CONTROL SYSTEMS [PDF]

OBJECTIVES. In this chapter : ▫ A short introduction to the steady state response of control systems to p y p y sinuso

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ME 304

CONTROL SYSTEMS Mechanical Engineering Department, Middle East Technical University

Radar Dish

Armature controlled dc motor

Outside

θD output

Inside

θr input p

θm Gearbox Control Transmitter

θD

dc amplifier ME 304 CONTROL SYSTEMS

Control Transformer

Prof. Prof Dr. Dr Y. Y Samim Ünlüsoy

Prof. Dr. Y. Samim Ünlüsoy

1

CH IX COURSE OUTLINE I. II. III.

INTRODUCTION & BASIC CONCEPTS MODELING DYNAMIC SYSTEMS CONTROL SYSTEM COMPONENTS

IV. V. VI. VII. VIII.

STABILITY TRANSIENT RESPONSE STEADY STATE RESPONSE DISTURBANCE REJECTION BASIC CONTROL ACTIONS & CONTROLLERS

IX.

FREQUENCY RESPONSE ANALYSIS

X. XI.

SENSITIVITY ANALYSIS ROOT LOCUS ANALYSIS

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

2

FREQUENCY RESPONSE - OBJECTIVES In this chapter : „

„

„

A short introduction to the steady state response p of control systems y to sinusoidal inputs will be given. Frequency domain specifications for a control system will be examined. Bode plots and their construction g asymptotic y p approximations pp using will be presented.

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

3

FREQUENCY RESPONSE – INTRODUCTION Nise Ni Ch Ch. 10 „

„

„

In frequency response analysis of control systems, the steady state response of the system to sinusoidal input is of interest. The frequency response analyses are carried out in the frequency q y domain, domain, rather than the time domain. It is to be noted that, time domain properties of a control system can be predicted from its frequency domain characteristics.. characteristics

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

4

FREQUENCY RESPONSE - INTRODUCTION „

For an LTI system the Laplace transforms off the th input i t and d output t t are related l t d to t each h other by the transfer function, T(s). Laplace Domain Input

„

R(s)

T(s)

C(s)

Output

In I the th frequency f response analysis, l i the th system is excited by a sinusoidal input of fixed amplitude and varying frequency. frequency.

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

5

FREQUENCY RESPONSE - INTRODUCTION „

Let us subject j a stable LTI system y to a sinusoidal input of amplitude R and frequency q y

ω in time domain.

r(t)=Rsin(ωt) „

The steady state output of the system will be again a sinusoidal signal of the same frequency,, but probably with a different frequency amplitude and phase. phase. c(t)=Csin(ωt+φ)

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

6

FREQUENCY Q RESPONSE - INTRODUCTION

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

7

FREQUENCY RESPONSE - INTRODUCTION „

To carry y out the same process p in the frequency domain for sinusoidal steady state analysis, one replaces the Laplace variable i bl s with i h s=jjω in the input output relation C(s)=T(s)R(s) C(s) T(s)R(s) with the result C(jω)=T(j ) T(jω)R(jω)

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

8

FREQUENCY RESPONSE - INTRODUCTION „

The input, output, and the transfer function have now become complex and thus they can be represented by their magnitudes and phases.

„

Input :

R(jω)= R(jω) ∠R(jω)

„

Output :

C(jω)= C(jω) ∠C(jω) C(jω)

Transfer Function :

T(jω)= (j ) T(jω) (j ) ∠Τ(jω) (j )

„

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

9

FREQUENCY RESPONSE - INTRODUCTION „

With similar expressions for the input and d th the ttransfer f ffunction, ti th the iinputt output relation in the frequency domain consists of the magnitude and phase expressions : C(jω)=T(jω)R(jω)

C(jω) = T(jω) (jω) R(jω) (jω)

∠C(jω)= ∠T(jω)+ ∠R ( jω)

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

10

FREQUENCY RESPONSE - INTRODUCTION „

For the input and output described by r(t)=Rsin(ωt) c(t)=Csin(ωt+φ) the amplitude and the phase of the output can now be written as

C =R T(jω) φ=∠ ∠T(jω) T(jω)

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

11

FREQUENCY RESPONSE „

Consider the transfer function for the general closed loop system. system C(s) G(s) T(s)= = R(s) 1+G(s)H(s)

For the steady state behaviour, behaviour insert s=jω.

C(jω) G(jω) T(jω) )= = R(jω) 1+G(jω)H(jω) T(jω) is called the Frequency Response Function (FRF) or Sinusoidal Transfer Function.. Function ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

12

FREQUENCY RESPONSE „

The frequency response function can be written in terms of its magnitude and phase.

T(jω) )= T(jω) ∠T(jω)

Since this function is complex, p , it can also be written in terms of its real and imaginary parts.

