ex9_16 - num2 =[1 0.9/0.9 num = conv(num1,num2 y2 =...

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% Example 9.16 % third order response % shows dominant poles % no zeros num1 = 25; den1 = [1 7 25]; num2 = 1; den2 = [1 1]; num = conv(num1,num2); % this multiplies the polynomials den = conv(den1,den2); t = 0:0.01:4; y1 = step(num,den,t); % with a zero at 0.9
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Unformatted text preview: num2 = [1 0.9]/0.9; num = conv(num1,num2); y2 = step(num,den,t); plot(t,y1,t,y2); xlabel('Time (sec)') ylabel('y(t)') title('Third order step response, no zeros') text(1.2,.5,'no zeros') text(.5,.8,'zero at -0.9')...
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This note was uploaded on 03/24/2010 for the course CENG 4331 taught by Professor Maryrandolph-gips during the Fall '09 term at UH Clear Lake.

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