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Convert the analog filter with system function, Ha(s)=s+0.5(s+0.5)2+16

Convert the analog filter with system function,

Ha(s)=s+0.5(s+0.5)2+16

into a digital IIR filter using the bilinear transformation. The digital filter should have a resonant frequency of ωr=π/2.

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Solution:

From the system function, we observe that Ωc=4. The sampling period T can be determined using the equation,

Ωc=2Ttanωr2

T=LΩctanωr2=L4tanπ4=0.5 s

H(z)=H(s)|s=2T(1z11+z1)=H(s)|s=4(1z11+z1)

H(z)=s+0.5(s+0.5)2+16|s=4(1z11+z1)

$ \begin{aligned} & =\frac{4\left(\frac{1-z^{-1}}{1+z^{-1}}\right)+0.5}{\left[4\left(\frac{1-z^{-1}}{1+z^{-1}}\right)+0.5\right]^2+16} \\ & =\frac{\left[4\left(1-z^{-1}\right)+0.5\left(1+z^{-1}\right)\right]\left[1+z^{-1}\right]}{\left[4\left(1-z^{-1}\right)+0.5\left(1+z^{-1}\right)\right]^2+16\left[1+z^{-1}\right]^2} \\\\ & =\frac{4.5+z^{-1}-3.5 z^{-2}}{36.25+0.5 z^{-1}+28.25 …

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