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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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written 2.3 years ago by |
Solution:
Given H(s)=2(s+1)(s+2)
Substitute s=2T[1−z−11+z−1] in H(s) to get H(z)
H(z)=H(s)|s=2T(1−z−11+z−1)=2(s+1)(s+2)|s=2T(1−z−11+z−1)
Given T=1sec
H(z)=2{2(1−z−11+z−1)+1}{2(1−z−11+z−1)+2}=2(1+z−1)2(3−z−1)(4)=(1+z−1)26−2z−1=0.166(1+z−1)2(1−0.33z−1)