chap16

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

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Unformatted text preview: 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 . ( sin e 155 . 1-0.5t Chapter 16, Solution 2. 8/s s s 4 2 + − + V x − 4 V ) t ( u ) e 2 e 2 4 ( v 3 8 j 3 4 s 125 . 3 8 j 3 4 s 125 . s 25 . 16 ) 8 s 8 s 3 ( s 2 s 16 V s 32 s 16 ) 8 s 8 s 3 ( V V s V ) s 4 s 2 ( s ) 32 s 16 ( ) 8 s 4 ( V s 8 4 V 2 V s s 4 V t ) 9428 . j 3333 . 1 ( t ) 9428 . 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 . 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 . − − 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 = →  + = 1 s : A 5- C C A 5 C B A 6 = →  + = + + = 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 ( ) ( ) A ) t ( u t 3229 . 1 sin 7559 . e or A ) t ( u e e e 3779 . e e e 3779 . e ) t ( i 3229 . 1 j 5 . s ) 646 . 2 j )( 3229 . 1 j 5 . 1 ( ) 3229 . 1 j 5 . ( 3229 . 1 j 5 . s ) 646 . 2 j )( 3229 . 1 j 5 . 1 ( ) 3229 . 1 j 5 . ( 2 s 1 ) 3229 . 1 j 5 . s )( 3229 . 1 j 5 . s )( 2 s ( s 2 Vs I ) 3229 . 1 j 5 . s )( 3229 . 1 j 5 . 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....
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chap16 - Chapter 16 Solution 1 Consider the s-domain form...

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