Ogata Root Locus

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S i m il ar l y, t h e p r o g r a m f o r t h e f o u r t h - o r d e r t r a n s f e r f u n c t i o n a p p r o x i m a t i o n w i t hT 0.1 sec is[num,denl = pade 0.1, 4);printsys num, den, st )

numlden =sA4 2O0sA3 + 180O0sA2 840000~+ 16800000sA4 + 200sA3 18000sA2 + 840000s + 16800000

N o t ic e t h a t t h e p a d e a p p r o x i m a t io n d e p e n d s o n t h e d e a d t i m e T a n d t h e d e s ir e d o r d e rf o r t h e a p p r o x im a t in g t r a n s f e r f u n c t io n .

EXAMPLE PROBLEMS AND SOLU TIONSA 6 1. Sketch the roo t loci for the system shown in Figure 6-39 a). The gain is assumed to be posi-tive.) Obse rve that for small or large values of the system is overdam ped and for med ium val-ues of K it is underdamped.

Solution The p rocedu re fo r plotting the ro ot loci is as follows:1 Locate the open-loop poles and zeros on the complex plane. Root loci exist on th e negative

real axis between 0 and -1 and between -2 and -3.2. The numb er of open-loop poles and that of finite zeros are the same.This means that there

are no asymptotes in the complex region of the s plane.

2)

Figure 6 39a) C ontrol system; b) root-locus plot.Chapte r 6 / Root-Locus Analysis


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3 De term ine the breakaway an d break- in poin ts.The character is tic equat ion fo r the sys tem IS

The breakaway a nd break- in poin ts are determined f romd K 2 s l ) s 2 ) s 3 ) s s 1 ) 2 > 5 )rls [ s 2 ) s 3 )1 2

as follows:

Not ice that bo th poin ts are on root loci. Therefore , they a re ac tual breakaway or break- inpoin ts . At poin t = -0.634, th e value of K is

Similarly, at = -2.36h

B ecause po in t s = -0.634 lies betw een two poles,it is a breakaw ay point, and because points = -2.366 l ies between two zeros, i t is a break -in point. )

4. Determ ine a suf f ic ient numbe r of po in ts t h d sa tisfy the angle condi tion . I t can he foundthat the root loci involve a circle with center at -1.5 tha t passes through the breakaway andbreak-in points .) The root-loc us plot for this system is shown in Figure 6-3Y h).

Note that this system is s table for m y positive value of K since all the ro ot loci lie in the left-half s plane.Small kalues of I : 0 K < 0.0718) correspond to an overdampcd sys tem. Medium value\

01 I< 0 .0 7 1 8 . : K ; 14) cor respond to a n und erda mp ed sys tem. Final ly . large values olK 14 = K ) correspond to an ov erda mp ed systern. With a large va lue of K , the s teady s tate canbe I-eachcd n much shorter t ime than with a \mall value o I


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A 6 2. Sketch the ro ot loci of the control system shown in Figure 6-40(a).Solution The op en-loop poles are located a t s = 0, s = -3 j4, and s = -3 4. root locusbranch exists on the real axis between the origin and -oo.There are three asymptotes for the root1oci.The angles of asy mptote s are

&18O0(2k 1Angles of asymptotes = 3 = 60, -60, 180

Referring to E quation (6-13), the intersection of the asymptotes and th e real axis is obtained as

Next we check the breakaw ay and bre ak-in points. For this system we have= - s s 2 6s 25)

Now we set

which yields

Figure 6-40(a) Con trol system; (b) root-locus plot.Chapter 6 Root Locus Analysis


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Notice that a t po in ts s = -2 ~2.0817he ang e co ndition is not satisf ied. He nce , they a re nei-ther b reakaw ay nor brea k-in points . In fact, if we c alculate the value of K , we obta in

(To be an actual breakaway or break- in poin t , the cor responding value of K must be rea l andpositive.)Th e angle of dep ar tu re f rom the complex pole in the u pper hal f s plan e is

Th e points where root-locu s branches cross the imaginary axis may b e found by substituting= j w in to the character is t ic equat ion and so lv ing the equat ion for and K as follows: Notingthat the character is t ic equat ion is

we have

which yields

Root-locus branches cross the imaginary axis at w 5 and w -S.The value of gain K at thecrossing points is 150. Also, the root-locus b ranch o n the real axis touches th e imag inary axis atw = 0 . Figure 6-40(b) shows a root-locus plot for the svstern.

