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Determine the system function $H(z)$ of the lowest order Chebyshev IIR digital filter with the following specifications:

Determine the system function $H(z)$ of the lowest order Chebyshev IIR digital filter with the following specifications: $ \begin{aligned} & 3 \mathrm{~dB} \text { ripple in passband } 0 \leq \omega \leq 0.2 \pi \\ & 25 \mathrm{~dB} \text { attenuation in stopband } 0.45 \pi \leq \omega \leq \pi \end{aligned} $

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Solution:

Let T = 1 and bilinear transformation is used

$ \varepsilon=\left[\frac{1}{A_1^2}-1\right]^{\frac{1}{2}}=\left[\frac{1}{0.707^2}-1\right]=1\\ $

The ratio of analog frequencies,

$ \frac{\Omega_2}{\Omega_1}=\frac{\frac{2}{T} \tan \frac{\omega_2}{2}}{\frac{2}{T} \tan \frac{\omega_1}{2}}=\frac{\tan \frac{0.45 \pi}{2}}{\tan \frac{0.2 \pi}{2}}=2.628\\ $

$ N \geq \frac{\cosh ^{-1}\left\{\frac{1}{\varepsilon}\left[\frac{1}{A_2^2}-1\right]^{-\frac{1}{2}}\right\}}{\cosh ^{-1}\left\{\frac{\Omega_2}{\Omega_1}\right\}}\\ $

$ \begin{aligned} & \geq \frac{\cosh ^{-1}\left\{\frac{1}{1}\left[\frac{1}{0.0562^2}-1\right]^{\frac{1}{2}}\right\}}{\cosh ^{-1}(2.628]} \\ & \geq \frac{3.569}{1.621} \geq 2.20 \approx 3 \\ & \Omega_c=\frac{\Omega_1}{\left[\frac{1}{A_1^2}-1\right]^{1 / 2 N}}=\frac{\frac{2}{T} \tan \frac{\omega_1}{2}}{\left[\frac{1}{0.707^2}-1\right]^{1 / 6}}=1.708 \end{aligned} $

Analog filter transfer function for N = 3.

$ \begin{gathered} H_a(s)=\frac{B_0 \Omega_c}{s+c_0 \Omega_c} \frac{B_1 \Omega_c^2}{s^2+b_1 \Omega_c s+c_1 \Omega_c^2} \\ y_N=\frac{1}{2}\left\{\left[\left(\frac{1}{\varepsilon^2}+1\right)^{\frac{1}{2}}+\frac{1}{\varepsilon}\right]^{\frac{1}{N}}-\left[\left(\frac{1}{\varepsilon^2}+1\right)^{\frac{1}{2}}+\frac{1}{\varepsilon}\right]^{\frac{-1}{N}}\right\} \\ y_N=\frac{1}{2}\left\{\left[\left(\frac{1}{1^2}+1\right)^{\frac{1}{2}}+\frac{1}{1}\right]^{\frac{1}{3}}-\left[\left(\frac{1}{1^2}+1\right)^{\frac{1}{2}}+\frac{1}{1}\right]^{\frac{-1}{3}}\right\}=0.5959 \end{gathered} $

$ \begin{aligned} & c_0=y_N=0.5959 \\\\ & b_1=2 y_N \sin \left[\frac{(2 \times 1-1) \pi}{2 N}\right]=2 \times 0.5959 \sin \frac{\pi}{6}=0.5959\\ \\ & c_1=y_N^2+\cos ^2 \frac{(2 \times 1-1) \pi}{2 N}=0.5959^2+\cos ^2 \frac{\pi}{6}=1.105\\ \end{aligned} $

$ \prod_{k=0}^{(N-1) / 2} \frac{B_k}{c_k}=1\\ $

$ \begin{aligned} \therefore \quad B_0=c_0 & =0.5959, \quad B_1=c_1=1.105 \\\\ H_a(s) & =\left(\frac{0.5959 \times 1.708}{s+0.5959 \times 1.708}\right)\left(\frac{1.105(1.708)^2}{s^2+0.5959 \times 1.708 s+1.105(1.708)^2}\right) \\\\ & \cdot\left(\frac{1.01}{s+1.01}\right)\left(\frac{3.223}{s^2+1.01 s+3.223}\right)\\ \end{aligned} $

Using bilinear transformation, H(z) is given by

$ \begin{aligned}\\ H(z) & =H_a(s)\left|s=\frac{21-z^{-1}}{1+z^{-1}}=\left(\frac{1.01}{s+1.01}\right)\left(\frac{3.223}{s^2+0.1 s+3.223}\right)\right|_{s=2 \frac{1-z^{-1}}{1+z^{-1}}} \\\\ & =\frac{3.25}{\left[2\left(\frac{1-z^{-1}}{1+z^{-1}}\right)+1.01\right]\left[2\left(\frac{1-z^{-1}}{1+z^{-1}}\right)^2+0.1 \times 2\left(\frac{1-z^{-1}}{1+z^{-1}}\right)+3.223\right]} \\\\ & =\frac{(3.25)\left(1+z^{-1}\right)^3}{7.423-1.554 z^{-1}+7.023 z^{-2}}\\ \end{aligned} $

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