hw06-f06-soln - EECS 20N: Structure and Interpretation of...

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Unformatted text preview: EECS 20N: Structure and Interpretation of Signals and Systems Department of EECS UNIVERSITY OF CALIFORNIA BERKELEY Problem Set 6 Issued: 30 November 2006 Due: 5 December 2006, 5pm HW6.1 (Discrete-Time FIR Filter Frequency Response) In this problem you will explore, among other things, the relationship between time scaling and frequency scaling. (a) A discrete-time LTI filter having impulse response g : Z R produces a two-point difference of the input, as described below: n Z , g ( n ) = 1 2 [ ( n ) ( n 1)] . Determine the frequency response G : R C of this filter. Sketch the mag- nitude response | G | and the phase response G over the frequency range < . Is this filter low-pass, band-pass, or high-pass? Explain. Solution G ( ) = summationdisplay n g ( n ) e in = summationdisplay n parenleftbigg 1 2 [ ( n ) ( n 1)] parenrightbigg e in = 1 2 parenleftBiggparenleftBigg summationdisplay n ( n ) e in parenrightBigg parenleftBigg summationdisplay n ( n 1) e in parenrightBiggparenrightBigg = 1 2 (1 e i ) = 1 2 e i 2 ( e i 2 e i 2 ) = 1 2 e i 2 parenleftBig 2 i sin parenleftBig 2 parenrightBigparenrightBig = e i 2 parenleftBig e i 2 sin parenleftBig 2 parenrightBigparenrightBig = e i ( / 2 / 2) sin parenleftBig 2 parenrightBig Hence, for < , | G ( ) | = | sin( 2 ) | and G ( ) = braceleftBigg 2 2 if 2 2 elsewhere. . 2 2 1 (a) | G ( ) | . 2 4 4 2 (b) G ( ) . Figure 1: Plots for G ( ) . Since the magnitude is highest when equals , and zero when equals 0, this is a high-pass filter. 1 (b) A discrete-time LTI filter is described by the following linear, constant-coefficient difference equation: y ( n ) = 1 2 [ x ( n ) x ( n 2)] . (i) Draw a delay-adder-gain block diagram representation (i.e., a signal flow graph implementation) of this filter. Solution The following figure shows the diagram: z- 2 +- + x ( n ) y ( n ) . 5 (ii) Determine the filters impulse response h : Z R . Solution The impulse response of the filter can be determined by setting x ( n ) to ( n ) : h ( n ) = 1 2 [ ( n ) ( n 2)] (iii) Determine the frequency response H : R C of this filter filter. Sketch the magnitude response | H | and the phase response H over the fre- quency range < . Is this filter low-pass, band-pass, or high- pass? Explain. Solution H ( ) = summationdisplay n g ( n ) e in = summationdisplay n parenleftbigg 1 2 [ ( n ) ( n 2)] parenrightbigg e in = 1 2 parenleftBiggparenleftBigg summationdisplay n ( n ) e in parenrightBigg parenleftBigg summationdisplay n ( n 2) e in parenrightBiggparenrightBigg = 1 2 (1 e i 2 ) = 1 2 e i ( e i e i ) = 1...
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This homework help was uploaded on 04/02/2008 for the course EE 20N taught by Professor Ayazifar during the Fall '08 term at University of California, Berkeley.

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hw06-f06-soln - EECS 20N: Structure and Interpretation of...

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