chap16

# chap16 - Chapter 16 Solution 1 Consider the s-domain form...

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Chapter 16, Solution 1. Consider the s-domain form of the circuit which is shown below. I(s) + 1 1/s 1/s s 2 2 2 ) 2 3 ( ) 2 1 s ( 1 1 s s 1 s 1 s 1 s 1 ) s ( I + + = + + = + + = = t 2 3 sin e 3 2 ) t ( i 2 t - = ) t ( i A ) t 866 . 0 ( sin e 155 . 1 -0.5t Chapter 16, Solution 2. 8/s s s 4 2 + + V x 4

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V ) t ( u ) e 2 e 2 4 ( v 3 8 j 3 4 s 125 . 0 3 8 j 3 4 s 125 . 0 s 25 . 0 16 ) 8 s 8 s 3 ( s 2 s 16 V s 32 s 16 ) 8 s 8 s 3 ( V 0 V s V ) s 4 s 2 ( s ) 32 s 16 ( ) 8 s 4 ( V 0 s 8 4 0 V 2 0 V s s 4 V t ) 9428 . 0 j 3333 . 1 ( t ) 9428 . 0 j 3333 . 1 ( x 2 x 2 x x 2 x 2 x x x x + + + = + + + + + = + + + = + = + + = + + + + + = + + + v x = V t 3 2 2 sin e 2 6 t 3 2 2 cos e ) t ( u 4 3 / t 4 3 / t 4 Chapter 16, Solution 3. s 5/s 1/2 + V o 1/8 Current division leads to: ) 625 . 0 s ( 16 5 s 16 10 5 s 8 1 2 1 2 1 s 5 8 1 V o + = + = + + = v o (t) = ( ) V ) t ( u e 1 3125 . t 625 . 0 0
Chapter 16, Solution 4. The s-domain form of the circuit is shown below. 6 s 10/s 1/(s + 1) + + V o (s) Using voltage division, + + + = + + + = 1 s 1 10 s 6 s 10 1 s 1 s 10 6 s s 10 ) s ( V 2 o 10 s 6 s C Bs 1 s A ) 10 s 6 s )( 1 s ( 10 ) s ( V 2 2 o + + + + + = + + + = ) 1 s ( C ) s s ( B ) 10 s 6 s ( A 10 2 2 + + + + + + = Equating coefficients : 2 s : -A B B A 0 = → + = 1 s : A 5 - C C A 5 C B A 6 0 = → + = + + = 0 s : -10 C -2, B , 2 A A 5 C A 10 10 = = = → = + = 2 2 2 2 2 o 1 ) 3 s ( 4 1 ) 3 s ( ) 3 s ( 2 1 s 2 10 s 6 s 10 s 2 1 s 2 ) s ( V + + + + + + = + + + + = = ) t ( v o V ) t sin( e 4 ) t cos( e 2 e 2 -3t -3t -t Chapter 16, Solution 5. s 2 2 s 1 + 2 I o s

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( ) ( ) A ) t ( u t 3229 . 1 sin 7559 . 0 e or A ) t ( u e e e 3779 . 0 e e e 3779 . 0 e ) t ( i 3229 . 1 j 5 . 0 s ) 646 . 2 j )( 3229 . 1 j 5 . 1 ( ) 3229 . 1 j 5 . 0 ( 3229 . 1 j 5 . 0 s ) 646 . 2 j )( 3229 . 1 j 5 . 1 ( ) 3229 . 1 j 5 . 0 ( 2 s 1 ) 3229 . 1 j 5 . 0 s )( 3229 . 1 j 5 . 0 s )( 2 s ( s 2 Vs I ) 3229 . 1 j 5 . 0 s )( 3229 . 1 j 5 . 0 s )( 2 s ( s 2 2 s s s 2 2 s 1 2 s 2 1 s 1 1 2 s 1 V t 2 t 3229 . 1 j 2 / t 90 t 3229 . 1 j 2 / t 90 t 2 o 2 2 2 o 2 = + + = + + + + + + + + + = + + + + = = + + + + = + + + = + + + = ° ° Chapter 16, Solution 6. 2 2 s 5 + I o 10/s s Use current division. t 3 sin e 3 5 t 3 cos e 5 ) t ( i 3 ) 1 s ( 5 3 ) 1 s ( ) 1 s ( 5 10 s 2 s s 5 2 s 5 s 10 2 s 2 s I t t o 2 2 2 2 2 o = + + + + + = + + = + + + + =
Chapter 16, Solution 7. The s-domain version of the circuit is shown below. 1/s 1 I x + 2s 1 2 + s Z 2 2 2 s 2 1 1 s 2 s 2 s 2 1 s 2 1 s 2 s 1 ) s 2 ( s 1 1 s 2 // s 1 1 Z + + + = + + = + + = + = ) 5 . 0 s s ( C Bs ) 1 s ( A ) 5 . 0 s s )( 1 s ( 1 s 2 1 s 2 s 2 s 2 1 x 1 s 2 Z V I 2 2 2 2 2 x + + + + + = + + + + = + + + + = = ) 1 s ( C ) s s ( B ) 5 . 0 s s ( A 1 s 2 2 2 2 + + + + + + = + B A 2 : s 2 + = 2 C C 2 C B A 0 : s = → + = + + = -4 B , 6 A 3 0.5A or C A 5 . 0 1 : constant = = → = + = 2 2 2 x 866 . 0 ) 5 . 0 s ( ) 5 . 0 s ( 4 1 s 6 75 . 0 ) 5 . 0 s ( 2 s 4 1 s 6 I + + + + = + + + + = [ ] A ) t ( u t 866 . 0 cos e 4 6 ) t ( i t 5 . 0 x =

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Chapter 16, Solution 8. (a) ) 1 s ( s 1 s 5 . 1 s s 2 2 ) s 2 1 ( s 1 ) s 2 1 //( 1 s 1 Z 2 + + + = + + + = + + = (b) ) 1 s ( s 2 2 s 3 s 3 s 1 1 1 s 1 2 1 Z 1 2 + + + = + + + = 2 s 3 s 3 ) 1 s ( s 2 Z 2 + + + = Chapter 16, Solution 9. (a) The s-domain form of the circuit is shown in Fig. (a). = + + + = + = s 1 s 2 ) s 1 s ( 2 ) s 1 s ( || 2 Z in 1 s 2 s ) 1 s ( 2 2 2 + + + 1 1 2 s 2/s 1/s s 2 (a) (b) (b) The s-domain equivalent circuit is shown in Fig. (b).
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