T(jω)= Re [ T(jω)] + jIm[ T(jω)]

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

13

FREQUENCY RESPONSE „

„

Remember R b that th t for f a complex l number b be b expressed in its real and imaginary parts : the magnitude is given by :

z= „

z = a+bj

( a+bj)( a - bj) =

the phase is given by :

2

2

a +b

-1 b

∠z ∠ z = tan

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

a 14

FREQUENCY Q RESPONSE „

The magnitude and phase of the frequency response function are given by : G(jω) G(jω) T(jω) = = 1+G(jω)H(jω) 1+G(jω)H(jω)

∠T(jω)= ∠G(jω) - ∠ [1+G(jω)H(jω)] These are called the gain and phase characteristics. ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

15

FREQUENCY RESPONSE – Example p 1a „

For a system described by the diff differential ti l equation ti

 +2x = y(t) x determine the steady state response xss(t) for a pure sine wave input

y(t)= 3sin(0.5t)

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

16

FREQUENCY RESPONSE – Example p 1b „

The transfer function is given by X(s) 1 T(s)= = Y(s) s ( s +2 )

 +2x = y(t) x Insert s=jω to get :

For

1 T(jω)= jω ( jω +2 )

ω=0.5 0 [rad/s]: [ d/ ] 1 1 T(0.5j)= = 0.5j ( 0.5j+2 ) -0.25 + j

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

17

FREQUENCY RESPONSE – Example p 1c „

Multiply and divide by the complex conjugate. ⎛ ⎞ ⎛ -0.25 - j ⎞ 1 -0.25 - j T(0.5j) = ⎜ ⎟⎜ ⎟= ⎝ -0.25 + j ⎠ ⎝ -0.25 - j ⎠ 1+0.0625 T(0.5j) = -0.235 - 0.941j

„

Determine the magnitude and the angle. T(0.5j) = cos - cos + sin + sin +

76o

cos sin -

cos + sin -

2

( -0.235 )

2

+ ( -0.941) = 0.97 -1 1

∠T(0.5j) T(0 5j) = tan t

-0.941 0.941 = -104 104o -0.235

-104o ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

18

FREQUENCY RESPONSE – Example p 1d „

The steady state response is then given by :

(

xss (t)= 3 ( 0 0.97 97 ) sin 0 0.5t 5t -104 104o

(

= 2.91sin 0.5t -104o

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

)

)

19

FREQUENCY Q RESPONSE – Example p 2a „

Express the transfer function (input : F, output : y) in terms of its magnitude and phase.

 +cy +ky =F my F k

c m F

ME 304 CONTROL SYSTEMS

y

G(s)=

1 2

ms +cs +k

Prof. Dr. Y. Samim Ünlüsoy

20

FREQUENCY Q RESPONSE – Example p 2b „

Insert s=jω in the transfer function to obtain the frequency response function. G(s)= T(jω)=

„

1 2

m ( ωjj) +c + ( ωjj) +k

=

(

1 ms2 +cs +k 1

)

k - mω2 +cωj + j

Write W i the h FRF iin a+bj bj form. f

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

21

FREQUENCY Q RESPONSE – Example p 2c „

Multiply and divide the FRF expression with the complex conjugate of its denominator.

( ) ( ) T(jω)= = 2 2 2 2 2 k mω +cωj k mω cωj ( ) ( ) k mω + cω ( ) ( ) k - mω2 - cωj

1

(

k - mω2 - cωj

)

⎡ ⎤ ⎡ ⎤ 2 k mω ⎢ ⎥ ⎢ ⎥ cω -cω T(jω)= ⎢ ⎥+⎢ ⎥j 2 2 ⎢ k - mω2 + ( cω )2 ⎥ ⎢ k - mω2 + ( cω )2 ⎥ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦

(

)

(

)

T(jω)=Re [ T(jω)] +Im[ T(jω)] j

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

22

FREQUENCY Q RESPONSE – Example p 2d „

Obtain the magnitude and phase of the frequency response function.

z = a2 +b2

T(jω) =

( (

k - mω

)

)

2

2

+ ( cω )

2 ⎡ 2⎤ 2 + ( cω ) ⎥ ⎢ k - mω ⎣ ⎦

b ∠z = tan-1 a

ME 304 CONTROL SYSTEMS

2

2

=

1

(

2

k - mω

-1

∠T(jω)= tan

Prof. Dr. Y. Samim Ünlüsoy

)

2

(

2

+ ( cω )

-cω k - mω2

23

)

FREQUENCY Q RESPONSE – Example p 3a „

„

„

The open loop transfer function of a control system is given as :

G ( s) =

300 ( s + 100 )

s ( s + 10 )( s + 40 )

Determine an expression for the phase angle of G(jw) (j ) in terms of the angles g of its basic factors.. Calculate its value at a frequency of factors 28.3 rad/s. Determine the expression for the magnitude of G(jw) in terms of the magnitudes of its basic factors . Find its value al e in dB at a frequency of 28.3 rad/s.

