control systems unit ii

8/6/2019 Control Systems Unit II

1/84

TI 'ME RESPONSE ANAL YSISCHAPTER 2

.~ J2.1 TIME RESPONSEThe time response of the system is the output of the closed loop system as a function of time.

denoted by c(t). The time response can be obtained by solving the differential equation governing" system. Alternativelytheresponse e(t) can be.obtained fromthe transfer function of the systemandinput to the system. ' . ,

Theclosedloop transfer function, C(s)= , O (s) , , M(s), . R(s)l+G(s)H(s)The-Output or Response in s-domain, 'C(s) is givenbytheproductof'the transfer function and

input, R(s).On taking inverse Laplace transform of this p roduc t the time domain response,c(t) can,obtained. ' ,Response ins-domain, C(s) ;;.' R{s)M(s)'Responsein time domain, c(t) = .c1{C(S)}= .c1{RCs)x M(s)}

, where, M(s) =' .o(s), ,1+G(s)H(s)The time response of a control system' consists oftwo parts: the transient and the steady "

response, The transient response is the response of the system when the input changes from one state 'another. The steady state response is the response astime, tapproaclies infinity.

Res) C(s) R(s) , ' "c(s)~M(s) = " O(s) ,l+G(s)H(s),

Input. Response(or Output)

The know.ledge of input signal is required to predictthe response of a system. - I n most o f 'systems the input signals are not known ahead oftime and also it is difficult to express the inputs'mathematically by simple equations. The characteristics of actual input signals are a sudden 'shock, 'sudden change, a constant velocity and a constant acceleration. Hence test signals which resembles ,characteristics are used as input signals to predict the performance of the system. The commonly

,Fig 2.1 :Closedl0;fPsystem.~- .~~..2.2' TEST SIGNALS'


2/84

Since.the test signals a r e simple functions for time, they can be easily generated in laboratories, The, maihematibl-and exper-imental analysisofcontrol systems using these signals can be earned o u t easily.f The use of the test signals can be justified because of a -correlation existingbetwe'e~ the responsef characteristics of a system, to a, test input signal and 'capability of the system to' cope, with actual input,:,:signals. -, - -,' - '- - S T E P S I G N A L

The step signal is a signal Whose value 'changes from zero to A at t =0_ and remains constant atA for t> 0 . _ The step signal resembles an actual steady- input to a system. A special case of step sig~1al is unit step in which A is unity.

The mathematicalrepresentation of the step signal' is,r(t) = ; t : : : : : 0=0 ; tcO ....(2.4)

" R A M P S I G N A L_ _The ramp signal is a signal whose value increases linearly with time, from an initial value of zero' at t ; = O. T4e ramp signal- resembles a constantvelocity input to the system. A special case of ramp signal is unit ramp signal in- -wh i ch the value of A is unity . " -

The mathematical representation of the ramp signalis,r{t)= At; t ~ 0= 0 ; t< - 0 . . . . . (25)

P A R A B O l iC S I G N A LIn parabolic signal,' the instantaneous value varies as square of th~

time from an initial value' of zero at t= O, The sketch of the signal with- respect to time resembles a parabola. TIle parabolic 'signal resembles aconstant accelerationinputto the system. A special case of parabolic signal,.isunit.parabolic signal.in which A is uni ty ;

-The mathematical representation of th e parabolic signal is,- ' Atir(t)=T 1;': 0= 0 , t


3/84

2.3~ince a : perfect impulse cannot he achieved in practice it is usually approximated by a pulse of small

width bu t W ith area, A. Mathematical ly an impulse signal is the derivative o f a step signal. Laplace t r ans fo rmoftheimpulse function is unity, .' .. .... .

TABLE 2-1: Standard Test S~als'.Name of the signal Time domamequatlon Laplace transform of

of signal, ret) the signal, R(s)A-Step - ;A -s..

Unit step 1 1-. sRamp At Aii

'.' ',1..Unitramp t 7\

At2A

Parabolic 2 53t2 . 1Unitparabolic .. -- 5 32 .'

Impulse o C t ) 1_ - . . - = .. ~ =' Jw n ; ''. . ' an d the sy s tem IS , u n d am p ed '

{ro o ts are teal an d equ al an d

When S =1 , SI' 5 : 2 = -w n, ; the s y stem is c r it ic a lly d am ped", is:': {ro o ts ar ,e real an d ,u nequ ,a l an dWhenl , ;>l ; s, 5 : 2 = - < ; 0 0 w "t;- -1; , , '1 " n, ? ' " th e s y s tem is o verd am ped

Wbeo0 < 'C ; n . n ' - ' : I

.....(2.17)"

.....(2.18)

.....(2.19)

= ~ C ; W w .;{ fO o t sare c ~m p lex co n ju gaten ,Jd , th e s ys tem IS u n d erd am p ed . ,

whe re , ffid = ' . 0 0 , , ) 1 - t ,2Here CiJdis called damped frequency of oscillation of the system anditsunit is tad /s ec .

.:...(2.20)..... (2.21)

2.7 .1 " RESPONSE O F U N D A M P E D S E C O N D ORDE~'SYSTEM F O R U N I T STEP I N P U T... ,. .... . . _. . _" .-.. .." -.The standard form of c losed loop.transfer f un c t i on of second order sys tem is;C(s) w~'R(s) = S2 +2 c;w ns +O ) ~

'Fo r u n d am ped sy s tem , Z =.0 .. C(s) 0 0 ;R(s) ;= 52 +002:.n .....(2.22)" , . I, , W hen the in p u t is u n it s tep ; r(t) = 1 an dR ts ) =~'"

.. , . Th " '. dorr .: ,(,),' (' ) , '00; 1 0 ) ;\:.e response m s- omam, C s =R 5 ~='- ~, . S + oon' S S + wn~y p ar tial f rac tio n ex pan sio n ,

".. ' " 00,2 A B" .. C(s )= n " ;:-c+-' ,-s (S2 +O)~,. S . S2 +O)~A'isobtained by multiplying C(s) by s an d let t in g s =O .

.....(2.23)

L\ \ \'Risobtained by m ultip ly in g C (s ) by (S2 + (0;) a nd le tt in g s" = -(f)/ o r s =jw u'


11/84

2.11

ret) c(t) 42 .; .;.;.-

o tFig 2.9.a : Input. Fig2.9;h: Response.Fig 2.9 : Response of undamped second order system for unit step input.

Using equation (2.24), the response of undamped second order system for unit step input is sketchedin fig 2.9, and-observed that the response is completely oscillatory. -

Note: Every practical system has some amount of damping. Hence undamped system does notexist inpractice.

The equation (2.24) is the response of undamped closed loop second-order system for unit stepinput. For step input of step value Ai the equation-(2.24) should-be multiplied by A... For closed loop undamped second order system,

Unit step-response = 1-cos matStep response,

2.7.2 RESPONSE O F U ND EROAM PED SECOND ORD ER S YST EM FORU NIT STEP INPU TThe standard form of dosed loop transfer function of second order system is,

C(s) _ w~R(s) = s2 +2i;;rons+ro~

For underdarnped system.D


12/84

2.12-A is obtained by multiplyingC(s)by sand letting s =O.

.~2 _ m2:. A =s x C(s)i_ 0 =s x ~ ~ a ) = - - - t = 1"S - _ s + 2t;:ronS + ro ~ - CO n

",,0-To solve for Band C,-cross multiply equation (225)and equate like power of s.On cross multiplication equation (2-;:25) aftersubstituting A = 1>we get,

r o~ = 5 2 + 21 ;r on 5+ ro ~ + (B s+ C )s2 2 2/' 2 2' ,ro n =_S + " ,ro n s t ro n ,+Bs + Cs

"-':Eqliatingcoefficients ofs2 w e get. 0 =1 +BEquating coefficient of s w e get, 0=2\;;0) + C, n -

:~B =-1:.C =-2c;O). , _ n _

...226)

Let us add an d subtract (,200-"'2o the d enominato r o f seco nd term in the equ atio n (2.26).n ". . ,

IWd=ronRI.....(2;27)

, -Let us multiply and divide byrodin the third term ofthe equation (2.27).

,---~~,--'-;-{} 1-1 "'-,SThe response in time domain is given by,, { I s+/'m /'m W }'C(t)"'Ll{C(S)}=Ll~_", ,,',"',n'" n '_ d ,- - 5 (5+/'ro )2 +m2 .W ( s+r ro )_2+m2" ~'n ' d 'd ",'" n ' d { } s+a I'L e-"'o/>srot = ',2 " ", '.(5+a) +00- I

,Let us express e(t) in a standard form as shown below.e-~"'nt . .


13/84

2: 13- - - ". .The equation (2.28) is the response of under damped closedloopsecond order system for unit step

input For step input of step value, A, the equation (2.28) should be multiplied by A.: r ~ 'orcios._ed.loop under damped s.econd order .system,, e--t;Olnt '. ~1-[/Unit stepresponse =1 - r:-::2 s in (c o d t+S ) ; 8=tan-1---_-.-. --/1 _(2 - s'I ' r e-~ " 'n t. . . ' . -t ~ 1 - S 2 ltep response =All---.--sm(COdt+s); 8=too -_--~1_(2 , S-Using equation (2.28) the response of underdamped second order system forunit step input is

sketched and observed that the response oscillates before settling to a final value. The oscillations dependson the value of damping ratio: '

r(t)

t ,_ _Fig2.10.a: Input.

c(t) i ' -1~

.' .' . . . .o - tFig 2.1 0 .h : Response.

Fig 2.10: Response of under damped s~cond order system for unit step input.---2.7.3 RESPONSE' OF CRIT ICAllY D AM PED SECOND ORD ER sYSTEM FOR UNIT STEP INPUT

The standard form of closed loop transfer function of second order system is,C(s )= co ~R(s) 52 +2t;con5+ co~

For critical damping l;= 1.. C(s ) f;l~R(s) s2+2rons+00; . . . . . (2 .29)

When input is unit step.rrt) = 1 and.Rfs) = lis.:.The response in s-domain,

co 2 1 00 2 00 2C(s) =Res) . n =_ n n(s+ COni S (s+ COni S (s + con)2By partial fraction expansion, we can write,002 A BCC(S)= n =~+ +--S (s+ooni s (s+roni -S+Wn

.....(2.30)

2.14


14/84

ABC 1, ' .C(s)=-+ +__ ::::__ OO nS (S+O O n)2 S+CO n S (s+oon )2The response i n time domain,

c(t):::: l-ront e-"'nl-e-''''n!c(t) = 1- e-wnt(l +ront )

.1 { ' - a t } 1I L e . ,=~'. . . . . (2.31)

The equation (2.31Jis the response of critically damped closed loop second order system for unit 0ep input., For step input of step value, A~the equation (2.31) shouldbe multiplied by A.., For closed loop critically damped second order system,

Unit step response =. 1- e'""'nt(l+cont) 0Step re~ponse = A [ 1- e-cont( l +O)o t ) J

Using equation (2.31), the response of critically damped second order system is sketched as shownfig 2.11 and observed that the response has no oscillations.

~ ~~

o o tFig 2 . 1 1 . a iInput. Fig 2.11.h : Response.Fig 2.1 1 : Response of critically damped second order system for unit step input.

RESPONSE OF OV ER D AM PED SECOND ORD ER SY STEM FOR UNIT ST EP INPUTThe standard form of closed loop transfer function of second order system is,

C(s) CO ~Res) 52 +2~cons+co!

For overdamped system t ; > I. The roots of the denominator of transfer 0 function are real and~~stjnct.Letthe roots of the denominator be sa' Sb'o

. . . .. (2.32)

'2. i5


15/84

For unit step inputr(t) '= I and R(s) = lis.00 2 _. 002:. C(s)=R(s) n - "O' n _, -(s+Sj}(s+Sz) S(5+51)(s +s.)

B y par t ia l fraction expansion we can write ,00 2 ABCC(s)= n, -+--+--5 (5:+51)(s+Sz) S S+SlS +52

00 2A=sxC(s)l_ =SX n.,-0 S (S7S1)(5+~) .= 0

2, ron - ro~ ,-Wn-2J~2-: 1 51S l[ -~ r o ll + ro ~ Jr ,2 :- 1+~ roC l+ (J) nJ~ 2 -1 ]

c=C(s) X (s-+~2)1 _ "" ro~' 1s --'2 5 (5+5) -"-t-..:. +,5\')'-_'I '=-'2 -~-~

,2 2,W n . ro r i~Sz [ - ( ,wn-ron~ t , z -1+('Wn - wn~r ,2 ' - I T ' " [2 roD~ r , 2 __ : ] S z , - r o n i2~r,2 -1 ~-

The response in time domain, c(t) is given by,

- ' 1 r o n , ,, 1 -'11 ron 1 -'1-1- --e + - e2 J ~ 2 - I 51 - 2~(/-1 S z,_ 'w n- ( e - S l l , ' e - S 2 1 )= 1--- --- -2~(,2 -:1 " 5 1 52

where, 51~r,ron-ronV r , 2 -1S z =(,ron +ronJ(/ -I

The equation (2.36) is the response of overdamped closed loop system for unit step.input, For sm

c{t )

c(t) .:...(2J

2 ..16


16/84

ret) c(t)

F;g 2.12 b : Response:o t.Fig 212.a : Input.Fig 2.12 : Response or over damped second order system forunitstep input.

