3 Time Domain Analysis - Advanced Control Engineering BURNS Pp 35-62

7/27/2019 3 Time Domain Analysis - Advanced Control Engineering BURNS Pp 35-62

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Time domain analysis

The m ann er in which a dynamic sys tem responds to an inpu t , expressed as a funct ionof t ime, is cal led the t ime response. T he theoret ical e valua t ion of th is response is saidto be under taken in the t ime domain, and is refer red to as t ime domain analysis . I t ispossible to compute the t ime response of a system if the fol lowing is known:9 the nature of the input(s) , expressed as a funct ion of t ime9 the math em at ica l m odel o f the sys tem.The t ime response o f any sys tem has two components :(a ) Transient response: This component of the response wil l ( for a s table system)

decay, usual ly exp onential ly , to zero as t ime increases . I t is a funct ion only of thesys tem dynamics , and i s independen t o f the inpu t quan t i ty .

(b ) Steady-state response: This is the response of the system af ter the t ransientcom pon ent has decayed a nd i s a funct ion o f bo th the sys tem dynamics and theinpu t quan t i ty .

Xo(t)TransientPeriod (t)

I ~ , I t ~ Steady-StaterrorI Transient /I Err~ / / f Steady-StatePeriod

Fig. 3.1 Transient nd steady-state eriodsof time esponse.


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36 Advanced Control EngineeringThe tota l response of the system is a lways the sum of the transient and steady-sta tecom pone nts. Figure 3.1 shows the transient and steady-sta te per iods of t imeresponse. Differences between the input function xi(t) (in this case a ramp function)and system response Xo(t) are cal led transient errors dur ing the transient per iod, andsteady-sta te errors du r ing the steady-sta te per iod. One of the major object ives ofcontrol system design is to minimize these errors .

In order to compute the t ime response of a dynamic system, i t is necessary to solvethe dif ferentia l equations (system mathematical model) for given inputs . There area number of analytical and numerical techniques available to do this , but the onefavoured by control engineers is the use of the Laplace transform.

This technique transforms the problem from the t ime (or t ) domain to the Laplace(or s) domain. The advantage in doing this is that complex t ime domain dif ferentia lequations become re la t ively simple s domain a lgebraic equations. When a suitablesolution is arr ived a t , i t is inverse transformed back to the t ime domain. The processis shown in Figure 3.2.

The Laplace t ransform of a func t ion of t ime f(t) is given by the integral& a [ f ( t ) ] - f(t)e-~tdt- F(s) (3.1)

where s is a complex variable cr + j~ and is called the Laplace operator.

LaplaceTransform

Z[f( t) ]= F(s)

Fig . 3 .2 The Laplace ransform process.

s Domain F(s)Algebraicequations

Time Domain f(t)Differentialequations

I nve rseLaplaceTransformS ' [F(s)]= f(t)


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T im e d o m a in a n a ly s is 3 73 .2 .1 L a p la c e t r a n s f o r m s o f c o m m o n f u n c t i o n sE x a m p l e 3 . 1

f ( t ) - 1 ( c a l l e d a u n i t s te p f u n c t i o n ) .S o l u t i o nF r o m e q u a t i o n ( 3 .1 )

E x a m p l e 3 . 2

f 0 ~~ [ f ( t ) ] - F ( s ) - 1 e - s t d t- [ - - s l ( e _ S t )

- - - - [ - - s l ( 0 - 1 ) ] - 1 - s ( 3 . 2 )

f ( t ) - - e - ~ t

2 ' [ f ( t ) ] - F ( s ) - - e - a t e - " d t

- f o ~ e - ( s + a ) t d t_ _ 1 ( e_ ( s + a ) t

s + a o

s --F a1= ~ ( 3 . 3 )s -F a

T a b l e 3 .1 g iv e s f u r t h e r L a p l a c e t r a n s f o r m s o f c o m m o n f u n c t i o n s ( c a l l ed L a p l a c et r a n s f o r m p a i r s ) .

3 . 2 .2 Pr o p e r t ie s o f t h e L a p la c e t r a n s f o r m( a) D e r i v a t i v e s : T h e L a p l a c e t r a n s f o r m o f a t i m e d e r i v a t i v e is

d nd t J ( t ) - s ~ F ( s ) - f ( 0 ) s ~ - l - f ' ( 0 ) s ~ -2 . . . . ( 3 .4 )w h e r e f l 0 ) , f ' ( 0 ) a r e th e in i ti a l c o n d i t i o n s , o r t h e v a l u e s o f f ( t ) , d / d t f ( t ) e t c . a t t - 0

( b ) L i n e a r i t y5 e [ f l ( t) - - E l ( S ) + F 2 ( s ) ( 3 . 5 )


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3 8 Ad van ced Contro l EngineeringTab l e 3 .1 C o m m o n L a p l a c e t r a n s f o r m p a i rsT i m e f u n c t i o n f ( t ) L a p l a c e t ra n s f o rm L ~ ' [ f ( t ) ] = F ( s )

1 u n i t i m p u l s e 6 ( 02 u n i t s t e p 13 u n i t r a m p t4 t n

5 e -at

6 1 - e -a t

7 s in w t

8 c o s w t9 e -a t s i n w t

1 0 e - a ' ( c o s w t - -

11 / s1 # 2n !

Sn+ !1

( s + a )g

s ( s + a )s 2 + w 2

Ss2 + ~

O3( s + a ) 2 + w

a . ss i n o 3 0w (s + a ) 2 + w

( c ) C o n s t a n t m u l t i p l i c a t i o n~ q ~ [ a f ( t ) ] - a F ( s ) (3 .6)

( d ) R e a l s h i f t t h e o r e m~ [ f ( t - T ) l = e - T S F ( s ) f o r T > 0

( e ) C o n v o l u t i o n i n t e g r a l(3 .7)

( f ) I n i t i a l va l ue t heo r emf o t f l ( T ) f 2 ( t - - T ) d T = F 1 ( s ) F 2 ( s ) (3 .8)

( g ) F i n a l v a l u e t h e o r e mf ( 0 ) , = lirr~ [ f ( t ) ] = s--,oolim s F ( s ) ] (3.9)

f ( c ~ ) = l i m [ f ( t ) ] - l i m [ s F ( s ) ]s---~O

(3 .10)

3 . 2 .3 I n v e rs e t r a n s f o r m a t i o nT h e i n v e r s e t r a n s f o r m o f a fu n c t i o n o f s i s g i v e n b y th e i n t e g r a l

1 f ~ + J ~- F ( s ) e S t d s( t ) = & a- l [F(s ) ] ~ ~o-j~ , (3 .11)


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Time domain analysis 39In pr ac t ice , inver se t r ans format ion i s mos t eas i ly ach ieved by us ing pa r t i a l f r ac t ionst o b r e a k dow n s o l u t i ons i n to s t a nd a r d c o m po ne n t s , a nd t he n u s e t a b le s o f L a p l a c et r ans form pa i r s , a s g iven in Table 3 .1 .