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

24

G ( s) =

300 ( s + 100 )

FREQUENCY RESPONSE – Example 3b

s ( s + 10 )( s + 40 )

∠G(jω) = ∠300 + ∠G(jω + 100) - ∠G(jω) - ∠G(jω + 10) - ∠G(jω + 40) -1 ⎛

ω ⎞ -1 ⎛ ω ⎞ -1 ⎛ ω ⎞ -1 ⎛ ω ⎞ = 0 + tan ⎜ ⎟ - tan ⎜ ⎟ - tan ⎜ ⎟ - tan ⎜ ⎟ ⎝ 100 ⎠ ⎝0⎠ ⎝ 10 ⎠ ⎝ 40 ⎠ -1 ⎛

ω ⎞ o -1 ⎛ ω ⎞ -1 ⎛ ω ⎞ = 0 + tan ⎜ ⎟ - 90 - tan ⎜ ⎟ - tan ⎜ ⎟ ⎝ 100 ⎠ ⎝ 10 ⎠ ⎝ 40 ⎠ o

o

-1 ⎛ 28.3 ⎞

∠G(28.3j) = 0 + tan ⎜

o

-1 ⎛ 28.3 ⎞

-1 ⎛ 28.3 ⎞

⎟ - 90 - tan ⎜ ⎟ - tan ⎜ ⎟ ⎝ 100 ⎠ ⎝ 10 ⎠ ⎝ 40 ⎠

= 0o + 15.8o - 90o - 70.5o - 35.3o = -180o ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

25

G ( s) =

300 ( s + 100 )

s ( s + 10 )( s + 40 )

G ( jω ) =

=

FREQUENCY RESPONSE – Example 3c

300 jω j + 100 jω jω + 10 jω + 40 2

2

2

ω + 40

300 ω + 100 2

ω ω + 10

2

2 2

300 28 28.3 3 + 100

G ( 28.3j ) =

2

28.3 28.3 + 10

2

2 2

28.3 + 40

( 300 )(103.9 ) = = 0.749 ( 28.3 )( 30.0 )( 49.0 ) ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

26

2

FREQUENCY RESPONSE „

Typical gain and phase characteristics of a closed loop system. |T(jw)| ∠T (jω) 0

Mr 1 0.707

ωr

ME 304 CONTROL SYSTEMS

BW

ω

Prof. Dr. Y. Samim Ünlüsoy

ω 27

FREQUENCY Q DOMAIN SPECIFICATIONS „

Similar to transient response specifications in time domain, frequency response specifications are defined. - Resonant peak, Mr, - Resonant frequency, ωr, - Bandwidth, BW, - Cutoff ff Rate.

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

28

FREQUENCY Q DOMAIN SPECIFICATIONS „

Resonant p peak, Mr : peak, This is the maximum value of the transfer function magnitude |T(jω)|.

|T(jω)| Mr 1

Mr depends on the damping ratio

ξ only and indicates the relative

stability of a stable closed loop system.

A large Mr results in a large overshoot of the step response. As a rule of thumb, Mr should be between 1.1 and 1.5. ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

ω

ωr

Mr =

1 2ξ 1 - ξ2 29

FREQUENCY Q DOMAIN SPECIFICATIONS „

Resonantt R frequency,, ωr : frequency

|T(jω)| Mr 1

This is the frequency at which the resonant peak is obtained.

ωr = ωn 1 -2ξ2 ωr

ω

Note that resonant frequency is different than both the undamped p and damped p natural frequencies! q ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

30

FREQUENCY DOMAIN SPECIFICATIONS „

Bandwidth,, BW : Bandwidth This is the frequency at which the magnitude of the frequency response function, |T(jω |T(jω)|, drops to 0 707 of its zero frequency 0.707 value.

„

|T(jω)| Mr 1 0.707

ωr

BW

ω

BW is directly proportional to ωn and gives an indication of the transient response characteristics of a control system. system The larger the bandwidth is, the faster the system responds.

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

31

FREQUENCY QU C DOMAIN O S SPECIFICATIONS C C O S |T(jω)|

„ „

Bandwidth,, BW : Bandwidth

Mr 1

It is also an indicator off robustness and noise filtering characteristics of a control system.