. Using equation (2.36), the response of overdamped second order system is sketched as shown ing 2.12 and observed that the response has no oscillations but it- takes longer time for the response ,to:reach the final steady value. . .-

TIME DOMAIN SPECrFICATIONS-The desired performance, characteristics of control systems are specified -in terms of time domain

ipecif icat ions . Systems with energy storage elements cannot respond instantaneously and will exhibit-ient responses, whenever-they are subjected to-inputs or disturbances. - .

The desired performance characteristics of a system of any order may be specified in terms of the.sient response toa unit step input signal. The r e sponse of a second order system for unit-step input

t h . var ious values of "damping ratio is shown mfig2,13.c(t)

o o t

ret)lr------...,._

tFig 2.13.a : lfi{n!t. . Fig2.13.b: Response.Fig 2.13 : Unitstepresponse of second order system.

__The transient response of a system -to a unit step input depends on the initial conditions. Therefore _!o compare-the time response of various systems it is necessary to start with standard initial conditions. Themostpractical standard is to start with the system at rest and ~6output and all time derivatives bef-oret~0 will be zero. The transient response of a practical controlsystemoften exhibits damped oscillation,efere reaching steady state. A typical damped oscillatory response of a system is shown i~ fig 2.14~

2.17


17/84

c(t)'-c(t,,)

1_1 -

Allowable error"- - 2% or 5%-

" ,td t, t" , 1 : . tFig 2.14 : Damped oscillatory response oj second order sys ternfor unit step input.

The time domain specifications are defined as follows. '1.DELAYTIl\fE (td)1 ,. RISE TIME(t )r

3. PEAK TIME (t)

4. PEAK OVERSHOOT(M).~ : '. . . ..

,5. SETTLING TIME (ts)

It is the time taken for response to reach 50% of the final value, fOIthe very first ~irne

'It is the time taken for response to raise from ,0 to 100% for the verfirst time. For underdamped system, the rise time is calculated from to 100%. But for overdamped system it is the time taken "by the-response to raise from 10% to 90%. For critically damped system, ~is the time taken for response t o raise from 5% to 95%. "Itis the time taken for the,response to reach the .peak value the veryfirst time. (or) It is the time taken for the response t o reach the peakovershoot, M .pIt is defined as the ratio of'the maximum peak value to the final value;where the .maximum'peak value:is measured from final value.Let, c(co) = Final valueofc(t).

c(tp) = Maximum value of cet).- - e(t }-c{co)Now, Peak overshoot, Mp =, P, 'c ( c o ) . . . . .(2.37Jc ( t ) - c (cO )% Peak overshoot, %M p = P _ x 100 ..... (2.38Jc(co) -

It is defined as the timetaken.by the response to reach and.stay withiila specified error. It is usually expressed as % affinal value, The usualtolerable error is 2 %or 5%-ofthefinal value.

EXPRESSIONS FORT IME DOMA IN SPEC IF ICAT IONS


18/84

Note: -On constructing rightangle triangle with sand ~1_~2 ,

) 1 ~ r : , 2tan6=--(, .

. -e-~"'n!j". ..: . ~ sm (mA+E l)"=Ovl-~2 .Since '- e~t;"ntr * 0, the term-sin (m dtr +8) = 0

W h e n , . 4 > = O ,1 t " ,2 1 t ,3 1 t. .. , s in 4 > ==0:.Wdtr+El==1tmd t , = = 1 t - E l

:we get

R T .' 1 t-81.:. use nne, t, =_._. . J. r o d~ 1- t} .r:-::1Here, e =tan1--t;,-; Dainped frequenc~ofosci1latioJ:l, ~d=fiDv1-t;,2

. ta _1~1-t;,21 t- n ---_~--r"=",;,C;_ in secc o ~ 1_r2.o .. ~:. Rise time, t, =

. h - c , , 2Note: I J or tan-'-""-- should be measured in radians.C ;

.....(2.39)

(refer note)

...... (2:40)

Peak time ( tp )To find the expression for peak time,tp' differentiate c(t) with respect to t and equate to O.

. d I .i.e.,-c(t) t= =0dt -tpThe unit step response Of under damped second order system is given by ,


19/84

= . 4 r o 'n . ' e c "" n '[ sin ( ( ro ~ t + e) -e )l.1~ 1;;2 .. . d'.. at t= t , -e(t) =0P dt

. ro n ---J' , 8 I' , C a s e =S . . . ' - - .r-e

I : . Peak: time, tp =.!E_ ' II ' . ' ro d IThe damped frequency o f osclllation.j, '=roil~I-(/'

- ..

.....(2:41)

:. Peak time,tp = g, (On 1 - , 1 ; ;

....;(2Al)

P~k- overshoot (M) .', e(t }-c(tO),%Peak Qversnoot,%Mp-= p.. . x 100

L- __~ ' ~ __~ __~...c~(w~)~ ~_,.;,...(2.43)

where, e(t) =Peak t:esponse at t =\ 'c(ob)=Final steady state' value.The unit step response of second order system is given by,

, e-C ;O>nr e(t)=l- ~sm(rodt+8), , 1 - 1 ; ; 2 ., _,e . ,c(t) = c(co) "" 1-, ~. sin(rodt,+9) = 1"'-0 = .1

] - ' - - t / " ,

Irod=ronRlI sin(1t+B)==-sin81

. " n : .---


20/84

' 1 ; 1 J :" 'C ( t J-C{o cy l+e- ~1_[ ,2 ,-1 ",Percentage PeakOvershoot, %Mp,= P C } x 100 = x lOO" co o 1,' ",

~~,.".,Percentage Peak Overshoot, %M P = e- 1=1 x 100 ......(2.45)

Settlingtime(t)The response of second order system has, two components. They ~e"

1. Decaying exponential component, JI-1:;2 .2~ Sinusoidal component, sin(ffiot + 8).

, '" In this the decaying exponential tenn dampens (or) reduces the oscillations produced by sinusoidal'component. Hence the settling time is decided by the" exponential component. The settling time can befound out by equating exponential component to percentage tolerance errors. '

,e~~"'"{' .-For 2% tolerance error band, at t= t" J l - r . ; , 4 , ,= 0.02

" For least values oIl:, , e-~"'nts:::0.02.On taking natural logarithm we get,

, 4t =--, r" ",ffirjFor the second order system. the iimecon~tant, T.,~

, , ' ' ~~, I':. Settling time,t, =-,-=4T" Sffi (for 2% error) .,..,(2.46)

For' 5% error, e-~" 'n t , =0.05 'O n taking natural logarithm we get,, 3t =--'Sffin

S 1 3:. ettmg nme, t, =(ill =3T (for 5% error) .....(2.47)" In general for a speCifi;d percentage error, Settling time can be eval11t~dusingequati;n (2A8}. "

2.71


21/84

EXAMPLE 2.1Obtain the response of unity feedback systemwhose open loop transterfunenon is G( s) = __ 4_. and wnen the in~ut. s(s + 5) . .is u n n step.

SOLUTIONThe closed loop System is shown in fig 1 .

C(s) G(s)The closed Ioopjranster function, -. - = -.--. . R(s) 1.+ G{sJ4 4

R ~ f ' ). Fig 1;' Closed loop system.

. C(s) '" s (s + 5)R() . 4 .s 1+---s{s+5)

S (s+5)s(s+.5)+4s (s +5)

.4 4 4" ' - S " " ' 2 + ' - , -5s--; 4-= (s+4) (s+1)s (s+5)+4

The response in s-dornam, C(s) =R(s) 4.. (s+1) (s+4)4:. C(s) = .s{s+1){s+4)Since the input is unit step, R(s) = _:l_s

By partial fraction expansion, we can write,4 ABCC(s)=. . =-+~+--. s (s+1) (5+4) . 5 s+1 s+4. '4 '4

A=C(s) x s[s=o= (5+1)(!5+4}s~o =';4 =1

B:::C(s) x.{s+1)1_ 1= _4 I ' 4. S--'. s{s+4), -It-l+4) . 3. . S - -1

C=C{s) X (S+4)ls~-4=S(S4+1lt __ 4 -. --4(~+1) 3The time domain responsectt) is obtained by taking inverse Laplace transform of C(s).Response in time domain; i:(t) = r'{C(s)} = r1{_:l_ _ _.1_ + _ :l _ - ' - r:}

. i s 3 s+1 3s+4

--4

RESULT. 1Response of unity feedback system, c(t) '" 1-_:[e-' - e-411.. . . 3. .

EXAMPLE 2.2A positional control system with velocity feedback is shown in fig 1. What is the response of the system for imit step

inputSOLUTION C(s)

2.22


22/84

Here (S2 + 12s + 1 00 ) is ch ara cte ris tic p o lyn om ia l. T he ro o ts o f the ch ara cte ris tic p o lyno m ia l a re ,...,12.J144-400 -12j16 6+'8S"S2~-= . 2 2 --J

T he T Oo ts a re co mplex co njuga te . T he sys tem is underdamped and so the respo nse o f the sys tem w ill have dampedOSC illa tio ns . .

T he respo nse in s - dom a in , C (s )= R{s ) 2 100. s +12s+100S in ce ' in p ut i s u nit s te p, R(5) =..!.. .' s

:.C (S)=..!. . 1 0 05 52+ 1 2::;+ 1 0 0100

5(52+125 + 100 )B y pa rtia l f ra ctio n "exp ans io n w e ca n write,C(5)= 100' =A+ Bs+C

S(52+12s+100 ) 552+125+100T he-res idue A is o bta ined by m ultip ly ing C (s } by 5a n d le ttin g s =0 .

1 0 0 100A = C(s ) x s is _ 0 = 52 + 12s + 1 00 s ~ 0 = 100 = 1T he res idue Band C are eva lua ted by cro ss m ultip lying the fo llo wing equatio n and equa ting the co eff ic ien ts o f like po wer o f s .

100 A Bs+C-+---,;----~5 (S2+ 12s +100) S 52+ 12s +100100 = = A(52 + 125+ 100)+(85 + C)s100 = AS2+ 12As+ 100A+Bs2 + C s

. On equating the co eff ic ien ts o f S '- w e ge t, 0 = A + B :. B= -A=-l

. O n equating ' co eff ic ien ts o f swe get, a = = 12A + C :. C= -12A=-121 -512.~.C(s) ;;;; ~+ -2 . =S . s +12s+1001 s+6s {s+ 6l + 82

s+12 s,+6+6(s +6)2+826 8

s6 _ . 1 s+6(s+9l+82 - s- (S+6)2+82

s

T he tim e do ma in respo nse is o bta ined by tak ing inverse lap lace trans to rm o f C (s ).T. "{ }. _ 1 { 1 "s+66 .8 }irne response, C(t)=L C{s) =L. '--22- . 2 2'. . '. S (5+ 6) +8 8(s+ 6) + 8

.=1- e--{; tci?s 8t-~ e -B t sin} t =1-e --{; '[ ~ s in8t + co s8 t]. "T he resu lt can be co nver ted to ano the r s tandard fo rm by co ns truc ting r ight ang le tr lanq le w ith ~ and J l - ~2 T he

" damp ing r a ti o l;is eva lua ted by co mpa r ing the c lo sed lo op trans fe r func tio n o f t he s ys tem with s tanda rd fo rm o f seco nd ~rde rtrans fe r func tio n . . .,

- 2.23


23/84

Consfructinq right angled triangleWilh t:;,and~1~_t:;,2' we~et,sine = 0.8 ; coss = 0.6; tans = 0.80.6.. e = tan-1 0.8 = 530 = 53 " x~rad =0.925 rad.'0.6 1800-

_~ ~ 1 __ t:;,2-=O.8 'C;~0.6c c s a =0:6I sine = 0.8

Tlmeresponse.icrt) = 1-e-6:[~sin8t + cos~t] =1- e~1~[ 1~Sin8t + 1~COS8tJ,- 10' ,- -,'= 1-"8e--{;t[Sin8t x 0 .6 + cos8t x 0.8] = 1-1.25 e-&[sin8tcos9+cos8t sine]- , -= 1-1.25 e--{;![sin (8t +9)]:: 1-125e- - - 0 1 Si,rl{8t+O.92S), I Note:B is~xpressed_ - ' In radians