3.2.4 Com mo n partial fraction expansions( i) Fac tore d roo ts

K A B= ~ (3.12)s(s + a) s (s + a)( ii ) Rep ea ted roo ts

K A B C. . . . t- - 4 - ~ (3.13)$2(S + a) s -~ (s + a)( ii i) Se co nd -or de r real roo ts (b 2 > 4ac)

K Ks(as 2 + bs + c) s(s + d)(s + e)

As

B C(s + d) ( s - t -e)

( iv ) Seco nd-o rder com plex roo ts (b 2 < 4ac)K A

s(as 2 + bs + c) sB s + C

t as 2 + b s + cC om pl e t i ng t he s qua r e g i ve s

A B s + CI (3.14)s (s + a) 2 + w2Note: In ( i i i ) and ( iv) the coef f ic ient a is usual ly factored to a uni ty value .

A t r ans f e r func t io n i s the Laplace t r ans form of a d i f f e ren t ia l equa t ion wi th ze roin i t i a l condi t ions . I t i s a ve ry easy way to t r ans form f rom the t ime to the s domain ,and a power fu l too l f o r the cont ro l enginee r .Example 3 .3Fin d the Laplace t r a ns fo rm o f the fo l lowing d i f fe r en t ia l equ a t io n g iven:(a) ini t ia l con di t i on s Xo - 4 , d x o / d t - 3(b) ze ro in i t i a l condi t ions

d2xod t 2

dxo+ 3 --7,. + 2Xo - 5f i t


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40 Ad vanc ed Cont r o l E ng ineeringX~(s)

G(s) X o ( S )

Fig . 3 .3 The t rans fer func t ion approach.

S o l u t i o n(a ) I n c l u d i n g i n i t i a l c o n d i t i o n s : T a k e L a p l a c e t r a n s f o r m s ( e q u a t i o n ( 3 . 4 ) , T a b l e 3 . 1 ) .

5( s 2 X o ( s ) - 4 s - 3) + 3 ( s X o ( s ) - 4) + 2X o(s) - -ss 2 X o ( s ) + 3 S X o ( S ) + 2Xo(S) 5- + 4 s + 3 + 1 2s

(S 2 -~- 3s + 2)Xo(s) - 5 + 4s 2 + 15s

4s 2 + 15s + 5S ~ - - S ( S 2 Jr- 3 s + 2) (3 .15)

(b ) Z e r o i n i ti a l c o n d i t i o n sA t t = 0 , Xo = 0, d x o / d t = O .

T a k e L a p l a c e t r a n s f o r m s5s Z X ' o ( S ) - ~ - 3SXo(S) + 2X o(s) = -s 5X o ( s ) - s ( s 2 + 3 s + 2) (3 .16)

E x a m p l e 3 . 3 (b ) is e a s il y so l v e d u s i n g t r a n s f e r f u n c t i o n s . F i g u r e 3 .3 s h o w s t h e g e n e r a la p p r o a c h . I n F i g u r e 3 . 39 X i (s ) i s t h e L a p l a c e t r a n s f o r m o f t h e i n p u t f u n c t i o n .9 X o ( S ) is t h e L a p l a c e t r a n s f o r m o f th e o u t p u t f u n c t i o n , o r s y s te m r e s p o n se .9 G ( s ) is t h e t r a n s f e r f u n c t i o n , i .e . th e L a p l a c e t r a n s f o r m o f th e d i f f e re n t i a l e q u a t i o n

f o r z e r o i n i ti a l c o n d i t i o n s .T h e s o l u t i o n i s t h e r e f o r e g i v e n b y

X o ( S ) - - G ( s ) X i ( s )T h u s , f o r a g e n e r a l s e c o n d - o r d e r t r a n s f e r f u n c t i o n

d 2 x o d xoa ~ + b - ~ + C Xo - K x i ( t )( a s 2 + b s + C ) X o (S ) - K X i ( s )

(3 .17)

H e n c eX o ( s ) - { K } X ~ ( s )a s 2 + b s + c (3 .18)


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Xi(s) ,.I K

"1 a~+bs+cFig. 3.4 Generalsecond-order transfer function.

Xo(S)

Time domain analysis 41

X~(S)=5/S 1s2+3s+2

Fig. 3.5 Example3.3(b) expressed as a transfer function.

Xo(S)v

Compar ing equa t ions (3 .17) and (3 .18) , the t rans fe r func t ion G (s) isKG (s) - (3.19)a s 2 + bs + c

which, us ing the form shown in Figure 3.3, can be expressed as shown in Figure 3.4.Re turn ing to Example 3 .3(b) , the so lu t ion , us ing the t rans fe r func t ion approach i s

shown in F igure 3 .5 . F rom F igure 3 .55

X~ -- s( s 2 -Jr- 3s + 2) (3.20)which is the same as equat ion (3.16) .

3.4.1 The impulse functio n- . . V T - ~ - ~ . . . . . . . . . . . . . .. . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . .. . . . ~ . . . . . . . . . . . . ~ ' - - - ~ ' : " " . . ~ ' : : : ' ~ - ' ~ ~ . ~ - - " - . - = ~ : - . . - : : ~ . ~ : ~ : ~ . . . . .. ~ : V . . . . .. . . . .. . . . .. . . . ~ . . . . .. . . . .. . . . .. . . . .. . . . ~ - . . . . ~: . - ~ : ~: . ~ . ~ .- . . . . .: = . - - - . . . . . . . . . . . . . . . . . . . . . . . : " ~ - - ~= - ~ - - -- - . ~ . . .- - = . ~. ~ = ~ = . . . . .An impulse is a pulse wi th a width At ---, 0 as show n in Figu re 3.6. The s t re ngth of animpulse is i t s area A, where

A -- heig ht h At. (3.21)The Laplace t ra ns for m of an impulse func t ion i s equa l to the a rea of the func t ion .The impulse funct ion whose area is uni ty is cal led a uni t impulse 6( t ) .

