Feedback Control System

Feedback Control System

THE ROOT-LOCUS DESIGN METHOD

Dr.-Ing. Erwin Sitompul

Chapter 5

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Frequency Response A linear system’s response to sinusoidal inputs is called the

system’s frequency response. Frequency response can be obtained from knowledge of the

system’s pole and zero locations.

( ) ( ),( )Y s G sU s

0( ) sin( ) 1( ),u t A t t

02 2

0

( ) .( )A

U s u ts

L

02 2

0

( ) ( ) .AY s G ss

Consider a system described by

where

With zero initial conditions, the output is given by

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Frequency Response

*0 01 2

1 2 0 0

( ) .... ,n

n

Y ss p s p s p s j s j

Assuming all the poles of G(s) are distinct, a partial fraction expansion of the previous equation will result in

1 21 2 0 0( ) .... 2 sin( ), 0,np tp t p t

ny t e e e t t

01

0

Im( )tan .

Re( )

where

Transient response Steady-state response

1( ) ,1

G ss

( ) sin(10 )u t t

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( ) ( )( )Y s G sU s

Frequency ResponseExamine

00( ) ,( ) s jG j G s

0 0 0( ) ( ) ( ),G j G j G j

2 2

0 0 0( ) Re ( ) Im ( )G j G j G j

010

0

Im ( )( ) tan

Re ( )G j

G jG j

0( ) 0.U j A

0 0 0( ) ( ) ( ),Y j G j U j

,M

.

But the input is

Therefore

0( ) ,Y j AM 0( ) sin( ).y t AM t • Meaning?

Magnitude

Phase

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Frequency Response A stable linear time-invariant system with transfer function

G(s), excited by a sinusoid with unit amplitude and frequency ω0, will, after the response has reached steady-state, exhibit a sinusoidal output with a magnitude M(ω0) and a phase Φ(ω0) at the frequency ω0.

The magnitude M is given by |G(jω)| and the phase Φ is given by G(jω), which are the magnitude and the angle of the complex quantity G(s) evaluated with s taking the values along the imaginary axis (s = jω).

The frequency response of a system consists of the frequency functions |G(jω)| and G(jω), which describe how a system will respond to a sinusoidal input of any frequency.

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Frequency Response Frequency response analysis is interesting not only because

it will help us to understand how a system responds to a sinusoidal input, but also because evaluating G(s) with s taking on values along jω axis will prove to be very useful in determining the stability of a closed-loop system.

As we know, jω axis is the boundary between stability and instability. Therefore, evaluating G(jω) along the frequency band will provide information that allows us to determine closed-loop stability from the open-loop G(s).

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Frequency Response

1( ) , 1.1

TsD s KTs

Given the transfer function of a lead compensation

(a) Analytically determine its frequency response characteristics and discuss what you would expect from the result.

1( ) ,1

TjD j KTj

21 1

2

1 ( )tan ( ) tan ( ).

1 ( )

TT TK

T

M = DΦ

• At low frequency, ω→0, M→K, Φ→0.• At high frequency, ω→∞, M→K/α, Φ→0.• At intermediate frequency, Φ > 0.

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Frequency Response(b) Use MATLAB to plot D(jω) with K = 1, T = 1, and α = 0.1 for

0.1 ≤ ω ≤ 100 and verify the prediction from (a).1( ) .

0.1 1sD ss

Using bode(num,den), in this case bode([1 1],[0.1 1]),

MATLAB produces the frequency response of the lead compensation

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y t

nt

2

2 2( ) ,2

n

n n

G ss s

Frequency ResponseFor second order system having the transfer function

we already plotted the step response for various values of ζ.

• The damping and rise time of a system can be determined from the transient-response curve.

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2

1( ) ,( ) 2 ( ) 1n n

G ss s

Frequency ResponseThe corresponding frequency response of the system can be found by replacing s = jω

This G(jω) can be plotted along the frequency axis, for various values of ζ.

