control systems compensator design via root locus.pdf

Assignment # 05 EE 324: Control Systems Semester: Spring: 2013

Due Date: 18.05.2013 at 8a.m.

The following concepts have been tested in this homework

• Lead, Lag and Lead-Lag Compensators

Note: Some Problems are provided with solutions to give you some insight.

Lead Compensator Design

Lag Compensator

1 Homework-5 Solution

Lead-Lag Compensator Design3 Homework-5 Solution

A-'7-6. Consider a system with an unstable plant as shown in Figure 7-34(a). Using the root-locus approach, design a proportional-plus-derivative controller (that is, determine the values of K, and T ~ ) such that the damping ratio of the closed-loop system is 0.7 and the undamped natural frequency w , is 0.5 radlsec.

Solution. Note that the open-loop transfer function involves two poles at s = 1.085 and s = -1.085 .. and one zero at s = -l/Td, which is unknown at this point.

Since the desired closed-loop poles must have w, = 0.5 rad/sec and f = 0.7, they must be located at

s = 0.5/180° f. 45.573"

(6 = 0.7 corresponds to a line having an angle of 45.573" with the negative real axis.) Hence, the desired closed-loop poles are at

The open-loop poles and the desired closed-loop pole in the upper half-plane are located in the diagram shown in Figure 7-34(b). The angle deficiency at point s = -0.35 f j0.357 is

Figure 7-34 (a) PD control of an unstable plant; (b) root-locus diagram for the system.

Example Problems and Solutions


JAMEEL AHMAD

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JAMEEL AHMAD

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Some Solved Problems on Compensators

Figure 7-35 Control system.

This means that the zero at s = - l /Td must contribute 11.939", which, in turn, determines the location of the zero as follows:

Hence, we have

The value of T, is

The value of gain Kp can be determined from the magnitude condition as follows:

or

Hence,

By substituting the numerical values of Td and Kp into Equation (7-12), we obtain

which gives the transfer function of the desired proportional-plus-derivative controller.

A-7-7. Consider the control system shown in Figure 7-35. Design a lag compensator G,(s) such that the static velocity error constant K, is 50 sec-' without appreciably changing the location of the original closed-loop poles, which are at s = -2 d~ j f i .

462 Chapter 7 / Control Systems Design by the Root-Locus Method


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Solution. Assume that the transfer function of the lag compensator is

Since K , is specified as 50 sec-', we have

Thus

Now choose K, = 1. Then

Choose T = 10. Then the lag compensator can be given by

The angle contribution of the lag compensator at the closed-loop pole s = -2 + jV6' is

v'6 v% /Gc(s) I = tan-' - - tan-' -

s=-2+jG -1.9 -1.995

which is small.Thus the change in the location of the dominant closed-loop poles is very small. The open-loop transfer function of the system becomes

The closed-loop transfer function is

To compare the transient-response characteristics before and after the compensation, the unit-step and unit-ramp responses of the compensated and uncompensated systems are shown in Figures 7-36(a) and (b), respectively.The steady-state error in the unit-ramp response is shown in Figure 7-36(~).

Example Problems and Solutions 463


JAMEEL AHMAD

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Figure 7-36 (a) Unit-step responses of the compensated and uncompensated systems; (b) unit- ramp responses of both systems; (c) unit-ramp responses showing steady-state errors.

Unit-Ramp Responses of Compensated and Uncompensated Systems

9 -

Uncompensated system has steady-state error of 0.4

Un~t-Step Responses of Compensated and Uncompensated Systems 1 2 . , , , , , , , , , -

J Compensated system

7----------- Uncompensated system

t Sec

(b)

0

0 4

0 2

Unit-Rarno Resoonse (35 < t < 40)

-

-

. . 40

39 5

39

3 3 8 5 Compensdted system 4 8 38 w

2 3 7 5

5 37 5 5 3 6 5 - Uncompensated system

36

35 5

3535 35 5 36 36 5 37 37 5 38 38 5 39 3 9 5 40

~ / i i i i i i i i ; , o t Sec

(a)

t Sec

( c )



A-7-44. Consider a unity-feedback control system whose feedforward transfer function is given by

Design a compensator such that the dominant closed-loop poles are located at s = -2 f @V"3 and the static velocity error constant K, is equal to 80 sec-'.

