Chapter 7 Frequency Response Method - WordPress.com...- Gain crossover frequency ΦM - Phase crossover frequency Note that, negative gain or phase margin means that the system is not

Chapter 7

Frequency Response Method

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Frequency Response Introduction

Concept of Frequency Response

Plotting Frequency Response

Asymptotic Approximations for Bode Plots

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1. Introduction.The Advantages of frequency response;

(1) When modeling transfer function from physical data

(2) When designing lead compensators to meet steady-state error

requirement and a transient response requirement

(3) When finding the stability of a non-linear system

(4) In settling ambiguities when sketching root locus

Approach:

(1) Introduce the concept of frequency response

(2) Define frequency response.

(3) Derive analytical expressions for frequency response

(4) Sketch and plot frequency response.

(5) Apply concept to control system analysis and design.

(6) Frequency response is an alternative way of analyzing and designing the

feedback control system.

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1.1 Concept of Frequency

Response. In linear systems at steady-state, sinusoidal inputs generates

sinusoidal output of the same frequency. However, they may

differ in amplitude and phase angle from the input. These

differences are functions of frequencies.

Representation of Sinusoids

Sinusoid can be represented by complex numbers called phasors.

Magnitude : amplitude of the sinusoid

Angle of complex number : phase angle of the sinusoid

can be represented by

where the frequency ω is implicit.

)cos( 11 tM

11 M

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(i) Magnitude frequency response

(ii) Phase frequency response

(iii) Frequency response

The above (i) – (iii) form the definition of frequency response.

The frequency response of a system whose transfer function is is

given by,

Cont’d…

)]()([)()()()( iioo MMM

)(

)()(

i

o

M

MM

)()()( io

)()( M

)()()( GGMjG

)(sG

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1.2 Analytical Expression for

Frequency Response.Plotting Frequency Response

can be plotted in several ways:

(1) Plot as a function of frequency with separate magnitude and

phase plots

(a) Magnitude curve (dB vs log ω, where dB=20logM)

(b) Phase angle vs logω

(2) Plot as a polar plot , where the phasor length is the

magnitude and the phasor angle is the phase.

Review;

Polar :

Rectangular :

Euler’s formula:

)()()( GGMjG

M

jBA

jMe

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First Order System,

Given a system as,

The frequency response,

Magnitude,

Phase angle,

Cont’d…

1)(

s

KsG

1

1)(

jjG

221

1)(

jG

)(tan1

tan 11

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Example 1: Frequency Response.

Solution:

Where

Respective magnitude and phase angle,

.

Determine the frequency response of the following transfer function,

2

2.0)(

ssG

15.0

1.0

2

2.0)(1

sssG

2221

25.01

1.0

5.01

1.0)(

jG

)5.0(tan)( 1

1 j

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2. Bode Plot. The log-magnitude and phase frequency response curves as

functions of log are called Bode Plots or Bode diagrams.

Consider a transfer function,

Magnitude frequency response,

Magnitude response in dB,(Log-Modulus=LM)

Phase Angle,

)())((

)())(()(

21

21

n

m

m

pspspss

zszszsKsG

jsn

m

m

pspspss

zszszsKjG

)()()(

)()()()(

21

21

jsn

m

pspsps

zszszsKjG

)(log20)(log20)(log20

)(log20)(log20)(log20log20)(log20

11

21

)(tan1

tan 11

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Normalized and Scaled Bode Plot.

(a.) G(s) = s;

(b.) G(s) = 1/s;

(c.) G(s) = (s + a);

(d.) G(s) = 1/(s + a)

Cont’d…Example

Figure 2: Normalized and

scaled Bode plots for

(a)G(s)=s; (b)G(s)=1/s; ©

G(s)=(s+a); (d)G(s)

=1/(s+a)

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Example 2: Bode Plot.

Solution:

Let s=jω. Then we have:

At low frequencies, ω approaches zero.

The magnitude response in dB is: 20logM=20loga

where M=|G(jω)| and is a constant. At high frequencies,

ω>>a. We then have:

The magnitude response is then loop transfer function,

Sketch a function G(s)=(s+a) for the logarithmic magnitude and phase

response.