3.4.2 The step functio nA step func t ion is descr ibed as x i ( t ) - B; Xi(s) - B / s for t > 0 (Figure 3.7). Fo r a uni ts tep funct ion x i ( / ) - 1, X i ( s ) - 1/s . This is somet imes referred to as a 'cons tantpos i t ion ' input .


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42 Ad vanc ed Cont r o l E ng ineering

x~(0Impu lse

Pu lse//7/777

Fig. 3.6 The impulse funct ion.

Fig. 3.7 The step funct ion.

x~(t)

3 .4 .3 T h e ra mp fu n c t io n. . . . . ~ . ..

A r a m p f u n c t i o n i s d e s c r i b e d a s x i ( t ) - Q t; X i ( S ) - - Q / s 2 f o r t > 0 ( F i g u r e 3 . 8 ) . F o r au n i t r a m p f u n c t i o n x i ( t ) - - t ; X i ( s ) - - 1/s 2 . T h i s i s s o m e t i m e s r e f e rr e d t o a s a ' c o n s t a n tv e l o c i t y ' i n p u t .

3 .4 .4 The parabo l ic funct ionA p a r a b o l i c f u n c t i o n is d e s c r i b e d a s x i ( t ) - K t 2 ; X i ( s ) - - 2 K / s 3 fo r t > 0 (F igu re 3 .9 ) .F o r a u n i t p a r a b o l ic f u n c t i o n x i ( t ) - - t 2 ; X i ( s ) - - 2/s 3 . T h i s i s s o m e t i m e s r e f e r r e d t o a sa ' c o n s t a n t a c c e l e r a t i o n ' i n p u t .


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xi(t)

Q

Fig. 3.8 The ramp function.


x~(t)

Fig. 3.9 The parabolic function.

3.5.1 Standar d formConsider a f i rs t -order different ial equat ion

dxo + bxo - cxi( t) (3.22)Take Laplace t ransforms, zero ini t ial condi t ions

asXo(s ) + bXo(s) - cXi (s )(as + b)Xo(s) = cXi (s )


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4 4 A d v a n c e d C o n t r o l E n g i n e e r in gT h e t r a n s f e r f u n c t i o n i s

G(s) Xo- x i ( s ) -

Ca s + b

T o o b t a i n t h e s t a n d a r d f o r m , d i v i d e b y ba ( s ) - al + ~ s

which i s wr i t t enKG ( s ) - 1 + Ts (3.23)

E q u a t i o n ( 3 .2 3 ) is t h e s t a n d a r d f o r m o f tr a n s f e r f u n c t i o n f o r a f ir s t - o r d e r s y st e m ,w h e r e K = s t e a d y - s t a t e g a i n c o n s t a n t a n d T - t i m e c o n s t a n t ( s e co n d s ) .

3 .5 .2 Impulse response o f f i rs t -order systemsE x a m p l e 3 . 4 (See a l so Append ix 1 , e x a m p 3 4 . m )F i n d a n e x p r e s s i o n f o r t h e r e sp o n s e o f a f i rs t - o r d e r s y s te m t o a n i m p u l s e f u n c t i o n o fa r e a A .

Solu t ionF r o m F i g u r e 3 . 1 0A____KK= A K / T (3 .24)X o ( s ) - 1 + ~ s ( s + 1 / 7 3

o r

X o ( s ) - X K ( ~ )- -- T - ( s + a ) (3.25)E q u a t i o n ( 3 . 2 5 ) i s i n t h e f o r m g i v e n i n L a p l a c e t r a n s f o r m p a i r 5 , T a b l e 3 . 1 , s o t h ei n v e r s e t r a n s f o r m b e c o m e s

A K e _ a t A K - t / r (3 .26)Xo(t) -- --T- = T eT h e i m p u l s e r e s p o n s e f u n c t i o n , e q u a t i o n ( 3 . 2 6 ) i s s h o w n i n F i g u r e 3 . 1 1 .

X~(s) = A . I K" [ l + T s

Fig . 3.1 0 Impulse response of a fi rst-order system.

X o ( S )r-


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Xo(O

A KT

Fig. 3.11 Response of a f i rst-order system to an impulse funct ion of area A.

Time dom ain analysis 45

3 . 5 .3 S t e p r e s p o n s e o f f i r s t - o r d e r s y s t e m sE x a m p l e 3 . 5 ( S ee a l s o A p p e n d i x 1 , e x a m p 3 5 . m )F i n d a n e x p r e s s i o n f o r t h e r e s p o n s e o f a f ir s t - o r d e r s y s t e m t o a st e p f u n c t i o n o fh e i g h t B .S o l u t i o nF r o m F i g u r e 3 . 1 2

X o ( s ) - s(1 + T s ) = B K s ( s + 1 T ) (3 .27 )E q u a t i o n ( 3 .2 7 ) is i n th e f o r m g i v e n i n L a p l a c e t r a n s f o r m p a i r 6 T a b l e 3 .1 , s o th ei n v e r s e t r a n s f o r m b e c o m e s

X o ( t) = B K ( 1 - e - t / r ) (3 .28 )I f B = 1 ( u n it s t e p ) a n d K = 1 ( u n i ty g a i n ) t h e n

X o ( t ) - ( l - e - t / T ) (3 .29 )W h e n t i m e t is e x p r e s s e d a s a r a t i o o f ti m e c o n s t a n t T , t h e n T a b l e 3 .2 a n d F i g u r e 3 .1 3c a n b e c o n s t r u c t e d .

Table 3.2 U ni t step response of a first-order systemt / T 0 0. 25 0.5 0.75 1 1.5 2 2.5 3 4Xo(t) 0 0 . 2 2 1 0 . 3 9 3 0 . 5 2 7 0 . 6 3 2 0 . 7 7 0 0 . 8 6 5 0 . 9 2 0 0 . 9 5 0 0 . 9 8 0


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X(s)=B/sI + Ts

Fig . 3 .12 S tep response o f a f i rs t -order sys tem.

X o ( t ) - - 1 - e - t / T

.~ Xo(S)

3 .5 .4 Exper im enta l de te rm ina t ion o f system t im e cons tan tusing step responseMethod one: T h e s y s t e m t im e c o n s t a n t i s t h e t im e t h e s y s t e m t a k e s to r e a c h 6 3 . 2 % o fi t s f ina l va lue ( see Table 3 .2) .Method two: T h e s y s t e m t i m e c o n s t a n t i s t h e i n t e rs e c t i o n o f t h e s lo p e a t t - 0 w i t ht he f i na l va l ue l i ne ( s ee F i gu r e 3 . 13 ) s i nce

d xodt

1.2

- - O - ( - 1 ) e - t / T - --~ l e - t / r (3 .30)

dxo 1d t ] t = o = ~ a t t - O (3.31)T h i s a l s o a p p l i e s t o a n y o t h e r t a n g e n t , s e e F i g u r e 3 . 1 3 .