RESULTThe response in time domain,

c(t) = 1- e - - - < i - t [ ~ sin8t + C O S 8 t J o r c(t);" 1-1.25e-61- sin{8t - + ' 0,925) .EXAMPLE 2.3

T he res po ns e of a se rvo mechan is m is , c(t) ~_1-+ 0 .2 ~ - 1 .2 e 1 O t w he n s ubje ct to a unit step input, Obtain anexp ress i on for c lo s ed lo o p tr an sfe r fu nctio n. D e te rm in e th e u nd am pe d n atu ra l fr eq ue nc y a nd d am p in g-r atio .". - _ ,. .SOLUTION

Given that, c(t) =1 + 0.2 e-OOl.--1.2e-10." On ta k in g L ap la c e transform of c(t) we get,

C(s) =..!.+0.2'_1 __ 12_1_- _ (5+ 60) (s+1{:i)+ 0.2 s{s+ 10),--1.2 5 (5+60), S (5+60) (5+1Q) s(5+'60){5+10)-, , S2 +70s+600+0.2 52-+25 -12s2-72$ .600 -' = = ' =~---,----:'-~-5(5-+60) (5+10) 5 (5+-60)(5+10)

1 ~----=:---'-..,...,..,..5 (5+60) (s+10),600

Sinceinputisunitstep, R(s):: 11s., 60 600-., qs) =.R(S ) (s+ 60)(s + 10)=R(s) g2 +70s+ 600

:. T he cI~ed loop transfer f~nc t io~ of the system, C(s) = 2 600 .._ . .. - R{s) s +70s + 600The damping ratio and natu ra l frequencyofosdJlation can beestimated by comparing theSYStem transfer function with

standar~ form o f s ec o nd ordertransfer func t ion .. C(sj ro~ -600. . R ( s ) =52 + 2 1 : ; m ns + r o ~ =S2 + 70s + 600

On comparing we get,ro ~ = 600 2 C ; o o n - = 70

2.24


24/84

EXAMPLE 2.4T h e u nity fe ed ba ck systemis cha ,ra cte r ized by an o pen lo o p tran sfe r func tio n G {s ) =K/s (s t1 0). D e ten nin e the ga in 'K ,s o that tile s ys tem will have a dam Ping ra tio o f 0 .5 fo r th is va lue o fK D ete rm ine peak o versho 6tand tim e a t peak o versho o t fo raunltstep.mput. -T he un ity.feedback sys tem is sho wn in f ig 1 .T h- - los d It' & f . C (s ) G {s }, ,ee e' co p rans,er unct i on -,-=---, - - - R{s ) 1+ G (s )G iven tha t, G (s )=K ls (s+10 )

K

RTI(~Fig 1.' Unity feedbacksystem.

SOLUTION

C (s ) s (s +1 0 ) K . K. R(s } - 1+ ,K _,;= s (s+10 )+K_ '" s 2+ i-Os+K

;s (s + 1 0 )T he va lU ~ o f Kcan be e~aluated by c o m pa rin g th e s ys te m tr an sfe r fu nc tio n Wfth s ta nd ard fo nn o ts ecen d o rde r tran sfe rfunc tio n . - . '

. C (s ) _ (\)~ KR(s ) s2+2 t;( \)n s+w~ s2+1 0s+~

On co mpar ing we get,2 - -

(\)n =: K-. .wn=-FK",

2t ;wn '= 1 dPut ~'" 0 .5 a rid r on=- . J K:. 2 x O .5 x . , J K '" 1 0

.jK => 10ro n = 1 0 r ad /s ecK=100

Thevalue ofqaln, K=lOO .Percen ta ge pea k o vers ho o t, % M . , _ =-e-c'~IFJ! xl 00

=e-- {J ,5~1~1-0_52100=-0:163xl00 =16.3%P kf 1t 1t, 1t 0 36' ..,eaktme, tp =-=, ~, '" ~, = - . .J sec_ wd ronV1-~2 10\'1.-0.52

RESULTT he va lue o f ga in ,P er ce nta ge p ea k o ve rs ho o t,P ea k tim e ,

K = - , 10 0= 16.3%

0 .3 63 s ec .EXAMPLE 2:5

2.25


25/84

Given that, G(s) = Kls{sT +1).: qs) K/s{sT+1) . K OK . KIT

o R C s ) 1+Kls(sT+1) s(sT+')+K"'S2T+s+K- 2 1K. . .. s +Ts+T .:Expression for ~ and wn can be obtained by comparing the transfer function with the sta ndard formofsecond order

Ira nster function. .. .

2I;ron '" 11 Tr=_1_", __1 '_._. " 2 w nT N - 2JKf2 - T. T

The peak overshoot; M p is reduced by increasing the damping ratiol;. The damping ratio l;is increased by reducing th egainK:

.:C(s) ro~~. R {s } = 52 + " 2t;ffinS + ro ~ 2 1 KS +-s+-T TOn comparing we get, .(()~'"KIT

KIT

When ~ = 0.75, Let ~ =';1 and K=K lWhen Mp = 0.25, Let.; = (2 and K = = K2Peak overshoot, M = e -r;;",R. p ~Taking n a n r r a l loqarnnrn on both sides, I n M p = = ~ -(;lC 21-1; .

(2 2On squaring we get, (tn ~)2 =4.1-';On crossing muttipli~tionweget,

(tn ~)2 = (1l? +s2(ln~l.(/n Mp)2 = r . , / [7 C2 +V n ~ )2 ]

(In ~)2 . ( ,2 = --;;--,---...;o.____".. . x 2 + {In ~)28 t r - . 1 . r2 _ 1u "'-2.,fKf' ." ~ - 4KT

Wh K K M 075 . K = 7 C 2 +(/nO.75)2en, = 1. P= . ,.. 1 4T (ln0:75f2 . 2When. K = I( M = 0.25 :. K = = 7 C + ( lnO.2S), '''2 'p , 2 4J(ln0.25)2

....{1)

A2)

On equating, equation (1)& (2) we get,1 -(In Mp)2

4KT = = 7 C 2+(ln Mp)"1 AT (In ~)2j(= 7 C 2 + (tn Mp)2

. 7 C 2 + (In ~)2K= .... 4T (In Mp)2

--=-7.68T T

9.952 30.06._-=--0.331T T11 .79 1.53

2.26EXAMPLE 2;6


26/84

A p os itio na l co ntro l s ys te m W ith ve lo city fe ed ba ck is s ho w n in f ig 1 : W ha t is th e re sp on se - c(t) to th ~ un it s te p in pu t G ive ntha t I ;; = 0 .5 . A ls o c alcu la te r is e tim e, p ea k tim e, m ax im um ove r sh o o t a nd s e ttlin g tim e .SOLUTION

. - C (s ) G (s )T he c lo sed lo op trans fe r func tio n , - - '= -~ . ......:.-cc--_. . '. R {s } 1+ G (5) H (5)G ive n th at G (s ) = 161s (5 + 0 .8 ) and H (s )= Ks +1

16 -

C ( s )

s .{s+0 .8 ) - 1 616 . -1 + (Ks+1 ) s ( s+0 .8 )+16 {Ks+1 )s (s +0 .8 )1 6 1 6= 52 + 0 .8s + 16Ks + 16 52+ (0 .8+ 16K) 5+16

T he va lues o f K and ro na re o bta ined I:lY co mpa ring the s ys te m trans fe r func tio n w ith s tand ard fo rm o f s eco ndo rde rtr a nsfer fu nc i ion .

_ . C (s )R(s)

_. C{s ) _ ro ~ _ 16"R (s ) - s2+2i;ro n5+ro ~ - S2 + (0.8+ 16K) s+16r .O n co mpa r in g we ge t.

- ro ~;= 16:. ron;= 4 r ad / sec

0 .8 + 1 6K = 2C;:ro n-:.K: 21;;ron-0.8:--2X 0 .5 x4. - 0 .8 = 0 .2

16 -1 616C( s )

R{s).16$2 + (0 .8+ 16 X 0.2) s+ 16

G ive n th at th e d am pin g ra tio , 1;;=0 .5 . H e nc e t h E ; ! sys tem is unde rdam ped and so the re spo nse o fthe s ys tem w ill ha ved am p e d- o s cilla tio n s . T h e r o o ts o f cha rac te r is tic pO lyn om ia l w il l b e co m p lex co n juga te : . .T he respo nse in 5 - do ma in , C (s ) =-R {s ) 2 16 .s +45+ 16F o r un it s te p input, R (s ) = lIs.

- 1 16 16:. C (s )=s s2+4s+16 s($2+4s+16)

B y p a rtia l f r ac tio n e xp an s io n we can wr i te ,C ( s ) ;= 16 . = A ~ Bs + C

s(52+4s+16).s s2+.45+16T he re sidueA is o bta ined by m ultip ly ing C (s ) by 5a nd le ttin g 5 = O .

16' 16A=C(s}x ,,1= . ",~=14. ~O 52+ 4s + 16 s ~O 16 _

. . " .. . -- c o d2.27


27/84

T he timedomain r esponse is obta lnedby taking inverse Laplace tranSfonn ofC('S}.

On equating the coefficients.of S2 we get, 0 '" A + B .'. B '" -A = ~1 ..' On.equating the coeffici.ents ot swe get, 0 . ' " 4A +C :. c=-4A =-:-4

1 -s-4. 1 s+4.'. C{s) = = ~ + ' "-;;-~.~--'--s s2+4s+16 s '52+45+4+12_ 1 s+ 2 + 2 _ 1 s+2' 2 &'-:-;-(5+2)2 +12 -; (s+ 2)2+12 .--..Ri{S+2)2 + 12

Theresponseln timedomain,C()'~L1{q n_.c1{1S+2 2. J 1 2 }. t - . s - $- {s+2Y-+12 :& (5+2)2+12_'. 2 ...= 1- e-2t cos.Ji2 t- M e-2tsin.J1'Z t2..,3

=1- e-21 [~sin(~t).+CoS(&t~]

T he result can- b e converted to another standard fo rm by ~nstructing rightangle triangle with (;and ..~ , '.'On c.onstructihg right angle triangle with I;and ~1- '(;2 .we get,sine", 0 . :866 = = J31 2; cose '" 0 . . 5 = 1 1 2'; ta n e '" 1732:. e = tan-11732 = 60"= 1047 rad

:. The response intime domain,eft) =1- e-2! [ ~ x2 x~irf.Ji2 t x..! -t-. ~ x cos . f i2 txJ3],,3 2..,3' 2. .

=1-e - 2 t ~ J s in .J12 t c o s e + c o s J 12 t Sine] I Note : ()is expressed in radians. I= 1- ~ e-21 [in(Ji2 t + " 8 ) ] =1- i s e-2 1[5in{J12t +:1047)]-

I~g. " . =0..8661;=0.5.. . '. IDamped trequenC :Y r ' . ~. ~ .' . .: .' 1 1 0 = ro n- v 1-(;- = 4-.11- 0 .. 5 - : : 3.464 rad I seco f OSCI ano n .' .' .

.. 7t:-.6 1C-t047:. Rlsetime,.t, =-~=O.6046secro d 3.464Peak time, tp = = ~= _It -. '"0 . . 9 0 . 7 secro d 3.464

. _~':[ --O;5x'l:: '%.MaXimUm} - ~ 2 J . ' ..0vershoot%Mp =e~ 1- -< !; x 10 0= e 1-0',5x1o.o.",-o.163x100",16.3% .. .

2.28


28/84

Rise time,Peaktime,.% Maximum overshoot, .Settling time,

t,.=0,6046sec~=O.907se~

%Mp=16.~%t.= 1.5 sec, for5% error= 2 sec, for 2% error

EXAMPLE 2.7. - . -

A unityfeedback controt system isctlaracterized by ihe fottowingopen loop transfer function G(s):: (0.4s +1}fs(s ;1"0.6) . 'Determin~ its transient response for unit step input and sketch ttie response. Evaluate themaximum overshoot andtheC(lrrespondiri~peaktime. " . .' ".'. ..SOLUTI()N

C(s)= G(s)The closed l o o p transfer function, .. R/.S.) 1 G () H ( )" + 5"' 5Given that, G(s) ="(0.45 + 1 )ls (s +0 .6 )F o r U I ;J it yf ee dba c k s y stem ,H (s ) = 1 .. '. . 0.4 . 5+1

". C(s) G(s) . . s(s'" 0~6}.', R(s) = l+G(s)'" 1 0.4s+1+ .5(5+0.6)0.4 5+ 1s(s+0.6) + 0.4 5+ 1

_ 0.45+1 0.45+1- 52 +0:65+ 0.48 + 1 - S2 + 5 +1

0.4 s+1The s ~domain response, C( s)= R(s) x S2+ s+1Fo r step input,H(s)== 115 ..'. C(s)=2 0.45+10.45+1

, S 52 +s+1 5(52 +s +1)B y partial fraction expansion C(s) can be . expressed as,

C{s)= 0.45+1 =~+ .Bs+Cs (s~ + s + 1) S S2+ 5+ 1The residue A issolved by multiplyinQC(s) by s andletting s =O." '0.4 5+11 .:. A=C{s) x 5Is=0= 2 .' =1

5 +s+1s'''0The residues 8 and Care solved by cross multiplying the.following equation andequating the coefficientsoflikepawers 015.

