ch13_ism - 13 The Laplace Transform in Circuit Analysis...

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13 The Laplace Transform in Circuit Analysis Assessment Problems AP 13.1 [a] Y = 1 R + 1 sL + sC = C [ s 2 +(1 /RC ) s /LC ) s 1 RC = 10 6 (500)(0 . 025) =80 , 000; 1 LC =25 × 10 8 Therefore Y = 25 × 10 9 ( s 2 +80 , 000 s +25 × 10 8 ) s [b] z 1 , 2 = 40 , 000 ± 16 × 10 8 25 × 10 8 = 40 , 000 ± j 30 , 000 rad/s z 1 = 40 , 000 j 30 , 000 rad/s z 2 = 40 , 000 + j 30 , 000 rad/s p 1 =0 rad/s AP 13.2 [a] Z = 2000 + 1 Y = 2000 + 4 × 10 7 s s 2 , 000 s × 10 8 = 2000( s 2 +10 5 s × 10 8 ) s 2 , 000 s × 10 8 = 2000( s +50 , 000) 2 s 2 , 000 s × 10 8 [b] z 1 = z 2 = 50 , 000 rad/s p 1 = 40 , 000 j 30 , 000 rad/s p 2 = 40 , 000 + j 30 , 000 rad/s 13–1
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13–2 CHAPTER 13. The Laplace Transform in Circuit Analysis AP 13.3 [a] At t =0 , 0 . 2 v 1 . 8 v 2 ; v 1 =4 v 2 ; v 1 + v 2 = 100 V Therefore v 1 (0 )=80 V = v 1 (0 + ); v 2 (0 )=20 V = v 2 (0 + ) I = (80 /s ) + (20 /s ) 5000 + [(5 × 10 6 ) /s ]+(1 . 25 × 10 6 /s ) = 20 × 10 3 s + 1250 V 1 = 80 s 5 × 10 6 s 20 × 10 3 s + 1250 ! = 80 s + 1250 V 2 = 20 s 1 . 25 × 10 6 s 20 × 10 3 s + 1250 ! = 20 s + 1250 [b] i =20 e 1250 t u ( t ) mA ; v 1 =80 e 1250 t u ( t ) V v 2 e 1250 t u ( t ) V AP 13.4 [a] I = V dc /s R + sL +(1 /sC ) = V dc /L s 2 +( R/L ) s /LC ) V dc L = 40; R L =1 . 2; 1 LC . 0 I = 40 s 2 +1 . 2 s
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Problems 13–3 [b] I = 40 ( s +0 . 6 j 0 . 8)( s . 6+ j 0 . 8) = K 1 s . 6 j 0 . 8 + K 1 s . j 0 . 8 K 1 = 40 j 1 . 6 = j 25 = 25/ 90 ; K 1 = 25/90 i =50 e 0 . 6 t cos(0 . 8 t 90 ) = [50 e 0 . 6 t sin 0 . 8 t ] u ( t ) A [c] V = sLI = 160 s s 2 +1 . 2 s = 160 s ( s . 6 j 0 . 8)( s . j 0 . 8) = K 1 s . 6 j 0 . 8 + K 1 s . j 0 . 8 K 1 = 160( 0 . j 0 . 8) j 1 . 6 = 100/36 . 87 [d] v ( t ) = [200 e 0 . 6 t cos(0 . 8 t +36 . 87 )] u ( t ) V AP 13.5 [a] The two node voltage equations are V 1 V 2 s + V 1 s = 5 s and V 2 3 + V 2 V 1 s + V 2 (15 /s ) 15 =0 Solving for V 1 and V 2 yields V 1 = 5( s +3) s ( s 2 +2 . 5 s +1) ,V 2 = 2 . 5( s 2 +6) s ( s 2 . 5 s [b] The partial fraction expansions of V 1 and V 2 are V 1 = 15 s 50 / 3 s . 5 + 5 / 3 s and V 2 = 15 s 125 / 6 s . 5 + 25 / 3 s It follows that v 1 ( t )= ± 15 50 3 e 0 . 5 t + 5 3 e 2 t ² u ( t ) V and v 2 ( t ± 15 125 6 e 0 . 5 t + 25 3 e 2 t ² u ( t ) V [c] v 1 (0 + )=15 50 3 + 5 3 v 2 (0 + 125 6 + 25 3 =2 . 5 V
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13–4 CHAPTER 13. The Laplace Transform in Circuit Analysis [d] v 1 ( )=15 V ; v 2 ( V AP 13.6 [a] With no load across terminals a b , V x =20 /s : 1 2 ± 20 s V Th ² s + ± 1 . 2 ³ 20 s ´ V Th ² =0 therefore V Th = 20( s +2 . 4) s ( s +2) V x =5 I T and Z Th = V T I T Solving for I T gives I T = ( V T 5 I T ) s 2 + V T 6 I T Therefore 14 I T = V T s 5 sI T V T ; therefore Z Th = 5( s . 8) s [b] I = V Th Z Th +2+ s = 20( s . 4) s ( s + 3)( s +6)
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Problems 13–5 AP 13.7 [a] i 2 =1 . 25 e t 1 . 25 e 3 t ; therefore di 2 dt = 1 . 25 e t +3 . 75 e 3 t Therefore di 2 dt =0 when 1 . 25 e t =3 . 75 e 3 t or e 2 t ,t . 5(ln 3) = 549 . 31 ms i 2 ( max )=1 . 25[ e 0 . 549 e 3(0 . 549) ] = 481 . 13 mA [b] From Eqs. 13.68 and 13.69, we have ∆ = 12( s 2 +4 s + 3) = 12( s + 1)( s +3) and N 1 = 60( s +2) Therefore I 1 = N 1 = 5( s ( s + 1)( s A partial fraction expansion leads to the expression I 1 = 2 . 5 s +1 + 2 . 5 s Therefore we get i 1 =2 . 5[ e t + e 3 t ] u ( t ) A [c] di 1 dt = 2 . 5[ e t e 3 t ]; di 1 (0 . 54931) dt = 2 . 89 A / s [d] When i 2 is at its peak value, di 2 dt Therefore L 2 di 2 dt !
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ch13_ism - 13 The Laplace Transform in Circuit Analysis...

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