Challenge_20-1 - 108.5 C=1.58 71.5 D=0.707 45 At test...

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EEL 3135: Discrete-Time Signals and Systems Dr. Fred J. Taylor, Professor Lesson Title: Convolution Theorem Lesson Number: 20 [Sections 7.6 and 7.7] Challenge A filter is clocked at a rate f s =1k Sa/s and has the pole-zero distribution shown in Figure 10 and is known to have DC gain of 0dB. What is the filter’s magnitude frequency and phase response at f=0.25kHz? Pole-zero distribution. Response The transfer function is given by: H(z) = k (z+0.5+j0.5) (z+0.5-j0.5)/ ((z- 0.5+j0.5) (z- 0.5-j0.5)) =k(z 2 +z+0.5)/(z 2 -z+0.5) H(z=1)=1=k(2.5)/(0.5)=5k. Therefore k=0.2. A=0.707 135 ° B=1.58
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Unformatted text preview: 108.5 C=1.58 71.5 D=0.707 45 At test frequency is at = /2: H(z=j)=0.2CD)/(AB) = 0.2 (1 -127 ) Vrification: b=[.2, .2, .2*.5]; a=[1, -1, 0.5]; x=[1,zeros(1,100)]; y=filter(b,a,x); 1 X 0.5+j0.5 X 0.5-j0.5 0 -0.5+j0.5 0 -0.5-j0.5 X 0.5+j0.5 X 0.5-j0.5 0 -0.5+j0.5 0 -0.5-j0.5 A B C D EEL 3135: Discrete-Time Signals and Systems Dr. Fred J. Taylor, Professor yf=fft(y); plot(abs(yf(1,1:100))) plot(360*angle(yf(1,1:100))/(2*pi)) Magnitude Frequency Response Phase Response 2 1.0 0.2 0.0 -127...
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This note was uploaded on 04/15/2010 for the course EEL 3135 taught by Professor ? during the Spring '08 term at University of Florida.

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Challenge_20-1 - 108.5 C=1.58 71.5 D=0.707 45 At test...

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