lecture 16 - Summary of Lecture #16 9/29/2010...

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1 1 Summary of Lecture #16 9/29/2010 Classification of responses Complete response = ZIR+ZSR = TSR + SSR = NTR + FCR Principle of superposition Impulse, step and frequency responses Examples 2 Classification of responses Complete response = ZIR+ZSR = TSR + SSR = NTR + FCR Zero state response (ZSR) Zero input response (ZIR) Transient response (TSR) Steady state response (SSR) Natural response (NTR) Forced response (FCR) 9 Ex 3. Principle of superposition Given v a (t) = 3 u(t) V, v b (t) = 2 sint u(t) V, and v c (0 - ) = 2 V . Find v c (t) for t > 0. V ) t ( u )] t sin t (cos 2 1 e 5 . 0 5 . 1 e 5 . 1 e 2 [ v t t t c - - + + - = - - - ZSR due to v a ZIR SSR, FCR v c1 v c2 v c3 ZSR due to v b TSR, NTR 10 ZIR: Keep initial conditions (I.C.) only. Short out v a and v b . V ) t ( u e 2 ) t ( v , 1 s 2 ) s ( V t 1 c 1 c - = + = V c1 11 ZSR due to v a Keep v a . Short out v b . Set I.C. to 0. V ) t ( u e 5 . 1 ) t ( u 5 . 1 ) t ( v , 1 s 5 . 1 s 5 . 1 ) 1 s ( s 5 . 1 )) 5 . 0 s /( 1 ( 2 ) 5 . 0 s /( 1 ) s ( V ) s ( V t 2 c a 2 c - - = + - = + = + + + = s 3 ) s ( V a = V c2 5 . 0 s 1 ) s / 1 ( 2 ) s / 1 ( 2 s 1 // 2 + = + 12 ZSR due to v b Short out v a . Keep v b . Set I.C. to 0. V ) t ( u ) t sin t (cos 2 1 ) t ( u e 2 1 ) t ( v , 1 s 5 . 0 1 s s 5 . 0 1 s 5 . 0 ) 1 s )( 1 s ( 1 ) s ( V t 3 c 2 2 2 3 c - - = + + + - + = + + = - 1 s 2 ) s ( V 2 b + = V c3 )) 5 . 0 s /( 1 ( 2 ) 5 . 0 s /( 1 ) s ( V ) s ( V b 3 c + + + =
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2 13 If input current or voltage is δ (t), then the ZSR output is the impulse response , h(t). V in (s) =1 [ or I in (s) = 1 ] V out (s) = H(s) [ or I out (s) = H(s) ] L - 1 [V out (s)] = L - 1 [H(s)] = h(t) Impulse response is the inverse Laplace transform of the transfer function H(s). + -
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lecture 16 - Summary of Lecture #16 9/29/2010...

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