hw3_sol.pdf - ECE 202 PROBLEM SET 3 SOLUTIONS 1 1(a Zin(s =...

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ECE 202: PROBLEM SET 3: SOLUTIONS 1. (a) Z in ( s ) = R 1 + Ls + R 2 || 1 sC = R 1 + Ls + R 2 R 2 sC +1 Y in ( s ) = 1 Z in ( s ) = 1 R 1 + Ls + R 2 R 2 sC +1 (b) H ( s ) = v R 1 v in = R 1 R 1 + Ls + R 2 R 2 sC +1 = R 1 L s + 1 R 2 C s 2 + s L + R 1 R 2 C R 2 CL + R 1 + R 2 R 2 CL (Voltage Divider) (c) Plugging in R 1 = 4Ω, L = 4 H , R 2 = 2Ω, and C = 1 8 F , we get H ( s ) = ( s +4) ( s +2)( s +3) Impulse Response: H ( s ) = V out ( s ) V in ( s ) = L{ v R 1 ( t ) } L{ δ ( t ) } ; which means v R 1 ( t ) = L - 1 { H ( s ) } Partial Fractions: H ( s ) = ( s +4) ( s +2)( s +3) = - 1 ( s +3) + 2 ( s +2) v R 1 ( t ) = - e - 3 t u ( t ) + 2 e - 2 t u ( t ) V Step Response: H ( s ) = V out ( s ) V in ( s ) = L{ v R 1 ( t ) } L{ u ( t ) } ; which means v R 1 ( t ) = L - 1 { H ( s ) s } Partial Fractions: H ( s ) = ( s +4) s ( s +2)( s +3) = 2 3 s + 1 3( s +3) - 1 s +2 v R 1 ( t ) = 2 3 u ( t ) + 1 3 e - 3 t u ( t ) - e - 2 t u ( t ) V (d) One possible representation of the circuit with initial conditions as voltage sources is shown in Fig. 1. An alternate representation with the initial capacitor voltage represented as a current source is shown in Fig. 2. Other representations are also possible. + - v in ( t ) R 1 Ls + - Li L (0 - ) R 2 + v R 1 - 1 sC - + v C (0 - ) s Z in Figure 1: Circuit with initial conditions represented as voltage sources.
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+ - v in ( t ) R 1 Ls + - Li L (0 - ) R 2 + v R 1 - 1 sC Cv C (0 - ) Z in Figure 2: Circuit with initial conditions represented as a mix of voltage and current sources. (e) We will use the circuit in Fig. 2 to answer this question (the other circuit can also be used). The voltage v R 1 ( t ) is the sum of its voltage drops from each source considered independently. (Start with 1 source, then find v in ( t ) while removing the other 2 sources. Repeat for each source). In other words, v R 1 ( t ) = v R 1 ,in ( t ) + v R 1 ,L ( t ) + v R 1 ,C ( t ) , where v R 1 ,in ( t
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