Solution. The static velocity error constant of the uncompensated system is K, = # = 0.625. Since K, = 80 is required, we need to increase the open-loop gain by 128. (This implies that we need a lag compensator.) The root-locus plot of the uncompensated system reveals that it is not possible to bring the dominant closed-loop poles to -2 -+ j 2 s by just a gain adjustment alone. See Figure 7-37. (This means that we also need a lead compensator.) Therefore, we shall employ a lag-lead compensator.

Let us assume that the transfer function of the lag-lead compensator to be

where Kc = 128. This is because

Root-Locus Plot of G(s) = 10/[s(s+2)(s+8)]

Figure 7-37 Root-locus plot of G ( s ) = 101 [ S ( S + 2 ) ( s + 8)). Real Axis

10

8

6

4

v, 2 -

4 Do 0 z - -2

-4

-6

-8


-

- Des~red closed-loop

- '----la

-

-

-

-

- lo l0 -5 0 5 10


JAMEEL AHMAD

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JAMEEL AHMAD

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Figure 7-38 Graphical determination of the zero and pole of the lead portion of the compensator.

and we obtain Kc = 128. The angle deficiency at the desired closed-loop pole s = -2 + j 2 f l is

Angle deficiency = 120" + 90" + 30" - 180" = 60"

The lead portion of the lag-lead compensator must contribute this angle.To choose Tl we may use the graphical method presented in Section 7-5.

The lead portion must satisfy the following conditions:

and

The first condition can be simplified as

By using the same approach as used in Section 7-5, the zero ( s = 1 1 ~ ~ ) and pole ( s = PIT,) can be determined as follows:

See Figure 7-38. The value of P is thus determined as

/3 = 14.419



JAMEEL AHMAD

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For the lag portion of the compensator, we may choose

Figure 7-39 (a) Root-locus plot of compensated system; (b) root- locus plot near the origin.

Then

Noting that

the angle contribution of the lag portion is -1.697" and the magnitude contribution is 0.9837.This means that the dominant closed-loop poles lie close to the desired location s = -2 & j 2 G . Thus the compensator designed,

is acceptable. The feedforward transfer function of the compensated system becomes

A root-locus plot of the compensated system is shown in Figure 7-39(a).An enlarged root-locus plot near the origin is shown in Figure 7-39(b).

Real Axis

(a)

Root-Locus Plot of Compensated System


60

40

20 V) .-

8 - -20

-40

-60 -60

- ............ i. . . . . . . . .;. .... . . . . . . .

........ ..;. ........... .;. ...........

q o x ; i . . . . . . . . . . . . . . . . . . . . . . - .:.

-. . . . .:. . . . . . . . .;, . . . . . . . .

-40 -20 0 20 40 60


Figure 7-39 (Continued)

Figure 7-40 (a) Unit step responses of compensated and uncompensated systems; (b) unit- ramp responses of both systems.

To verify the improved system performance of the compensated system, see the unit-step responses and unit-ramp responses of the compensated and uncompensated systems shown in Figures 7-40 (a) and (b), respectively.

Root-Locus Plot of Cqmpensated System near the Origin

Unit-Step Responses of Compensated and Uncompensated Systems

10

8

4

2 a ? , z " -2

-4

-6

-8

t Sec

(4

Chapter 7 / Control Systems Design by the Root-Locus Method

-5 0 5 10 Real Axis

(b)

-

- I - Desired closed-Io ,p -

-

-

-

-

-

poks



Un~t-Ramp Responses of Compensated and Uncompensated Systems

9 -

8 -

7 -

6 -

Uncompensated system -

t Sec

('J)


A-7-9. Consider the system shown in Figure 7-41. Design a lag-lead compensator such that the static velocity error constant K, is 50 sec-' and the damping ratio 5 of the dominant closed-loop poles is 0.5. (Choose the zero of the lead portion of the lag-lead compensator to cancel the pole at s = -1 of the plant.) Determine all closed-loop poles of the compensated system.

