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Unformatted text preview: LECTURE NOTES 102 Fall ' 90 Review Oct 28 H Midterm Monday Nov 2 OPEN BOOK L IPOP d n v H t L dt n + 1 n- 1 a k d k v H t L dt k = m b k d k u H t L dt k LAPLACE TRANSFORM : L H f' L = s L H f L- f H L L H f'' L = s 2 L H f L- f' I M- s f H L L I f H k L M = s k L H f L- j = k- 1 s H k- 1- j L f H j L H L Of Left Side : s n V H s L + k = n- 1 a k s k V H s L- j = n- 1 s H n- 1- j L v H j L H L- k = n- 1 a k j = k- 1 s H k- 1- j L v H j L H L n = 2 m = 1 fi v'' H t L + a 1 v' H t L + a v H t L = b u H t L + b 1 u' H t L WHAT IS THE INPUT TO THE SYSTEM THE RIGHT SIDE ALL OF IT Laplace Transform s 2 V H s L- v' H L- s v H L + a 1 H s V H s L- v H LL + a V H s L = b U H s L + b 1 H sU H s L- u H LL Collect terms containing V H s L and U H s L fi I s 2 + a 1 s + a M V H s L- v' H L- H a 1 + s L v H L = H b + b 1 s L U H s L- b 1 u H L ' SOLVE FOR V H s L " V H s L = TRANSFORM OF RIGHT SIDE I s 2 + a 1 s + a M @ IPOP TRANSFER FUNCTION D + H a 1 + s L v H L + v' H L I s 2 + a 1 s + a M @ INITIAL STATE TRANSFORM D Define : H H s L = 1 s 2 + a 1 s + a which we shall call the ' System Transform Function' because this depends only on the system. H H s L = ¥ e- st W H t L t W H . L is the System Weighting Function W H L = fi ¥ e- st W' H t L t = s H H s L Then we can write V H s L = H H s [email protected] a 1 + s L v H L + v' H L D + H H s L @H b + b 1 s L U H s L- b 1 u H LD The First term is the Transform of the Initial Condition H OR zero Input Response L Inverse Transform : W' H t L v H L + W H t L H v' H L + a 1 v H LL The SECOND term is the H Zero State L Input Response- The STEADY STATE TRANSFER FUNCTION Here it depends on what we call the ' Input' : u H t L or b u HL + b 1 u' H t L , t ‡ Let W...
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## This note was uploaded on 01/16/2012 for the course EE 102 taught by Professor Levan during the Spring '08 term at UCLA.

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view - LECTURE NOTES 102 Fall ' 90 Review Oct 28 H Midterm...

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