. . "

0.4 s+lA Bs+C. =-+----;;"~~S{$2 + s.'; 1 ) S S2 + S+ 1

On cross multiplication we get;0.4s+1 =A(S2+S+1)+(Bs+C)s0.4s +1 =A$2+ As +A+ 8s2+ Cs

2.29. . -. ' -


29/84

T he tim e do main res po ns eis o bta ined bytak fng invers e Lap lace trans fo rm of C (s ) ..'. T he res po ns e in tim e do main ,

c(t) = .c1C(5)}= . c . . ; { . ! _ - . 5'+0 .5 '. 0.1-.jfj35 }'. 5 {s+ 0,5)2 + 0 .75 JO.75 (s + 0.5)2 + 0.75'"1-e---{J 5tco sJO.75 t - ~ e-0 5 1 s in JO.75 t- JO .75 -'' " 1 7 e. .Q51[.1155sin( JO.75 t ) - f'" co s ( JO.75 \ l l

T he trans ien trespo nse is the pa rt o fthe o utput which van is hes as t tends to in fin ity. Here as t te nd s.to in fin ity th ee xpo nen tia l co mpo nent e ~5ltend s to ze ro . He nce the trans ien t respo nse is g iven by the damped s inuso ida l compo nen t '. 'T he trans ien t respo nse o f c(t) =e-o.5 1 [0 .1155 s ln (.J0 .75 t) + Cos( . J~ .75 t } lT he va lue o f ~and ff in can be eS!im ated by co rnpa r tnq th e c ha ra cte ris tic e qu atio n o f the sys tem with-s tandard fo rm o fseco nd o rder charac te r is tic equa tio n . . < : ( t )

:. Sl+ 2!;f f inS +ff in2:: s2+s +1On c ompa rin g we get,

/ ff i~ =1 ,I 2l; f f in = 1;'. (J) n '" 1 rad I sec I . . I;= _1_. = = . . ! . = 0.52r on 2