0 707 0.707

ωr

ωBW = ωn ME 304 CONTROL SYSTEMS

(

BW

ω

)

1 − 2ξ 2 + 4ξ 4 − 4ξ 2 + 2 Prof. Dr. Y. Samim Ünlüsoy

32

FREQUENCY DOMAIN SPECIFICATIONS „

Cut--off Rate : Cut

|T(jω)|

This is the slope of the magnitude of the frequency response function, f i | |T(j (jω)|, )| at higher (above resonant) frequencies. frequencies „

„

It indicates the ability of a system t to t distinguish di ti i h signals from noise.

Mr 1 0.707

ωr

BW

ω

Two T systems having h i the h same bandwidth b d id h can have different cutoff rates.

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

33

BODE PLOT Dorf & Bishop Ch. Ch 8, 8 Ogata Ch Ch. 8 „

„

The Bode plot of a transfer function is a useful graphical hi l tool t l for f the th analysis l i and d design d i off linear control systems in the frequency domain. The Bode plot has the advantages that - it can be sketched approximately using straightline segments without using a computer. - relative stability characteristics are easily determined, and - effects of adding controllers and their parameters are easily visualized. ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

34

BODE PLOT

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

35

BODE PLOT Nise Section 10 10.2 2 „

The Bode plot consists of two plots drawn on semisemi-logarithmic paper. paper 1. Magnitude of the frequency response f function ti i decibels, in d ib l i.e., i 20 log|T(j g| (jω)| on a linear scale versus frequency on a logarithmic scale. scale. 2. Phase of the frequency response function on a linear scale versus frequency on a logarithmic scale. scale.

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

36

BODE PLOT

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

37

BODE PLOT „

„

„

It is possible to construct the Bode plots of the open loop transfer functions, but the closed loop frequency response is not so easy to plot. It is also possible, however, to obtain the closed loop p frequency q y response p from the open loop frequency response. Thus, it is usual to draw the Bode plots of the open loop transfer functions. functions. Then the closed loop frequency response can be evaluated from the open loop Bode plots.

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

38

BODE PLOT „

„

It is p possible to construct the Bode p plots by y adding the contributions of the basic factors of T(jω T(jω) by graphical addition. Consider the following general transfer function. P

( p=1 1

K ∏ 1+Tps T(s)=

)

⎛ 2 ⎞ s s ⎟ ⎜ N s ∏ (1+ τms ) ∏ ⎜1+2ξ q + 2 ⎟ ωnq ω ⎟ m=1 q=1 ⎜ nq ⎠ ⎝ M

ME 304 CONTROL SYSTEMS

Q

Prof. Dr. Y. Samim Ünlüsoy

39

P

(

K ∏ 1+ Tp s T(s)=

p=1

)

⎛ s s2 ⎜ N s ∏ (1+ τms ) ∏ ⎜1+2ξ q + ωnq ω2 m=1 q=1 ⎜ nq ⎝ Q

M

„

BODE PLOT

⎞ ⎟ ⎟ ⎟ ⎠

The logarithmic magnitude of T(jω T(jω) can be obtained by summation of the l logarithmic ith i magnitudes it d off individual i di id l terms. P

log T ( jω ) =logK + ∑ log 1+ jωτp p

N

-log ( jω )

ME 304 CONTROL SYSTEMS

⎛ jω - ∑ log 1+ jωτm - ∑ log 1+ jω+ ⎜ ωnq ⎜ ωnq m q ⎝ M

Q

Prof. Dr. Y. Samim Ünlüsoy

2ξ q

2

⎞ ⎟ ⎟ ⎠

40

P

(

K ∏ 1+ Tp s T(s)=

p=1

)

⎛ s s2 ⎜ s ∏ (1+ τms ) ∏ ⎜1+2ξ q + ωnq ω2 m=1 q=1 ⎜ nq ⎝ N

Q

M

„

BODE PLOT

⎞ ⎟ ⎟ ⎟ ⎠

Similarly, the phase of T(jω T(jω) can be obtained by simple summation of the phases of individual terms.

⎛ 2ξ ω ω ⎞ q nq ⎟ -1 o -1 -1 ⎜ φ = ∠ T ( jω ) = ∑ tan ωτp -N 90 - ∑ tan ωτm - ∑ tan ⎜ 2 2 ⎟ ω ω p m q ⎜ nq ⎟ ⎝ ⎠ P

ME 304 CONTROL SYSTEMS

(

)

M

Prof. Dr. Y. Samim Ünlüsoy

Q

41

BODE PLOT „

Therefore, any transfer function can be constructed from the four basic factors : 1. Gain Gain,, K - a constant, 2. Integral Integral,, 1/jω, or derivative factor factor,, jω – pole or zero at the origin, 3. First order factor – simple lag, 1/(1+jωT), or lead 1+jωT (real pole or zero), zero) 4. Quadratic factor – quadratic lag or lead. 2 2 ⎡ ⎡ ⎛ ω ⎞ ⎛ ω ⎞ ⎤ ⎛ ω ⎞ ⎛ ω ⎞ ⎤ 1 ⎢1+2ξ ⎜ j ⎟ +⎜ j ⎟ ⎥ or ⎢1+2ξ ⎜ j ⎟ +⎜ j ⎟ ⎥ ⎢ ⎢ ⎝ ωn ⎠ ⎝ ωn ⎠ ⎥⎦ ⎝ ωn ⎠ ⎝ ωn ⎠ ⎥⎦ ⎣ ⎣

