Modern Control - Lec 05 - Analysis and Design of Control Systems using Frequency Response

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


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


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


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


Consider the type-1


system,

31


Consider the type-2


system,

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Modern Control - Lec 05 - Analysis and Design of Control Systems using Frequency Response

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Transcript of Modern Control - Lec 05 - Analysis and Design of Control Systems using Frequency Response