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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=1−2+0.6=−57
B=(s+0.6)×1(s+2)(s+0.6)|s=−0.6=1−0.6+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
1s−pi=zz−epiT
∴ Transfer function of the digital filter
H(z)=−5z/7z−e−2T+5z/7z−e−0.6T
=5z7(−1z−e−2+1z−e−0.6)( Assume T=1)
=5z7[−(z−e−0.6)+(z−e−2)(z−e−2)(z−e−0.6)]
=5z(e−0.6−e−2)/7z2−(e−2+e−0.6)z+e−2.6
∴ Transfer function of the digital filter
H(z)=0.2953zz2−0.6841z+0.0743
Case 2: By BLT Method
Put s=2(z−1)T(z+1)=2(z−1)1(z+1) in (1)⋯(2)
∴ Transfer function of the digital filter
H(z)=1(2z−2z+1+2)(2z−2z+1+0.6)
=1(2z−2+2z+2z+1)(2z−2+0.6z+0.6z+1)
=(z+1)24z(2.6z−1.4)
=(z+1)24z×2.6(z−0.5385)
∴ Transfer function of the digital filter
H(z)=0.0962(z+1)2z(z−0.5385)
(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(z−e−aTcosbT)z2−2ze−aTcosbT+e−2aT
Here, a=0.1,b=3 and T=1
∴ Transfer function of the digital filter
H(z)=z[z−e−0.1×1cos(3×1)]z2−2ze−0.1×1cos(3×1)+e−2×0.1×1
=z(z−e−0.1cos3)z2−2ze−0.1cos3+e−0.2
∴ 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)=2z−2z+1+0.6(2z−2z+1+0.6)2+9
=2z−2+0.6z+0.6z+1(2z−2+0.6z+0.6z+1)2+9
=2.6z−1.4z+1(2.6z−1.4)2(z+1)2+9
=2.6z−1.4(z+1)(2.6z−1.4)2+9(z+1)2(z+1)2
=2.6z−1.4(z+1)×(z+1)26.76z2−7.28z+1.96+9(z2+2z+1)
=(2.6z−1.4)×(z+1)215.76z2−10.72z+10.96
=2.6(z−0.5385)×(z+1)215.76(z2−0.6802z+0.1244)
∴ Transfer function of the digital filter
H(z)=0.1650(z−0.5385)(z+1)2z2−0.6802z+0.1244