ENGINEERING BACKGROUN D 7 iut\ _ Lf0\fT _ K8) and o f the outpu t i s L(BeluJt) = b(s){LeJut\ luJt ) = IUJ s Apply th e invers e Laplac e transfor m t o ge t [Be^ir) = -L f K+l °° -M- e +ST ds. 27r JK-ioo S-IUJ As an exampl e tak e 6(5 ) = 0/(s + c) . The n iu + c[ J v n J If 6 G #°°(R.H.P.) , tha t is , c 0 , the n e" ct 0 quickly , s o indee d afte r a "short" tim e th e outpu t look s lik e b{iu)e luJt a s advertised . I f c 0 , the n th e output ha s a n exponentiall y explodin g par t (whic h i s consisten t wit h th e fac t that b is unstable). Note fro m th e sam e computatio n tha t inpu t e( tw +r)* gives outpu t Be (iu,+r)t = b ^iu) + f )[ c (tw+r)t + e -ct^ Thus b(iu H - r) describe s ho w £ handle s a comple x frequenc y iu) + r . EXAMPLE 1 . L R v.Ct) ] ^ ^ j ^ c j v 2 (t) FIGURE 1.3 . Question. Wha t t 2 is produced b y v\l Tha t is , take v\ a s input,t 2 as output , and find th e transfe r function . Solution. Le t c(t) b e th e curren t i n th e loo p a t tim e t. KirchoiT s Law s an d the definitio n o f inductor-capacitor, etc. , giv e T dc n 1 r , 1 f , L— + RcJr—lcdt = Vi, cdt = w 2 . Assume zer o initia l condition s an d tak e Laplac e transform s t o ge t Lsc{s) + Rc{s) + ^-c{s) = 1)1(5) , -p -c(s) = v2{s). G s C s Eliminate c(s) t o get tha t th e desire d transfe r functio n vzjs) = 1 ixis) LCs 2 + RCs + l' EXAMPLE 2 (se e Figure 1.4) . This i s a mechanica l syste m drive n b y forc e F(t) whos e heigh t y(t) w e wan t to measure . Tak e F(t) a s input , y(t) a s output . The n
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