Chap6StudentSolutions

One way is to think of this circuit as two voltage

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Unformatted text preview: ∗ comb( F )]e − j 20πF (b) H( F ) = j sin(2πF ) (d) H( jΩ) = H( F ) = 1 − e − j 4 πF (c) 8e jΩ 8 − 5e − jΩ Similar to Exercise 24, except for discrete time. 26. Find and sketch the frequency response of each of these circuits given the indicated excitation and response. (a) Excitation, v i ( t) - Response, vC 2 ( t) R1 = 1 kΩ R 2 = 10 kΩ + vi (t) + C1 = 1 µF C2 = 0.1 µF - v 2(t) C - The transfer function can be found in multiple ways. One way is to think of this circuit as two voltage dividers. The first voltage division is from the excitation, vi ( t ) , to the voltage across the first capacitor. The second voltage division is from that voltage to the response voltage, vC 2 ( t ) . Solutions 6-18 M. J. Roberts - 8/16/04 H ( jω ) = 1 Zπ ( jω ) 1 jω C2 = 1 R1 + Zπ ( jω ) jω R2C2 + 1 R2 + jω C2 VC 2 ( jω ) Zπ ( jω ) = Vi ( jω ) R1 + Zπ ( jω ) first voltage division second voltage division 1 1 R2 + jω C jω C1 2 Zπ ( jω ) = 1 1 + R2 + jω C1 jω C2 Substitute Zπ ( jω ) into the expression for the transfer function and simplify. H( jω ) = (b) 1 1 − ω R1R2C1C2 + jω (C1 + C2 ) R1 + R2C2 [ 2 Excitation, v i ( t) - Response, iC1 ( t) R1 = 1 kΩ R 2 = 10 kΩ + iC1(t) vi (t) C1 = 1 µF C2 = 0.1 µF Think of the transfer function as the transfer function from the excitation to the current in R1 times the transfer function from the current in R1 to iC1 ( t ) . 1 R2 + I ( jω ) IR1 ( jω ) 1 jω R2C2 + 1 jω C2 H ( jω ) = C1 = = 1 1 Vi ( jω ) Vi ( jω ) Zi ( jω ) jω R C + 1 + C2 + R2 + 22 jω C2 input impedance C1 first transfer function jω C1 second transfer function Zi ( jω ) = R1 + Zπ ( jω ) = R1 + second transfer function jωR2C2 + 1 jω (C1 + C2 ) − ω 2 R2C1C2 Combine expressions and simplify to yield H( jω ) = (c) jωC1 ( jωR2C2 + 1) 1 − ω R1R2C1C2 + jω R1 (C1 + C2 ) + R2C2 [ 2 Excitation, v i ( t) - Response, v R 2 ( t) Solutions 6-19 M. J. Roberts - 8/16/04 C1 = 1 µF C2 = 1 µF + + vi (t) R 1 = 10 kΩ vR2(t)...
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