I t is note d that if the or de r of the nu me rato r of G s ) H s ) s lower than tha t of the denomi-nato r by two or m ore, and if so me of the closed -loop poles move o n the root locus tow ard th e r ightas gain K i s increased , then o the r c losed- loop poles mus t m ove toward the lef t as gain K is in-crea sed.T his fact can be se en clearly in this prob lem . If the gain K is increased from K = 34 toK = 68, the complex-conjugate closed-loop poles are move d from = -2 13.65 to = 1 + j4:the th i rd pole i s mov ed f rom = -2 (which cor responds to K = 34) to = -4 (which corre -s ponds to K = 68).Thus, the m ovem ents of two com plex-conju gate closed-loop poles to th e r ightby on e unit cause the remaining closed-loo p pole (real pole in this case) to mo ve to the left by twounits.

A 6 3. Consider the system shown in Figure 6-41(a). Sketch the ro ot loci for the system. Ob serve tha tfor smal l o r large values of K t he s y s t em i s unde rdamp ed and fo r med ium va lues o f K it isove rdamped .Solution A root locus ex is ts on the rea l ax is between the or ig in and m. Th e angles of asymp-to tes of the root- locus branches are o bta ined as

+180(2k + 1)Angles of asymptotes = 3 60, -60 -180Th e intersection of the asymptotes and t he real axis is located o n the real axis at

xample Problems and Solutions


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Figure 6-41a) Co ntrol system;bj root-locus plot. b)

The breakaway and break-in points are found from d K / d s = 0. Since the characteristic equatio n iss3 + 49' + 5s + K = 0we haveK = - s3 + 4s2 + 5 s )

Now we set

which yields s = -1, s = -1.6667Since these points are on root loci, they are actual breakaway or break-in points. At po int s = -1,the value of K is 2 , and at point s = -1.6667, the value of K is 1.852.)The angle of dep arture from a complex pole in the upper half s plane is obtaine d from

= 1800 153.430 go0or6 = -63.43

The root-locus branch from the complex pole in the upp er half s plane breaks in to the real axisat s = -1.6667.Next we de term ine the poin ts where roo t-locus branche s cross the imaginary axis. By substi-tuting s = jw into the characteristic equation , we have( j ~ ) ~4(jw) ' + 5 ( j w ) + K = 0

or( K 4w2)+ jo(5 w2)=

from which we obtainw = r t f l , K = 2 0 or w = O , K = O

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Root-locus branches cross the imaginary axis at fi nd w fl.he root-locus branchon the real axis touches the w axis at w 0. A sketch of the root loci for the system is shown inFigure 641 b).

Note that since this system is of third order, there are three closed-loop poles.The nature ofthe system response to a given input depends on the locations of the closed-loop poles.

For 0 < K < 1.852, there are a set of complex-conjugate closed-loop poles and a real closed-loop pole. For 1.852 K < 2 , there are three real closed-loop poles. For example, the closed-loop poles are located at

s -1.667, s -1.667, s -0.667, for K 1.852-1, s - 1 s -2, for K 2

For 2 < K , there are a set of complex-conjugate closed-loop poles and a real closed-loop poleThus, small values of K ( 0 < K < 1.852) correspond to an underdamped system. Since the realclosed-loop pole dominates, only a small ripple may show up in the transient response.) Mediumvalues of K (1.852 K < 2 ) correspond to an overdamped system. Large values of K ( 2 < K )correspond to an underdamped system. With a large value of K , he system responds much fasterthan with a smaller value of K .Sketch the root loci for the system shown in Figure 6-42(a).Solution The open-loop poles are located at 0 , s -1, s -2 j3, and -2 3. A rootlocus exists on the real axis between points s 0 and s -1. The angles of the asymptotes arefound as follows:

+180(2k 1 )Angles of asymptotes 45 , - 4 j 0 , 135 ,4

4Figure 6 42a) Control system; b) root-locus plot.

Example Problems and Solutions


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Th e intersection of the asym ptotes and the real axis is found from

Th e breakaway and break- in poin ts are found f rom d K / d s 0 . Noting thatK - s s l ) s 2 4 s 13 - s4 5s3 17s2 13s)

we have

d = - 4s3 15s2 34s 13 0dsfrom which we get

Point s -0.467 i s on a root locus .T l~erefore ,t is an actual breakaway p oint.T he gain values Kcorresponding to poin ts -1.642 2.067 are com plex quant i t ies . S ince the gain values a renot real positive, these points are neither breakaway no r break-in points .The angle of de par tu re f rom the com plex pole in the upper half s plane is