0.8

" ~ 0 . 6

0 .4

0 .2

0 0.5

T

46 A d van ced C o n t r o l E n g i n eeri n g

T! j

/ /

1.5 2 2.5 3 3.5Num ber o f T im e Cons t an t s

4 .5

Fig. 3.1 3 Un it s tep response of a f i rs t -order system.


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T im e d o m a in a n a ly s is 4 73.5 .5 Ram p respon se of f irs t -order systemsE x a m p l e 3 . 6F i n d a n e x p r e s s i o n f o r t h e r e s p o n s e o f a f i r s t - o r d e r s y s t e m t o a r a m p f u n c t i o n o fs l o p e Q .So l u t i onF r o m F i g u r e 3 . 1 4

O KX o ( s ) - s2(1 + T s) Q K / T A B C. . . . l- + (3 .32 )s 2( s + 1 / T ) s -ff ( s + 1 / T )( S e e p a r t i a l f r a c t i o n e x p a n s i o n e q u a t i o n ( 3 . 1 3 ) ) . M u l t i p l y i n g b o t h s i d e s b ys2 (s + l / T ) , w e g e t

TOK A B= A s 2 + = ~s + B s + ~ + C s 2 (3 .33).e . T 1 1

E q u a t i n g c o e f f ic i e n ts o n b o t h s i d e s o f e q u a t i o n ( 3 .3 3 )( $ 2 ) " 0 - - A + C

A( s l ) 9 0 - uQ K B(sO)" X - Y

(3.34)(3 .35)(3 .36)

F r o m ( 3 . 3 4 )C - - - . , 4

F r o m ( 3 . 3 6 )B = Q K

S u b s t i t u t i n g i n t o ( 3 . 3 5 )

H e n c e f r o m ( 3 . 3 4 )X = - Q K T

C - - Q K T

X~(s)=QIs2 . . I K"1 l + T sF i g . 3.1 4 Ram p esponse of a f i rst-order system (see also Figure A l l ) .

Xo(S)


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48 Advanc ed Control Engineering

v 4X

>r176 3".

.

o o 1

J. f JJ

2 3 4 5 6 7N u m b e r o f T ime C o n s ta n ts

Fig. 3.1 5 Uni t ramp response of a f i rst-orde r system.

In se r t ing va lues o f A , B and C in to (3 .32 )X o(s ) Q K T Q K Q K T. . . . ~- + (3 .37 )s ~ ( s + l / T )

I n v e r s e t r a n s f o r m , a n d f a c t o r o u t K QX o( t) = K Q ( t - T + T e - t / r ) (3.38)

I f Q - 1 (un i t r am p) an d K = 1 (un i ty ga in ) t henX o ( t ) - t - T + T e - t / T (3.39)

T h e f i r s t t e r m i n e q u a t i o n ( 3 . 3 9 ) r e p r e s e n t s t h e i n p u t q u a n t i t y , t h e s e c o n d i s t h es t e a d y - s t a t e e r r o r a n d t h e t h i r d i s t h e t r a n s i e n t c o m p o n e n t . W h e n t i m e t i s e x p r e s s e da s a ra t i o o f t i m e c o n s t a n t T , t h e n T a b l e 3 . 3 a n d F i g u r e 3 .1 5 c a n b e c o n s t r u c t e d . I nF i g u r e 3 . 1 5 t h e d i s t a n c e a l o n g t h e t i m e a x i s b e t w e e n t h e i n p u t a n d o u t p u t , i n t h es t e a d y - s t a t e , i s th e t i m e c o n s t a n t .

Table 3.3 U nit ram p response of a first-order systemt /T 0 1 2 3 4 5 6 7xi(t)/T 0 1 2 3 4 5 6 7Xo(t)/T 0 0 .36 8 1 .135 2 . 05 3 .018 4 .007 5 6


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3.6.1 Standard formCons ider a second-order d i f f e ren t i a l equat ion

d 2 x o dxoa dt 2 + b - -~ -+ CXo - ex i ( t ) (3.40)Take Lap lace t r ansfo rms , zero in i t i a l cond i t i ons

The t r ansfer func t ion i s

a s Z X o ( S ) + b s X o ( s ) + c X o ( s ) - e X i ( s )(a s 2 + b s + c ) Xo ( s ) - eX i ( s )

X o eG (s) - -~ i ( s ) - a s 2 + b s + c

To ob ta in the s t andard fo rm, d iv ide by ce

a ( s ) - a S 2 -b+ c S + 1which i s wri t ten as

(3.41)

K9 (3.42)G (s) - s 2 + 2__~ + 10.;2 ~n

This c an also be no rma l ized to ma ke the s 2 coeff icient uni ty , i .e .KJnG( s) - s2 -a- 2ff60nS + 6o2 (3.43)

Equ at ions (3 .42) and (3 .43) a re t he s t an dard fo rms o f t r ansfer func t ions fo r a second-o r d e r sy s t em , w h e r e K - - s t ead y - s t a t e g a i n co n s t an t , COn = u n d a m p e d n a t u r a lf r equency ( rad / s ) and ~ - - dam ping ra t io . Th e mean ing o f t he parame ter s ~ n an dare explained in sect ions 3 .6 .4 and 3 .6 .3 .

3.6.2 Roots of the characteristic equation and theirrelationship to damping in second-order systemsAs discussed in Sect ion 3 .1 , the t ransie nt re sponse of a system is inde pen den t of theinpu t . Thus fo r t r ans i en t r esponse ana lys i s , t he sys t em inpu t can be cons idered to bezero , and equat ion (3 .41) can be wr i t t en as

(as 2 -+- bs + c) Xo (s ) - 0I f Xo( S) r O, t hen

as 2 + bs + c - 0 (3.44)


16/28

5 0 A d v a n c e dCon trol EngineeringTab l e 3.4 Tran sient behaviour of a second-order systemD i s c r i m i n a n t R o o t s T r a n s i e n t r e s p on s e t y p eb 2 > 4ac s l and s2 r e a l O v e r d a m p e d

and unequal Transient(-ve) Responseb 2 = 4ac s l and s2 real Criticallyand equal Damped Transient(-ve) Responseb 2 < 4ac s l a nd s2 c o m p l e x U n d e r d a m p e dconjugate of the Transientform: sl, s2 = -cr + jw Re spo nse

T h i s p o l y n o m i a l i n s is c a l le d t h e C h a r a c t e r i s t i c E q u a t i o n a n d i ts r o o t s w i ll d e t e r m i n et h e s y s t e m t r a n s i e n t r e s p o n s e . T h e i r v a l u e s a r e

- b + ~ / b 2 - 4 a cs l , s 2 = ( 3 . 45 )2 aT h e t e r m ( b 2 - 4 a c ) , c a l le d t h e d i s c r i m i n a n t , m a y b e p o s i ti v e , z e r o o r n e g a t i v e w h i c hw i ll m a k e t h e r o o t s r e a l a n d u n e q u a l , r e a l a n d e q u a l o r c o m p l e x . T h i s g i ve s r is e to t h et h r e e d i f f e r e n t t y p e s o f tr a n s i e n t r e s p o n s e d e s c r i b e d i n T a b l e 3 .4 .

