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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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Solution:

S=2T(1211+21)=2(1z11+z1)[2(1z11+z1)+0.1]2+9=2(1z1)+0.1(1+z1)(1+z1)(2(1z1)+0.1(1+z1))2(1+z1)2+9

H[z]=[2(1z1)+0.1(1+z1)](1+z1)[2(1z1)+0.1(1+z1)]2+9(1+z1)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)=e0.1tcos(3t) for t0=0 otherwise. 

Sampling the function gives,

h[nT]=e0 nT cos(3nT) for n0=0 otherwise. 

H[2]=n=0ea e.nT cos(3nT)zn=n=0e0.1nT[ej3nT+ej3nT2]zn=12[n=0e(0.1j3)nTzn+n=0e(0.1+j3)nTzn]

=12[11e(0.1j3)Tz1+11e(0.1+j3)τ1z]=12[1e(0.1+j3)Tz1+1e(0.1j3)Tz1(1e(0.1j3)Tz1)(1e(0.1+j3)Tz1)]

=12[2e0.ITcos(ej3T+ej3T)z112e0.1Tcos(3T)z1+e0.2Tz2]

H[z]=1e0.1Tcos(3T)z112e0.1Tcos(3T)z1+e0.2Tz2

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