# Hw-8 - The University of Akron Engineering Faculty...

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Unformatted text preview: The University of Akron Engineering Faculty Electrical & Computer Department Digital Signal Processing I Dr. Roberto J. Acosta Course # 4400:646 Homework # 8 DSP I Student Name: Sriram Yarlagadda Student Number: 2196944 Due Date GRADE: Jan 30 2009 PROBLEM-1 Given: Required: Solution: For Analog system the time domain impulse response and Frequency response (amplitude and phase) are obtained as follows using a MATLAB program Time Domain impulse Response %Problem -1a %Given analog filter Transfer function Ha(s)=OmegC/(s+OmegC) %Required --to obtain the time domian impulse response %Applying the inverse laplace transforms to obtain the %time domain impulse response %on Applying inverse laplace transforms %time domain impulse response is Ha(t)=OmegC*exp(-OmegC*T) clc; clear all ; close all ; OmegC=10; %radians/sec N1=5; Ha=zeros(N1,1); x=1; for t=0:0.1:N1 Ha(x)=OmegC*exp(-OmegC*t); x=x+1; end figure(1) plot(Ha) title( 'Impulse response of analog filter' ); xlabel( 'Magnitude of impulse response' ) ylabel( 'time' ) Frequency Response %obtaining the frequency response for the analog filter %replace S-->jw w--rad/sec N2=1000; num=OmegC; mag1=zeros(N2,1); phase1=zeros(N2,1); for w=1:1:N2 den=i*w+OmegC; mag1(w)=20*log10(abs(num/den)); phase1(w)=angle(num/den); end figure(2) subplot(2,1,1) plot(mag1) title( 'magnitude response of the analog filter' ) xlabel( 'Frequency --rad/sec' ) ylabel( 'Magnitude in dB' ) subplot(2,1,2) plot(phase1) title( 'Phase response of an analog filter' ) xlabel( 'Frequency --rad/sec' ) ylabel( 'Phase---angle/dec' ) For Digital system the time domain impulse response and Frequency response (amplitude and phase) are obtained as follows using a MATLAB program Time Domain Impulse Response %obtaining the time domain impulse response of the digital filter %the difference equation obtained is %y[n]=alpha*y[n-1]+a*(x[n]+x[n-1]) N3=100; alpha =(1-OmegC)/(1+OmegC); a=(1-alpha)/2; x=zeros(N3,1); y=zeros(N3,1); %x[n] is an impuls input x(1)=1; y(1)=a*x(1); for n=2:1:N3 y(n)=alpha*y(n-1)+a*(x(n)+x(n-1)); end figure(3) plot(y),grid title( 'Impulse response of the digital filter' ) xlabel( 'Magnitude of impulse response' ) ylabel( 'no of samples(time)--n' ) Frequency Response %obtaining the frequency response of the digital filter %replace z-->exp(jw) w--rad/sec w=-2*pi:0.01:2*pi; num1=a.*(1+exp(-i.*w)); den1=1-alpha.*exp(-i.*w); figure(4) subplot(2,1,1) plot(w,20*log10(abs(num1./den1))),grid title( 'magnitude response of the digital filter' ) xlabel( 'Frequency --rad/sec' ) ylabel( 'Magnitude in dB' ) subplot(2,1,2) plot(w,angle(num1./den1)),grid title( 'Phase response of an digital filter' ) xlabel( 'Frequency --rad/sec' ) ylabel( 'Phase---angle/dec' ) Comparison of Amplitude Response: In Analog system the amplitude response Magnitude goes to -40 dB finally. Similarly in digital System the amplitude response has a peak reaching the -40dB....
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## This note was uploaded on 02/20/2011 for the course EE EE 646 taught by Professor Dr.robertojacousta during the Spring '10 term at The University of Akron.

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Hw-8 - The University of Akron Engineering Faculty...

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