DeCarlo Ch14a Solutions

# Linear Circuit Analysis: The Time Domain and Phasor Approach

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2/23/02 page P14.1 © R. A. DeCarlo, P. M. Lin PROBLEM SOLUTIONS CHAPTER 14 S OLUTION 14.1. (a) Z ( s ) = R ( Ls + 1 Cs ) R + Ls + 1 Cs = Cs ( RLCs 2 + R ) Cs ( RCs + LCs 2 + 1) = RLC ( s 2 + 1 LC ) LC ( s 2 + R L s + 1 LC ) = R ( s 2 + 1 LC ) s 2 + R L s + 1 LC (b) Z ( s ) = R + ( Ls )( 1 Cs ) Ls + 1 Cs = R + LCs C ( LCs 2 + 1) = R + Ls LCs 2 + 1 = RLCs 2 + Ls + R LCs 2 + 1 = RLC ( s 2 + 1 RC s + 1 LC ) LC ( s 2 + 1 LC ) Hence, Z ( s ) = R s 2 + 1 RC s + 1 LC s 2 + 1 LC SOLUTION 14.2. (a) Z in ( s ) = V s ( s ) I s ( s ) = (10 + 0.2 s )( 80 s ) 10 + 0.2 s + 80 s = s (800 + 16 s ) s (0.2 s 2 + 10 s + 80) = 800 s + 4000 s 2 + 50 s + 400 (b) If i s ( t ) = 3 e - 20 t u ( t ) A then I s ( s ) = 3 s + 20 and V s ( s ) = Z in ( s ) I s ( s ) = 800 s + 4000 s 2 + 50 s + 400 - 3 s + 20 = 2400 s + 12,000 ( s + 10)( s + 20)( s + 40) = K 1 s + 10 + K 2 s + 20 + K 3 s + 40 Here K 1 = 2400 s + 12,000 ( s + 20)( s + 40) s =- 10 = + 12,000 - 24,000 (10)(30) = - 40 K 2 = 2400 s + 12,000 ( s + 10)( s + 40) s =- 20 = 12,000 - 48,000 ( - 10)(20) = 180