Solution. Let us employ the lag-lead compensator given by

where p > 1. Then

K, = lim sG,(s)G(s) s-0

K,(T,S + 1)(T2s + 1) = lims

1

'-' (5, + 1)(pT2s + 1) S ( S + 1) ( s + 5)

- Kc - - 5

The specification that K, = 50 sec-' determines the value of Kc, or

Kc = 250



JAMEEL AHMAD

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We now choose Tl = 1 so that s + ( 1 1 ~ ~ ) will cancel the (s + 1 ) term of the plant. The lead portion then becomes

For the lag portion of the lag-lead compensator we require

where s = s1 is one of the dominant closed-loop poles. For s = sl , the open-loop transfer function becomes

Noting that at s = sl the magnitude and angle conditions are satisfied, we have

where k = 0,1,2,. . . . In Equations (7-13) and (7-14), P and s, are unknowns. Since the damping ratio 5 of the dominant closed-loop poles is specified as 0.5, the closed-loop pole s = s1 can be writ- ten as

where x is as yet undetermined. Notice that the magnitude condition, Equation (7-13), can be rewritten as

Noting that Kc = 250, we have

The angle condition, Equation (7-14), can be rewritten as



We need to solve Equations (7-15) and (7-16) for P and x. By several trial-and-error calculations, it can be found that

Thus

The lag portion of the lag-lead compensator can be determined as follows: Noting that the pole and zero of the lag portion of the compensator must be located near the origin, we may choose

That is,

With the choice of T2 = 6.25, we find

and

( 3.3002 ) ( 3.3002 ) = tan-' --- - tan-' ----- = -1.937" (7-18) -1.74515 -1.89054

Since

our choice of T2 = 6.25 is acceptable. Then the lag-lead compensator just designed can be writ- ten as

Therefore, the compensated system has the following open-loop transfer function:

A root-locus plot of the compensatedsystem is shown in Figure 7-42(a). An enlarged root-locus plot near the origin is shown in Figure 7-42(b).

Example Problems and Solutions 47 1


Real Axis

(a)


The closed loop transfer function becomes

Root-Locus Plot of Compensated System near the Origin

The closed-loop poles are located at

1

0.8

0.6

0.4

0.2

0

-0.2

-0.4

-0.6

-0.8

-1


-I -0.5 0 0.5 1 Real Axis

(b)

- .............. i . .

-. . . . . . . . . . . . j . . . . . . . . . . . . . . . . . . . . . . . . . .

- . . . . . . . . . . . . ;

- . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .:.. ................ -

....: ............

................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . :

.:. ..................

. . . . . . . . . . . . . . : .

- . . . . . . . . . . j . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . -. .:

. . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .i ................. -

: ................ .-

. . . . . . . . . . . . . . . . . . . . . . . . . . . .I...

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ...-


Unit-Step Response of Compensated System 1.4

Figure 7-43 (a) Unit-step response of the compensated system; (b) unit-ramp response of the compensated system.

t Sec

(a)

Unit-Ramp Response of Compensated System

t Sec

(b)

Notice that the dominant closed-loop poles s = -1.8308 f j3.2359 differ from the dominant closed-loop poles s = hs, assumed in the computation of P and T,. Small deviations of the dominant closed-loop poles s = -1.8308 h j3.2359 from s = f s, = -1.9054 + j3.3002 are due to the approximations involved in determining the lag portion of the compensator [See Equations (7-17) and (7-18)].

Figures 7-43(a) and (b) show the unit-step response and unit-ramp response of the designed system, respectively. Note that the closed-loop pole at s = -0.1684 almost cancels the zero at s = -0.16025. However, this pair of closed-loop pole and zero located near the origin produces a long tail of small amplitude. Since the closed-loop pole at s = -17.205 is located



very much farther to the left compared to the closed-loop poles at s = -1.8308 + j3.2359, the effect of this real pole on the system response is very small. Therefore, the closed-loop poles at s = -1.8308 f j3.2359 are indeed dominant closed-loop poles that determine the response characteristics of the closed-loop system. In the unit-ramp response, the steady-state error in following the unit-ramp input eventually becomes 1/K, = $ = 0.02.

A-7-10. Figure 7-44(a) is a block diagram of a model for an attitude-rate control system.The closed-loop transfer function for this system is

The unit-step response of this system is shown in Figure 7-44(b). The response shows high- frequency oscillations at the beginning of the response due to the poles at s = -0.0417 & j2.4489. The response is dominated by the pole at s = -0.0167. The settling time is approximately 240 sec.