Next we shall f ind the points where root loci may cross the w axis . Since t he characteris ticequation is

by substituting s jw in to i t we obtain

f rom which we o btain.6125, K 37.44 o r 0, K 0

The root- locus branches that ex ten d to the r ight-half p lan e cross the imaginary axis a tw 11.6125. Also, the root-locus branch o n the real axis touches the imaginary axis at 0 Fig-ure 6-42 b) shows a sketch of the root loci for the system. Notice that each root-locus branch thatexte nds to th e r ight half s plane crosses its own asymptote.Chapter Root Locus Analysis


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Ad-5. Sketch the root loci for the system shown in Figure 6-43 a).Solution. A root locus exists on the real axis between points s -1 and s -3.6. The asymp-totes can be de term ined as follows:

+180 2k + 1Angles of asymptotes 90 -903 - 1The intersection of the asymptotes and the real axis is found from

Since the characteristic equ ation iswe have

The breakaway and break-in points are found fromd (3s + 7 . 2 s ) ( s + 1) ( s 3 + 3.6s ) 0s S +

Figure 6 43a) Control system; b) root-locus plot.Example Problems and Solutio ns


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from which we get

Point s 0 corresponds to the actual breakaway point. But points s 1.65 0.9367 are neitherbreakaway nor break-in points, because the corresponding gain values K become complexquantities.

To check the points where root-locus branches may cross the imaginary axis, substitute jwinto the characteristic equation, yielding.( j ~ ) ~3 . 6 ( j ~ ) ~Kjw K 0

Notice that this equation can be satisfied only if w 0, 0. Because of the presence of a dou-ble pole at the origin, the root locus is tangent to the jw axis at 0. The root-locus branches donot cross the jw axis. Figure 6-43(b) is a sketch of the root loci for this system.

A 6 6. Sketch the root loci for the system shown in Figure 6-44(a).Solution. A root locus exists on the real axis between point s -0.4 and -3.6. The angles ofasymptotes can be found as follows:

*180(2k 1)Angles of asymptotes 90, -903 - 1

Figure 6-44(a) Control system; (b) root-locus plot.

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Th e intersect ion of the asymptotes a nd th e real axis is obtain ed from

Next we shal l find th e breakaw ay points. Since the characterist ic equation is

we have

The breakaway an d break-in points a re found from

from which we get

Thus, the breakaway or break-in points a r e a t 0 and -1.2. Note tha t -1.2 is a do ubleroot . Wh en a double root occurs in dK /ds a t point s -1.2, d2K /(ds 2) 0 at this poin t .Thevalue of gain K at point -1.2 is

This means that w ith K 4.32 the characterist ic equation has a t riple root at po in ts -1.2.Thiscan b e easily verified as follows:

Hen ce , thr ee root - locus bran ches mee t a t point -1.2. The angles of depa r tures a t point-1.2 of the root locus branches that appro ach the asymptotes ar e 80/3, that is, 60 and-60 . (See Problem A-6-7.Finally, we shall examin e if root-locus b ranch es cross the imaginary axis. By su bstituting s jwinto the ch aracteristic equa tion, we hav e

This eq uat ion ca n be satisfied only if w 0 0.A t po i n t w 0 the roo t locus is tangent tothe j o axis because of the presence of a doub le pole at the origin. Th ere ar e no points that roo t-locus branche s cross the imag inary axis.

A sketch of the root loci for this system is shown in Figure 6-44(b).xample Problems and So lutions


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A-6-7. Referring to Problem A-6-6 obtain the equations for the root-locus branches for the systemshown in Figure 6-44(a). Show that the root-locus branches cross the real axis at the breakawaypoint at angles 0 .Solution The equations for the root-locus branches can be obtained from the angle condition

which can be rewritten as/ s 0.4 b s + 3.6 *180(2k 1)

By substituting u jw we obtain

By rearranging, we havetan- -) tan- tan- + tan- L) *l8O0(2k 1 )u + 0.4 u 3 6Taking tangents of both sides of this last equation, and noting that

we obtain

which can be simplified to

which can be further simplified to

For u -1.6 we may write this last equation as

Chapter Root-Locus nalysis


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which gives the equat ions for the roo t-locus as follows:w o

T h e e q u a t i o n w = 0 represe nts th e real axis. Th e roo t locus for 0 K o is between points= -0.4 a nd = -3.6. (The rea l axis oth er than this l ine segment and the or igin = 0 corre-sponds to the roo t locus for w K 0.The equa t ions

represent the complex branches for 0 K m These two branches lie between = -1.6 a nd= 0 [See Figure 6-44(b).] The slopes of th e complex root-locus branch es at the breakaw aypoint a = -1.2) can be found by evaluating d o l d a of Equat ion 6-21) a t point = -1.2.

