Chap6StudentSolutions

# J roberts 81604 r 1 k ict c 1 f vi t in

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Unformatted text preview: onse, iC ( t) Solutions 6-5 M. J. Roberts - 8/16/04 R = 1 kΩ + iC(t) C = 1 µF vi (t) - In this case the transfer function is the reciprocal of the input impedance. (c) Excitation, v i ( t) - Response, v R ( t) + vR(t) - R = 1 kΩ + vi (t) C = 1 µF L = 1 mH (d) Excitation, ii ( t) - Response, v R ( t) + ii (t) R = 100 Ω vR(t) L = 1 mH C = 1 µF Divide the excitation current between the two branches and multiply the current in the right branch by R to get the response voltage. Then solve for the ratio of the response voltage to the excitation current. 10. Classify each of these transfer functions as having a lowpass, highpass, bandpass or bandstop frequency response. 1 1 + jf (a) H( f ) = (c) H( jω ) = − (b) j10ω 100 − ω 2 + j10ω Solutions 6-6 H( f ) = jf 1 + jf M. J. Roberts - 8/16/04 (d) H( F ) = sin( 3πF ) sin(πF ) This case is not as “pure” as the previous ones. It is generally lowpass because the transfer function magnitude at lower frequencies is generally greater than at high frequencies. But there are nulls in the transfer function that make it look somewhat like a bandstop filter or a multiple bandstop filter. (e) H( jΩ) = j[sin(Ω) + sin(2Ω)] This case is also not perfectly clear. The response at zero frequency is zero and the response at Ω = π is also zero. These criteria fit a bandpass filter. But the response 2π is also zero at Ω = . So it might again look like a bandstop in some ways. 3 11. Plot the magnitude frequency responses, both on a linear-magnitude and on a logmagnitude scale, of the systems with these transfer functions, over the frequency range specified. 20 (a) H( f ) = , −100 < f < 100 22 20 − 4π f + j 42πf (b) 2 × 10 5 H( jω ) = , −500 < ω < 500 (100 + jω )(1700 − ω 2 + j 20ω ) 12. Draw asymptotic and exact magnitude and phase Bode diagrams for the frequency responses of the following circuits and systems. (a) An RC lowpass filter with R = 1 MΩ and C = 0.1 µF . 1 1 1 1 jωC H( jω ) = = = = 6 −7 1 + R jωRC + 1 jω10 10 + 1 j 0.1ω + 1 jωC Solutions 6-7 M. J....
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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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