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2/23/02 page P14.2 © R. A. DeCarlo, P. M. Lin K 3 = 2400 s + 12,000 ( s + 10)( s + 20) s =- 40 = 12,000 - 96,000 ( - 30)( - 20) = - 140 V s ( s ) = 180 s + 20 - 40 s + 10 - 140 s + 40 and for t > 0, v s ( t ) = 180 e - 20 t - 40 e - 10 t - 140 e - 40 t V SOLUTION 14.3. (a) Y p ( s ) = Cs + 1 R = 2 × 10 - 3 s + 1 0.10 = 2 × 10 - 3 ( s + 50) Then Z p ( s ) = 500 s + 50 and Z in ( s ) = 1.25 s + 500 s + 50 = 1.25 s 2 + 62.5 s + 500 s + 50 and Y in ( s ) = I s ( s ) V s ( s ) = 1 Z in ( s ) = s + 50 1.25 s 2 + 62.5 s + 500 = 0.80 s + 40 s 2 + 50 s + 400 With v s ( t ) = 90 e - 40 t u ( t ) , then V s ( s ) = 90 s + 40 and I s ( s ) = V s ( s ) Y in ( s ) = 90 s + 40 0.80 s + 40 ( s + 10)( s + 40) = 72 s + 3600 ( s + 10)( s + 40) 2 = K 1 s + 10 + C 1 s + 40 + C 2 ( s + 40) 2 Here K 1 = 72 s + 3600 ( s + 40) 2 s =- 10 = 3600 - 720 (30) 2 = 2880 900 = 3.20 and with p ( s ) = 72 s + 3600 s + 10 p ( - 40) = 720 - 30 = - 24 p ( s ) = ( s + 10)(72) - (72 s + 3600) ( s + 10) 2 p ( - 40) = - 30(72) - ( - 2880 + 3600) ( - 30) 2 = - 2160 - 720 900 = - 3.20 Then C 2 = p ( - 40) 0! = - 24 , C 1 = p ( - 40) 1! = - 3.20 ,
2/23/02 page P14.3 © R. A. DeCarlo, P. M. Lin I s ( s ) = 3.20 s + 10 - 3.20 s + 40 + 24 ( s + 40) 2 and for t > 0 i s ( t ) = 3.20 e - 10 t - 3.20 e - 40 t + 24 te - 40 t A SOLUTION 14.4. (a) Find Z in ( s ) vis Y in ( s ) Y in ( s ) = Cs + 1 Ls + 20 + 1 10 = 200 Cs + 10 LCs 2 + 10 + 20 + Ls 10( Ls + 20) = 10 LCs 2 + (200 C + L ) s + 30 10 Ls + 200 and Z in ( s ) = 1 Y in ( s ) = 10 Ls + 200 10 LCs 2 + (200 C + L ) s + 30 With C = 10 - 3 F and L = 0.05 H Z in ( s ) = 0.50 s + 200 0.0005 s 2 + 0.25 s + 30 = 1000 s + 4 × 10 5 s 2 + 500 s + 60,000 (b) If i s ( t ) = 0.3 u ( t ) A, then I s ( s ) = 0.30 s and V in ( s ) = Z in ( s ) I s ( s ) = 300 s + 400 s ( s 2 + 500 s + 60,000) = 300 s + 400 s ( s + 2000)( s + 300) = 300 K 1 s + K 2 200 + K 3 300 It follows that K 1 = s + 400 ( s + 200)( s + 300) s = 0 = 400 200(300) = 1 150 K 2 = s + 400 s ( s + 300) s =- 200 = 200 ( - 200)(100) = - 1 100 and K 3 = s + 400 s ( s + 200) s =- 300 = 100 ( - 300)( - 100) = 1 300 Thus

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2/23/02 page P14.4 © R. A. DeCarlo, P. M. Lin V in ( s ) = 300 1 150 s - 1 100 s + 200 + 1 300 s + 300 = 2 s - 3 s + 200 + 1 s + 300 and for t > 0 v in ( t ) = 2 - 3 e - 200 t + e - 300 t V SOLUTION 14.5. Z ( s ) = s + 20 s + 40 and the network is “at rest” (a) If v in ( t ) = 20 u ( t ) V in ( s ) = 20 s then I in ( s ) = V in ( s ) Z ( s ) = 20 s s + 40 s + 20 = 20 s + 40 s ( s + 20) Using a partial function expansion I in ( s ) = 20 s + 40 s ( s + 20) = 20 K 1 s + K 2 s + 20 in which case K 1 = s + 40 s + 20 s = 0 = 40 20 = 2, K 2 = s + 40 s s =- 20 = 20 - 20 = - 1 Thus I in ( s ) = 20 2 s - 1 s + 20 and i in ( t ) = 20(2 - e - 20 t ) u ( t ) A (b) Note that v in ( t ) = 20 e - 40 t V in ( s ) = 20 s + 40 Then I in ( s ) = V in ( s ) Z ( s ) = 20 s + 40 s + 40 s + 20 = 20 s + 20 in which case i in ( t ) = 20 e - 20 t u ( t ) A (c) Note that v in ( t ) = 20 e - 20 t V in ( s ) = 20 s + 20 Then
2/23/02 page P14.5 © R. A. DeCarlo, P. M. Lin I in ( s ) = V in ( s ) Z ( s ) = 20 s + 20 s + 40 s + 20 = 20 s + 40 ( s + 20) 2 Using a partial fraction expansion I in ( s ) = 20 s + 40 ( s + 20) 2 = 20 C 1 s + 20 + C 2 ( s + 20) 2 Here p ( s ) = s + 40 p ( - 20) = 20 p '( s ) = 1 p '( - 20) = 1 and C 1 = p '( - 20) 1!

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