2

1( ) .( ) 2 ( ) 1n n

G jj j

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Frequency Response

• The damping of a system can be determined from the peak in the magnitude of the frequency response curve.

• The rise time can be estimated from the bandwidth, which is approximately equal to ωn.

The transient-response curve and the frequency-response curve contain the same information.

?

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Frequency Response Bandwidth is defined as the maximum frequency at which

the output of system will track an input sinusoid in a satisfactory manner.

By convention, the bandwidth is the frequency at which the output is attenuated to a factor of 0.707 times the input.

The maximum value of the frequency-response magnitude is referred to as the resonant peak Mr.

2 0.707

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Bode Plot TechniquesAdvantages of working with frequency response in terms of Bode plots:1. Dynamic compensator design can be based entirely on Bode plots.2. Bode plots can be determined experimentally.3. Bode plots of systems in series can be simply added, which

is quite convenient.4. The use of a logarithmic scale permits a much wider range

of frequencies to be displayed on a single plot compared with the use of linear scales.

Bode plot of a system is made of two curves, The logarithm of magnitude vs. the logarithm of frequency

log M vs. log ω, or also Mdb vs. log ω. The phase versus the logarithm of frequency

Φ vs. log ω.

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Bode Plot Techniques

1 2

1 2

( )( ) ( )( ) .( )( ) ( )

m

n

s z s z s zKG s K

s p s p s p

For root locus design method, the open-loop transfer function is written in the form

For frequency-response design method, s is replaced with jω to write the transfer function in the Bode form

1 20

( 1)( 1) ( 1)( ) .( 1)( 1) ( 1)

m

a b n

j j jKG j K

j j j

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Bode Plot Techniques

12

1

( )( ) ,( )s zKG s K

s s p

To draw the Bode plot of a transfer function, it must be rewritten in magnitude equation and phase equation. For example

10 2

( 1)( ) .( ) ( 1)a

jKG j Kj j

0 12

( ) ( 1) ( ) ( 1),a

KG j K jj j

Then

0 12

( 1),( )

( 1)( ) a

K jKG j

jj

0 12

log log log ( 1)( ) log ( 1) ,( ) a

K jKG jjj

0 1db2

20log 20log ( 1)( ) 20log 20 ( 1) .( ) a

K jKG jjj

Phase Equation

Magnitude Equation

(log)

(db)

Magnitude Equation

Magnitude Equation

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Bode Plot TechniquesExamining all the transfer functions we have dealt with so far, all of them are the combinations of the following four terms:

( )nj

12( ) 2 ( ) 1n nj j

1( 1)j

0K • Gain

• Pole or zero at the origin

• Simple pole or zero

• Quadratic pole or zero

Once we understand how to plot each term, it will be easy to draw the composite plot, since log M and Φ are the additive combination of the magnitude logarithms and the phases of all terms.

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Bode Plot Techniques0K

Magnitude

Phase

Gain

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Bode Plot Techniques( )nj

Magnitude

Phase90n

log log( )n n jj

Pole or zero at the origin

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Bode Plot TechniquesAs example, Bode plot of a zero (jω) at origin will be as follows:

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Approximation Magnitude Phase

Bode Plot Techniques1( 1)j Simple pole or zero

1

1

1

1( 1)j

2( 1) 1j j

( 1)j

1 0

( 1) 45j

( ) 90j

( 1) 1j

( 1) 1j j

( 1)j j

• The point where ωτ = 1 or ω = 1/τ is called the break point.

( 1)j

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Bode Plot TechniquesAs example, Bode magnitude plot of a simple zero (jωτ+1) is given below, with τ = 10.

• The break point lies at ω = 1/τ = 0.1.

db

10( 1)j

db

120log( 2) 3 db( 1)j

db

120log( )( 1)j

• Correction of Asymptote

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Bode Plot TechniquesThe corresponding Bode phase plot of a simple zero (jωτ+1) is given as:

1 1 0

1( 1) 45j

1( ) 90j

• Corrections of Asymptotes by 11°,at ω = 0.02 and ω = 0.5.

