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Convert the following filters with system functions. H(s)=(s+0.1)(s+0.1)2+9
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written 2.4 years ago by |
Solution:
S=2T(1−2−11+2−1)=2(1−z−11+z−1). [2(1−z−11+z−1)+0.1]2+9=2(1−z−1)+0.1(1+z−1)(1+z−1)(2(1−z−1)+0.1(1+z−1))2(1+z−1)2+9
H[z]=[2(1−z−1)+0.1(1+z−1)](1+z−1)[2(1−z−1)+0.1(1+z−1)]2+9(1+z−1)2
By Impulse Invariant method:
H(s)=s+0.1(s+0.1)2+9
The inverse L.T. of the given function is
h(t)=e−0.1tcos(3t) for t⩾0=0 otherwise.
Sampling the function gives,
h[nT]=e−0 nT cos(3nT) for n⩾0=0 otherwise.
H[2]=∞∑n=0e−a e.nT cos(3nT)z−n=∞∑n=0e−0.1nT[ej3nT+e−j3nT2]z−n=12[∞∑n=0e−(0.1−j3)nTz−n+∞∑n=0e−(0.1+j3)nTz−n]
=12[11−e−(0.1−j3)Tz−1+11−e−(0.1+j3)τ−1z]=12[1−e−(0.1+j3)Tz−1+1−e−(0.1−j3)Tz−1(1−e−(0.1−j3)Tz−1)(1−e−(0.1+j3)Tz−1)]
=12[2−e−0.ITcos(ej3T+e−j3T)z−11−2e−0.1Tcos(3T)z−1+e−0.2Tz−2]
H[z]=1−e−0.1Tcos(3T)z−11−2e−0.1Tcos(3T)z−1+e−0.2Tz−2
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