Chap6StudentSolutions

# J roberts 81604 solve for the transfer function 2

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Unformatted text preview: R 2 = 10 kΩ - - Similar to (a). (d) Excitation, ii ( t) - Response, v R1 ( t) R 2 = 10 k C1 = 1 µF i i(t) + C2 = 1 µF vR1(t) R 1 = 10 kΩ - (e) Excitation, v i ( t) - Response, v RL ( t) R 2 = 10 kΩ + + C1 = 1 µF vi (t) R 1 = 10 kΩ C2 = 1 µF vRL(t) RL = 1 kΩ - - Write two nodal equations and solve for the transfer function. Summing currents to zero at the middle node and the right-hand node, VR1 ( jω )[ jωC1 + jωC2 + G1 ] − Vi ( jω ) jωC1 − VRL ( jω ) jωC2 = 0 VRL ( jω )[ jωC2 + GL + G2 ] − Vi ( jω )G2 − VR1 ( jω ) jωC2 = 0 Solutions 6-20 M. J. Roberts - 8/16/04 Solve for the transfer function, −ω 2 R1R2C1C2 + jωR1 (C1 + C2 ) + 1 H( jω ) = R + R2 R −ω 2 R1R2C1C2 + jω (C1 + C2 )1 + 2 R1 + R2C2 + L RL RL |H( jω )| 1 ω -25000 25000 Phase of H( jω ) π ω -25000 25000 -π 27. Find and sketch versus frequency the magnitude and phase of the input impedance, V(f) V (f) Z in ( f ) = i and transfer function, H( f ) = o , for each of these filters. Ii ( f ) Vi ( f ) + i i(t) 1 µF vi (t) (a) - + vo(t) - i i(t) 100 Ω 50 mH vi (t) (b) 1 kΩ + - 10 nF + vo(t) - 28. The signal, x(t), in Exercise 23 is the excitation of an RC lowpass filter with R = 1kΩ and C = 0.3 µF. Sketch the excitation and response voltages versus time on the same scale. From Exercise 23, x( t) = 500 rect (1000 t) ∗ comb(500 t) Solutions 6-21 M. J. Roberts - 8/16/04 1∞ n X( f ) = ∑ sinc δ ( f − 500 n ) 2 2 n =−∞ The transfer function is H( f ) = 1 j 2πfRC + 1 Therefore the output is ∞ 1 1 n Y( f ) = ∑ sinc 2 δ ( f − 500n) 2 j 2πfRC + 1 n =−∞ 1∞ 1 n δ ( f − 500 n ) Y( f ) = ∑ sinc 2 j1000πnRC + 1 2 n =−∞ Converting to the time domain, n sinc ∞ 2 1 y( t) = ∑ e j1000πnt 2 n =−∞ j1000πnRC + 1 or ∞ e j1000πnt e − j1000πnt 1 n 1 + ∑ sinc y( t) = + 2 j1000πnRC + 1 − j1000πnRC + 1 2 n =1 − j1000πnt ∞ − e j1000πnt ) + e j1000πnt + e − j1000πnt 1 n j1000πnRC (e y( t) = 1 + ∑ sinc 2 2 2 n =1 (1000πnRC) + 1 ∞ 1 n 2000πnRC sin(1000πnt) + 2 cos(1000πnt) y( t) = 1 + ∑ sinc 2 2 2 n =1 (1000πnRC) + 1 1 -2 ms 2 ms t 29. Dra...
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## This note was uploaded on 06/19/2013 for the course ENSC 380 taught by Professor Atousa during the Spring '09 term at Simon Fraser.

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