application to control design

Lecture 10Feedback Control Systems

President University Erwin Sitompul FCS 10/1

Dr.-Ing. Erwin SitompulPresident University

http://zitompul.wordpress.com2 0 1 3


Application to Control Design( )Y s

+–

( )R sK 2

1( 1)s s

Root locus of G(s) = 1/[s(s+1)2]

Bode plot for KG(s) = 1/[s(s+1)2], K = 1.

Chapter 6 The Frequency-Response Design Method


Application to Control Design

I

II

III

IVI

II

III

IV

The detour of C1 excludes the pole on the imaginary axis.

The detour of C1 includes the pole on the imaginary axis.already done last time

will be done now



Application to Control Design

I

II

III

IV

In this new detour of C1, only the section IV-I is changed.

As in previous evaluation, this path can be evaluated by replacing

,js r e 90 90 , clockwise.

2

1( )

( 1)j j js r eG s

r e r e

Performing it,

1jr e

jR e

• Evaluation of G(s) forms a half circle with a very large radius.

• The half circle starts at 90° and ends at –90°, routing in counterclockwise direction.



Application to Control Design Combining all four sections, we

will get the complete Nyquist plot of the system.

III

IV

I

II



Application to Control Design There is one encirclement of –1 in counterclockwise direction (

N = –1 ). There is one pole enclosed by C1 since we choose the contour

to enclose the pole at origin ( P = 1 ). The number of closed-loop pole in RHP

Z = N + P = –1 + 1 = 0 No unstable closed-loop pole (the same result as when

encircling the pole at origin to the right).



Application to Control Design If we choose K > 2, the plot will encircle –1 once in clockwise

direction ( N = 1 ). Since P = 1, Z = N + P = 2

The system is unstable with 2 roots in RHP.• Thus, the matter of choosing

a path to encircle poles on the imaginary axis does not affect the result of analyzing Nyquist plot.



Stability Margin A large fraction of control systems behave in a pattern

roughly similar to the system we just discussed: stable for small gain values and becoming unstable if the gain increases past a certain critical point.

Two commonly used quantities that measure the stability margin for such systems are gain margin (GM) and phase margin (PM).

They are related to the neutral stability criterion described byIf |KG(jω)| < 1 at G(jω) = –180°, then the system is stable.“

”



Stability Margin Gain margin (GM) is the factor by which the gain can be

raised before instability results. A GM ≥ 1 is required for stability.

Phase margin (PM) is the amount by which the phase of G(jω) exceeds –180° when |KG(jω)| = 1. A positive PM is required for stability.

• PM and GM determine how far the complex quantity G(jω) passes from –1.

• Within stable values of GM and PM, as can be implied from the figure, there will be no Nyquist encirclements.



Stability Margin Crossover

frequency, ωc, is referred to as the frequency at which the gain is unity, or 0 db.



Stability MarginChapter 6 The Frequency-Response Design Method

PM is more commonly used to specify control system performance because it is most closely related to the damping ratio of the system.

The relation between PM, ζ, and furthermore Mp for a second-order system can be summarized in the following two figures.

2

2 2( ) ,2

n

n n

T ss s


Stability Margin Conditionally stable system:

a system in which an increase in the gain can make it stable.

Several crossover frequencies exist, and the previous definition of gain margin is not valid anymore.



Example on Stability Margin As one example, determine the stability properties as a

function of gain K for the system with the open-loop transfer function

2

3

( 10)( ) .K sKG ss

Real Axis

Imag

inar

y A

xis

-1 -0.8 -0.6 -0.4 -0.2 0 0.2

-1

-0.5

0

0.5

1



Example on Stability Margin We again choose path IV-I as a half circle with a very

small radius, routing from negative imaginary axis to positive imaginary axis in counterclockwise direction.

This path can be evaluated by replacing

,js r e 90 90 , counterclockwise.

I

II

IIIIV

2

3

( 10)( )

( )j

j

js r e

r eG s

r e

3 3

100jr e

3 3jR e • Evaluation of G(s) forms an one-and-

a-half circle with a very large radius.• Inserting the value of θ, the half

circle starts at 270° and ends at –270°, routing in clockwise direction.


Performing it,


Example on Stability MarginChapter 6 The Frequency-Response Design Method

• The Nyquist plot was drawn for a stable value K = 7.• There is one cc encirclement and one ccw

encirclement of the –1 point• Net encirclement equals zero the system is stable.


CompensationChapter 6 The Frequency-Response Design Method

As already discussed before, dynamic compensation is typically added to feedback controllers to improve the stability and error characteristics when a mere proportional feedback alone is not enough.

In this section we discuss several kinds of compensation in terms of their frequency-response characteristics.

To this point, the closed-loop system is considered to have the characteristic equation 1 + KG(s) = 0.

With the introduction of compensation, the closed-loop characteristic equation becomes 1 + KD(s)G(s) = 0.