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

42

BODE PLOT Some useful definitions : „

The magnitude is normally specified in decibels [dB]. The value of M in decibels is given by : M[dB]=20logM [ ] g

„

Frequency ranges may be expressed in terms of decades or octaves. Decade : Frequency band from ω to 10ω. Octave : Frequency band from ω to 2ω 2ω..

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

43

BODE PLOT Gain Factor K. „

„

The gain factor multiplies the overall gain by a constant value for all frequencies. It has no effect on phase. M[dB]

G(s)= ( ) K G(jω)=K

20logK

M = 20log G(jω) = 20log(K) [dB]

φ =0

0 φ[o]

M : magnitude, φ : phase. ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

0

ω ω

44

BODE PLOT Integral Factor 1/jω – pole at the origin. „

„

Magnitude is a straight line with a slope of -20 dB/decade becoming zero at ω=1 [rad/s]. Phase is constant at -90o at all frequencies. M[dB]

1 1 1 G(s) = , G ( jω ) = =- j s jω ω

20 0

M= 20log G(jω) ⎛1⎞ = 20log ⎜ ⎟= -20logω ⎝ω⎠ o

φ = -90

ω 0.1

1

10

-20 -20 dB/decade slope Im

-1/ω ME 304 CONTROL SYSTEMS

decade

φ

Re

φ[o] 0 -90

Prof. Dr. Y. Samim Ünlüsoy

ω

45

-1/ω2

BODE PLOT

Re

φ

Double pole at the origin. „

Im

Simply double the slope of the magnitude and the phase, i.e., -40 dB/decade becoming zero at ω=1 1 [rad/s] and -180o phase. G(s)=

1 s2

, G ( jω ) =

1

( jω )2

=-

1 ω2

M = 20log G(jω) ⎛ 1 = 20log ⎜ ⎝ ω2

φ = -180

o

ME 304 CONTROL SYSTEMS

M[dB]

decade

40 0

ω 0.1

1

10

-40 40

⎞ ⎟= -40logω ⎠

-40 dB/decade slope

φ[o] 0 -180

Prof. Dr. Y. Samim Ünlüsoy

ω

46

BODE PLOT Derivative Factor jω – zero at the origin. „

„

Magnitude is a straight line with a slope of 20 dB/decade becoming zero at ω=1 [rad/s]. Phase is constant at 90o at all frequencies. M[dB]

G(s)= s , G ( jω ) = ωj

20 0

M 20log M= 20l G(jω) G(j ) = 20log ( ω )

φ = 90o ME 304 CONTROL SYSTEMS

decade

ω 0.1

1

10

-20

φ[o]

20 dB/decade slope

90 0 Prof. Dr. Y. Samim Ünlüsoy

ω

47

BODE PLOT Double zero at the origin. „

Simply double the slope of the magnitude and the phase, i.e., 40 dB/decade becoming zero at ω=1 1 [rad/s] and 180o phase. M[dB]

G(s)= s2 , G ( jω ) = -ω2

M = 20log G(jω) = 40log ( ω )

φ =180o

40 0 0

ω 0.1

1

10

-40

φ[o]

40 dB/decade slope

180 0

ME 304 CONTROL SYSTEMS

decade

Prof. Dr. Y. Samim Ünlüsoy

ω

48

BODE PLOT – First Order Factor Simple lag (Real pole) 1/(1+jωT). 1 G(s)= 1+ Ts

G(jω)=

1 1 - jωT 1 ωT = j 2 2 2 2 1+ jωT 1 - jωT 1+ω T 1+ω T

⎛ 1 M = 20log G(jω) = 20log ⎜ ⎜ 2 2 ⎝ 1+ω T

⎞ ⎟ ⎟ ⎠

M= -20log 1+ω2 T2 [dB] φ = tan-1 ( -ωT ) = -tan-1ωT

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

49

BODE PLOT – First Order Factor Simple lag (Real Pole) 1/(1+jωT). M = -20log 1 + ω2 T2 [dB]

M[dB] 0

0.1 T

1 T

10 T

100 T

ω F For ω >

-40

ME 304 CONTROL SYSTEMS

1 T 1 T

M ≅ -20log l ω T [dB] [d ]