Since tan- 60, the root-locus bran ches intersect the real axis with angles +60A 6 8. Consider t he system shown in Figure 6-45(a), which has a n unstable feedforw ard transfer func-tion. Sketc h the root-locus plot an d locate th e closed-loop poles. Show that, a lthough the closed-loop po les lie o n the n egative real axis and t he system is not oscillatory, the un it-step respon se curvewill exhibit overshoot.

Solution. Th e root-locus plot for this system is shown in Figure 6-45(b).The closed-loop poles arelocated at s = -2 a nd s = 5.

Th e closed-loop transfer function becomes

a)Figure 6 45:a) Con trol system; b) root-locus plotExample Problems and Solutions

losed-loop zero


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Figure 6 46Unit-step responsecurve for the systemshown in Figure6-45 (a).

The unit-step response of this system is

The inverse Laplace transform of C ( s )givesc(t) 1 + 1 . 66 6~ -~ ' 2 .666e- , for t 0

The unit-step response curve is shown in Figure 6-46. Although the system is not oscillatory, theunit-step response curve exhibits overshoot. (This is due to the presence of a zero at s -1.)

A 6 9. Sketch the root loci of the control system shown in Figure 47(a) . Determine the range of gainK for stability.Solution. Open-loop poles are located at s 1 s -2 + j d nd -2 d . A root locusexists on the real axis between points s 1 and 03. The asymptotes of the root-locusbranches are found as follows:

*180 2k + 1 )Angles of asymptotes 3 60, -60, 180The intersection of the asymptotes and the real axis is obtained as

The breakaway and break-in points can be located from d K / d s 0 . SinceK - r l ) ( s 2+ 4s + 7) - (s3 3s2 + 3s 7)

we have

which yieldss I ) ~ 0

Chapter 6 Root Locus Analysis


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(a)igure 6-47(a) C ontro l system; (b) root-locus plot.

Thus the equat ion d K / d s = 0 has a double root t 1. (This means that the characteris ticequat ion has a t r iple root a t s = -1.) T h e breakaway point is located at = -1. Thre e root- locusbranches meet a t th is breakaway point .Th e angles of depar ture of the branches a t the breakawaypoint are ilX0/3, that is. 60 and -60 .We shall next det erm ine the po ints where root-locus branches may cross the imaginary axis.Not ing that the character is tic equat i on is

.r l ) . s2 4s + 7 + K 0o r

r 3 , ~ ~3 . ~ 7 K = owe substitute j w in to i t and ob tain

( jw ) 3 ( j ~ ) ~3 ( j w ) 7 + K -By rewriting this last equati on, we have

( K 7 3w2) ,043 w2 0This equatio n is satisfied when

= K = 7 + 3 w l 6 o r w = 0Example Problems and Solutions


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Th e root-locus branches cross the imaginary axis at w = h d where K = 1 6 ) and w = 0 (whereK = 7). Since the value of gain K at th e origin is 7, th e range of gain value K for stability is

Figure 6-47(b) shows a sketch of th e root loci for the system. Notice th at al l branches consist ofpar ts of straig ht lines.The fact that th e root-locu s branch es consist of straight lines can b e verified as follows: Sincethe angle condit ion is

we have- 1s 1 / s 2 + j f l - s + 2 d = h 1 8 0 2 k + 1 )

By subst i tut ing s = a + jw into this last equatio n,

u + 2 + j w +d / a + 2 + j w d / a 1 + jw 80 2k + 1 )which can b e rewri t ten as

w + a w - v 3t a n ) t ) -tan- (*) L W 2 k + 1 )Taking tangents of b oth sides of this last equation , we o btain

2 w u + 2 ) wq 2 + 4 C T + 4 - w 2 + 3 a - 1

which c an be simplified to2 w u + 2 ) u 1 ) = - w a 2 + 4 a + 7 W2

o rw 3 a 2 + 6 u + 3 w2) =

Fu rthe r simplification of this last equ ation yields

which defin es three lines:

98 hapter Root i.o


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Thus th e root-locus branches consist of three l ines. Note that th e roo t loci for > 0 consist ofport ions of th e straight l ines as shown in Figure 6-47(b ). (No te that each straight l ine starts froman op en-loo p pole and extends to infini ty in the direction of 180, 60, or -60 measu red fr om thereal axis.) The rem aining port ion of each straight l ine corresponds to K 0.