It is desired to speed up the response and also eliminate the oscillatory mode at the beginning of the response. Design a suitable compensator such that the dominant closed-loop poles are at s = -2 + j 2 a .

t Hydraulic servo Aircraft I -

Rate gyro

(a)

Unit-Steo Res~onse of Uncom~ensated System

Figure 7-44 (a) Attitude-rate control system; (b) unit-step response.

b I t

50 100 150 200 250 300 Time (sec)

(b)



. .., .• • •--

--

•,1

• " !

I r-,1

II

•/

1/

)00

..

...

•

8-7-7. 'lb9 801\21;ion to such a problem is not unique. "We"imzill _t tlI<>801~lCl11B to the problem in wI1at follows. Nots that frem the requ1remsntstated in the problem. the daD1nant clossd-100p po1BB mat; have ~ • 0.5 and Wn:= 3, or

S=-I.S"±j 2. Nel

Notice that the angle deficiency is . '. . .Angle deficiency = 180 - 120 - 100.894 • - 40.894

Method 1, If 1I'B c:booes the zero ~" the lead """I"""sator at a • -1 80 that itwill cancel the~ pole at a • -1, tIlen the ..............tor pole _ be l"""teclat 8 • -3, or

or

(j. (s) = I<

5+/S+3

)_.Dl..-'Tz

5+1S-f-~

'lb9 value ~ It can be cIetem1ne<1 by Use ~ the magnitude condition.

K = IS(S+.I)30

or

31< 5+/S+ .3

10

S(S+I) =J

$+15i-3

'lb9 open-100p trlInafer _ion is

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jameel

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jameel

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jameel

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,(j,,{r) t7/f) = s (s+ 3)

The closec!-loop transfer fun6tion C(s)!R(s) teoromI8 as follows:

crs-J _ --=",,,,,,,,,,,P__K(s) - S I. + 3 SO +.f

Method 2: Referring to the figure shown below, if we bisect angle OPA and' take20.447· each side, thep the locations of the zero and pole are found as follows:

zero at s • -1.9432pole at s • -4.6458

Thus, Ge(S) can be given as

(iris) = I< 7iS-+/ =k I!... s+/.fP-.1Z =2. 'II K S+f.,,..1Z ,72 S T I T2. S T ,. *~rr .7 S'rf4/N"f

.P

j3

jz

o I % ()-

-it

The value of J( can be determined by 1188 at ~ EgnitueSe condiUCIl.

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jameel

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jameel

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jameel

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'DIe closed-loop tnDsfer functiClll is

eM _ S-./39 (P•.J'/~S+I)

!?(.) - SfSrl)(p.2/S-Z.S.,.. /) T'.S'-/$~(",J'"/Il~.s+/)

= =2,..:'t;.:.,c.;..I'...;S:.....:T--=.S':.;.,;./.~'3:.:r:...- .....,...po2/J'z sJ +-/. Z/,sZ s" + 3.p~5<-s + s: nF

It is interssting to ccmpare the ststic velocity error constants for thetwo systems designed above.

For the system designed by MethDd 1.

K~ =.Ii- s f = 3S ..p S(St3)

For the system designee! by IIet!lod 2.

o.$"/ fZp.s+ I

P·2/S-ZS+/

It)S(S+I)

'DIe system designed by MethDd 2 gives a 1s<ger value ~ the static velocityerror ccnstant. '!his means that the. system designed by _ 2 will giveSIIIS1ler steady-state errors in following ramp inputs t:ban the system designedby_l.

In what follows., we c:ompare the unit-step respcnses Of the three systems:the original unc:mpensated system, the system designed by _ 1, and thesystem designed by _ 2. 'DIe HATLl\Il _am used to obtain the unit-stepresponse curves is given below. 'DIe resulting unit-step response curves areshown on the next page.

" ........ Comparison of unlt-step responses for three systems ........

num - [0 0 10):den - [1 1 10/:num1 - (0 0 9]:denl - [1 3 91:num2 - (0 0 2.644 5.1381:den2 - [0.2152 1.2152 3.644 5.138);I - 0:0.02:8:c - steplnum,den,tJ:c1 - atep(num1.den1,t);c2 - step(num2,den2,t);plotlt,c, ',',t,c1,'.' ,t,c2,'-.')gridtille('Comparlaon of Unlt-Slep Responses for Three Svotemo')xleball'l sec',ylabel('Outputo') .textn.5,1.6,'Uncompensated system')leXlI1.1.0.5,'CompensetedOVOlem with K - 0.5138. Tl - 0.5148, T2 - 0.2152'),exl(1.1,O.3,'Compenseted ovotem with K - 0;3, T1 - 1, T2 - 0.3333')

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jameel

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jameel

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•,•• • •,..••

•• . I\... -, .••

·· .• "'\. I \· ,- r-..·, · \.V· \ 7\ / .... Ul• .0.814 ..... io'.... ..n • ....~ po.