1 .1631

~~~ -O_5~M axim um o versho o t, M p = eP:~1-O,52=0 .163 o . t , ,=3.628 sec t .Fig 1Response of under damped system.% M axim um o versho o t, %M p = M p x 100:: 0 .163 ~dOO'" 16 .3%

. . . 1 1 : 11:.. r cPeak tim e, tp = - = r;-;:2 - . ~ - 3 .628 s ec, {I)o ff in V 1-!;2 1x ,,1- 0 .52 .T he respo nse o f the sys tem is underda tnped and it is sho wn in -f ig 1 .

RESULTT rans ien trespo nse o fth! s ys tem ,c{t) - e -O$l[ O .1155s in (.J0 ,75 t)+ cos(JO,75 t ) ]% M a xlm um pe ak ovEu "Sh oo t, %M p - 16 .3%Peak tim e, tp ' = 3 .628sec

EXAMPLE 2.8A un ity fe ed ba ck c on tr ol s ys te m hasen ampl i f ier with ga in KA :: 10 and ga in ra tio , G (s ):: 1 /5(5 +2) in the feed f o rwa rd

path. A d er iva tive feed back , H (s ):: s Ko is in tro duced as a m ino r lo op aro un d G(s ). D eterm in e th e der iva tive feed back co ns tan t,Ko so tha tthe sys tem damping fac to r is 0 .6 . . - - . '

1.30Step 1 : R educ ing the inne r feedback lo op .


30/84

1G(s)

1+G(s} H(s)s (s + 2) _ 1 _ --;;-__ ~

1+-.-' -'- SK - s(s+2)+sKo - s2+2s+sKo5(5+2) ,0 .1

C(s)

Step 2:C om bin ing blo cks in cascade10 C(s)

Step 3 : R educ ing th e un ity fee dback path10 R(s} 10 C(s)

52+ (2 +K;)s + 10

, , ." C(s) 10'T he c lo sed lo o p tran s fe r func tio n , _ ' '" -2;O-----~, R (s ) 5 +(2+Ko)s+10T he given s ys tem is a seco nd o rde r sys tem . T he va lue o f K g can be de te rm ined by co mpa r ing the sys tem trans fe r

func tio n w ith s tanda rd fo rm o f seco nd o rde rna r is fe r functo n g ive~ be lo w . . ,

. .. .. ( 1 )

S tanda rd formof } c{S) , , ro~~nd o rde r tra ns fe r functio n R(s ) ,'" 52 + 2l;;ro ns + ro ~On o ornpar ing equatio n (1 ) & (2 ) w e get ,

(r)~"" 1 0.'. ro n = = . f I C i ' " 3.162 rad Isec

, . ... . ( 2 )

:. Ko = 2~wn":2=2xO.6x3.162-2=1.7944 '

RESULTT he vatue o f c o n sta n t, Ko'" 1.7944 ,

2.31 _ , : : _ . "T he un ity feed ba ck sys tem is sho wn in f ig 1 .


31/84

C(5} 5 (5+2) 10 10.. R(s('l+ 10 = = s (5+2)+10'= s2+2S+10

s (s+ 2)T he va lu es o f d am pJ ng' ra tio I;a nd n atu ra l fr eq ue nc y o f o s cilla tio ~ (f ) n a re o bta in ed by co m par in g the sys tem tran sfunctio n w ith s tanda rd fo rm o f seco nd o rde r trans te r functio n. ' 1t . , - : ~

. S tanda rd .fo rm o f -}, C (s ) ,ro~ "Seco nd o rde r tran s fe r func tio n R(s ) = $2 + 2i;wns + w~ - ., .. :(

T he c lo sed lo o p tran sfe r func tio n, C(s) - ~R(s) - 1+ G{s) -'_ ' G (s )

I'h e -c lo s e d lo o p tr a ns fs r fu n cu o n ,G ive n th at, G (5 ) = lOis (5+2)10 Fig 1:Unltyfeedback system:

On compar i ng e qua tio n (1 ) & (2 ) we get,w~=10

:. ron:: J W =3.162 ra d I sec 2i;w ~ : ; 2- 2 - 1:.1;;=-' -=--'=0.316-2w;, 3.16-2-1~1-[: -1Jl-d.3162e = tan '-- = = tan 1249 ra d(," 0.316

Wd= r.'on~1-(,2 = 3.162JhO.3162 = 3 ra dis ec ,1t - e IT -1249R is e tim e , ti =--,' = = ~ ' - 0.63 s ec

(l)d .)_~ -O,316~,

_P er ce nta ge o v er sh o o t, % M p = eR100 = ekO_31fj2, xl 00= 0.3512 x 100 = 35.12%

Peak o ve r s h o o t = 35.12 x 1 2 u nits = 4.2~44 un'i ts100 y-Peak- t ime, 'IT IT' ,t~:; _'-'=- -:; t047 sec~ O J d 3 , "T. 1 1Ime co ns tan t, T = = - : ; . ; _ - - : - ~ - -C : ; w n O.316x3.162 1sec:. For 5% error, Settling time, ts= 3T = 3 s9f

Fo r 2% error, Settling time, t. = 4 T = 4 s ecRESULT

2.32EXAMPLE 2.10


32/84

Ac losed I? o p s elY o is represented by th e d if fe re n tia l e qu a tio n ::~ + 8:~ =64 eWhere e is the d isp lacem en t o f the o utput s ha ft, r is thed is p laeem en t.o f:the inputsha ft and e = r - c . D ete rm ine

und am ped na tu ra l f re quen cy, da mpin g ra tio an d pe rcen ta ge m axim um o vers ho o t fo r un it s tep in putSOLUTION

The ma t hema ti ca l equanons go ve rn in g the s ys tem a re ,'d 2e de-+8~_=>64edt~ d te = r-c

Put e-= r - c in equa tio n (1),--d2e de '--:. dt2 +8"dt;= 64{r - e )

Let L{e} = C (s ) a nd L{r } = R(s)O n tak ing Lap lacetra ns fo nn o f e quatio n(3 ) we ge t. . __

S2 C(s ) + 8s C (s ) = 6 4 [R(s )- G (s )]: . _ - g 2 C(s ) + a s C(s ) + 64 C(s ) = 64 R{s)

( S 2 + 8s + 6 4) C {s ) = 64 R(sj'qs} 64. . R ( s f ; : : : ; ~2 + 8s+64

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

. .. .. ( 3 )

..... (4)~era tio C (s )JR(s )is the clo sed I~p trans fe i: functio n of the sys tem . On ~m pa r fng the sys tem trans fe r functio n w ill i

s ta nd ard to rm o ts ec on d o rd er tr an sfe rfun ctlo n , we G 8 n est imate the va lues o f i ; and ( ) ) n ' - ,Standard form o f -} C (s } - , (J)~

__Se co nd order trans fe r func!io n _- 'R(s) 0 = ' S2 + 2 i ; - { ) ) n s + ro~On compa li ngequa tio n (1) & (2)we ge t,..... (5)

ro ~ =64 2i;ro n = 8i; =__ =_8_ = 0.5_ 2ro r , 2x8:. ffin =8 rad I sec-I;~ - _ - - - ' O . s ~

Percentage .peak overshoot, % M . , = eR100 .= e h-o x 100= 16.3%ft:ESULT

..~.

- Undamped na tJ . ir a lir e quency o f osci l la t ion, ro n = 8 r ed /s e c

2.332.9 TYPE NUMBER OF CONTROL_SYSTEMS


33/84

The type number is specified for loop transfer function G(s)H(s). The number or poles of the 1 0transfer function lying at the origin decides the type number of the system. In general, ifN is the.num _'of poles at the: origin then the type. number is N. -The loop transfer function can be expressed asa ratio o f two polynomials in s.

G(s\H(s)=K P(s) =K (HZ,) (S+Z:!) (S+z.;_}, .. Q(s) SN (s+pd (s+PZ)(S+P3) .where, zl' ~, ~, ~.....are zeros of transfer function

Pl' P z > P3' are polesof transfer functionK =ConstantN =Number of poles at the origin

The value of N in the denominator polynomial of loop transfer function shown in equation (2. )decides the type number of the system.

If N =0, then the system is type .; 0 systemIf N = 1, then the system is type - 1 systemIfN'" 2, then the system is type - 2 systemIf N =3, then the system is type - 3 system and s o on.

. . . . . ( 2A~-

2.10 STEADY STATE ERRORThe steady state error is the value of error signal e(t), when t tends to infinity, The steadys .

erroris a measure of system accuracy. These errors arise from the nature of inputs, type of systemfrom non linearity of system components. The steady state performance of a stable control system ' ..generally judged by its steady state error to step,ramp and parabolic inputs. -

Consider a closed loop system shown in fig 2.15.Let, R(s) = Input signal

E(s} =Error signalC(s) H(s) =Feedback signalC(s) =Output signal or response

The error signal, E(s) ;' R{s) - C(s) H(s)The output signal, C(s) = E(s)G(s)On substituting for C(s)from equation (2.51)in equation (2.50) we get, ..

.F ig 2.15......(2.5 .....~(2.5.1.....

2.34

:.e(t):=.cI{E(S)}=:.c1{ R(s).. }. .....(2.53)


34/84

. 'I+O(s}H(s)Let, e55 = steady state error.The steady state error isdefmed as the, value of e(t) when t tends to infinity.

:. ess = Lt e{t)t-e-eo ....:(2.54) .The final value theorem of Laplace transform states that,

If,'F(s)= L{f(t}} then, Lt f(t) =Lt s F(s)t-+oo .-+0 .....(2.55)Using final value theorem, .

sR(s). The steady state error, ess = . Lt e(t) = Lt sEes ) = .Lt . . -- . . : . .. . ;. . . .-H


35/84

K = Lt O(s) H(s) = = Lt K (s+ z,)(s+Z:2)(s+ Z3) 00P HO HO S(S+Pl) (S+P2) (S+P3)....... . 1 1

:.ess=--_=- ~ =O1+ Kp 1+00in systems with type number 1 and above, for unit step input the value of'K, is infinity and so the'-steady state error is zero. .

. .2 .13 STEADY STATE ERROR WHEN THE INPUTISUNrrRAMP SIGNAL

sR(s)Steady state error,e,s =Lt. 'HO l+G(s)H(s)When the input is unit ramp,' Res) ~ ~. s

~. 152 1

:. e ss::: Lt . s . = LtHO 1+O(s) H(s), ....os +sG(s) H(s)1 ..... (2:61)----:::-.Lt sG(s)H(s)s. . . . 0 K, v

where.K, =Lt sG(5) H(s). , s....o

The constant K, is called v elo city e rr or c on sta nt. T ype -Osys tem -,

'x,= Lt sG(s)H(S)= Lt sK (s+zt)(s+Z:2)(S+Z3)""". 0< - - 'l o O . H O (S+Pl) (S+P2) (S.+P3) ......

:. CS1 = 1i K, =110 = 00Hence in type-O systems when the input is unit ramp, .the steadystate error is infinity.

. T y p e.;.l s y st emK ~ L G() H( ) = L K (s+. z,) (s+ Z:2)(5+ Z3)".''''v ts s S ts. .

HO HO S(S+P')(S+Pi)(S+P3) .......',e,s':'"11K, = = constant

Hence in type-I systems when the input is unit ramp there will be a constant steady state error.

= K_ZI.Z:2Z3......constantPIP2P3 ..

, T ype -2sys tem'K ='L G()H()=L K ' (S+ZI) (S+Z2) (S+Z3.) . . . . . . . .v t s. sst 5'2 .' . 00""'0. .....0 s (s+p,) (s+pz) (s+pj) .....,

~


36/84

Steady state error, ess =Lt _ sR(s), , HO 1+0'(5) H(s).

,- ~: }-When the.input is unit parabola, R(s)=3'- . S

1s-L S3, , 'Lt, 1_:.es = t ' " ---'-;2;:----;:-2--...__- ._..o 1+ G{s) H(s) .-+0 s ~s G(s) H(s) 1u s2G(S}H(s) = K .. cs~ o - , . _ ..(2 .62 )

- where, K, = Lt S20(5) H(s). HO.The constant K is called accele ratio n erro r, co nstan t -

- T y pe -a syst~mx, =Lt s2G(s)H(S')= Lt s2K(s+Zl) (s+ZzJ (S+z3) 0

HO '_ ,HO (s+Pt){S+P2)(S+PS) ......1 _ 1

.r . ess ::;'-=-=,0:>, Ka 0 -Henceintype-O systems for unit p~bolic input, thesteady state " e r r o r is inf ini ty,

Type-l system-. K =Lts2G(s)H(s)~ u s2K (s+zD(s+~)(s+Zjj ..... 0-

a $40 - HOS (s+P~)(S+P2)(-s+P3) ...... -1 1.e=-=-=oo. s s Ka 0

Hence in type-lsystems for unitparabolic input, the steadystate error is infinity... ..~. _.. . . .T ype-2 sys tem _

. K. = Lt s20(s) H(s) = Lt s2K ;s +Zl} (8 +zzHs+ ~) .._.... = K ZIZ2:Z3..'.. constant.-0 . _ . . o S (s+ Pl).(S+Pi) (5+P:i).PIP2P3:1:. e ss = - =cons tan tK a '

H-ence in type-2 system when the input is unitparabolic signal there will be a-constant steadystate error.Type-3 system

K, = Lts2G(S) H(s) = Lt S2K(S+Zl) (s+Zi) (s+~) .5-+0 .HO S3 (S+Pl) (S+P2)(S+P3)";'" < X l -

2.37TABLE-2.2 : Static Errot Constant for TABLE-2.3:Steady State Error for


37/84

Various Type Number of SystemsError Type n um be .. of sys temConstant 0 1 2 3

K cons t an t 00 00 00pK 0 cons t an t 00 00vK. 0 0 cons t an t 00

Various Types of InputsInput Type number-of sys temSigmd 0 1 2 3

1Unit Step -- 0 0 0r-x,1U nH Ram p 00 K 0 0

1UnitParabo1ic 00 00 - 0K,.2 .15 GENERALIZED ERROR COEFFIC IENT

. . ~The drawback in static error coefficients is that it does not show the variation of error with time.. - -and input should be it standard input. The generalized error coefficients gives the steady stateerror as afunction of time. Also using the generalized error coefficients, the steady state error can be found for anyt y p e of input.

The error signal in s-domain, E(s) can be expressed as a product of two s-domain functions.R(s) IE(s} = = 1+ G(s) H(s) i+G(s) H(s) R(s) = = F(s) R(s) .....(2.63)

where,F(s)=- 1. 1+G(s) R(s)Let, e(t) L-I{E(s)} (error sighal in time domain). \'f(t) = L-I{F(s)}

ret) = L-I{R(s)} (input signal in time domain)The convolution theorem of Laplace transform states that the Laplace transform of the convolution

of two time domain signals is equal to the product of their individual Laplace transform.-i.e., L{f{t) * r(t)}= F(s) R(s)where * is the symbol for .convolution operation.: .c-'{F(S) R(s)}'; fl.t) * r(t) ..... (2.64)

From equation (2.63) & (2.64) We can 'wri te ,e(t) "" f{t). r(t)

Mathematically the convolution of -f(t) and ret) is defined as,+


38/84

( T ') . ( ) . () T2 . 'f( ) T3 'H()+ ()n r- n.(r t - ""r t -T t + - . , t-- r t ....,.+ -I - r t)...2! 3! nlwhere, (t ) = IS ! derivative ofr(t)

r e t ) =2nd derivative of r( t )n.- ret) =nth derivative ofr(t)

On substituting the Taylor's series expansion - o f r(t - T), the error e(t) can be written as,eft) = f l . f(T) [ r ( t ) - T i(t) + r2 i'(t) - T3r (t)+ ....+(_1)0 , m ~(t)..] dT21 3! .. n ! ' .o . , .! t " t 2 "e(t) = J leT) r(i)dT - f f(T) T_ret) dT + f f(n~l r e t ) dT

o ,0 o - 1 f(T)~ rm-,..1 f(T) (_:1)n Tn ~'(t)dT....ooo 3! 0 n l .

Since ret), i(t), T( t ) , . . . . . ~'(t) are constants when the integration is done -with respect to T, the errorsignal can be written as,, ,e(t) =ret) J f(T) dT - ret) J Tf(T) dt + f ; : ) J T2f(T) dt

o ,0 0

- r(t)J! Tf(T) dt+....+(-lt ~'(t)fl.T O f(T)dt...31 n t, 0 ' 0!

Let, Co =+f f(T) dTo