A 6 10. Consider th e system shown in Figure 6-48(a). Sketch the root lociSolution The op en-loo p zeros of the system are located at . The open-loop poles a re lo-ca ted a t 0 and = -2. This system involves two poles an d two zeros. Hence, there is a possi-bility that a circular root-lo cus bra nch exists. In fact, such a circular ro ot locus exists in this case,as shown in the fol lowing. The angle condit ion is

By subst i tut ing w into this last equation, we o btain

Taking tangents of both sides of this equation an d no ting that

Figure 6 48(a) Contro l system; (b) root-locus plot .xample Problems and Solutions


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we obtain

which is equivalent to

These two equations are equations for the root 1oci.The first equation corresponds to the root locuson the real axis. The segment between s = 0 and = -2 corresponds to the root locus for0 K < m The remaining parts of the real axis correspond to the root locus for < 0.) Thesecond equation is an equation for a circle. Thus, there exists a circular root locus with center at= i = 0 and the radius equal to a / 2 . The root loci are sketched in Figure 6-48 b). [That

part of the circular locus to the left of the imaginary zeros corresponds to K > 0 The portion ofthe circular locus not shown in Figure 6-48 b) corresponds to K < 0.1

A 6 11. Consider the control system shown in Figure 6-49. Plot the root loci with MATLAB.Solution. MATLAB Program 6-11 generates a root-locus plot as shown in Figure 6-50.The rootloci must be symmetric about the real axis. However, Figure 6-50 shows otherwise.

MATLAB supplies its own set of gain values that are used to calculate a root-locus plot. It doesso by an internal adaptive step-size routine. However, in certain systems, very small changes in thegain cause drastic changes in root locations within a certain range of gains.Thus,MATLAB takes toobig a jump in its gain values when calculating the roots, and root locations change by a relatively largeamount. When plotting, MATLAB connects these points and causes a strange-looking graph at thelocation of sensitive gains. Such erroneous root-locus plots typically occur when the loci approach adouble pole or triple or higher pole), since the locus is very sensitive to small gain changes.

M TL B Program 6 1

Figure 6 4 9Control system.

num = [O 0 1 0.41;den = [ 3.6 0 01;rlocus(num,den);v = [-5 1 -3 31; axis(v)gridtitle( Root-Locus Plot of G(s) = K s 0.4)/[sA2(s 3.6)) )

Chapter 6 / Root Locus Analys is


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Figure i s0Root-locus lot.

Root Locus Plot of G s) K s+0.4)/[s2 s+3.6)]

Real xis

In the probl em considered here, the critical region of gain K is betwe en 4.2 and 4.4.Thus weneed to set the ste p size small enough in this region. We may divide the region for K as tollows:

Entering MATLAB Prog ram 6-12 into the com put er, we obrain the plot as shown in Figure 6-51,f we change the plot comma nd plot(r, o ) n MATLAB P r ~ g r a m-12 to plot(r, - 1, we o bt ain Fig-ure 6-52. F igures 6-51 an d 6-52 respect ive ly, show sa t i s fa~ tc~ ryoot-lo cus plots.

MATLAB Program 6 1 2A, -- - - - -- - Root-locus plot ----------n u m = [O 0.41;den = [ 3.6 01;K1 = [0:0.2:4.21;K2 = [4.2:0.002:4.4];K3 = [4.4:0.2:10];K4 = [ 0:5:200];K = [KI K2 K3 K4];r = rlocus(num,den,K);plot(r, ol)v = [-5 1 -5 51; axis(v)gridtitIe( Root-Locus Plot of G(s)= K(s 0.4)/[sA2(sxlabel( Rea1Axis )ylabel( lmag Axis )

Example Problems and olutions


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Figure 6 5 1Root locus plot.

Figure 6 5 2Root locus plot.

Root-Locus Plot o f G s) =K s+O 4) / [ s 2 s+3.6 ) ]

-5 -5 4 -3 -2 1 0Real AXIS

Root-Locus Plot o G s)= K s+0.4) / [ s2 s+3.6) ]

Real AxisA 6 12. Consider the system whose open loop transfer function G s ) H s ) s given by

Using MATLAB plot root loci and their asymptotes.Solution. We shall plot the root loci and asymptotes on one diagram. Since the open loop transfer function is given by

G s ) H s )= s s l ) s 2 )K

s3 3s2 2sthe equation for the asymptotes may be obtained as follows: Noting that

lim = lim =3 m s 3~~ 2~ S- m S~ 3 ~ 2 3~ S q3

Chapter 6 / Root Locus nalysis


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the equ ation for the asymptotes may be given by

num = [O Oden = [ I 2 1

and fo r the asympto tes ,numa = [ O O 11dena = [ I 11In using the following root- locus and plot com man ds

the num ber of rows of r and tha t o f a must be the same.To ensure this, w e include the gain con-s tan t K in the commands. For example,