Prof. Dr. Y. Samim Ünlüsoy

50

BODE PLOT – First Order Factor It is clear that the actual magnitude curve can be approximated pp by y two straight g lines. M[dB]

0.1 0 T

1 T

10 T

100 T

ω

-3 M ≅ -20log1 20l 1 = 0 [dB]

-20 M ≅ -20log ω T [dB]

-40 For ω >

1 T

Prof. Dr. Y. Samim Ünlüsoy

51

BODE PLOT – First Order Factor ωc=1/T is called the corner (break) frequency. frequency. Maximum error between the linear approximation and the exact value will be at the corner frequency. M[dB] 0

M= -20log 1+ω2 T2 [dB]

0.1 T

1 T

10 T

100 T

ω

-3

1⎞ ⎛ M ⎜ ω = ⎟ = -20log 2 T⎠ ⎝ ≅ -3[dB]

-20

-40

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

52

BODE PLOT – First Order Factor ω

0.1ωc 0.5ωc

ωc

2ω c

10ωc

Error [dB]

0.04

3

1

0.04

M[dB] 0

0.1 T

1 1 T

10 T

100 T

ω

-3 -20

-40

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

53

BODE PLOT – First Order Factor „

„

„

„

Transfer function G(s)=1/(1+Ts) is a low pass filter. filter. At low frequencies the magnitude ratio is almost one,, i.e.,, the output p can follow the input. For higher frequencies, however, the output cannot follow the input because a certain amount of time is required to build up output magnitude (time constant!). Thus, the higher the corner frequency the faster the system response will be.

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

54

BODE PLOT – First Order Factor Simple lag 1/(1+jωT). φ[o] 0

φ = tan-1 ( -ωT ) = -tan-1ωT 0.1 T

1 T

10 T

100 T

ω For ω >

10 T

φ ≅ -90 [o ]

-90

ME 304 CONTROL SYSTEMS

0.1 T

Prof. Dr. Y. Samim Ünlüsoy

55

BODE PLOT – First Order Factor It is clear that the actual phase curve can be approximated by three straight lines. φ[o] 0

o

φ≅0[ ]

0.1 T

1 T

10 T

-5.7

100 T

ω Linear variation in the range 0.1 10 ≤ω≤ T T

-45

-90

φ ≅ -90 [o ]

In this case corner frequencies q are : 0.1/T and 10/T ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

56

BODE PLOT – First Order Factor ω

0 01ωc 0.01

0 1ω c 0.1

ωc

10ωc

100ωc

φ [o]

-0.57

-5.7

-45

-84.3

-89.4

Error [o]

0.6

5.7

0

-5.7

-0.6

Thus the maximum error of the linear approximation is 5.7o.

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

57

BODE PLOT – First Order Factor Simple lead (Real zero) 1+jωT. G(s)=1+ Ts G(j ) 1 G(jω)=1+ωTj Tj

M = 20log G(jω) = 20log ⎛⎜ 1+ω2 T2 ⎞⎟ ⎝ ⎠

M=20log 1+ω2 T2 [dB] φ = tan-1 ( ωT ) = tan-1ωT

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

58

BODE PLOT – First Order Factor Simple lead (Real zero) 1+jωT. M = 20log 1 + ω2 T2 [dB]

M[dB]

For ω >

20

0

1 T

M ≅ 20log ω T [dB]

ω 0.1 T

ME 304 CONTROL SYSTEMS

1 T

10 T

100 T

Prof. Dr. Y. Samim Ünlüsoy

59

BODE PLOT – First Order Factor It is clear that the actual magnitude curve can be approximated pp by y two straight g lines. 1 For ω >

1 T

40 M ≅ 20log 20l ω T [dB]

20 M ≅ 20log1 = 0 [dB]

0

ω 0.1 T

ME 304 CONTROL SYSTEMS

1 T

10 T

100 T

Prof. Dr. Y. Samim Ünlüsoy

60

BODE PLOT – First Order Factor Simple lead 1+jωT. φ = tan-1 ( ωT )

φ[o]

For ω >

45

10 T

φ ≅ 90 [o ]

0

ω 0.1 0 1 T

ME 304 CONTROL SYSTEMS

1 T

10 T

100 T

Prof. Dr. Y. Samim Ünlüsoy

61

BODE PLOT – First Order Factor It is clear that the actual phase curve can pp by y three straight g lines. be approximated φ[o] φ ≅ 90 [o ]

90

Linear variation Li i ti in the range 0.1 10 ≤ω≤ T T

45 5.7 φ≅0[ ] 0 o

ME 304 CONTROL SYSTEMS

0.1 T

1 T

10 T

Prof. Dr. Y. Samim Ünlüsoy

100 T

ω

62

BODE PLOT – Quadratic Factors As overdamped systems can be replaced by two first order factors,, only y underdamped p systems are of interest here. G(s)=

ω2 n

s2 +2ξωns +ω2 n

G(jω)=

A set of two complex conjugate poles.