IC1 = " ' " f ,Tf(T) dTo

I ,C2 =+J rZf(T) dT ,

o

f ' 3 .C3=- Tf(T)dToI

,'C_n=(_I)n f TorCf) dTo

eft) =ret) Co + ret) c1 + 1(t) ~~ +'r(t) ~:+. . .~(t) ~~h..., C () C .() c2";.{) c3.,,() . cn n.() ,= 0 r t + 1rt +_ .\t +- r t +....+- r t .....21 '3! n !

The equation(2.65) is the general equation for error signal, e(t)......(2.65)

2.39 ..-:~'lijm~~~" )2.16 EVALUATION ,OFGENERALIZED I:RRORCOEFFICIENTS


39/84

T he generalized error coefficient is given by;te n =(-ItJ -T"f(T)_dT;o

, 1 -where F(s)= , - ._. l+G(s)H(s)We know that L{ f(T)} =F(s), hence -b y the definition of Laplace transform,

t....F(s)=J fCT) e-sTdT ..... (2.67) ..

oOn taking , ~ . t oon both sides of equati on (2',67) we get,

tLt F(s) =Lt J f(I) e--'sTdIS40 _. '40 .0,t t

= f fCT) ~ to e-sTdT= f fCT) dT= Co .o . . ~. . : . 1 Co = .~~F(s) I

On differentiating equation (2.68) with respect to s we get, -~ F(s) =~Jtf(T) e""sTdTd s ds .o

t ~. . = J f(1)!e-ST)dT = J (1) (-1)e~rdT0- 0t= : - J TfCT) e~sTdTo

..... (2.68)

..... (2.69)

On taking ~~on both sidesofequation '(2.69) we get,t

Lt ~F(s)= Lt - J . Tf(T) e-sTdT, - - - t o d s s-so o 't . i_= . o . . f TfCT) Lte _:sTT =- J Tf(T) dT=C]. . - - - t oo 0. d.. C]= Lt ~F(s)

. - - - t o d s . , .On dif ferent ia t ing equation (2.68) on both sides with r e s pec t to s we get,..... (2.70)

i_ r i_ (F(S )] =~[- j T f ( T ~ e - ' ; d T ] .


40/84

s. 41- EXAMPLE 2.11

, , '10(s+2)


41/84

For~ uhity feedback control system the open loop transferfunction. G(s) = S2(s + 1)" . Finda) th e position. velocity and acceleration error Constants.b) the steady state error when the input is R(s), where R( s) = 1_ ~ + ~- , s s 3s

SOLUTIONa') To find static error constants

Fora unity feedback syStem H(s)::;1Position error constant, K p '" Lt G(s)H(s): Lt G(s) : Lt 1~ s + 2i '" 00540 5-+0 _ '5--+0 S (s+ 1, -

." - 10(5+2)Velocity error constant. K . "'5~O SG(s)H(s) = .~os G(s) '" ~o s S2(5+ 1 )Acceleration errorconstant, K;,:Lt s2G(S)H(S)= .Lt S2G(S)S40 .- S40

: Lts2 10(s+2)", 10~2=20HO 52(S + 1) 1b Y T o find steady state errorMethod-I

Steady state error for non-standard input isobtained usinggeneralized error series. given below.'The error signal, e(t) = r(t)Co + f(t)C1 + f(t) C2 +....: . +~ (t) Cn +.;.., '2! n !_. 3- 2 1 -GIVen that,R(s}=--~+-_s S2 3s3

Input signal in time domain, r(t)";' . c 1 {R(S)}= r 1 { ~ _ ~ -+ ~ 3 ' } ,_' s s 3s, 1 t2, t2= 3-2t+-- '" 3-2t+-32!, _6

. . r e t ) = i.r(t) = -2 +22t = -2 + . . ! . .dt 6 3"'() d2 () d.() 1rt =dt2rt = dtrt ='3~'() d3,() d ..() 0' r t = dt3 r t'" t r t '"

The derivatives of ret)is zeroalter second derivative. Hence we have to evaluate only three constants Co.C1 and C 2The generalized error conStantsare given by.

2.42c , = Lt AF(5) '" Lt A[ 53+52 ]2


42/84

....Od5 s.... ds 53+5 + 105+ 20= Lt f(53 +52 +10s+20X3s2 +25)-(53 +252)(352+2S+10}].... (53+52 + 105+ 20), .:::Lt[35S+254 +354+253 +3053 +20s2 +6052 +40s-3t -254 ~ 1053-354 -253 -10s2]HO, . (53+52+105+20):;: Lt r , 2053 +7052 +4052] '" 0H 1 (53 + 52 +lOs + 20) .

C2 = Ltd22 F(5)'" Lt A[~F(5)] = Lt A[ 2053 + 7052 + 4052]....od5 5->0 ds ds ....0 ds (53+52+ 105 + 20)

(S3+ 52+ 1O s + 20r(6052 + 1405+40)':;'(20s3+ 7052 +405) 2:':(53 +52+105+20) (352~25+10)202 x40 1;Lt =(53+52+105+20f 204 . '10,

Error signal, e(t)::: r(t)eo + f(t)C, + r e t ) C2 = (3- 2t+ f ) x 04- ( - 2 + _ ! _ ) x 0 + . : ! . x J... x _ ! _ : : : _ ! _.' . 21 6. ' . 3 3 10 2! 60, 1 1Steady state error, es s = = Lt e(t) = = Lt - = -. " H", . 1-+", 60, 60

M~thod -IIThe error signal in s -domain, E(5) =1+~~;H(5)Given that, R(s):::~_2.+_1~. G(s) = 10(5+2). 'H(5)::1. S 52 353' , 52{S+ 1 ) ,

3 ,2 1 3 2 i--2+-3 --2+---J:.E(5) = 55 3s = = ,5 5 35 ', 1+ 10(5+2) 52(5+1)+10(5+.2)52(5 + 1 ) , 52(5 + 1 )= ~ [5 2 ( 5 : : ) ( : ~~s+ 2 ) ] - , ~ [ 5 2 ( 5 : ~ ( : ~ ~ s +)]+ 3 : 3 [ 5 2 ( 5 : : ) ( : ; ~ i 5 +2)]

The Steadystate error ess c an b e o b ta in ed f rom f ina l value the~rem:

,2 .43Method .~ III


43/84

Errorsignal in s ~domain,E(s) - R(s) .. . 1+G(s)H(s) ,. E(s) .= 1 .R(s) 1+ G(s)H(s).10(s+ 2), G iven tha t. G (5) : : '. ' .H(s)::: 1, s2(s+1) ,

E(5) , 1 S2(5 +1)., R(s) = 1+ '10($ + 2) S2(5+ 1 )+ 1 O (s + 2),52(S+1)

53 + S2 52 +53 52 53= 53+52+105+20 = 20+105+52+53 - 20 + 40+"", ' . . [ S 2 S3 ] _ 1 2' ,,1, 3E(5) = R(5) 20 + 40 +... - 20 s R(s) -r 40 S R(5)+".

On ta kin g in ve rs e Laplace trans fo rm o f the abo ve equatio n we ge t,

20 - 1 ' - 1O s +52.+S3 S2+$3, '. $3 S4 S5. S2+_+_.:,_+.......,.( -) (-) 2 (-) 20(-) . 20

$3 54 S5'

S2 S3 .-+-"'"20 40

e() l ..() 1,~(), t =-rt +-r t+...,20 40,. '. 3 2 .. 1Given that, R(s) '" ---+--s 52 3s3

.'

S3 S4 55 S6~+-+-+-( f ( _ )4 (-) 40 (-) 40. 3 s 4 3s5 . . s f:------10 40 402 2020

Div id ing numerato r p olyno m ia l "by denomina to rpo lyno rn ia l ,\.

r e t ) = !!.r(t) = -2 + ..!2t = = -2 +.!.dt. .' 6 3..( ) d2 () d '( ) 1't :::-2 r t =, . . . . . . . . r t ::-dt dt, 3,

3 ."let) = : t 3 ret)~ : t f(t) = a

E' I' .. do , '() 1"{)"1(1)\ 1:. rror.signa In time omain, e t = -r t =~ - ::--'- " 20 203' 60, '. ' 1 1Steady s ta te e rr or , ess':: Lt e(t) = = Lt - =-I....o \-.'" 60 60

. . . .

,RESULT(a) Po s it io n ' e r ro r c o n s ta n t,

V e lo city e rr o r c on sta nt,A cc ele ra tio n e rr o r c on sta nt,

K=co~K , . : : coK.=20

1

,~',__ '({~~i$is )SOLUTION

2.44


44/84

~... a) G{5):: 20(5 + 2),~, s(s +1)(5 + 3)'

Let us assume unity feedback system, :; H(s )=1The open loop system has a pole atortgio. Henceitis atype-1system. In systems with type oumber-1, the velOCity (iclillP)input wil l give a constants teady.s tate error.' , .The steady state error with uoit velocity input. ess '" ~Velocity error constant, K " = = 5~0 SG(5) H($l "'s~o 5G{s)

= Lt 5 . 20(5+ 2). 20 x 2 = = 40$-)0 s(s + 1)(s + 3)1 x 3 3,

S dvs '. 1'3' O'07' tea y state error, ess '" - = = -- ' " . . 5" 'K v 40). . 10b .G(s}= (S+2)(5+3)Let us assume unity feedback system. :. H(5}=1. .. .The open 190P system has no pole atorigin. Hence it is a type-O system, In systems with type number-D. the step input ., w ill give a constant Steady state error. .The steady state error With unit step inp.ut,e~ '" ._1_, . . . ., '1 +Kp, ." . . . 10 'Position error constant, Kp = Lt 'G{s)H(s) = Lt G(5) = Lt -:- - - -- ,- ,-. s-.O " S~O 5->0 (s+ 2)(s +3)

Steady state error, 1 13 3eS s ",._-=_ -=-=.,.-::0.3751+~ 1+~ 3+5 83c} G(5) = = ,10

52(S+1)(s+2) .Let us assume unity feedback system, .: H(s)::1.

The open loop system has tw o poles atorigin. Hence it isatype-2 system. In systems w i t h type. number~2, the, acceieration (parabolic) inputwill give a c o n s ta n t steady state error. .The steady state error with unit acceleration input, eS s " " _ ! _. . . ' ~Acceleration error constant, Ka"'.Lt S2 G(s)H(s) = Lt S2 G{s} '" Lt S2 ' 10. . ::10 '"50->0, ' S... O s-.O s~($+1)(s+2) 1x2

1 1 'Stecidystate error, ess = _ - ' " - ' " 02Ka5


45/84

2.46

.C " '" L t ~F(S1'" Lt ~[..:.F(S)Jl.", Lt ~[ 2s+10 ]s_,.ods2s-->ods ds - . _ 5':"0 ds. (0.1s2+S+10)2


46/84

= Lt j_[(0.1s2+S+1O)" X 2-(25+10) X 2(~.152+S+10}(0.25+1)]', -s- 0 ds (0.152+5+10)-

:.C2'" 102X2-10:2x10x1=O10Error signal, ett) '" r(t)Co + r(t)Cf+ ret) C~ '" r{t)C,+ 0 + 0 = ( a , + a21 ) 0.1_ . 2. - _

:.' Steady state error, e.. '" Lt e(t):::. Lt [(a, + a2t) 0.1] :::


47/84

,Given that, r(t) =a1) +a1 + i e. d, ":. r= -r(t) = 81-t:821dt.., d .-.r(t) = = dt ret) = a2...() d ..() 0r t ==-r t ;,

dt:. Err~r signal in time domain"e(t) = 1~ r e t } = 1~ (81 + 82t)

0+5+0.152 5+0.15252 535+-+-,

(-) (-)10 (-)100, 53-100

S354 55_; ) 1 5 0 ; ) 1 0 0 0 ; : 1 1 0 0 0 0 'S4 ,551000 +1 OOO(}

Steady state error;

, RESULT"(a) Position emir constant,(b ) Velocity error constant,,(cl Acceleration error constant, ,

, , a t2(d) . When input, r{~)= = 80 + 81t+T'K = eo, pK , . = 10K.= aSteady state error, es s = O C J

EXAMPLE 2.14Consider a unity feedback system with a closed loop tr an sfe r fu nc tio n , q R ( S = 2 Ks + b. Determine open IOOp'S S -s as r b

transf~rfunctiOnG(~). Show that steady state errorwith unit ramp input is given by (a ~ K) .SOLUTION

" F o r unity feedback system, H(s)=lThe closed loop transfer fun~on. M(s)::: C(s) ,,;G{s) ,, ';, ,R(5) 1+G{5)H(S)

G(s) , ": ' l+G(S) = = M(s)

~l+G(s)

On cro ss mu!tiplica~on o f the abo ve equa tio n we get, ,G(s) =M(s)f1 +G{S)] ==M(s) +M(s) G(s):. G{s) -M(s) G(5) .,; M(s) =?, ,G{s)j1 _M(s)] = M(s ) =? M(s) = Ks + b


48/84

2.492.19 COMPONENTS OF AUTOMATIC CONTROL SYSTEM ':

The basic components of an automatic control system are Error detector, Amplifier and Controller;


49/84

Actuator (Power actuator), Plant and Sensor or Feedback system. The block diagram of an automatie'control system is shown in fig 2.16. '

(Inputsignal):. ~. u ,;.~, ,., ,., :

~t

I1+----------ensor orFeedback signal FeedbacksystemFig 2.16: Block diagram of automatic control system.

, The plant is the open loop system whose output is automatically controlled by closed loop sy s t em ,The comhined unit of error detector, amplifier and controller is called automatic controller, because,without this unit the system becomes open loop system.

In automatic control systems the reference input wil l bean inputsignal proportional to . desiredoutput. The feedback signal is a signal proportional to current output of the system. The error detectorcompares the reference input and feedback signal and if there is a difference it produces an error signalAn amplifier can be used to amplify the error signal and the controller'modifies the error signal for bettercontrol action.

The actuator amplifies the controller output and converts to: the required form of energy-that is'acceptablefor the plant; Depending on the input to the plant, the output.will change..This process continues as long as there is a difference betweenreference input and feedback signal. If the,difference is zero, then there is no error signal and the output settles at the desired value.

Generally, the error signal will be a weak signal and so it has to be amplified and then modified forbetter control action. In most of the .system the controller itself amplifies the error signal arid integrates'..or differentiates to produce a control signal (i.e., modified error signal). The different types of controllersare P, PI, PO and PID controllers. '.2.20 CONTROLLERS

A controller is a device introduced in the system to modify the error signal and to produce a co~trolsignal'. The manner in which the controllerproduces the control signal is called the controlaction. Thecontroller modifies the transient response of the system. The electronic controllersusingoperationalamplifiers are presented in this section. , '

The following six basic control actions are very common among industrial analog controllers.

2.50-Depending on the control actions provided the controllers can be classified a s follows.-I. Two position or ONOFF controllers.2.


50/84

Proportional controllers.3. Integral controllers.4. Proportional-plus-integralcontrollers.5. Proportional-plus-derivative controllers.6. Proportional-plus-integral-p Ius-derivative controllers.

ON~OFF (OR) T W O POSmON CONTROLLER 'The ON-OFF or two positi~n controller has only two -fixedpositiots. They are either on or off.

The on-off control system is' very simple in construction and hence 'less expensive. _For this reason, it is 'very widely used in both industrial and domestic control systems.

The ON-OFF control action may be provided by a relay. There are different types.of relay. The most-popuiar ~me is electromagnetic relay. It is a device which has NO (Normally Open) and NC(Normally Closed) contacts, whose opening and closing are controlled by the relay.coil, When the relaycoil is excited, the relay operates and the contacts change their positionsIi.e., NO-t NC and NC -4NO).

- 'Let the output signal from the controller be u(t) and the actuatirig error signal be e(t). In thiscontroller, u(t) remains at either amaximum or minimum value.