M TL B rogram 6 1 3

num = [ O O I ] ;den = [ I 3 2 01;numa = [O 0 0 1dena = [ I 3 3 1K1 = 0:0.1:0.3;K2 = 0.3:0.005:0.5;K3 = 0.5:0.5:10;K4 = 1O:S:I 00;K = [Kl K2 K3 K4];r = rlocus(num,den,K);a = rlocus(numa,dena,K);y = [r a];plot(y, - 1v = [-4 4 -4 41; axis(v)gridtitle( Root-Locus Plot of G s) = K/[s(s ) s 2)) and Asymptotes )xlabel( Rea1 Axis )ylabeU1lmagAxis )

Manually draw open-loop poles in th hard copy

Example Problems nd Solutions


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Root-Locus Plot of G s) = Ki[ .s s+l) s+2)] nd Asymptotes

Figure 6 53Root-locu s plot. Real Axis

Including gain K in rlocus comm and ensures that the r matrix and a matrix have the same num ber ofrows. MATLAB Program 6 13 will genera te a plot of root loci and their asymptotes. See Figure 6 53.Drawing two or m ore plots in one diagram can also be accomplished by using the hold com-mand. MATLAB Program 6-14 uses the hold comm and. The resulting root-locus plot is shownin Figure 6-54.

M TL B Program 6 1 41~- - -- - - Root-Locus Plots ------------

num = [O 0 1 I;den = [ I 2 01;numa = [O 0 0 11;dena = 1 3 3 1IK1 = 0:0.1:0.3;K2 = 0.3:O.OOS:O.S;K3 = O.5:0.5:10;K4 = 10:5:100;K = [ K l K2 K3 K4];r = rlocus(num,den,K);a = rlocus(numa,dena,K);plot(r, ol)holdCurrent plot heldplot(a, - 1v = [-4 4 -4 41; axis(v)gridtitle( Root-Locus Plot of G(s) = K/[s(s+l ) s+2)1and Asymptotes )xlabel( Rea1 Axis )ylabel( lmag Axis )

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Root-Locus Plot of G s) = Ki[.s s+l) s+2)] nd Aysmptotes

Figure 6-54Root-locus plot. Real xis

Consider a unity-feedback system with the following feedforward transfer function G ( s ) :K ( s 2)C ( s ) s' 4 ) ( s 5)'

Plot root loci for the sy stem with MATLAB.Solution. A M AT LA B program to p lo t the root loci i s g iven as MA TL AB Program 6-15. Theresulting root-locus plot is shown in Figure 6-55.Notice th at this is a special case where no ro ot locus exists on t he real axis .This mean s thatfor any value of K > 0 the closed-loo p poles of the system ar e two sets of complex-co njugatepoles. N o real closed-loop poles e xis t .) For example, with K 25, the characteris tic equationfor the system becom es

s4 10s' 54s' 140 s 200sL 4s 10) ( s2 6s 20)S 2 j2.4495)(s 2 2.44 X)(s 3 ;3.3166)(s 3 3.3166)

M TL B Program 6 1% -- - - -- - - -- Root-Locus Plot ------------num = [O 0 4 41;den = [ I 10 2 9 40 1001;r = rlocus(nurn,den);plot(r,'ol)holdcurrent plot heldplot(r,'-'1v = [-8 4 -6 61; axis(v); axis('squarel)gridtitle('Root-Locus Plot of G(s) = s 2) 2/[(sA2 4 ) s 5) 211)xlabel('Rea1 Axis')ylabel('lmag Axis')

xample Problems and Solutions


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Root Locus Plot of G s )= ~ + 2 ) ~ / [ ~ ~ + 4 ) s + 5 ) * ]

Figure 6-55Root-locus plot. Real Axis

Since no closed-loop poles exist in the right-half s plane the system is stable for all values ofK 0.

A 6 14. Consider a unity-feedback control system with the following feedforward transfer function:

Plot a root-locus diagram withMATLAB Superimpose on the plane constant lines and con-stant w circles.Solution. MATLAB Program 6 16 produces the desired plot as shown in Figure 6-56.

