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The normalized transfer function of an analog filter is given by Ha(sn)=1s2n+1.414sn+1

The normalized transfer function of an analog filter is given by Ha(sn)=1s2n+1.414sn+1

Convert the analog filter to a digital one with a cut-off frequency of 0.4π using the bilinear transformation.

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Solution:

Ωc=2Ttanωc2=2tan(0.2π)=1.45Ha(s)=1(s1.45)2+1.414(s1.45)+1=(1.45)2s2+2.055s+(1.45)2H(z)=Ha(s)s=2T(1z11+z1)

H(z)=2.14(1z1)2(1+z1)2+(2.055)2(1z1)1+z1+2.1=2.1(1+z1)24(1z1)2+4.11(1z2)+2.1(1+z1)2=2.1(1+z1)24(12z1+z2)+4.11(1z2)+2.1(1+2z1+z2)=2.1(1+z1)210.213.8z1+1.99z2=0.2(1+z1)210.37z1+0.195z2

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