Step and Impulse Response using Halijak Substitution method

# Step and Impulse Response using Halijak Substitution method...

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1. Step and Impulse Response using Halijak Substitution method clc; clear all ; close all ; syms z ; T=0.0094; a1=-100*T; b1=200*T+3333.33*T^2+666665*T^3; c1=-100*T-3333.33*T^2+666665*T^3; a2=2+370.666*T; b2=-6-741.332*T+24266.6*T^2+693333*T^3; c2=6+370.666*T-24266.6*T^2+693333*T^3; Nr=[a1 b1 c1 0]; Dr=[a2 b2 c2 -2]; sys=tf(Nr,Dr,0.0094) nr=[-50 33.3333*50 13333.3*50]; dr=[1 185.333 12133.3 693333]; x=ones(150,1); y=zeros(150,1); y(1)=-0.171*x(1); y(2)=1.869*y(1)-0.171*x(2)+0.4974*x(1); y(3)=1.869*y(2)-1.443*y(1)-0.171*x(3)+0.4974*x(2)-0.1241*x(1); for n=4:150 y(n)=1.869*y(n-1)-1.443*y(n-2)+0.365*y(n-3)-0.171*x(n)+0.4974*x(n-1)-0.1241*x(n-2); end figure() plot(y) title( 'Step response using Halijak Substitution method' ); xlabel( 'No of samples' ) ylabel( 'y(n)' ) % figure(); % DSTEP(Nr,Dr); y=zeros(150,1); p=zeros(150,1); p(1)=1;

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0 50 100 150 -0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 Step response using Halijak Substitution method No of samples y(n) y(1)=-0.171*p(1); y(2)=1.869*y(1)-0.171*p(2)+0.4974*p(1);
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Step and Impulse Response using Halijak Substitution method...

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