T h e t r a n s i e n t r e s p o n s e o f a s e c o n d - o r d e r s y s t e m i s g i v e n b y t h e g e n e r a l s o l u t io nX o ( t ) - A e s ' ' + B e s2 t (3 .46 )

T h i s g i v e s a s t ep r e s p o n s e f u n c t i o n o f th e f o r m s h o w n i n F i g u r e 3 . 16 .

Xo(t) Und erdam ping (s l and s2 complex)

Cr i t ical damping(sl and s2 rea l and equal)Overdamping(s, an d s2 rea l and unequa l)

Fig. 3.1 6 Effe ct hat roots of the characteristic equ ation have on the dam ping of a second-order system.


17/28

Time dom ain analysis 513 .6 .3 C r it ic a l d a m p i n g a n d d a m p i n g r a t ioC r i t i c a l d a m p i n gW hen the dam pin g coef f ic ien t C o f a second-o rder sys tem ha s i ts c ri t ica l va lue Co, thes y s te m , w h e n d i s tu r b e d , w i ll re a c h i t s s t e a d y - s ta t e v a lu e i n th e m in im u m t im e w i th o u tover s hoo t . As ind ica ted in Tab le 3 .4 , th i s is wh en the roo ts o f the Chara c te r i s t icE q u a t io n h a v e e q u a l n e g a t iv e r e a l r o o t s .D a m p i n g r a t i oT h e r a t i o o f t h e d a m p in g c o e ff ic i e nt C in a s e c o n d - o r d e r s y s t e m c o m p a r e d w i th t h eva lue o f the da m pin g coef f ic ien t Cc r equ i r ed fo r c r it i ca l dam pin g i s ca l led theD a m p i n g R a t i o ~ ( Z et a ). H e n c e

C~ = - - (3.47)CcT h u s

= 0 N o d a m p i n g< 1 U n d e r d a m p i n g= 1 C r i t ic a l d a m p in g> 1 O v e r d a m p i n g

E x a m p l e 3 . 7Fin d the va lue o f the c r it i ca l dam pin g coef f ic ien t Cc in te rms o f K and m fo r thes p r i n g - m a s s - d a m p e r s y s te m s h o w n i n F i g u re 3 .1 7.

X o ( t )

/ / / / / / / / / / / / / /

\ \ \ \ c

KF(t)

Lumped Param eter Diagram( a )

K X o C x o

T T

F ( t )

F r e e - B o d yDiagram( b )

Xo(t)Xo(t)~o(t) ~ + v e

F ig . 3 . 1 7 S p r in g - m a s s -d a m p e r y ste m .


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5 2 Ad van ced Control EngineeringS o l u t i o nF r o m N e w t o n ' s s e c o n d l a w

~ F x - m3~oF r o m t h e f l e e - b o d y d i a g r a m

F ( t ) - K x o ( t ) - C J co (t) - m 3 ~ o(t)T a k i n g L a p l a c e t r a n s f o r m s , z e r o i n i t i a l c o n d i t i o n s

F ( s ) - K X o ( s ) - C s X o ( s) - m s ZX o (s )o r

( m s 2 + C s + K ) X o ( s ) - F ( s )C h a r a c t e r i s t i c E q u a t i o n i s

m s 2 + C s -+- K - 0

a n d t h e r o o t s a r eC Ki.e. S 2 nt- - - --] - - 0m m

{ 21 C - 4~ , ~ 2 - ~ mF o r c r i ti c a l d a m p i n g , t h e d i s c r i m i n a n t i s z e ro , h e n c e t h e r o o t s b e c o m e

Also , f o r c r i t i ca l dampingS 1 - - $ 2 Cc2m

Cc 4Km 2 m

giv ing2 ~

4 K m 2m

(3.48)

(3.49)

(3.50)

Cc - 2x / 'K m (3.51)

3 .6 .4 G enera l ized secon d-order system response to a un i t s tepi n p u tC o n s i d e r a s e c o n d - o r d e r s y s t e m wh o s e s t e a d y - s t a t e g a i n i s K , u n d a m p e d n a t u r a lf r e q u e n c y is COn a n d wh o s e d a m p i n g r a t i o i s ( , wh e r e ( < 1. F o r a u n i t s te p i n p u t , t h eb l o c k d i a g r a m i s a s s h o wn i n F i g u r e 3 . 1 8 . F r o m F i g u r e 3 . 1 8

K~2nX ~ - - s ( s 2 + 2 f fC O n S - +- C O 2 ) ( 3 . 5 2 )


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T im e d o m a in a n a ly s is 53

X (S)= 1/s K~ 29 s 2 + 2 ~ n S + s ~ 2

Xo(S)y

Fig. 3.1 8 Ste p response of a general ized second-order system for f f < 1 .E x p a n d i n g e q u a t i o n ( 3 . 5 2 ) u s i n g p a r t i a l f r a c t i o n s

A B s + CX o ( S ) = - +s ( s 2 + 2 f f co n S + c o 2 )2E q u a t i n g ( 3 . 5 2 ) a n d ( 3 . 5 3 ) a n d m u l t i p l y b y S ( S 2 - Jr -2ffconS + C O n )

(3 .53)

E q u a t i n g c o e f f i c i e n t s

g iv ing

K c o 2 n - A ( s 2 - + - 2 f f c o n S t - c o2 ) + B s 2 - 1- C s

($ 2): 0 = A + B( s 1 ) : 0 - - 2 ff c o n A - J r -C( s O ) 9 KJ n - ~nA

A = K , B = - K a nd C = - 2 f fc o n KS u b s t i t u t i n g b a c k i n t o e q u a t i o n ( 3 . 5 3 )

I ! { S -1 - 2 ffc O n ) ]X o ( s ) - K - s 2 - Jr-2ffconS -Jr-CO2nC o m p l e t i n g t h e s q u a r e

s + 2ff~n

/ s + 2 , / 1( S + ~ c o n ) 2 + ( c o n V / l _ ~ 2 ) 2T h e t e r m s i n t h e b r a c k e t s { } c a n b e w r i t t e n i n t h e s t a n d a r d f o r m s 1 0 a n d 9 i nT a b l e 3 . 1 .

