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Apply bilinear transformation to H(s)=2(s+1)(s+2) with T=1 sec and find H(z).
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Solution:

Given H(s)=2(s+1)(s+2)

Substitute s=2T[1z11+z1] in H(s) to get H(z)

H(z)=H(s)|s=2T(1z11+z1)=2(s+1)(s+2)|s=2T(1z11+z1)

Given T=1sec

H(z)=2{2(1z11+z1)+1}{2(1z11+z1)+2}=2(1+z1)2(3z1)(4)=(1+z1)262z1=0.166(1+z1)2(10.33z1)

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