HW 03_examples

# HW 03_examples - 1 ME261 Solved Examples for Homework#3...

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Unformatted text preview: 1 ME261 - Solved Examples for Homework #3 Problem 1: For each circuit in Figure 1., derive the transfer function using equivalent impedance method and impulse responses using partial fraction expansion and inverse Laplace transform. Figure 1. Solution: Z C = 1 Cs ; Z L = Ls; Z R = R Circuit 1: loop equations: (1) Y ( s ) = I & 1 sC ¡ (2) X ( s ) = Z R I + Z C I = IR + I 1 Cs | {z } Y ( s ) = I 1 Cs CsR + I 1 Cs = Y ( s ) ( RCs + 1) transfer function: H ( s ) = Y ( s ) X ( s ) = 1 RCs +1 = 1 RC h 1 s + 1 RC i de&ne: & = RC , impulse response: h ( t ) = L & 1 f H ( s ) g = 1 & & e & t=& ¡ u ( t ) (see Laplace Transform Table 11.1 Ambardar) Circuit 2: loop equations: (1) X ( s ) = i 1 Z R 1 + ( i 1 & i 2 ) Z R 2 (2) Y ( s ) = i 2 Z c (3) Y ( s ) + Z R 2 ( i 2 & i 1 ) = 0 i 2 = Y ( s ) Z c and ( i 1 + i 2 ) = Y ( s ) Z R ) i 1 = Y ( s ) ¢ 1 Z R & 1 Z c £ now sub i 1 and i 2 into X ( s ) equation: X ( s ) = Y ( s ) ¢¢ 1 + Z R Z c £ + 1 £ transfer function: H ( s ) = Y ( s ) X ( s ) = 1 2+ Z R Zc = 1 2+ RCs = 1 RC ¢ 1 2 RC + s £ impulse response: h ( t ) = L & 1 f H ( s ) g = 1 RC & e & 2 t=RC ¡ u ( t ) (see table) Circuit 3: loop equations: (1) X ( s ) = i 1 Z R 1 + ( i 1 & i 2 ) Z R 2 (2) Y ( s ) = i 2 Z L = ( i 1 & i 2 ) Z R (3) i 2 = Y ( s ) Z L (4) ( i 1 & i 2 ) = Y ( s ) Z R now sub i 1 and...
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## This note was uploaded on 04/18/2010 for the course ME 360 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.

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HW 03_examples - 1 ME261 Solved Examples for Homework#3...

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