Chap6StudentSolutions

Chap6StudentSolutions - M J Roberts Chapter 6 Fourier...

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M. J. Roberts - 8/16/04 Solutions 6-1 Chapter 6 - Fourier Transform Analysis of Signals and Systems Selected Solutions (In this solution manual, the symbol, , is used for periodic convolution because the preferred symbol which appears in the text is not in the font selection of the word processor used to create this manual.) 1. A system has an impulse response, h u LP t t e t ( ) = ( ) 3 10 , and another system has an impulse response, h u HP t t t e t ( ) = ( ) ( ) δ 3 10 . (a) Sketch the magnitude and phase of the transfer function of these two systems in a parallel connection. H , H LP HP j j j j ω ω ω ω ( ) = + ( ) = + 3 10 1 3 10 H P j j j ω ω ω ( ) = + + + = 3 10 1 3 10 1 ω -40 40 |H P ( j ω )| 1 ω -40 40 Phase of H P ( j ω ) - π π (b) Sketch the magnitude and phase of the transfer function of these two systems in a cascade connection. 2. Below are some pairs of signals, x t ( ) and y t ( ) . In each case decide whether or not y t ( ) is a distorted version of x t ( ) .
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M. J. Roberts - 8/16/04 Solutions 6-2 t 2 x( t ) -2 2 (a) t 2 y( t ) -2 2 t 2 x( t ) -2 2 (b) t 2 y( t ) -2 2 (a) Inverted and amplified, undistorted (b) Time shifted, undistorted t 2 x( t ) -2 2 (c) t 2 y( t ) -2 2 t 2 x( t ) -1 1 (d) t 2 y( t ) -1 1 (c) Clipped at a negative value, distorted (d) Time compressed, distorted t 2 x( t ) -2 2 (e) t 2 y( t ) -2 2 t 2 x( t ) -2 2 (f) t 2 y( t ) -2 2
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M. J. Roberts - 8/16/04 Solutions 6-3 (e) Constant added, distorted (f) Log-amplified, distorted 3. Classify each of these transfer functions as having a lowpass, highpass, bandpass or bandstop frequency response. f -10 10 |H( f )| 1 (a) F -2 2 |H( F )| 1 (b) -4 π 4 π |H( j )| 1 (c) ω -100 100 |H( j ω )| 1 (d) (a) Lowpass (b) Bandpass (c) Lowpass (d) Bandpass F -2 2 |H( F )| 1 (e) -4 π 4 π |H( j )| 1 (f) (e) Highpass (f) Bandstop 4. Classify each of these transfer functions as having a lowpass, highpass, bandpass or bandstop frequency response. (a) H rect f f ( ) = 1 100 10 (b) H rect comb F F F ( ) = ( ) ( ) 10 (c) H rect rect comb j ( ) = + + 20 4 20 4 2 π π π π π Bandpass 5. A system has an impulse response, h rect . . t t ( ) = 10 0 01 0 02 .
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M. J. Roberts - 8/16/04 Solutions 6-4 What is its null bandwidth? 6. A system has an impulse response, h u n n n [ ] = [ ] 7 8 . What is its half-power DT-frequency bandwidth? Using α α n j n e u [ ] ← → F 1 1 the transfer function is H j e e j j ( ) = = 1 1 7 8 8 8 7 . This is a DT lowpass filter. Its maximum transfer function magnitude occurs at Ω = 0. The -3 dB point must be the first frequency at which the square of the magnitude of the transfer function is one-half of its maximum value (the “half-power” bandwidth). The low-frequency gain is H 0 8 ( ) = The -3 dB point occurs where H j dB ( ) = = 3 2 2 8 2 32 .
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