All previous discussions pertaining to the frequency response of KG(s) applies now directly to the compensated case, where the response of KD(s)G(s) is of interest.

We call this quantity L(s), the “loop gain” or open-loop transfer function, L(s) = KD(s)G(s).


PD CompensationChapter 6 The Frequency-Response Design Method

The PD compensation transfer function is given by

( ) ( 1)DD s T s

• Frequency-Response of PD Compensation

The stabilizing influence is apparent by the increase in phase and the corresponding 20 dB/dec- slope at frequencies above the break point 1/TD.

We use this compensation by locating 1/TD so that the increased phase occurs in the vicinity of crossover (that is, where |KD(s)G(s)| = 1 Effect: PM is increased.


PD CompensationChapter 6 The Frequency-Response Design Method

Note: The magnitude of the compensation continues to grow with increasing frequency.

This feature is undesirable because it amplifies the high-frequency noise.

• Frequency-Response of PD Compensation


Lead CompensationChapter 6 The Frequency-Response Design Method

In order to alleviate the high-frequency amplification of the PD compensation, a first order pole is added in the denominator at frequencies substantially higher than the break point of the PD compensator.

Effect: The phase lead still occurs, but the amplification at high frequency is limited.

This resulting lead compensation has a transfer function of

1( ) , 11

TsD sTs

where 1/α is the ratio between the pole/zero break-point frequencies. • Frequency-Response of

Lead Compensation



Note: A significant amount of phase lead is still provided, but with much less amplification at high frequencies.

A lead compensator is generally used whenever a substantial improvement in damping of the system is required.

The phase contributed by the lead compensation is given by1 1tan ( ) tan ( )T T

It can be shown that the frequency at which the phase is maximum is given by

max1

T

The maximum phase contribution, i.e. the peak of D(s) curve, corresponds to

max1sin ,1

max

max

1 sin1 sin



For example, a lead compensator with a zero at s = –2 (T = 0.5) and a pole at s = –10 (αT = 0.1, α = 0.2) would yield the maximum phase lead at

max1

T

1 4.47 rad s(0.5) 0.2



The amount of phase lead at the midpoint depends only on α and is plotted in the next figure.

We could increase the phase lead up to 90° using higher values of the lead ratio, 1/α.

However, increasing values of 1/α also produces higher amplifications at higher frequencies.

A good compromise between an acceptable PM and an acceptable noise sensitivity at high frequencies must be met.

• How?

• How?


First Design Using Lead Compensation Chapter 6 The Frequency-Response Design Method

Find a compensation for G(s) = 1/[s(s+1)] that will provide a steady-state error of less than 0.1 for a unit-ramp input. Furthermore, an overshoot less than 25% is desired.

ss 20

1 1lim ( ), ( )1 ( ) ( )s

e s R s R sKD s G s s

ss 0

1lim( ) 1 ( 1)s

e ss KD s s

1 0.1(0)KD

0lim ( )v s

K s L s

(0) 10KD

0lim ( ) ( )ss KD s G s

(0)KD

ss1

v

eK

• KD(0) must be greater than 10, so we pick K = 10.



• Sufficient PM to accommodate overshoot requirement (25%) is taken to be 45°.

• From the read of Bode Plot, the existing PM is 20° additional phase of 25° at crossover frequency of ω = 3 rad/s

45°



• However, adding a compensator zero would shift the crossover frequency to the right also requires extra additional PM or gain attenuation.

• To be save, we will design the lead compensator that supplies a maximum phase lead of 40°.

5 • As can be seen, 1/α = 5 will accomplish the goal.

• The maximum phase lead from the compensator must occur at the crossover frequency.

• This requires trial-and-error in placing the compensator’s zero and pole.

• The best result will be obtained when placing zero at ω = 2 rad/sec and the pole at ω = 10 rad/s.

12110

1( ) 101

sKD s

s

K = 10


Design Procedure of Lead Compensation Chapter 6 The Frequency-Response Design Method

1. Determine the open-loop gain K so that the steady-state errors are within specification.

2. Evaluate the PM of the uncompensated system using the value of K obtained above.

3. Allow for extra margin (about 10°) and determine the needed phase lead φmax.

4. Determine α from the graph.5. Pick ωmax to be at the wished crossover frequency, thus the

zero is at 1/T and the pole is at 1/αT.6. Draw the frequency response of the compensated system

and check the new PM.7. Iterate on the design if the requirement is not met. This is

done by changing the position of the pole and zero.


Homework 10 No.1, FPE (6th Ed.), 6.25. No.2

Design a compensation for the system

Due: Wednesday, 11.12.2013.


2

8( )(1 )(1 3)

G jj j

which will yield an overall phase margin of 45° and the same gain crossover frequency ω1 as the uncompensated system.Hint: Use MATLAB to draw the Bode plots before and after the compensation. Check the PMs. Submit also the two Bode plots.

application to control design

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