1 2

⎛ ω ⎞ ⎛ ω ⎞ ⎜j ⎟ +2ξ ⎜ j ⎟ +1 ⎝ ωn ⎠ ⎝ ωn ⎠ 2

2 ⎡ ⎛ ω ⎞2 ⎤ ⎛ ⎞ ω M=20log G(jω) = -20log 20log ⎢1 - ⎜ ⎟ ⎥ + ⎜ 2ξ ⎟ [dB] ωn ⎠ ⎢ ⎝ ωn ⎠ ⎥ ⎝ ⎣ ⎦

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

63

BODE PLOT – Quadratic Factors 2

2 ⎡ ⎛ ω ⎞2 ⎤ ⎛ ω ⎞ ⎥ M 20log G(jω) = -20log M= 20log ⎢1 - ⎜ + 2ξ ⎟ ⎜ ⎟ [dB] ωn ⎠ ⎢ ⎝ ωn ⎠ ⎥ ⎝ ⎣ ⎦

Low frequency asymptote, asymptote ωωn : 2

⎛ ω ⎞ ⎛ ω ⎞ M ≅ -20log ⎜ ⎟ = -40log ⎜ ⎟ [dB] ⎝ ωn ⎠ ⎝ ωn ⎠

Low and high frequency asymptotes intersect at ω=ωn, i.e. corner frequency is ωn. ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

64

BODE PLOT – Quadratic Factors Therefore the actual magnitude curve can be approximated by two straight lines. lines M[dB] 20

LF Asymptote

ξ (increasing)

0

HF Asymptote

-20 40 -40 -40dB/decade slope

-60

ωn/100

ωn/10

ωn

10ωn

100ωn

Frequency ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

65

BODE PLOT – Quadratic Factors φ = ∠G(jω)= -tan-1

ω 2ξ ωn

2

⎛ ω ⎞ 1-⎜ ⎟ ω ⎝ n⎠

o φ ≅ 0 [ ] At low frequencies, frequencies ω→0 : o φ ≅ − 90 [ ] At ω=ωn :

At high frequencies, ω→ :

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

φ ≅ -180 [o ]

66

BODE PLOT – Quadratic Factors Thus, the actual phase curve can be approximated by three straight lines. lines 0

φ[o] ξ (increasing) -90o/decade slope

-90

-180

ωn/10

ωn

10ωn

100ωn

Frequency

Corner frequencies q are : ωn/ /10 and 10ωn. ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

67

BODE PLOT – Quadratic Factors „

„

It is observed that,, the linear approximations for the magnitude and phase will give more accurate results f for damping d i ratios i closer l to 1.0. 10 The peak magnitude is given by : Μr =

„

1 2ξξ 1− ξ 2

The resonant frequency : ωr = ωn 1 -2ξ2

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

68

BODE PLOT – Quadratic Factors „

For ξ=0.707 : (or M=20log1=0 g dB). ) Mr=1 ( Thus, there will be no peak on the magnitude plot.

„

Note the difference that in transient response for step input, input there will be no overshoot for critically or overdamped p systems, y , i.e.,, for ξ ≥ 1.0.

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

69

BODE PLOT – Example 1a „

Sketch the Bode plots for the given open loop transfer function of a control system. 100000 (1+s ) T(s)=

„

(

s ( s +10 ) 0 0.1s 1s2 +14s +1000

)

First convert to standard form. 100000 (1+ jω ) T(jω)=

⎡⎛ ω ⎞2 ⎤ ⎛ ω ⎞ ( jω )(10 )(1+0.1ωj)(1000 ) ⎢⎜ j ⎟ +1.4 ⎜ j ⎟ +1⎥ ⎝ 100 ⎠ ⎢⎣⎝ 100 ⎠ ⎥⎦ T(jω)= T(jω)

ME 304 CONTROL SYSTEMS

10 (1+ jω ) ⎡⎛ ω ⎞2 ⎤ ⎛ ω ⎞ ( jω )(1+0.1ωj) ⎢⎜ j ⎟ +1.4 ⎜ j ⎟ +1⎥ ⎝ 100 ⎠ ⎢⎣⎝ 100 ⎠ ⎥⎦

Prof. Dr. Y. Samim Ünlüsoy

70

BODE PLOT – Example 1b T(jω)=

„

10 (1+ jω ) ⎡⎛ ω ⎞2 ⎤ ⎛ ω ⎞ 1+0 1ωj) ⎢⎜ j +1 4 ⎜ j ( jω )(1+0.1ωj ⎟ +1.4 ⎟ +1⎥ ⎝ 100 ⎠ ⎢⎣⎝ 100 ⎠ ⎥⎦

