

Page 23: Control chap10
Page 24: Control chap10
Page 25: Control chap10

Introduction When checking the stability of a system,

the shape of contour that we will use is a counter that encircles the entire right half-plane.

Page 26: Control chap10

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)

Page 27: Control chap10

Introduction Examples to determine the stability of a

system0, 0,

0, the system is stable

P N

Z P N

0, 2, ( ' ')

0 ( 2) 2,system unstable

P N clockwise ve

Z

Page 28: Control chap10

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.

Page 29: Control chap10

Sketching the Nyquist Diagram Example 10.4

Sketch a nyquist diagram based on the block diagram below.

Page 30: Control chap10

Sketching the Nyquist Diagram

Solution The open loop transfer function G(s),

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

500( )

( 1)( 10)( 3)G s

s s s

2 3

2 2 3 2

500( )

( 1)( 10)( 3)

( 14 30) (43 )500

( 14 30) (43 )

G jj j j

j

Page 31: Control chap10

Sketching the Nyquist Diagram

Magnitude frequency response

Phase frequency response

2 2 3 2

500( )

( 14 30) (43 )G j

31 1

2

(43 )( ) tan tan

14 30

BG j

A

Page 32: Control chap10

Sketching the Nyquist Diagram

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 ∞.

Page 33: Control chap10

Sketching the Nyquist Diagram

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

2 3

2 2 3 2

2 3

2 2 3 2 2 2 3 2

( 14 30) (43 )500

( 14 30) (43 )

( 14 30) (43 )500

( 14 30) (43 ) ( 14 30) (43 )

j

j

real imaginary

3

2 2 3 2

(43 )0

( 14 30) (43 )

Page 34: Control chap10

Sketching the Nyquist Diagram

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

Second

2 2 3 2

00

( 14 30) (43 )

3(43 )0

Page 35: Control chap10

Sketching the Nyquist Diagram

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

3

2 2 3 2 2 2 3 2

(43 ) 0

( 14 30) (43 ) ( 14 30) (43 )

3

1

2

3

(43 ) 0

0

6.5574

6.5574

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

Page 36: Control chap10

Sketching the Nyquist Diagram

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

3 3

2 2 3 2

(43 ) (43 )

( 14 30) (43 )

2 2 3 2( 14 30) (43 )

Page 37: Control chap10

Sketching the Nyquist Diagram

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

2 3

2 2 3 2

2 3

2 2 3 2 2 2 3 2

( 14 30) (43 )500

( 14 30) (43 )

( 14 30) (43 )500

( 14 30) (43 ) ( 14 30) (43 )

j

j

real imaginary

2

2 2 3 2

( 14 30)0

( 14 30) (43 )

Page 38: Control chap10

Sketching the Nyquist Diagram

There are two conditions that could produce zero real value.

First

Second

2 2 3 2

00

( 14 30) (43 )

2( 14 30)0

Page 39: Control chap10

Sketching the Nyquist Diagram

Calculate the frequency values for the first condition.

Calculate the frequency values for the second condition

2

2 2 3 2 2 2 3 2

2

( 14 30) 0

( 14 30) (43 ) ( 14 30) (43 )

( 14 30) 0

1.4639

2 2

2 2 3 2

( 14 30) ( 14 30)

( 14 30) (43 )

Page 40: Control chap10

Sketching the Nyquist Diagram

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

Page 41: Control chap10

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

Sketching the Nyquist Diagram

0

1.4639

6.5574 AC

Page 42: Control chap10

Sketching the Nyquist Diagram 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).