TL B Program 6 1 6num = [0 0 1 21;den = [ 9 8 01;K = 0:0.2:200;rlocus(num,den,K)v = [ 10 2 6 6 ; axis(v1; axis( squarel)sgridtitle( Root-Locus Plot with Constant \zeta Lines and Constant \omega-n Circles )gtext( \zeta = 0.9 )gtext( 0.7 )gtext( 0.5 )gtext( 0.3 )gtext( \omega-n = 10 )gtext( 8 )gtext( 6 )gtext( 4 )gtext( 2 )

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Root-Locus Plot with Constant ( Lines and Constant o Circles

Figure 6 56Roo t-locus plot withconstan t lilies andconstant w c:ircles. Real Axis

A 6 15. Consid er a unity-feedback con trol system with the following feedforward tran sfer function:

Plot root loci for the system with M AT LA B. Show that th e system is stable for all values of K > 0Solution. MAT LAB Program 6-17 gives a plot of roo t loci as show n in Figure 6-57. Since the rootloci are entirely in the left-half s plane the system is stable for all K > 0

M TL B rogram 6 1 7num [O 1 0 25 01;den = [ 0 404 0 16001;K 0:0.4:1000;rlocus(num,den,K)v [-30 20 -25 251; axis(v); axis( square )gridtitle( Root-Locus Plot of G(s) K(sA2 25)s/(sA4 404sA2 1600) )

A 6 16. A simplified form of the ope n-loop tran sfer function of an a irplane with an auto pilot in the lon-gitudinal mo de is

Such a system involving an open -loo p pole in the right-half plane may be conditionally stable.Sketch the roo t loci when a = b = 1, ( = 0.5 and w = 4 Find the range of gain K for stability.E x am p l e P r o b l em s and So l u t i o n s 407


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Root-Locus Plot of G s) = ~ s ~25)s/ s4+ 404s2+ 1600)

Figure 6 57Root-locus plot . Real Axis

Solution. The o pen- loop transfer function for the system is

To sketch the root loci , we fol low this procedure:1 Loca te the open -loop poles and ze ro in the complex plane. Root loci exist o n the real axis

between 1and 0 and be tween -1 and m.2. Determ ine the asymptotes of the root loci .There are three asymptotes whose angles can be

determined as180 2k 1)Angles of asymptotes = = 60, -60, 1804 1

Referr ing to Equat ion 6-13), the abscissa of the intersect ion of the asy mptotes an d th e realaxis is

3 Determ ine the breakaw ay and break-in points. Since the characteristic equa tion is

we obtain

By differentiating K with respect to s, we get

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Th e num erator can be factored as follows:3s4 10s3 21s2 24s 16

Points 0.45 and -2.26 are o n root loci on the real axis. Henc e, these points a re actu-al breakaway and break-in points, respectively. Points 0.76 2.16 d o not satisfy theangle condition. Hen ce, they ar e neither breakawav nor break- in points.

4 Using R outh's stability criterion, determ ine th e value of K at which the r oot loci cross theimaginary axis. Since the characteristic equ ation is

the Rou th a rray becomes

The values of K t h a t m a k e t h e s t e rm in the f i r st c o lumn e qua l z e ro a re K 35 7 a ndK 23.3.

The crossing points on the imaginary axis can be foun d by solving the auxiliary equationobta ined f rom the s2 ow, that is, by solving the following eq uat ion fo r s:

Th e results arekj2.56, for K 35.7i j1 .56 , for K 23.3

The crossing points on th e imaginary axis are thus ~ t j 2 5 6 nd s ij1.56.5 Find the angles of dep arture of t he root loci from the complex poles. For th e open-l oop pole

a t s -2 j 2 d , he angle of depar ture 8 is

o r-54.5

(The angle of depar tu re f rom the open- loop pole a t -2 12 f l is 54S0. )6 Choose a test point in the broad neighborhood of the jw axis and the origin, and apply the

angle conditio n. If the test po int do es not satisfy the angle condition , select anoth er test pointuntil i t does. Continu e the sam e process a nd locate a sufficient num ber of points that satisfythe angle condition.

xample Problems and Solutions 4 9


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Figure 6-58Root-locus plot.Figure 6-58 shows the ro ot loci for this system. From s tep 4 the system is stable for23.3 K 35.7. Otherwis e, it is unstable.Thus, th e system is conditionally stable.

