Chap9_Practice - Problem 9.3 For the circuit shown in Fig....

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Unformatted text preview: Problem 9.3 For the circuit shown in Fig. P9.3, determine (a) the transfer function H = V o / V i , and (b) the frequency o at which H is purely real. R C L 2 L 1 + _ V i + _ V o V 1 Figure P9.3: Circuit for Problem 9.3. Solution: (a) KCL at node V 1 gives: V 1 V i Z C + V 1 Z L 1 + V 1 R + Z L 2 = , where Z C = 1 / j C , Z L 1 = j L 1 , and Z L 2 = j L 2 . Also, voltage division gives V o = V 1 R R + j L 2 . Solving for the transfer function gives H = V o V i = 2 RL 1 C R ( 1 2 L 1 C )+ j ( L 1 + L 2 2 L 1 L 2 C ) . (b) We need to rationalize the expression for H : H = 2 RL 1 C R ( 1 2 L 1 C )+ j ( L 1 + L 2 2 L 1 L 2 C ) R ( 1 2 L 1 C ) j ( L 1 + L 2 2 L 1 L 2 C ) R ( 1 2 L 1 C ) j ( L 1 + L 2 2 L 1 L 2 C ) = 2 R 2 L 1 C ( 1 2 L 1 C )+ j 3 RL 1 C ( L 1 + L 2 2 L 1 L 2 C ) R 2 ( 1 2 L 1 C ) 2 + 2 ( L 1 + L 2 2 L 1 L 2 C ) 2 . The imaginary part of H is zero if = 0 (trivial solution) or if = radicalbigg L 1 + L 2 L 1 L 2 C . Problem 9.13 Generate Bode magnitude and phase plots for the following voltage transfer functions: (a) H ( ) = j 100 10 + j (b) H ( ) = . 4 ( 50 + j ) 2 ( j ) 2 (c) H ( ) = ( 40 + j 80 ) ( 10 + j 50 ) (d) H ( ) = ( 20 + j 5 )( 20 + j ) j (e) H ( ) = 30 ( 10 + j ) ( 200 + j 2 )( 1000 + j 2 ) (f) H ( ) = j 100 ( 100 + j 5 )( 100 + j ) 2 (g) H ( ) = ( 200 + j 2 ) ( 50 + j 5 )( 1000 + j ) Solution: (a) H ( ) = j 100 10 + j = j 100 10 ( 1 + j / 10 ) = j 10 1 + j / 10 . Constant factor 10 = + 20 dB Zero @ origin Simple pole with c = 10 rad/s M [ dB ] = 20log | H | = 20log10 + 20log 20log | 1 + j / 10 | Magnitude Phase (rad/s) () - 90 o 0.1 1 10 100 o 90 o j 1 1 + j /10 (rad/s)- 20 0.1 1 10 100 20 40 dB M [dB] 20 log 20 log |1 + j /10| 20 log 10 = 20 dB (b) H ( ) = . 4 ( 50 + j ) 2 ( j ) 2 = . 4 2500 ( 1 + j / 50 ) 2 2 = 1000 ( 1 + j / 50 ) 2 2 . M [ dB ] = 20log | H ( ) | = 20log1000 + 40log | 1 + j / 50 | 40log . Line starts at 60 dB at = 1 rad/s, and has slope of 40 dB/decade Constant factor 1000 = 60 dB Pole @ origin of order 2 Simple zero with c = 50 rad/s, of order 2 (rad/s) ()- 180 o- 180 o- 20- 40- 40 log - 60- 8 dB 1 10 50 100 1000 Magnitude 20 40 60 M [dB] dB 40 log |1 + j /50| 20 log 1000 = 60 dB (rad/s) 1 10 100...
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This note was uploaded on 02/21/2011 for the course MSE 231 taught by Professor Bedford during the Spring '10 term at University of Michigan.

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Chap9_Practice - Problem 9.3 For the circuit shown in Fig....

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