•....U

,

,,

8-7-8. The closed-loop transfer ~ion C(s)/l!(s) is given by

we obtain

or

whic!h results in

J<=K

8-7-9. The an;le deficiency at the closed-l""" pole e'. -2 + j2 .[3ie

1SO-- 120· - 90-. - .30

The lead O.I........tor _ CXlDtri_ 30'.

IAIt U8 chooee -the zero of the lead. C'liJi'Iensatar at 8 • -2. 'ft8l, t1wpate .of the cc ili&lAter III18t be located at a = -4. 1bu8,

~ {r' = K Si"2...,."/ 'S't ,.

The gain II: ie cIetem1nac1 fraa the _i_ CCIldition.

- 1-1)7 -


jameel

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jameel

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K = IS(S+ 9-)I .' = /.,. /0 S=-2~}2.fi

or

Hence,

s+zl

s-t- SL s- I .= IS (o•.s S +1) .5"- -2"j,g

~~(.J) =/.6 s+z$+ 9-Next, we shall obtain unit-step responses of the original system and the

. canpensated syst8D. '!be original system has the following Closed-loop· trans- __ter t1mctiOlu

efT) _ /t)

RfI) - st~2s +/0

'!be cuapensated system has the following Closed-loop transfer :f1.mCticn:

C{!) ._ /61«S) - $&+¥-s+16

The un!t-step respmse curves of the original syst.em and canpensated system areshown below.

Ir-,

"-.r'- \ I...m<, \ L--- -! "~V

n#l..!J

0.2

1.4

r-0.1

1

1.2

0.4

1 Ui 2 2..5 3 3.15SIc

4.5 5

B-7-10. The aDg1e deficiency is

ISO- - 135- -135- ~ - go-

A lead. CUIIlensatar can· CXIIltribute 90-. Let us choose tbe zero of the la4

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jameel

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jameel

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UnIt_....-of"'_......'1''''-", ....,.,....

00 , ,.. to "

~7-11. The original uncaapensated system bas the follawiilg closed-locp traDefer fUnCtiau

..£f:2.. _ /~

I?CS) - s· +~S-t-/'The two closed-locp' poles are located at B • -2 ± j2,J3. Chcoee a lag~-Bator of tI1e follawiilg fonu

The pole and zero of tI1e lag 0 "censator _ I1e located cloee to tI1e origill.Let us chooee T • 20. .'Iben, the lag Cillillensatar tecaa_

=

NoUoe t:bat

Is+(),f~rI = (J. 'P~&JS+-D.() I ... -Z+jUl'

S+,.DJ"j = !-UJ"+-jz-/3 _ /-/.ff+;"25S+D.DI - -. S~_Z.foJ2.a

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jameel

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Un~t-Ramp Responses of Compensated and Uncompensated Systems

9 -

8 -

7 -

6 -

Uncompensated system -

t Sec

('J)


A-7-9. Consider the system shown in Figure 7-41. Design a lag-lead compensator such that the static velocity error constant K, is 50 sec-' and the damping ratio 5 of the dominant closed-loop poles is 0.5. (Choose the zero of the lead portion of the lag-lead compensator to cancel the pole at s = -1 of the plant.) Determine all closed-loop poles of the compensated system.

Solution. Let us employ the lag-lead compensator given by

where p > 1. Then

K, = lim sG,(s)G(s) s-0

K,(T,S + 1)(T2s + 1) = lims

1

'-' (5, + 1)(pT2s + 1) S ( S + 1) ( s + 5)

- Kc - - 5

The specification that K, = 50 sec-' determines the value of Kc, or

Kc = 250



jameel

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jameel

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We now choose Tl = 1 so that s + ( 1 1 ~ ~ ) will cancel the (s + 1 ) term of the plant. The lead portion then becomes

For the lag portion of the lag-lead compensator we require

where s = s1 is one of the dominant closed-loop poles. For s = sl , the open-loop transfer function becomes