T e r m ( 1 ) = - - S~ S + ~ n ~ 2 + ( ~ n r 2

T er m (2) - - v/1 _ ff2 )2~ n ( s ~ + ~ n + ( ~ n , / i - ~ ) ~


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5 4 Ad vanc ed Contro l EngineeringI n v e r s e t r a n s f o r m , )o ( t ) - - K I 1 - e - ~ W n t(c o s ( c o n V /l-~ 2 )t- c o n V /l_ ~ 2 s in ( c o n V /1 - ~ 2 )

2~ (s in COnV /1- - { V /1 _ i f 2 } { e - f f ~ n t ( - - f f2 ) ) } ] (3 .55)E q u a t i o n ( 3 . 5 5 ) c a n b e s i m p l i f i e d t o g i v e

[ { ( ~ ) V /1 ~ 2 ) t } ] ( 3 .56 )o ( t) -- K 1 - e -~ w n t o s ( C O n V / l - f f 2 ) t + V / l - i f 2 s in (C O n -W h e n ( - 0

X o ( t ) - K[1 - e~ + 0}]= K[1 - cos C O n] (3 .57)

F r o m e q u a t i o n ( 3 . 5 7 ) i t c a n b e s e e n t h a t w h e n t h e r e i s n o d a m p i n g , a s t e p i n p u t w i l lc a u s e t h e s y s t e m t o o s c i l l a t e c o n t i n u o u s l y a t C O n r ad / s ) .D amped na t u ra l f r equency - O dF r o m e q u a t i o n ( 3 .5 6 ) , w h e n 0 < ff > 1 , t h e f r e q u e n c y o f t r a n s i e n t o s c i l la t i o n isg i v e n b y

C O d - c o n V /1 _ i f 2 ( 3 . 5 8 )w h e r e COd i s c a l le d t h e d a m p e d n a t u r a l f r e q u e n c y . H e n c e e q u a t i o n ( 3 .5 6 ) c a n b ew r i t t e n a s [ {o ( t ) - K 1 - e - ~ ~ c o s c o d t + ~ / 1 ( 2 s in c oa t ( 3 .5 9 )

e - ( ~ O n t= K 1 - --1-------~ in ( c o o t + 4 ) ) (3 .60)w h e r e

t an 05 x / ' 1 _ ( 2 ( 3 . 6 1 )W h e n ( - 1 , t h e u n i t s te p r e s p o n s e is

X o ( t ) - - K [1 - e-~~ + COnt)]a n d w h e n ~ > 1 , t h e u n i t s t e p r e s p o n s e f r o m e q u a t i o n ( 3 .4 6 ) is g i v e n b y

(3 .62)

(3 .63)


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0.8

1.6

1.4

1.2

0.6 \

0.40.2 Y/# f f = l . 0

~ = 0 . 2

~ = 0 . 4~ = 0 . 6~=o.8

= 2.0

0 0.5 1 1 .5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5

COnt rad)Fig. 3.1 9 Unit s tep response of a second -order system.

The genera l i zed sec ond-order sys t em response to a un i t s tep inpu t i s shown in F igure3.19 for the condi t ion K - 1 (see also Ap pen dix 1, sec_ord.m).

3.7.1 Step response analysisI t is poss ib l e t o i den ti fy t he math ema t i ca l model o f an und erd am pe d second -ordersystem from i t s s tep response funct ion.

Cons ider a un i ty -ga in (K = 1) second-o rder un der dam ped sys t em respon d ing toan inpu t o f t he fo rm

x i ( t ) - B ( 3.64 )The resu l t i ng ou tpu t Xo(t) would be as shown in F igure 3 .20 . There a re two methodsfo r ca l cu l a t ing the damping ra t io .Method (a)" Percen tage Ove rshoo t o f f ir s t peak

a l% O ve rs ho ot - -=_ x 100B (3.65)N o w

al - B e - r


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56 Ad vanc ed Cont r o l E ng ineering

X o ( t ) ~ ( e-~Wntwith reference to f inal value)F ig . 3 .2 0 S tep response ana lys is .

T h u s ,Be-(W.(r/2)% O v e r s h o o t - 1 00B (3 .66 )

S i n c e th e f r e q u e n c y o f t r a n s i e n t o s c i l l a t io n i s Odd, t h e n ,27rOdd

27rOdnV/1 _ i f2

S u b s t i t u t i n g ( 3 . 6 7 ) i n t o ( 3 . 6 6 )% O v e r s h o o t - e -2~r(wn/2wn 1~ ~-~ 2 x 1 0 0% O v e r s h o o t - e -r162

(3 .67)

(3 .68 )M e t h o d ( b ) " L o g a r i t h m i c d e c r e m e n t . C o n s i d e r t h e r a t i o o f su c ce s si ve p e a k s a l a n d a 2

H e n c e

a l - - B e - < ~ "( r/ 2) (3 .69 )a2 - B e - f fwn(3r /2 ) (3 .70)

a l e - ~ w n ( r / 2 )a2 e -(wn(3r/2) = e { - @ n ( T / 2 ) + @ n ( 3 r / 2 ) }

= e


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T i m e d o m a i n a n a ly s is 5 7E q u a t io n ( 3. 71 ) c a n o n ly b e u se d i f t h e d a m p in g i s li g h t a n d t h e r e i s m o r e t h a n o n eo v e r s h o o t . E q u a t io n ( 3 . 6 7 ) c a n n o w b e e m p lo y e d to c a l c u l a t e t h e u n d a m p e d n a tu r a lf r e q u e n c y

27rWn = (3.7 2)x /1

3 .7 .2 S te p response per forman ce spec ifica tionThe th ree parameter s shown in F igu re 3 .21 a re u sed to spec i fy per fo rmance in thet im e d o m a i n .(a ) R i s e t i m e t r : The sho r tes t t ime to ach ieve the f ina l o r s teady- s ta te va lue , fo r the

f i r s t t ime . Th is can be 100% r i se t ime as shown , o r the t ime taken fo r examplef r o m 1 0 % to 9 0 % o f t h e fi n a l v a lu e, t h u s a l l o w in g f o r n o n - o v e r s h o o t r e s po n s e .(b) O v e r s h o o t : T h e r e l a t i o n s h ip b e tw e e n th e p e r c e n t a g e o v e r s h o o t a n d d a m p in g

r a t i o i s g ive n in e q u a t io n ( 3. 68 ). F o r a c o n t r o l s y s t em a n o v e r s h o o t o f be tw e e n0 an d 10% (1 < ~ > 0 .6) is gene ral ly acceptab le .

