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

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EECS 20N: Structure and Interpretation of Signals and Systems Department of EECS U NIVERSITY OF C ALIFORNIA B ERKELEY 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 [ δ ( n ) δ ( 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 π ω < π . Is this filter low-pass, band-pass, or high-pass? Explain. Solution G ( ω ) = summationdisplay n g ( n ) e iωn = summationdisplay n parenleftbigg 1 2 [ δ ( n ) δ ( n 1)] parenrightbigg e iωn = 1 2 parenleftBiggparenleftBigg summationdisplay n δ ( n ) e iωn parenrightBigg parenleftBigg summationdisplay n δ ( n 1) e iωn parenrightBiggparenrightBigg = 1 2 (1 e ) = 1 2 e i ω 2 ( e i ω 2 e i ω 2 ) = 1 2 e i ω 2 parenleftBig 2 i sin parenleftBig ω 2 parenrightBigparenrightBig = e i ω 2 parenleftBig e i π 2 sin parenleftBig ω 2 parenrightBigparenrightBig = e i ( π/ 2 ω/ 2) sin parenleftBig ω 2 parenrightBig Hence, for π ω < π , | G ( ω ) | = | sin( ω 2 ) | and G ( ω ) = braceleftBigg π 2 ω 2 if ω 0 π 2 ω 2 elsewhere. . 0 π 2 π π 2 π 1 (a) | G ( ω ) | . 0 π π π 2 π 4 π 4 π 2 (b) G ( ω ) . Figure 1: Plots for G ( ω ) . Since the magnitude is highest when ω equals π , and zero when ω equals 0, this is a high-pass filter. 1
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(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 ) 0 . 5 (ii) Determine the filter’s impulse response h : Z R . Solution The impulse response of the filter can be determined by setting x ( n ) to δ ( n ) : h ( n ) = 1 2 [ δ ( n ) δ ( 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 π ω < π . Is this filter low-pass, band-pass, or high- pass? Explain. Solution H ( ω ) = summationdisplay n g ( n ) e iωn = summationdisplay n parenleftbigg 1 2 [ δ ( n ) δ ( n 2)] parenrightbigg e iωn = 1 2 parenleftBiggparenleftBigg summationdisplay n δ ( n ) e iωn parenrightBigg parenleftBigg summationdisplay n δ ( n 2) e iωn parenrightBiggparenrightBigg = 1 2 (1 e i 2 ω ) = 1 2 e ( e e ) = 1 2 e (2 i sin( ω )) = e ( e i π 2 sin( ω )) = e i ( π/ 2 ω ) sin( ω ) Hence, π ω < π , | H ( ω ) | = | sin( ω ) | and H ( ω ) = braceleftBigg π 2 ω if ω 0 π 2 ω elsewhere.
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