u(t) = ul; for e(t) _< - 0= u2; for e(t) > 0

R(s) ,.,,~u,U,~ "-

U(s)

I E(S)=L{e(~)};' ,- V e s ) =L{u(t)} I Fig 2.17 : Blockdiagram of on-off controller.Feedback signalPROPORT IONAL CONTROLLER ( P :- CONTROLLER)

The proportional controller is a device that produces a control signal, urt) proportional to the-inputerror signal,e(t). .

In P-controUer, uCt) a: e(t).. ti(t) = Kp e(t)

where, Kp_= Proportional gain o r constant -.On taking Laplace transform of equation (2.81) we get,

U(s) = KpE(s)

.....(2.81)

.......2.82). . U(s) . It.', -Transfer function ofP - controller, -- = Kp ,. . . E(s) - ..... (2.83)


51/84

-atBPl~4i~ii~sis, ' ).", . . - 2.52ANALYSIS OF P~(ONTROllER SHOWN IN:HG 2,20

The assumption made inop-arnpcircuitanalysis are,


52/84

1. "The voltages at both inputs are equal2. The input current is zero.

Based on the above assumptions, th~ equivalent circuit of op-amp amplifier 'and sign changer areshown in fig 2.22 and 2.23~ .'

i, -- It,- - - . .

I

i 2 - R-++~

+ uet )

_Fig 2.22 : Equivalent circuit a/amplifier. "Fig 2.23 : Equivalent circuit of sign changer.fl" 2 2' ' . eft)From rg z, 2, e(t) "" i.R, ; :_1 : ""R .....2~88)Iuj(t) ""-il~ ,:...2.89)

Substitute fori, from equation (2.88) in equation (2.89).. ''_ e{t) , -.. u1(t) - -R R2I .....(2~90)

. - , u(t)From fig 2.23, u(t) = = -ci.R; :.12==-Rul(t) = i 2 R

Substitute for ~ from equation (2.91) in equation (2.92) .. ' _ u(t) _ ".. u](t)- -T=,-u(t) -

On equating the equations (2.90) arid (2.93) we get,, R2u(t).: - e(t) -" ,Rl

.....(2.91)

.....(2.92)

_.....(2.93)

.....(2.94)On taking Laplace transform of.equation (2.94) we get,

RUes) =_2 E(s)R]Ues) R2--=-E(s) It]

- . .....(2.95).....(2.96) -


53/84


54/84

2:55On taking Laplace transform of equation (2.111) with zero initial conditions we get,. . .

K E(s)U (s )=K E (s -)+___c _l'_ --. P . lj S .....(2.112)


55/84

U{s) . .( . - 1 J .:. Transfer function of PI - controller, ~~ - K 1+ -E(s) _ .1 ': . 'T is ..... (2.113)

Theequarion (2.112) gives the output of the Pl-controller.for the-input E(s) and equation (2.113) isthe transfer function-of the Plcontroller, The block diagram of Pl-controlleris shown in fig 2.28,

Feedback signal'

~(s)

I -.Fig 2.28 : Block diagram of PI-don troller.

-The _advantages of both P-controller and I-controller are combined in Pl-controller, The proportionalaction increases the loop gain and make s the systemless sensitive t o variations of system parameters. T h eintegral action eliminates or reduces the steady state error.

The integralcontrol action is' adjusted by varying the integral time. The changein. value of Kpaffects both the proportional and integral partsofcontrol action. The Inverse of the Integral time T; iscalled the reset rate.

Feedback signal

. .' ".. -. . E X A M P L E O F E lE C T R O N I C P I - C O N T R O L L E R

. The PI-controller can be realized by anop-amp integrator with gain followed by a sign e(t)changer-as shown in fig 2.29:

By deriving the transfer function of thecontroller shown in fig (2.29) and comparing withthe transfer function of Pi-controller defined byequation (2.1_14),it can be proved that the circuitshown in.fig 2.29, work as PI-controller.ANALYSIS O F PI~ CONTROLLER SHOWN ' IN fIG 2.2.9

R

"i' \Fig 2.29 sPl-controller using op-amp.

The assumptions made.inop-amp circuit analysis ate,1 " The voltages at both inputs are equaL2. The inputcurrentis zero. .

Based on the above assumptions the equivalent circuit of op-amp. integrator and sign changer .areshown in fig 2.30. and 2.31.

... e{t).. ]]=-


56/84

2.57:. Transfer function ofPD ~controller; U(s) '" K (l +T s)E(s) p d ....:(2.124)

The equation (2.123) gives the output of the PD-controller for the input E(s) and equation (2. 124Y


57/84

is the transfer function of PO-controller.The block diagram of PD-controUer is shown in fig 2.32.

Feedback signalFig 2.32 : B lock d iagram of P D - con tro ller.

Thederivativecontrol acts on rate ofchange of error and not on the actual error signal. Thederivative con t r o l action is effective only during transient periods and .s o if does not produce correctivemeasures for any constant error. Hence t he -d er iv at iv e c o n tr o lle r is never used- alone, butit is employed inassociation with propor t iona l and integral controllers. The.derivative controller does 110 taffect the steady-state error directly but anticipates the error, initiates an early corrective action and tends to increase thestability of the system. While derivative control action has an advantage of being anticipatory it has the 'disadvantage that it amplifies noise signals and may cause a saturation effect in the actuator,

The derivative control action is adjusted by varying the derivative time. The change in the value ofK,affects both the proportional and derivative parts of control action. The derivative control is also called'rate control

Kp ( 1 +Td s )Feedback signal.

E X A M P L E O F E L E C T R O N IC P D ~ C O N T R O L L E RThe Plf-controller c a n be realized by an

op-amp differentiator with gain followed by a signchanger as shown in f ig 2.33. e(t)

By deriving the transfer function of thecontroller shown in fig 2.33 and comparing withthe transfer function ofPD-_controllerdefined byequation (2.124) it can be proved that the circuitshown in fig 2.33 will work as PD-controller.A N A L Y S I S OF PD -CONTROLLER SHOW-N I N F IG 2 . 3 3

~Fig 2.33 : P D co ntro lle r U Sin g o p-am p.

The assumptions made in op-amp circuit analysis are,1 : The voltages at both inputs are equaL2. The input current is zero.

Based on the above assumptions the equivalent circuit of op-amp differentiatorand sign changer.are shown in Hg 2.34 and 2.35.


58/84

In PID - controller, u(t) a [e(t) + f e(i) dt +_!e(t)].. . dt. . K f . d .:. u(t) = Kp e(t) +_P e(t)dt +Kp T d ....;,-e(t). . T dt

~ . . .~.59'j.J. . .. ;.(2 .1 3 ,-~


59/84

. Iwhere, K =Proportional' gain.: pT;= Integral time

.~

~1"d =Derivative timeOn taking. Laplace transform of equation (2.1'34) With zero initial conditions we get,- .

. . KpE(s),U(s).=KpE(s)+-::r,- -s-+KpTd ~E(s)I

.--~1

.

:. Transfer function of PID -Controller, U(s). ~ Kp ( 1 + 2 _ . + ~ d S ) 'E(s) v . 'lis ....;(2.13~The equation (2.135) gives the output of the PID-controller forthe inputEts) andequation (2.136)1

is' the transfer function of the PID-controller. The block diagram of PID--contoller is sho-wn in' fig' 2.36 ...iU(s)E(s) ( 1 J. Kp 1+-'-+Tds....lis.

Feedback signalFig 236: Block diagramof Plls- controller.

The combination of proportional control action, integral control action and derivative control action,is called.Plfr-corrtrol action. Thiscombined action has the advantages of the' each of the three individual:control actions. .

. Theproportional controller stabilizes the gain but produces a steady state error. The integral controllerreduces or eliminates the. steady state error.The derivative controller reduces the rate of change of error.:- . - .E X A M P L E O F E L E C T R O N IC P ID --C O N T R O L L E R

The, PID-controil,er can be realizedbyop -a r np amplifier with integral and derivativeaction followed by sign changer as shown in eft)fig 2.37.

By deriving the transfer function of the'controller shown infig (2.37) and comparing withthe transfer function of PID-controller.defined . =


60/84

2.61.


61/84

..:..(2.146)

.The equation (2.146) is the transfer function of op-ampPllf-controller, 'On comparing equation(2.146) with equation (2.136) we. get,'~ .. .Propo r t i ona l gain 'K-~~R2', P RIDerivative time, TJ",,-R1C1 ; Integral time; 1 ; =R2CZ 'Also' , RICI +R2C2 .1., 'R2C2By varying the values ofR, and ~the values ofK", Tdand T; are adjusted.

2.21 RESPONSE WITHP, PI, PO AND PIC CONTROLLERSIn feedback control systems a controller may be I n t r oduced to ' modi f y the error signal-and to

achieve better control action. The introduction of controllers will modifY the jransient response and thesteady 'stateerror of the system. The effects due to introduction ofP, PI, PDand PlD controllers arediscussed in th is s ec tio n , .H F E C T O F P fW P O R T IO N A L C O N T R O L L E R ( P - C O N T R O lL E R ) .

The proportional controller produces anoutputsignalwhich is proportional 10 errorsignal.Thetransfer function o f proportional controller is given below. (Refer equation 2~83). .. . . U(s) .Transfer function of P-controller, E(s) "" K p

. Thet~ Kp in the transfer functionofprbportional controller is called the gainofthecontroller, .'..Hence the proportional controller amplifies t he-e r ro r signal and increases the l oop gain of the system. The .following aspects of system behaviour are improved by increasing loopgain ..* Steady state trackingaccuracy.* .Disturbance signal rejection ..* Relative stability.In addition toincrease in loop gain it decreases the.sensitivity of the system to parametervariations .'The drawback in proportional control action is that it produces a constant steady state error, .

2.62The block diagram of unity feedback system with Pl-controlleris shown in fig 2.40,

C(~)


62/84

Fig2.40: Block diagram offeedback system with PI-controller._ Let the open loop -transfer 'function G(s) be a second order Systyffi with transfer function, asshown in equation' (2.148).' - - , ,

2- Open loop transfer function,O(s) = = C O n ", '_ ,- '_ S,(S+2Sff in)Now, loop transfer function = G c{s ) G (s ) H (s ) = = G c(s ) G (s )K ( I + T ; s ) , f f i '~= ." - lis ) X ,s{s +2Smn}Nowthe closed loop transferfunction is given by ,

....J2~147)

I H(s)=ll.. K pw ; (I+T;s)- - g2T,( s +2s r o , ,) .....(2.143)

_ C ( _ s ) = . , . . . 0 , ; . . . . : ,, , - , - ( S , . : - ) O _ _ ; ( : . . : : . s ) ~R ( s ) 1+O c ( s ) G ( s )

- - 2Kpw; , ( l + TiS )s2T; (S+~(On)~(j):(1 +T;s )1+ ----:o:-"'----"--~-

's2T;(s+2sro,,) -Kp(O~.(l + TiS) -

s j +2,ro S 2 + K ro 2 -s + Kp ro ~n p "T; 0Kiro; (I+Tis)

32r2'K2 K2s + ~(j)nS .;--, P(Oll;;+ i ron_From the' closed loop transfer function (equation (3.149 it is observed that the Pl-controller

mtroduces a zero in the system -apd increases the order by one. The increase in the, order of the' systemresults i n a'lessstable system than the original one because higher order systems are less stable than lowererder systems. .. - -'F rom the loop transfer function (equation (3 .148 it is observed that-the PI-controller increase theltypehunberbyone. Theincrease in 'type number results in reducing the steady state error. For example

i I f . thesteady state-error of the original system is constant, thenthe integral controller win reduce the error

.....(2.149)

i I D zero.


63/84


64/84


65/84

rROGRAM 2.1Consider the standard closed loop transfer function of the secoiid order systemgiven' below.

M(s)=())n2/ (s2+2~())nS+())n2),Wr;'te a MATLABrogralil, to find-the unit step response for various values of;_


66/84

damping ratiof~' Take, natu'ral frequency of osc+Tlat ion.. (On=1 rad/sec.;-~~-~. %un; t step response fo rvari ous va..ues of dampi 1 \9 r atto ; zeta.%The natural frequency' of oscillation, wn..l.

cTct=O:0.2:12:c=zeros(61,6)izeta=[OO.2 0.4 O.~ 0.8.1];.' for n=1:6;nUlILcof=[Oall;den~cof';'rl 2*zeta(ri) 1]:c (1 :-61,n)=step(num_cof; den_cof, t) ;, \

%specify a time' vector'%initialize "respons.e array, as zero%store zeta as a n array _%forloop to compute cet) 6 times

end,pJ ot(t, c " k') i gri dx1abel(' ti me,t in sec') ;yl abel( 'unit step response. c (t)');text(2.8,l~g6,\zeta=0~)-_text:(2,8,1.58, '\zeta=0.2')texte2.8, 1. ~O, \ze,t:a"O. 4')texf(2.8,1.12,'\zeta.=O.6')text(i. 8,0.95, '\i:e'ta=O. 8')text(2.8.0.7Z,\~et~=1.0')

, The output waveForms are shown in fig p2.1-,

1.81.61.42e= - 1_2c:oQ;!~1 ;) 0:8

'2:


67/84


68/84


69/84

-:.~~ ~sis .OUTPUT 2.7(JUnit step response of the syst:einl is,51 = '.