1

1)()(

ajaajjG

ajG )(

00 9090)(

aa

a

jajG

20logM = 20loga + 20log(ω/a) = 20logω

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where a < ω < . Notice that the high frequency approximation is

equal to the low frequency approximation when ω = a.

These straight line approximations are called asymptotes. a is called

the break frequency.

For the phase response:

(i) at break frequency, the phase is 450 ( ).

(ii) at low frequencies, the phase is 00 ( ).

(iii) at high frequencies, the phase is 900 ( ).

Cont’d…Example

)()( ajjG

ajG )(

090)( jG

Figure 1: Bode plots of (s + a): a.

magnitude plot; b. phase plot.

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System.

Low frequency.

High Frequency.

)1()(

)(

ajaajjG

assG

0

)(

M

jG

90

90)(

M

a

jajG

Cont’d…Example

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System.

Low frequency.

High Frequency.

Cont’d…Example

)1()(

)(

ajaajjG

assG

aM

jG

log20log20

)(

log20log20log20log20

90)(

aaM

a

jajG

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Example 3: Bode Plot for Ratio of First Order Factors.

Solution:

(1) Convert G(s) to show normalized components,

unity low frequency gain.

The break frequencies are at 1, 2 and 3.

Magnitude plot starts a decade below the lowest break frequency and

extend a decade above the highest break frequency, (0.1 rad to 100 rad).

Draw the Bode plot for system shown in figure below, where,

)2)(1(

)3()(

sss

sKsG

12

1

132

3

)(s

ss

sK

sG

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Cont’d…Example

(2) Magnitude Bode Plot.

(1) At =1; K will moves up or down by 20logK.

(2) At =2; begins at -20dB/decade slope

(3) At =3; begins at +20dB/decade slope, changes slope from -

60dB/decade to -40dB/decade.

-20dB/decade slope is drawn from 23.52dB at =0.1, to 3.52dB (a

20dB decrease) at =1.

Figure 3: Bode Magnitude

plot for G(s)

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Cont’d…Example

(3) Phase Bode Plot.

(1) Slope -45dB/decade from frequency of 0.1 to 0.2.

(2) Increase slope -90dB/decade from frequency of 0.2 to 0.3.

(3) Return to -45dB/decade from frequency of 0.3 to 10rad/s.

(4) +45dB/decade from frequency of 20 to 30rad/s.

(5) 0dB/decade from frequency of 30 to infinity.

Figure 4: Bode Phase Plot for

G(s).

GAIN & PHASE MARGINS IN BODE PLOT

GM – gain margin

ΦM – phase margin

GM- Gain crossover frequency

ΦM- Phase crossover frequency

Note that, negative gain or phase margin means that the system is not stable

GAIN & PHASE MARGINS IN BODE

PLOT

542

sss

KsG

Example: If K=200, find the gain margin and phase

margin .

GAIN & PHASE MARGINS IN BODE

PLOT

GAIN & PHASE MARGINS IN BODE

PLOT

6.02dB

15º

-6.02dB

-165º

7rad/sec5.5rad/sec

PID CONTROLLER

pc KG

Proportional compensator (P).

Use to improve steady state error type 0.

Consider P-compensator transfer function as:

High Kp gives better steady state but poor transient response.

Too high Kp can cause instability.

)()( sEKsA P

cG pG

)(sH

R(s) E(s) A(s) Y(s)

B(s)

+

-

PID CONTROLLER

)()( sEs

KsA I

Integral compensator (I).

Use to improve steady state error type 0.

Consider the I-compensator and actuating signal of :

Slow response, can be used with P-compensator to remedy this

problem.

s

KG I

c

cGpG

)(sH

R(s) Y(s)

+

B(s)-

A(s)E(s)

PID CONTROLLER

Dc sKG

Derivative compensator (D). Consider the D-compensator as and actuating signal as:

Quick response. No effect at steady state because no error signal. Useful for controlling type 2 together with a P-controller. Response only to rate of change and no effect to steady state.

)()( sEsKsA D

cGpG

)(sH

R(s) Y(s)

+

B(s)-

A(s)E(s)

PID CONTROLLER

d

i

pc sTsT

KsG1

1)(

Proportional-integral-derivative compensator (PID) .

Involves three separate parameters; the proportional, the

integral and derivative values.

di

pc sKs

KKsG )(