(c ) Se t t l i ng t ime t s " This i s the t ime fo r the sys tem ou tpu t to se t t le down to wi th in ato l e r a n c e b a n d o f t h e fi n al va lu e, n o r m a l ly b e tw e e n + 2 o r 5 % .

U s in g 2 % v a lu e , f r o m F ig u r e 3 . 2 10.02B = Be -~n t s

I n v e r t50 - - e CaJ"ts

X o ( t ) e e - ( W n t" ' . . . ~ ~ (with reference to final value)

Rise- ,q m-Time trS e t t l in g T i m e ts

F i g . 3 . 2 1 S t e p r e s p o n s e p e r f o r m a n c e s p e c i f i c a ti o n .


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58 Advanced ontrolEngineeringTake na tu ra l l ogs

givingIn 50 = ff~nts

ts - ln 50 (3.73)The t e rm (1/ffCOn) i s somet imes cal led the equivalent t ime constant Tc for a second-ord er system. N ote that In 50 (2% tolerance) i s 3 .9 , and ln20 (5% tolerance) i s 3 .0 .Thus the t ransient per iod for both f i rs t and second-order systems i s three t imes thet ime cons t an t t o wi th in a 5% to l erance band , o r four t imes the t ime cons t an t t owi th in a 2% to l erance band , a usefu l ru l e -o f - thumb.

Trans fer func t ion t echn iques can be used to ca l cu l a te t he t ime response o f h igher -o rder sys t ems .Exam ple 3.8 (See also Appendix 1 , examp38.m)Figure 3 .22 show s , i n b lock d i agr am fo rm, t he t r ansfer func t ions fo r a res i s tancethermometer and a va lve connected toge ther . The inpu t xi(t) i s t empera tu re and theo u t p u t Xo(t) i s valve posi t ion. Find an expression for the uni t s tep response funct ionwhen there are zero in i t ial condi t ions.SolutionF r o m F i g u r e 3 .2 2

25X o ( s ) s(1 -+- 2s) (s 2 + s + 25) (3 .7 4)12.5

S(S -Jr-0. 5) (S 2 -Jr- S -q- 25) (3.75)A B C s + D= t + (3.76 )s (s + 0.5) (s + 0.5) 2 + (4.97) 2

Resistance Thermometer

I 1i ( s )= l / s 9 1 + 2sValve

25s 2 + s + 2 5

Fig. 3.2 2 Block diagram representation of a resistance thermometer and valve.

~Xo(S)


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T i m e d o m a i n a n a l y s is 5 9N o t e t h a t t h e s e c o n d - o r d e r t e r m i n e q u a t i o n ( 3 .7 6 ) h a s h a d t h e ' s q u a r e ' c o m p l e t e ds i n ce i t s r o o t s a r e c o m p l e x ( b 2 < 4 a c ) . E q u a t e e q u a t i o n s ( 3 . 7 5 ) a n d ( 3 . 7 6 ) a n d m u l t i -p l y b o t h s id e s b y s ( s + 0 .5 ) ( s 2 + s + 25) .

1 2 . 5 - ( s 3 - +- 1 .5 s 2 - t - 2 5 .5 s + 1 2 .5 ) A + ( s 3 - - I - s 2 n - 2 5 s ) B-t- (S3 + 0.5 s2 )C -+- (s 2 -~- 0 . 5s )D

E q u a t i n g c o e f f i c i e n t s( $ 3 ) : O = A + B + C(s 2) : 0 = 1 .5 A + B + 0 . 5 C + D(s 1) " 0 = 2 5 . 5 A + 2 5 B + 0 . 5 D(sO) : 12 .5 = 12 .5A

( 3 .7 7 )

S o l v i n g t h e f o u r s i m u l t a n e o u s e q u a t i o n sA - l , B - - 1 . 0 1 , C - 0 .0 1 , D - - 0 . 5

S u b s t i t u t i n g b a c k i n t o e q u a t i o n ( 3 . 7 6 ) g i v e s1 1 .0 1 0 . 0 1 s - 0 .5

X o ( s ) - s - ( s + 0.5----~+ ( s + 0 .5 ) 2 + ( 4 .9 7 ) 2 ( 3 .7 8 )I n v e r s e t r a n s f o r m

X o ( t ) - 1 - 1 . 01 e - ~ - 0 . 0 1 e - ~ s i n 4 . 9 7 t - c o s 4 . 9 7 t ) ( 3 .7 9 )E q u a t i o n ( 3 .7 9 ) s h o w s t h a t t h e t h i r d - o r d e r t r a n s i e n t r e s p o n s e c o n t a i n s b o t h f i rs t-o r d e r a n d s e c o n d - o r d e r e l e m e n t s w h o s e t i m e c o n s t a n t s a n d e q u i v a l e n t t i m e c o n s t a n t sa r e 2 s e c o n d s , i .e . a t r a n s i e n t p e r i o d o f a b o u t 8 s e c o n d s . T h e s e c o n d - o r d e r e l e m e n th a s a p r e d o m i n a t e n e g a t i v e s in e t e r m , a n d a d a m p e d n a t u r a l f r e q u e n c y o f 4 .9 7 r a d / s.T h e t i m e r e s p o n s e i s s h o w n i n F i g u r e 3 . 2 3 .

1.2

0 .8

>r176.6

0 .4 /~

m

fJ/o 2 /

o o 1 2 3 4 5 6T ime (s)F ig . 3 .23 T ime response o f th i rd -o rde r sys tem.


26/28

60 Advanced Control Engineering

Example 3.9A ship has a mass m and a res is tance C t imes the forward veloci ty u(t). I f the thru s tfrom the p ropel ler i s K t imes i t s angu lar veloci ty ~o(t) , determine:(a) The f i rs t -order di fferentia l equ at io n and hence the t ransfer funct ion re la t ing U(s)

and ,~(s).W he n the vessel has the p aram eters : m = 18 000 103 kg, C = 150000 Ns/ m, andK = 96 000 Ns /rad , f ind,(b) the t ime constant .(c) an express ion for the t ime respon se of the ship whe n there is a s tep change of ~( t )

fro m 0 to 12.5 rad/s. Ass um e th at the vessel is init ial ly at rest .(d) What i s the forward veloci ty af ter( i) one min ute(i i) ten minutes.

