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Convert the following filters with system functions (i) H(s)=1(s+2)(s+0.6) (ii) H(s)=(s+0.1)(s+0.1)2+9 into a digital filter by means of impulse invariant and BLT method.
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(i) H(s)=1(s+2)(s+0.6)(1)

Let 1(s+2)(s+0.6)=As+2+Bs+0.6

A=(s+2)×1(s+2)(s+0.6)|s=2=12+0.6=57

B=(s+0.6)×1(s+2)(s+0.6)|s=0.6=10.6+2=57

H(s)=5/7s+2+5/7s+0.6

Here, s=0.6 and s=2 are the poles in s-plane.

Case 1: By Impulse Invariance Transformation

1spi=zzepiT

Transfer function of the digital filter

H(z)=5z/7ze2T+5z/7ze0.6T

=5z7(1ze2+1ze0.6)( Assume T=1)

=5z7[(ze0.6)+(ze2)(ze2)(ze0.6)]

=5z(e0.6e2)/7z2(e2+e0.6)z+e2.6

Transfer function of the digital filter

H(z)=0.2953zz20.6841z+0.0743

Case 2: By BLT Method

Put s=2(z1)T(z+1)=2(z1)1(z+1) in (1)(2)

Transfer function of the digital filter

H(z)=1(2z2z+1+2)(2z2z+1+0.6)

=1(2z2+2z+2z+1)(2z2+0.6z+0.6z+1)

=(z+1)24z(2.6z1.4)

=(z+1)24z×2.6(z0.5385)

Transfer function of the digital filter

H(z)=0.0962(z+1)2z(z0.5385)


(ii) H(s)=(s+0.1)(s+0.1)2+9

H(s)=(s+0.1)(s+0.1)2+32(3)

Case 1: By Impulse Invariance Transformation

s+a(s+a)2+b2=z(zeaTcosbT)z22zeaTcosbT+e2aT

Here, a=0.1,b=3 and T=1

Transfer function of the digital filter

H(z)=z[ze0.1×1cos(3×1)]z22ze0.1×1cos(3×1)+e2×0.1×1

=z(ze0.1cos3)z22ze0.1cos3+e0.2

Transfer function of the digital filter

H(z)=z(z+0.8958)z2+1.7916z+0.8187

Case 2: By BLT Method

Put (2) in (3)

Transfer function of the digital filter

H(z)=2z2z+1+0.6(2z2z+1+0.6)2+9

=2z2+0.6z+0.6z+1(2z2+0.6z+0.6z+1)2+9

=2.6z1.4z+1(2.6z1.4)2(z+1)2+9

=2.6z1.4(z+1)(2.6z1.4)2+9(z+1)2(z+1)2

=2.6z1.4(z+1)×(z+1)26.76z27.28z+1.96+9(z2+2z+1)

=(2.6z1.4)×(z+1)215.76z210.72z+10.96

=2.6(z0.5385)×(z+1)215.76(z20.6802z+0.1244)

Transfer function of the digital filter

H(z)=0.1650(z0.5385)(z+1)2z20.6802z+0.1244

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