Noting that at s = sl the magnitude and angle conditions are satisfied, we have

where k = 0,1,2,. . . . In Equations (7-13) and (7-14), P and s, are unknowns. Since the damping ratio 5 of the dominant closed-loop poles is specified as 0.5, the closed-loop pole s = s1 can be writ- ten as

where x is as yet undetermined. Notice that the magnitude condition, Equation (7-13), can be rewritten as

Noting that Kc = 250, we have

The angle condition, Equation (7-14), can be rewritten as



We need to solve Equations (7-15) and (7-16) for P and x. By several trial-and-error calculations, it can be found that

Thus

The lag portion of the lag-lead compensator can be determined as follows: Noting that the pole and zero of the lag portion of the compensator must be located near the origin, we may choose

That is,

With the choice of T2 = 6.25, we find

and

( 3.3002 ) ( 3.3002 ) = tan-' --- - tan-' ----- = -1.937" (7-18) -1.74515 -1.89054

Since

our choice of T2 = 6.25 is acceptable. Then the lag-lead compensator just designed can be writ- ten as

Therefore, the compensated system has the following open-loop transfer function:

A root-locus plot of the compensatedsystem is shown in Figure 7-42(a). An enlarged root-locus plot near the origin is shown in Figure 7-42(b).



Real Axis

(a)


The closed loop transfer function becomes

Root-Locus Plot of Compensated System near the Origin

The closed-loop poles are located at

1

0.8

0.6

0.4

0.2

0

-0.2

-0.4

-0.6

-0.8

-1


-I -0.5 0 0.5 1 Real Axis

(b)

- .............. i . .

-. . . . . . . . . . . . j . . . . . . . . . . . . . . . . . . . . . . . . . .

- . . . . . . . . . . . . ;

- . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .:.. ................ -

....: ............

................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . :

.:. ..................

. . . . . . . . . . . . . . : .

- . . . . . . . . . . j . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . -. .:

. . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .i ................. -

: ................ .-

. . . . . . . . . . . . . . . . . . . . . . . . . . . .I...

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ...-


Unit-Step Response of Compensated System 1.4

Figure 7-43 (a) Unit-step response of the compensated system; (b) unit-ramp response of the compensated system.

t Sec

(a)

Unit-Ramp Response of Compensated System

t Sec

(b)

Notice that the dominant closed-loop poles s = -1.8308 f j3.2359 differ from the dominant closed-loop poles s = hs, assumed in the computation of P and T,. Small deviations of the dominant closed-loop poles s = -1.8308 h j3.2359 from s = f s, = -1.9054 + j3.3002 are due to the approximations involved in determining the lag portion of the compensator [See Equations (7-17) and (7-18)].

Figures 7-43(a) and (b) show the unit-step response and unit-ramp response of the designed system, respectively. Note that the closed-loop pole at s = -0.1684 almost cancels the zero at s = -0.16025. However, this pair of closed-loop pole and zero located near the origin produces a long tail of small amplitude. Since the closed-loop pole at s = -17.205 is located



very much farther to the left compared to the closed-loop poles at s = -1.8308 + j3.2359, the effect of this real pole on the system response is very small. Therefore, the closed-loop poles at s = -1.8308 f j3.2359 are indeed dominant closed-loop poles that determine the response characteristics of the closed-loop system. In the unit-ramp response, the steady-state error in following the unit-ramp input eventually becomes 1/K, = $ = 0.02.

A-7-10. Figure 7-44(a) is a block diagram of a model for an attitude-rate control system.The closed-loop transfer function for this system is

The unit-step response of this system is shown in Figure 7-44(b). The response shows high- frequency oscillations at the beginning of the response due to the poles at s = -0.0417 & j2.4489. The response is dominated by the pole at s = -0.0167. The settling time is approximately 240 sec.

It is desired to speed up the response and also eliminate the oscillatory mode at the beginning of the response. Design a suitable compensator such that the dominant closed-loop poles are at s = -2 + j 2 a .

t Hydraulic servo Aircraft I -

Rate gyro

(a)

Unit-Steo Res~onse of Uncom~ensated System

Figure 7-44 (a) Attitude-rate control system; (b) unit-step response.

b I t

50 100 150 200 250 300 Time (sec)

(b)



control systems compensator design via root locus.pdf

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