• Corresponds to 1/5ωbreak and 5ωbreak

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Bode Plot TechniquesQuadratic pole or zero

12( ) 2 ( ) 1n nj j

• Asymptotes can be used for rough sketch.• Afterwards, correction must be made

according to the value of damping factor ζ.

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Bode Plot Techniques

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Bode Plot Techniques

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Bode Plot: ExamplePlot the Bode magnitude and phase for the system with the transfer function

2000( 0.5)( ) .( 10)( 50)

sKG ss s s

0.5

10 50

2000 0.5( 1)( )

10( 1) 50( 1)

j

j jKG jj

Convert the function to the Bode form,

0.5

10 50

2( 1).

( 1)( 1)

j

j jj

• ωb1 = 0.5, ωb2 = 10, and ωb3 = 50

• One pole at the origin

5 terms will be drawn separately and finally composited1 1 1

0.5 10 502, ( ) , ( 1), ( 1) , ( 1) .j j jj

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Bode Plot Techniques

40

60

20

0

–20

–40

db

2

1( )j

0.5( 1)j

110( 1)j

150( 1)j

ωb1 = 0.5 ωb2 = 10 ωb3 = 50

: Rough composite

–20 db/dec

–20 db/dec–40 db/dec

0 db/dec

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Bode Plot Techniques

40

60

20

0

–20

–40

db

ωb1 = 0.5 ωb2 = 10 ωb3 = 50

: Rough composite: 3-db-corrected composite

+3 db–3 db

–3 db

Final Result

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Bode Plot Techniques

2

1( )j

0.5( 1)j

110( 1)j

150( 1)j

ωb1 = 0.5 ωb2 = 10 ωb3 = 50

0.1

2.52

5010

250

+11°

–11° –11°

+11° –11°

+11°

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Bode Plot Techniquesωb1 = 0.5 ωb2 = 10 ωb3 = 50

Final Result

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Neutral Stability Root-locus technique determine the stability of a closed-

loop system, given only the open-loop transfer function KG(s), by inspecting the denominator in factored form.

Frequency response technique determine the stability of a closed-loop system, given only the open-loop transfer function KG(jω) by evaluating it and performing a test on it.

The principle will be discussed now.

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Neutral StabilitySuppose we have a system defined by

• System is unstable if K > 2.• The neutrally stable points lie on

the imaginary axis, K = 2 and s = ±j.

1,( )( ) 180 .

KG sKG s

All points on

the locus fulfill:1,( )

( ) 180 .KG jKG j

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Neutral Stability The frequency response of the

system for various values of K is shown as follows.

At K = 2, the magnitude response passes through 1 at the frequency ω = 1 rad/sec (remember s = ±j), at which the phase passes through 180°.

After determining “the point of neutral stability”, we know that: K < Kneutral stability stable, K > Kneutral stability unstable.

At (ω = 1 rad/sec, G(jω) = –180°), |KG(jω)| < 1 stable K, |KG(jω)| > 1 unstable K.

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Neutral Stability A trial stability condition can now be stated as follows

If |KG(jω)| < 1 at G(jω) = –180°, then the system is stable.

This criterion holds for all system for which increasing gain leads to instability and |KG(jω)| crosses magnitude = 1 once.

For systems where increasing gain leads from instability to stability, the stability condition is

If |KG(jω)| > 1 at G(jω) = 180°, then the system is stable.

For other more complicated cases, the Nyquist stability criterion can be used, which will be discussed next.

While Nyquist criterion is fairly complex, one should bear in mine, that for most systems a simple relationship exists between the closed-loop stability and the open-loop frequency response.

“ ”

“ ”

8/35Erwin Sitompul Feedback Control System

Homework 8 No.1, FPE (5th Ed.), 6.3.

Hint: Draw the Bode plot in logarithmic and semi-logarithmic scale accordingly.

No.2.

(a) Derive the transfer function of the electrical system given above.

(b) If R1 = 10 kΩ , R2 = 5 kΩ and C = 0.1 μF, draw the Bode plot of the system.