Hw-9-10 - 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 # 1 DSP I Student Name: Sriram Yarlagadda Student Number: 2196944 Due Date: GRADE: PROBLEM-1 Required: To obtain a BPF which meet the given specifications Solution %Problem 1-Home work # 9 clc clear all ; close all ; % given initial conditions Fph=460*10^3; %Fph=460 KHz Fsh=470*10^3; %Fsh=470 KHz Fpl=450*10^3; %Fpl=450 KHz Fsl=440*10^3; %Fsl=440 KHz Rp=1; % Maximum pass band deviastion in dB alphas=60; %Maximum stop band attenuation in dB %chosing the sampling frequency %we have the center frequency in the pass band as 455KHz %selecting sampling frequency to be 5 times of center frequency Fs=5*(Fpl+Fph)/2; %Fs=5*455KHz=2275 KHz %normalizing all the digital frequencies Wph=2*pi*Fph/Fs; %Wph=1.27044625991323 rad/sample Wsh=2*pi*Fsh/Fs; %Wsh=1.29806465686787 rad/sample Wpl=2*pi*Fpl/Fs; %Wpl=1.2428278629586 rad/sample Wsl=2*pi*Fsl/Fs; %Wsl=1.21520946600396 rad/sample %Converting the digital frequencies in to analog frequencies %applying bilinear (prewrap) technique for conversion Omegph=tan(Wph/2); %Omegph=0.737145173427723 rad/sec Omegsh=tan(Wsh/2); %Omegsh=0.758678621716582 rad/sec Omegpl=tan(Wpl/2); %Omegpl=0.716045729309657 rad/sec Omegsl=tan(Wsl/2); %Omegsl=0.695359489039528 rad/sec %Calculating Omeg0 Omeg0=sqrt(Omegph*Omegpl); %Omeg0=0.726518859572239 rad/sec %Converting the digital band pass filter in to proto type analog low pass %filter PBW=Omegph-Omegpl; %PBW--pass band width %PBW=0.0210994441180664 rad/sec %Omegs -stop band frequency of low pass filter %Omegp -pass band frequency of low pass filter Omegs=-(-1*(Omegsh^2)+(Omeg0^2))/(Omegsh*PBW); Omegp=-(-1*(Omegph^2)+(Omeg0^2))/(Omegph*PBW); %Omegs=2.9837894352411 %Omegp=1.00000000000001 %Calculating the filter order E=sqrt((10^(Rp/10))-1); %E=0.508847139909588 A=10^(alphas/20); %A=1000 K=Omegs/Omegp; %K=2.98378943524108 %1/K=0.335144292753756 K1=sqrt(A^2-1)/E; %K1=1965.22574574666 %1/K1=0.000508847394333348 N=log10(K1)/log10(K); N=ceil(N); %order=N=7 %Calculating the cut off frequency so that pass band specifications are met Omegc=Omegs/((10^(alphas/10)-1)^(1/(2*N))); %Omegc=1.11223555560011 rad/sec %obtaining the transfer function %calculating the poles of butterworth %z-- zeros, p--poles , k--normalization constant [z,p,k]=buttap(N); %poles p= % -0.222520933956314 + 0.974927912181824i % -0.222520933956314 - 0.974927912181824i % -0.623489801858733 + 0.78183148246803i % -0.623489801858733 - 0.78183148246803i % -0.900968867902419 + 0.433883739117558i % -0.900968867902419 - 0.433883739117558i % -1 %plotting the poles figure(1) plot(real(p),imag(p), '*' ) title( 'Pole locations' ) xlabel( 'real axis' ) ylabel( 'imaginary axis' ) % Above obtained is a normalized transfer function %transforming the normalized transfer function to unnormalized one %obtaining the magnitude and phase response for the Butterworth filter l1=length(p); a=0.01; b=ceil(Omegs)+1;...
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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-9-10 - The University of Akron Engineering Faculty...

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