Solution(a ) m(du/dt) + Cu = K~o(t)

u K/C- ( s ) =~o 1 + (m /C ) s(b) 120 secon ds(c) u(t) = 8(1 - e -0"00833')(d) (i) 3.148 m /s

(ii) 7.946 m /sExample 3.10(a) Dete rmi ne the t ransfer funct io n re la t ing Vz(s) and V1 (s) for the passive e lectr ical

ne twork shown in F igure 3 .24 .(b) W he n C = 2 ~tF and R] = R2 = 1 Mgt , de termin e the s tead y-s ta te gain K andt ime cons tan t T .(c) Find an express ion for the uni t s tep response.

R1

Fig. 3.24 Passiveelectrical network.

I

v~(t C R2 v2(t)


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Time dom ain ana lys is 61Solut ion

V2 R z / R 1 + R 2( a ) ~ - [ ( s ) - 1 + ( R m R 2 C / R 1 + R 2 ) s(b) 0.5

1 . 0 s econds(c ) Vo(t) = 0.5(1 - e - t )E x a m p l e 3 .1 1D e t e r m i n e t h e v a l u e s o f COn a n d ff a n d a l s o e x p r e s s i o n s f o r t h e u n i t s t e p r e s p o n s e f o rt h e s y s t e m s r e p r e s e n t e d b y t h e f o l l o w i n g s e c o n d - o r d e r t r a n s f e r f u n c t i o n s

X o 1(i) ~ -i (s) - 0.25s 2 4- s 4- 1

X o _ 10(ii) - -X i ( S ) $ 2 4 - 6 s 4 - 5Xo _ 1(iii) ~ ( s ) s 2 + s + l

Solut ion( i ) 2 . 0

1 . 0 ( C r i t i c a l d a m p i n g )Xo(t) = 1 - e - 2 t ( 1 + 2 t)

( i i ) 2.2361 . 3 4 2 ( O v e r d a m p e d )Xo(t) - - 2 - 2 .5e - t + 0 .5e -St

(iii) 1.00 . 5 ( U n d e r d a m p e d )Xo(t) = 1 - e - ~ 0 . 866 t + 0 . 577 s i n 0 . 8 66 0

E x a mp l e 3 .1 2A t o r s i o n a l s p r i n g o f s ti ff n e ss K , a m a s s o f m o m e n t o f in e r t i a I a n d a f lu i d d a m p e rw i t h d a m p i n g c o e f f i c i e n t C a r e c o n n e c t e d t o g e t h e r a s s h o w n i n F i g u r e 3 . 2 5 . T h ea n g u l a r d i s p l a c e m e n t o f th e f r ee e n d o f t h e s p r i n g is 0 i( t) a n d t h e a n g u l a r d i s p la c e -m e n t o f t he m a s s a n d d a m p e r i s Oo(t).

K- 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . _ ~ _ e o ( t )

CF i g . 3 . 2 5 T o r s i o n a l s y s te m .


28/28

62 Ad vanc ed Con trol Engineering( a ) Deve l op t he t r ans f e r f unc t i on r e l a t i ng 0 i ( s ) and Oo(s).( b ) I f t he t i m e r e l a t i ons h i p f o r 0 i( t) i s g i ven by 0 i (t ) = 4 t t hen f i nd an exp r e s s i on f o r

t h e t i m e r e s p o n s e o f Oo(t) . A s s u m e z e r o i n i t i a l c o n d i t i o n s . W h a t i s t h e s t e a d y -s t a t e e r r o r be t ween 0 i ( t ) and 0o ( t ) ?

So l u t i onOo 1

( a ) ~ -i ( s ) - ( / ) s 2 + ( C) s + 1(b ) Oo(t ) = 4 t - 0 . 2 + e - 2s t ( 0 . 2 cos 9 . 682 t - 0 . 361 si n 9 . 6 820

0 . 2 r a d i a n s .E x a m p l e 3 . 1 3W h e n a u n i t y g a i n s e c o n d - o r d e r s y s t e m i s s u b j e c t t o a u n i t s t e p i n p u t , i t s t r a n s i e n tr e s p o n s e c o n t a i n s a f ir s t o v e r s h o o t o f 7 7 % , o c c u r r i n g a f t e r 3 2 .5 m s h a s e l a p se d . F i n d( a) t h e d a m p e d n a t u r a l f r e q u e n c y( b) t h e d a m p i n g r a t i o(c ) t h e u n d a m p e d n a t u r a l f r e q u e n c y( d ) t h e s y s t e m t r a n s f e r f u n c t i o n( e) t h e t im e t o s e t tl e d o w n t o w i t h i n + 2 % o f th e f i n al v a lu eSo l u t i on( a ) 96 . 66 r ad / s(b) 0 .083( c) 96 . 99 r ad / s(d) G ( s ) = 0.106 x 10-3s 2 + 1 .712 x 10-3s + 1( e) 0 . 486 s econd sE x a m p l e 3 . 1 4A s y s t e m c o n s i s t s o f a f ir s t - o r d e r e l e m e n t l i n k e d t o a s e c o n d - o r d e r e l e m e n t w i t h o u ti n t e r a c t i o n . T h e f i r s t - o r d e r e l e m e n t h a s a ti m e c o n s t a n t o f 5 s e c o n d s a n d a s te a d y -s t at e g a in c o n s t a n t o f 0 .2 . T h e s e c o n d - o r d e r e l e m e n t h a s a n u n d a m p e d n a t u r a lf r e q u e n c y o f 4 r a d / s , a d a m p i n g r a t i o o f 0 . 25 a n d a s t e a d y - s t a t e g a i n c o n s t a n t o fu n i t y .

I f a s t e p i n p u t f u n c t i o n o f 1 0 u n i t s i s a p p l i e d t o t h e s y s t e m , f i n d a n e x p r e s s i o n f o rt h e t i m e r e s p o n s e . A s s u m e z e r o i n i t i a l c o n d i t i o n s .So l u t i onXo(t) - 2 . 0 - 2 . 0 4 6e - ~ + e - t ( 0 . 0 4 6 c o s v / ~ t - 0 . 0 9 4 s in v / ~ t )\ /

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