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Chap9_Practice

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

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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 ϖ 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 = 0 , where Z C = 1 / j ϖ C , Z L 1 = j ϖ L 1 , and Z L 2 = j ϖ L 2 . Also, voltage division gives V o = V 1 R R + j ϖ L 2 . Solving for the transfer function gives H = V o V i = ϖ 2 RL 1 C R ( 1 ϖ 2 L 1 C )+ j ϖ ( L 1 + L 2 ϖ 2 L 1 L 2 C ) . (b) We need to rationalize the expression for H : H = ϖ 2 RL 1 C R ( 1 ϖ 2 L 1 C )+ j ϖ ( L 1 + L 2 ϖ 2 L 1 L 2 C ) × R ( 1 ϖ 2 L 1 C ) j ϖ ( L 1 + L 2 ϖ 2 L 1 L 2 C ) R ( 1 ϖ 2 L 1 C ) j ϖ ( L 1 + L 2 ϖ 2 L 1 L 2 C ) = ϖ 2 R 2 L 1 C ( 1 ϖ 2 L 1 C )+ j ϖ 3 RL 1 C ( L 1 + L 2 ϖ 2 L 1 L 2 C ) R 2 ( 1 ϖ 2 L 1 C ) 2 + ϖ 2 ( L 1 + L 2 ϖ 2 L 1 L 2 C ) 2 . The imaginary part of H is zero if ϖ = 0 (trivial solution) or if ϖ 0 = radicalbigg L 1 + L 2 L 1 L 2 C .

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Problem 9.13 Generate Bode magnitude and phase plots for the following voltage transfer functions: (a) H ( ϖ ) = j 100 ϖ 10 + j ϖ (b) H ( ϖ ) = 0 . 4 ( 50 + j ϖ ) 2 ( j ϖ ) 2 (c) H ( ϖ ) = ( 40 + j 80 ϖ ) ( 10 + j 50 ϖ ) (d) H ( ϖ ) = ( 20 + j 5 ϖ )( 20 + j ϖ ) j ϖ (e) H ( ϖ ) = 30 ( 10 + j ϖ ) ( 200 + j 2 ϖ )( 1000 + j 2 ϖ ) (f) H ( ϖ ) = j 100 ϖ ( 100 + j 5 ϖ )( 100 + j ϖ ) 2 (g) H ( ϖ ) = ( 200 + j 2 ϖ ) ( 50 + j 5 ϖ )( 1000 + j ϖ ) Solution: (a) H ( ϖ ) = j 100 ϖ 10 + j ϖ = j 100 ϖ 10 ( 1 + j ϖ / 10 ) = j 10 ϖ 1 + j ϖ / 10 . Constant factor 10 = + 20 dB Zero @ origin Simple pole with ϖ c = 10 rad/s M [ dB ] = 20log | H | = 20log10 + 20log ϖ 20log | 1 + j ϖ / 10 |
Magnitude Phase ω (rad/s) φ(ω) φ - 90 o 0.1 1 10 100 0 o 90 o j 1 1 + j ω/10 ω (rad/s) - 20 0 0.1 1 10 100 20 40 dB M [dB] 20 log ω 20 log |1 + j ω/10| 20 log 10 = 20 dB (b) H ( ϖ ) = 0 . 4 ( 50 + j ϖ ) 2 ( j ϖ ) 2 = 0 . 4 × 2500 ( 1 + j ϖ / 50 ) 2 ϖ 2 = 1000 ( 1 + j ϖ / 50 ) 2 ϖ 2 . M [ dB ] = 20log | H ( ϖ ) | = 20log1000 + 40log | 1 + j ϖ / 50 |− 40log ϖ . Line starts at 60 dB at ϖ = 1 rad/s, and has slope of 40 dB/decade Constant factor 1000 = 60 dB Pole @ origin of order 2 Simple zero with ϖ c = 50 rad/s, of order 2

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ω (rad/s) φ φ(ω) - 180 o - 180 o - 20 - 40 - 40 log ω - 60 - 8 dB 0 1 10 50 100 1000 Magnitude 20 40 60 M [dB] dB 40 log |1 + j ω/50| 20 log 1000 = 60 dB ω (rad/s) 0 1 10 100 1000 Phase (1 + j ω /50) - 180 o (c) H ( ϖ ) = 40 + j 80 ϖ 10 + j 50 ϖ = 40 ( 1 + j 2 ϖ ) 10 ( 1 + j 5 ϖ ) = 4 ( 1 + j ϖ / 0 . 5 ) ( 1 + j ϖ / 0 . 2 ) . Constant factor 4 = 12 dB Simple pole with ϖ c = 0 . 2 rad/s Simple zero with ϖ c = 0 . 5 rad/s M [ dB ] = 20log | H ( ϖ ) | = 20log4 + 20log | 1 + j ϖ / 0 . 5 |− 20log | 1 + j ϖ / 0 . 2 |
0 ω (rad/s) φ(ω) 0.01 0.1 1 10 - 90 o - 45 o - 5 - 10 90 o 45 o Magnitude Phase ω (rad/s) 15 20 12 10 5 0 0.01 0.1

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

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