1/,3*exp(-4*t)+1-4/'3*exp(-t)


70/84

unit step response of the system2. is,s2 =. .l-exp(-6*t) =eos (8*t:)-3/'4*exp(-s=t) rsin(8~t)

Unit step response of' zhe system3 7'S,s3 =. 1+1/'FexpC-60*t:)6/'5*exp(-J.O*t)the output waveform ..is .showni n fig p2. 4

! s~,~--~~-. ~ O .B /r~(t) .. L . . : ! . . ; - ; . . , - : - .: 3 - , I., :I .I.6 / .... ~ ~.: '~ :............ "' .. -

< > . , :.. ~. 0.4 .. / , : ---.~'2 I I

::::l J0.2/ _ .

II01

. :

-- .;." ,.: _ ;., ;. . . . i ! - _ ( r s2(t) ~1\

o .3 4 5 '6lime, t irrsec 7 9 toP ig P2.4: Unit-ste p re spo nse O J sy ste m s give n in pro gram .4.

PROGRAM 2.5' .consider. the closed loop .transfer function of the follQwi-ng second order system, .

M(S) =16/ (52+45+16).....te a MATlAB Prag ram tof; nd the rise time, peak t;me, m~ximum"peak overshoot,and se:ttlingtime from the unit Step response of the 5ysteQJ.c1 .


71/84

plot(t,cl, '--k' ,t,c2, '-.k' ,t,c3,'-k'); gridxlabel('time,t in sec');ylabel('lesponses,cl(t),c2(t),c3(t)');text(O.25,1.1S. 'cl(t) ')text(l. 45,1. 5. c2 (t)')text(1.35,O.7,'c3(t)')


72/84

OUTPUTThe output waveform is !ihown in .fig p2.6.

1 .B -.- - ....... - - - - _- ..~- , .. ..: ...- . - - .. _ - - _" ,, ,.:

..~:,-----.!....,.. - - _ :- _ : --; .._ .

""'''' ./,/,/. ..":7"c2(tl/

':"'I'1.6. .......;.. "

:./'~1.~ .. , ... -.. - ; : "..-:.: .t: . :~(~C~.~; t>--L~.~t / -4~-- .~..-.~.-~t : /: g O.B:-r ..... : ..; . > . - . :&, ~.~ O.6 J ,'- --' ., -. /- ..c: : I: jI :0.4 ""I: .; .../~..

f0..2 .. / . ... : .I ./o I ~o . 0.2 0.4 0.6 O.B

. - - '.' ,," - - .~ ~ -: -".".: ... ---~. . ...~.. , - - - .- . " , . . . . - ..1.2 VI til 1.8 2time,t in sec

Fig Pl.6 : Step; ramp and parabolic response of system given in program 2.6.PROGRAM 2.7

consider the c'losed 'Ioop transfer function of' the fol1ow;ng second order system,M e s) =5/ (s'+s+5)

Write a MATLA!;lrogram to find the response for the input signal, r(t)=2-2t+t2.

%prag ram to fi nd res pons e fo r 9iveil input.cl c .num_cof=[OO 5J;den_cof=[l i 5];t""O:O.005:3; %spedfy a time -veccorr=2-2*t+t. "2;. %input si gnal


73/84

2.74rQ2S Define step' sign~L .. .: . . . .. . ' . " r(t )The step signal ISa signal Whose value changes, from 0 to A and remains '.

constant at A for t:> O.The mathematical representation of step signal is, A-----r{t)"" A , t;::O

= O,!<


74/84

Define ramp signalQ2.6

Q2.7

A ramp signal is a signal whose value increases linearly with time froman initial value of zero at t= O. Mathematical representation otrarnpSignal is, .r{t) = A t, t:


75/84

2.76.The closed l ~ , ! p transfer function of second order -system is given by -2 - : : ; - - _ ; 2 ~ " - - - '- - '- 2 O f } ~ . , Determines-+ ...S+.the damping ratio and natural frequency of osdlkulon.Let us compare the given transfer "function with the standard form of second .order transfer function

. C(s). m~ 200-: R(s) = = s2+2I;mn5+(j)~ - 52+205+200-


76/84

.-23

.. m -~ = 200 I 2I;m;, ",20. .. . ... '. 20 20ron=mo=14,14rad/sec -~=-. -=- 4 :.;-~0.7(}7. _ 2Xffin 2x1 .14 .Damping ratio, s : : 0.707Natural frequency of oscillation,ffin = 14.14 red/sec,A second order system has a-itamping~atio of 0.6 and natural frequencyof osCilladon is 10rad/sec.-Determine the damped frequency of oscillation: .Damp~dfrequencyofosciHation, md=(j)n~1-tl = 10J1-(O.6)2 =10xD.8=8rad/secThe open loop-transfer functlon.of a unity feedback system is G(s) = 20 . What is Ihenatureof- . .. - - s(s + 10) -response of closed loop system for unit step input ._ ..The closed loop transfer function,C(s) G(s) 20fs(s+10} 20 20R(s) "" 1+G(s) = 1+_ _ _ _ 1 Q _ _ = 5(5+10)+ 20"" S2 +10s;.205(S+10)The standard form of second order transfer function i s , C(s) = . {J}~ .-- - - R(s) s2+2I;mns+ro~-On comparing system transfer function with standard form of second o rde r transfer function we get,2 . - . - -(J)n=20 21;(J)n=10

""'20 447 dl ./,=~=_1_0_-=_t12:.(J)n=V'::U= .. ra sec '> 2x(J)n 2x4.47Since damping ratio, I; >1 , the system is overda rnped and the response will be exponentially rising.List the time domain specifications.The ,time domain sp~cjficatjons are,(i) . Delay time (ii)(iv) Maximum overshoot (v)

" f D . : 2 . 2 4

Rise timeSettling time.

(iii) Peak time

i');25 Define delay time.ltis me time taken for response to reach 50% of the final value, tne very first time.

!2.26_.Define rise time. .. _-It is the time taken for response to raise from 0 to 100% ,the very flfSt time. For underdamped system,the risetime is Calculated from 0 to 100%. But for overdamped system it ls the time taken by theresponse to raise from 10%to 9 0% . For critically dampedsystem, it isthe_ time taken far response toraise from 5% to 95%.

2 ..77Q2.29

Q).30

Define settling time, ,It is defined as the time taken by the response to reach and stay within a specified error and the erroris usually specified as % offinal value. The, usual tolerable.error is 2% or 5% of the final value.The damping ratio of a system is 0.75 and the natura/frequency of oscillation is 12 rad/sec -.Determine tile peak overshoot and the peak time.

-C1t 0.75",


77/84

Q2.31

Peak overshoo~) M p = ePe ~H075_l2 = 0 .028 ; :. %M p= 0 .Q28 x 1 00" 2.8%Damped f requency of oscillation, ro d =r o nJ 1': 'c,2 "1 2J 1- (0.75)2 = 7.94 rad I sec~ .P k ti rt 1Cea time, Ip=- =-=0.396 secro d 7.94The damping ratio of system is 0.6,and the natural frequency of oscillation is 8 rod/sec; Determinethe rise time.R' " 11 :-9use time, t, =~-ro 'd

e-I _lJl-C,2- an -.- C,rod'" fDnJ1-c,2 ~ SJ1"':'(0,6)2= 6.4 rad / sec

,R' f 1t - 0.927 0 . 3 4.. Ise Ime, t, = 6.4 = secQ2.32 Wha: is type number of a system? What is its significance?'

The type number is given by number of poles of loop transfer function at the origin. The type numberof the system decides the steady state error.

Q2.33 . Distinguish betweentypeand order of a system.. .(i) Type number is specified for loop transfer function but order can be specified for any transfer'function. (open loop. or closed loop transfer function).(ii) The type number is given by number of poles - o f loop transfer function lying at origin of s-plane but ..

the order is given by the number of poles of transfer function. .Q2.34 For the system with following transfer/unction, determine type and order of the system.

K . .. 20 (s+2). (i) G(s) H(s) = ) 2.6 8 (ii) G(s) H(s):::= 2(: 3!~(:.' ~~s(s+1(s + s+ ') S s+ :t :.'+0.5/

(s+4) 10(iii) G(s) H(s) = (s _ 2) (s +0,15) (iv) G(s) H(s) = s3(s2 +2s + 1).Ans: (i} Type - 1.order -4 (ii) Type -2,order - 4

What is steady state error? .'.'The steady state error is the value of error Signa! e(t), when t tends to infinity The steady state erroris a measure 'of system accuracy. These errors arise from the nature of inputs, type of system and!from non-linearity of system components.

. Q:i.35

.a

J

{iii) Type - 0, order - 2 (iv) Type - 3, order - 5.

2.78Define velocity error constant.The velocity error constant Kv-= Lt sG(s) H(s). The steady state error in type-1 system for unit rampinput is given by 1 / 1 < , . . s ->o - _, 'Define' acceleration error constant.The acceleration error' constant 'K a = = It S2 G(s) H{s). The steady state error in type-2 system for unit", $--70 ,


78/84

parabolic input is_given by 11K.A unity feedback system hasa open loop transfer/unction-of G(s)= - 10 . Determine the, '- (s+ 1) (s + 2)steady state error for unit step input

- , " ,- - 1The steady state error for unit step input; ess = -r--,where, Kp = Lt G(s) H(s) ., "-1+Kp 5--70' _ 'For unity feedback system H(s) = 1.10 -:.K = Lt G(s)= Lt. ' , Sand _e =-'-=,-, -=-p S~>O s -s- o (5+1)(5+-2) _ ss 1+Kp 1+5 6-'. fi -d"'b'k has a ooen loootransti fi . if G{:'l- 25 (5+ 4) D .A umty ee ac system as a open oop trans.)er uncuon 0 S, = ' - - ',.- setermtne .. '''' -' ,s(s+0.5) (s+2)thesteady state error for.unu.ramp input

- 1 hThe steady state error for unitramp input is, ess =-K' were, K v = Lt s G(s) H(s). for unity feedback_ v - - s---tOsystem H(s) = 1., .[ 25 (s+4) ] 25x4 1 1,-.K.= It sG(s)=U s ' '=~-,-, =100 and ess="'-=-=O.Q1HO' ',S--70_ 5(5+0.5)(5+2)" 0.Sx2 -, / Ky 100 '. fi 'db' k - h 1 - - ,F. ' fi .' if G('l 20(5+5) D .A Unity ee ac system as a open oop transfer unction 0 $-,=', ' '. etermine, .' s(s+O.l)(~+ 3)the steady state error for parabolic input.

The steady state error for unit ramp input is ess = K l - ,a

where, Ka= Lt s2G(S) H(s) . For unity$-->0feedback system H(s) = 1... Ka =, U S2[' 20 (s+S) ] = = 20xS = 100 = 333.33. HO s2(s+0.1) (s+3) 0.1x 3 0.3 and

1 1 'ess = = -= -- =.003Ka 333.33What ar.e generalized error coefficients?They are the coefficients of generalized error series.The generalized error series is given by,- (t) 'C (t) C'() C2() .C3 ...().. Cn"-()e t = o r t + l r t +-r t +~r t +.....+~r t ...... 2! 3] n! -. The coefficients Co' C, ,Cr'" .C" are called gener~!ized error coefficients or dynamic error coefficients,The nth coefficient, C; '" It ~ F(5) , where, F(s) = 1+ G'~) H(s) ., . ~o ~n .

. .Gil/e.the relation -between generalized and static error coefficients.


79/84


80/84


81/84

2.822;24 EXERCISESE2.1 What is .the unit-step response of the system shown in fig E2.l'

~", ~. .~,IO C(s)

+ f ~_ ~ _ , _ S _ + _ 1 _ , _ _ , _ _ ~S_2_,__


82/84

~~j~Fig H2.l.~2' Obtain the unit-step response of a unity-feedback system

h 'I" fe fu .,. O(s) = ' 5(s+ 20) ,w ose open- oop trans er notion IS ( 459) (2 341 16....) ." ' s s+ ' s + . s+ oJ"

E2;5

E2.3 'The open loop transfer function of an unity feedback control system is given by G(s) = 100 ,s(s +2)~(s+ 5), For unit step input.find the time response of the closed loop system and determine % over shoot and the'~ti~ ", ' ,A Servomechanism has it s moment' of inertia J ~ 1 0 1 0-6 K g-m ', retarding friction, B = = 4 00x I 0-6 N -IIl/(rad/sec) and elasticity coefficient,K-== 0.004 N-mlrad. Find the natural.frequencyand damping factor ofthe system.For a second order system whose open loop transfer function G(s) = __ 4_, determine the maximum, , ,'. ,- s(s+ 2) , ,over shoot, the time to reach the maximum overshoot when a stepdisplacement of 18 is .given to thesystem. Find the rise time,' time constant and the settling time for an error of 7% .

: E 2 . 4

, "25'Consider the: unity feedback closed loop system where the forward transfer function is G(s) = ---, .s{s + 5)Obtain the rise time; Peak time, Maximum overshoot and the settling time when the syst.em is subjected to

'a unit-step input. 'E2.7 Consider t h e system shown in f ig E2.7, where c : =0:6 and : 0.5 rad/sec, Determine the rise time, peak

time, .maximum overshoot and Settling time, when the system is subjected to a unit-step I npu t .

E2.6

E{s ) C(s)

'-------'---..;.,...-~_,____JFig E27:E2.8 For the system shown in fig E2.8, determine the values of K and K" so that the maximum overshoot in the

unit step response is 0.2 and.the peak time isI sec. With these values of K and K " , 'obtain rise time and.settllng time ..{ C(s)


83/84

2.84:'ANSWER. FOR EXERCISE PROBLEMSE2.1

. ~'.'. " .;0-: '

E2.2E2_.3


84/84

!fE2.6

2.7- I1 : : 2 . 8 '

i t E 2 . ~: 1 - ' "~2.10'~ -

Maximum overshoot =0.16, when -input is 18%(Mp= 2.88%Peak time, tp=1.81 sec. Rise time, t,=1.21 seeTime constant, T=l sec, Settling tjme{or]%:error = 2.66 sec." Rise time, t,=0.55sec, %peakOvElrsli'0otM~=c 9.5%p~ak time, t p = O .7S5.sec,S_ettling:time, tS )";'" 1.33sec _(for2% error): . t 5 = 1 , sec( for 5%etror)- RiseUme;t, = 0.55sec, Maximum overshoot, Mp= 0.095- " '_ '., _ "1.-Peaktime, tp = 0.785sec, Settl!~g time, t,;=1 sec (for 5% criterion)- K";12.5, Rise: time, 1,=_O.65sec' '.',~ = 0,178 ; Settling time,_ts';' 2.48 sec (for2% error) ; ts= 1.86 sec (for 5% error)K=lA2, T=1.09' ': E2,.10 K=2,95N~m T:=0.471sec

J1C(b). Rise time, tr = 1.91 secPeak time., - tp= 2.79 secPeak overshoot- M p =:0.1265Se;tling t ime,. Is'" S.4sec

=0

The total steady state error is

control systems unit ii

Documents