Identify the basic factors and corner frequencies : -Constant gain K : K=10, 20log10=20 [dB] -First order factor (simple lead – real zero) : T=1 (ωc1=1/T=1) - for magnitude plot -Integral factor : 1/jω -First order factor (simple lag – real pole) : T=0.1 T=0 1 (ωc1 1=1/T=10) - for magnitude plot -Quadratic factor (complex conjugate poles) : ωn=ωc1=100, ξ=0.7 - for magnitude plot

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

71

BODE PLOT – Example 1c T(jω)=

„

10 (1+ jω ) ⎡⎛ ω ⎞2 ⎤ ⎛ ω ⎞ 1+0 1ωj) ⎢⎜ j +1 4 ⎜ j ( jω )(1+0.1ωj ⎟ +1.4 ⎟ +1⎥ ⎝ 100 ⎠ ⎢⎣⎝ 100 ⎠ ⎥⎦

Identify the basic factors and corner frequencies : -Constant gain K : K=10, 20log10=20 [dB] -First order factor (simple lead – real zero) : T=1 (ωc2=0.1/T=0.1, ωc3=10/T=10) – for phase plot -Integral factor : 1/jω -First order factor (simple lag – real pole) : T=0.1 (ωc2=0.1/T=1, ωc3=10/T=100) – for phase plot -Quadratic factor (complex conjugate poles) : ωn=100 (ωc2=ωn/10=10, ωc3=10ωn=1000) – for phase plot ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

72

BODE PLOT – Example 1d 1+jω

M[dB] 40

K=10

20 0

1

10

100

1000

ω[rad/s]

-20

Quadratic factor

-40

1/(1+0.1jω) 1/jω

-60

Bode (magnitude) plot ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

73

BODE PLOT – Example 1e 90

0

1+jω

φ[o]

1

10

100

1000 K=10

ω 1/(1+0.1jω)

-90 90

1/jω Quadratic factor

-180 180

-270 Bode (phase) plot ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

74

BODE PLOT – Example 1f Matlab plot: full blue lines

num=[100000 100000] den=[0 1 15 1140 10000 0] den=[0.1 bode(num,den) grid

Approximate plots: dashed lines ME 304 CONTROL SYSTEMS

40 Magnittude (dB)

(just 4 lines to plot !)

Bode Diagram

60

20 0 -20 -40 -60 60 0

Phase ((deg)

„

-90

-180

-270 -1 10

10

0

1 10 Frequency

Prof. Dr. Y. Samim Ünlüsoy

10

2

10

75

3

STABILITY ANALYSIS Nise Sect Sect. 10 10.7, 7 pp pp.638 638-641 638-

„

„

„

Transfer functions which have no poles or zeroes on the right hand side of the complex plane are called minimum phase transfer funtions. Nonminimum p phase transfer functions,, on the other hand, have zeros and/or poles on the right hand side of the complex plane. The major disadvantage of Bode Plot is that stability of only minimum phase systems can be determined using Bode plot.

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

76

STABILITY ANALYSIS „

From the characteristic equation : 1 + G(s)H(s) = 0 or G(s)H(s)= -1 Then the magnitude and phase for the open loop transfer function become :

20log G(jω)H(jω) =20log1= 0 dB

∠G(jω)H(jω)= -180

o

Thus, when the magnitude and the phase angle of a transfer function are 0 dB and -180 80o, respectively, i l then h the h system iis marginally stable. ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

77

STABILITY ANALSIS „

„

„

If at the frequency, for which phase becomes equal to -180o, gain is below 0 dB, then the system is stable (unstable otherwise). otherwise) Further, if at the frequency, for which gain i becomes b equall to t zero, phase h is i above -180o, then the system is stable (unstable otherwise). otherwise) Thus, relative stability of a minimum phase system can be determined according to these observations.

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

78

GAIN and PHASE MARGINS Nise Ni pp. 638638-641 „

„

Gain Margin : Additional gain to make the system marginally stable at a frequency q y for which the p phase of the open loop transfer function passes through -180o. Phase Margin : Additional phase angle to make the system marginally stable at a frequency for which the magnitude of the open loop transfer function is 0 dB.

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

79

GAIN and PHASE MARGINS Mag gnitude (dB))

50

Gain G i Margin

0 -50 -100

Phase ((deg)

-150 -90 -135 -180

Phase M i Margin

-225 -270 -1 10

10

0

10

1

10

2

10

3

Frequency (rad/sec)

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

80

BODE PLOT „

Can you identify the transfer function approximately if the measured meas red Bode diagram is available ?

ME 304 CONTROL SYSTEMS

Prof. Dr. Y. Samim Ünlüsoy

81

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