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

Let $\frac{1}{(s+2)(s+0.6)}=\frac{A}{s+2}+\frac{B}{s+0.6}$

$\therefore A=(s+2) \times\left.\frac{1}{(s+2)(s+0.6)}\right|_{s=-2}=\frac{1}{-2+0.6}=\frac{-5}{7}$

$B=(s+0 .6) \times\left.\frac{1}{(s+2)(s+0.6)}\right|_{s=-0.6}=\frac{1}{-0.6+2}=\frac{5}{7}$

$\therefore H(s)=\frac{-5 / 7}{s+2}+\frac{5 / 7}{s+0.6}$

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

Case 1: By Impulse Invariance Transformation

$\frac{1}{s-p_{i}} =\frac{z}{z-e^{p_{i} T}}$

$\therefore$ Transfer function of the digital filter

$H(z)=\frac{-5 z / 7}{z-e^{-2 T}}+\frac{5 z / 7}{z-e^{-0.6 T}}$

$=\frac{5 z}{7}\left(\frac{-1}{z-e^{-2}}+\frac{1}{z-e^{-0.6}}\right)(\text { Assume } \mathrm{T}=1)$

$=\frac{5 z}{7}\left[\frac{-\left(z-e^{-0.6}\right)+\left(z-e^{-2}\right)}{\left(z-e^{-2}\right)\left(z-e^{-0.6}\right)}\right]$

$=\frac{5 z\left(e^{-0.6}-e^{-2}\right) / 7}{z^{2}-\left(e^{-2}+e^{-0.6}\right) z+e^{-2.6}}$

$\therefore$ Transfer function of the digital filter

$H(z)=\frac{0.2953 z}{z^{2}-0.6841 z+0.0743}$

Case 2: By BLT Method

Put $s=\frac{2(z-1)}{T(z+1)}=\frac{2(z-1)}{1(z+1)}$ in $(1) \cdots (2)$

$\therefore$ Transfer function of the digital filter

$H(z)=\frac{1}{\left(\frac{2 z-2}{z+1}+2\right)\left(\frac{2 z-2}{z+1}+0.6\right)}$

$=\frac{1}{\left(\frac{2 z-2+2 z+2}{z+1}\right)\left(\frac{2 z-2+0.6 z+0.6}{z+1}\right)}$

$=\frac{(z+1)^{2}}{4 z(2.6 z-1.4)}$

$=\frac{(z+1)^{2}}{4 z \times 2.6(z-0.5385)}$

$\therefore$ Transfer function of the digital filter

$H(z)=\frac{0.0962(z+1)^{2}}{z(z-0.5385)}$

(ii) $H(s)=\frac{(s+0.1)}{(s+0.1)^{2}+9}$

$H(s)=\frac{(s+0.1)}{(s+0.1)^{2}+3^2} \cdots(3)$

Case 1: By Impulse Invariance Transformation

$\frac{s+a}{(s+a)^{2}+b^{2}}=\frac{z\left(z-e^{-a T} \cos b T\right)}{z^{2}-2 z e^{-a T} \cos b T+e^{-2 a T}}$

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

$\therefore$ Transfer function of the digital filter

$H(z)=\frac{z\left[z-e^{-0.1 \times 1} \cos (3 \times 1)\right]}{z^{2}-2 z e^{-0.1 \times 1} \cos (3 \times 1)+e^{-2 \times 0.1 \times 1}}$

$=\frac{z\left(z-e^{-0.1} \cos 3\right)}{z^{2}-2 z e^{-0.1} \cos 3+e^{-0.2}}$

$\therefore$ Transfer function of the digital filter

$H(z)=\frac{z(z+0.8958)}{z^{2}+1.7916 z+0.8187}$

Case 2: By BLT Method

Put $(2)$ in $(3)$

$\therefore$ Transfer function of the digital filter

$H(z)=\frac{\frac{2 z-2}{z+1}+0.6} {\left(\frac{2 z-2}{z+1}+0.6\right)^{2}+9}$

$=\frac{\frac{2 z-2+0.6 z+0.6}{z+1}} {\left(\frac{2 z-2+0.6 z+0.6}{z+1}\right)^{2}+9}$

$=\frac{\frac{2.6 z-1.4}{z+1}}{ \frac{(2.6 z-1.4)^{2}}{(z+1)^{2}}+9}$

$=\frac{\frac{2.6 z-1.4}{(z+1)}}{\frac{(2.6 z-1.4)^{2}+9(z+1)^{2}}{(z+1)^{2}}}$

$=\frac{2.6 z-1.4}{(z+1)} \times \frac{(z+1)^{2}}{6.76 z^{2}-7.28 z+1.96+9\left(z^{2}+2 z+1\right)}$

$=\frac{(2.6 z-1.4) \times(z+1)^{2}}{15.76 z^{2}-10.72 z+10.96}$

$=\frac{2.6(z-0.5385) \times(z+1)^{2}}{15.76\left(z^{2}-0.6802 z+0.1244\right)}$

$\therefore$ Transfer function of the digital filter

$H(z)=\frac{0.1650(z-0.5385)(z+1)^{2}}{z^{2}-0.6802 z+0.1244}$