written 2.3 years ago by | • modified 2.3 years ago |
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.
written 2.3 years ago by | • modified 2.3 years ago |
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.
written 2.3 years ago by | • modified 2.3 years ago |
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(1−z−11+z−1)=H(s)|s=4(1−z−11+z−1)
H(z)=s+0.5(s+0.5)2+16|s=4(1−z−11+z−1)
$ \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 …