Control chap10

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CONTROL SYSTEMS THEORY

Sinusoidal Tools

CHAPTER 10STB 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)

( )( )

( ) ( ) ( )

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

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.

1( )M j A •B

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.

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

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

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

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 )( )

( 4) ( 4)

j jG j

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

(2 )( )

1( ) tanB

1( ) tan2

Introduction We can plot the magnitude frequency response and

phase frequency response

220log ( ) 1 4 vs. log M

1( ) tan 2 vs. log

Introduction We can also plot the polar plot

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

Introduction Exercise 10.1

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

Solution

( 2)( 4)G s

( 2)( 4)

G jj j

Introduction

2 2 4 3 3 2 2

8 6 8 6

64 8 48 8 6 48 6 36

j j j j j

Introduction

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 half-plane.

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

system0, 0,

0, the system is stable

0, 2, ( ' ')

0 ( 2) 2,system unstable

P N clockwise ve

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.

Sketching the Nyquist Diagram Example 10.4

Sketch a nyquist diagram based on the block diagram below.

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

2 2 3 2

500( )

( 1)( 10)( 3)

( 14 30) (43 )500

( 14 30) (43 )

G jj j j

Magnitude frequency response

Phase frequency response

2 2 3 2

500( )

( 14 30) (43 )G j

(43 )( ) tan tan

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,

2 2 3 2

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 )

real imaginary

2 2 3 2

(43 )0

( 14 30) (43 )

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.

Second

2 2 3 2

( 14 30) (43 )

3(43 )0

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

2 2 3 2 2 2 3 2

(43 ) 0

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

(43 ) 0

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, ∞.

2 2 3 2

(43 ) (43 )

( 14 30) (43 )

2 2 3 2( 14 30) (43 )

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

2 2 3 2

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 )

real imaginary

2 2 3 2

( 14 30)0

( 14 30) (43 )

There are two conditions that could produce zero real value.

Second

2 2 3 2

( 14 30) (43 )

2( 14 30)0

Calculate the frequency values for the first condition.

Calculate the frequency values for the second condition

2 2 3 2 2 2 3 2

( 14 30) 0

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

( 14 30) 0

1.4639

2 2 3 2

( 14 30) ( 14 30)

( 14 30) (43 )

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.

1.4639

6.5574 AC

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).

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