Consider the system shown in Figure 6-59, where the dead time T is 1 sec. Suppose that we ap-proximate the dead time by the second-order pad e approxim ation. The exp ression for this ap-proximation can be obtained with MA TLA B as follows:

[num,den] = pade(1, 2);printsys(num, d en)numlden =

s 2 6s 12

Hence

Using this approximation, determine the critical value of K (where K > 0 for stability.Solution Since the characteristic equation for the system is

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Figure 6 59A control systemwith dea d time.

by substituting Equ ation (6-22) into this characteris tic equation, we obtain

Applying th e R ou th s tabili ty criterion, we ge t the R out h table as follows:

Hence, for s tabili ty we requ ire-6K2 36K 114 > 0

which can be written as( K 8.2915)(K 2.2915) < 0

o rK < 2.2915

Since K must be positive, the range of K for stability is0 < K < 2.2915

Notice th at according to the present analysis , the upper l imit of K for stability is 2.2915.Thisvalue is great er than t he exact upp er l imit of K (Earl ier , we obta ined t he exact up per l imit of Kto be 2, as shown in Figure 6-38.) This is because we appro xim ated e- by the second-order padeapproximation.A higher-order p ade approximation will improve the accuracy. How ever, the com -putations involved increase considerably.

A 6 18. Con sider the system shown in Figure 6-60.The plant involves the dead time of T sec. Design a suit-able controller G,(s) for the sys tem.Solution. We shall present the Sm ith predictor ap proac h to design a controller . Th e f irs t s tep todesign the controller G ,( s) is to design a suitable controller G ( s ) when the s y st em has no deadt ime. Ot to J.M. Smith designed a n innovative controller scheme, now called the Smith predictor,Example Problems and Solutions 4


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Then the closed-loop transfer function C s ) / R s ) an be given by

Figure 6 62Step-responsecurves.

Hence, the block diagram of Figure 6-61 a) can be modified to that of Figure 6-61 b).The closed-loop response of the system with dead time e-TJis the same as the response of the system with-out dead time c - ~ , except that the response is delayed by T sec.

Typical step-response curves of the system without dead time controlled by the controller,(s) and of the system with dead time controlled by the Smith predictor type controller are

shown in Figure 6-62.It is noted that implementing the Smith predictor in digital form is not difficult, because deadtime can be handled easily in digital control. However, implementing the Smith predictor in ananalog form creates some difficulty.

Step Response

S m ~ t hredictor type controller

0 0 5 1 1 5 2 2 5 3 3.5 4 4.5 5Time sec)

PROBLEMSB 6 1. Plot the root loci for the closed-loop control system B 6 3. Plot the root loci for the closed-loop control systemwith with

B 6 2. Plot the root loci for the closed-loop control systemwith B 6 4 . Plot the root loci for the system with

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B 6 5. Plot the r oot loci for a system with

Dete rmin e the exact points w here the root loci cross the waxis.8 64. Show that the ro ot loci for a control system with

are arcs of th e circle cen tere d at the origin with radius equalto m .8 6 7. Plot the root loci for a closed-loop control systemwith

B 6 8. Plot the root loci for a closed-loop control systemwith

B 6 9. Plot the root loci for a closed-loop control systemwith

Locate the closed-loop poles on the root loci such that thedominant closed-loop poles have a damping ratio equa l to0.5. Dete rmin e th e corresponding value of gain K.B 6 10. Plot the root loci for the system shown in Figure6 63. Deter mine th e range of gain K for stability.

Figure 6 63Control system.R 6 11. Consider a unity-feedback c ontrol system with thefollowing feedforward transfer function:

Plot the roo t loci for the system. If th e value of gain K is setequal to 2 where are the closed-loop poles located?

B 6 12. Consider the system whose open-loop transferfunction G s ) H s ) s given by

Plot a root-locus diagram with MA TLA B.B 6 13. Consider the system whose open-loop transferfunction is given by

Show that th e equa tion for the asy mpto tes is given by

Using MA TL AB plot the root loci and asymptotes forthe svstem.B 6 14. Consider the unity-feedback system whose feed-forward transfer function is

Th e constant-ga in locus for the system for a given value ofK is defined by th e following equatio n:

Show that the constant-gain loci for 0 K omay begiven by

Sketch the constant-gain loci for K 1 2 5 1 0 and 20 onthe plane.B 6 15. Consider the system show n in Figure 6 64. Plot theroot loci with M AT LA B. Locate the closed-loop poles whenthe gain K is set eq ual t o 2.

Figure 6 64Control system.414 Chapter Root Locus Analysis


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B 6 16. Plot root-locus diagrams for the nonminimum- B 6 17. Consider the closed-loop system with transport lagphase systemsshown in Figures 6-65 a) and b), respectively. shown in Figure 6-66. Determine the stability range for

gain K

Figure 6 66Contro l system.

Figure 4 65a) and b) Nonminimum-phase systems.

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