Given the specifications $\alpha_p=3 \mathrm{~dB} ; \alpha_s=16 \mathrm{~dB} ; f_p=1 \mathrm{KH}_2$ and $f_s=2 \mathrm{KHz}$. Determine the order of the filter using the Chebyshev approximation. Find $H(s)$.

Solution:

From the given data we can find,

$ \begin{aligned} & \Omega_p=2 \pi \times 1000 \mathrm{~Hz}=2000 \pi \mathrm{rad} / \mathrm{sec} \\ & \Omega_s=2 \pi \times 2000 \mathrm{~Hz}=4000 \pi \mathrm{rad} / \mathrm{sec} \end{aligned} $

Step: 1

$ \begin{aligned} N \geq \frac{\cosh ^{-1} \sqrt{\frac{10^{0.1 \alpha_s}-1}{10^{0.1 \alpha_p}-1}}}{\cosh ^{-1} \frac{\Omega_s}{\Omega_p}} & =\cosh ^{-1} \frac{\sqrt{\frac{10^{1.6}-1}{10^{0.3}-1}}}{\cosh ^{-1} \frac{4000 \pi}{2000 \pi}} \\ & =1.91 \end{aligned} $

Step 2: Rounding N to the next higher value we get $N=2$.

For N even, the oscillatory curve starts from $\frac{1}{\sqrt{1+\varepsilon^2}}$.

Step 3: The values of the minor axis and major axis can be found below,

$ \begin{aligned} & \varepsilon=\left(10^{0.1 \alpha_p}-1\right)^{0.5}=\left(10^{0.3}-1\right)^{0.5}=1 \\\\ & \mu=\varepsilon^{-1}+\sqrt{1+\varepsilon^{-2}}=2.414 \\\\ & a=\Omega_p \frac{\left[\mu^{1 / N}-\mu^{-1 / N}\right]}{2}=2000 \pi \frac{\left[(2.414)^{1 / 2}-(2.414)^{-1 / 2}\right]}{2}=910 \pi \\\\ & b=\Omega_p \frac{\left[\mu^{1 / N}+\mu^{-1 / N}\right]}{2}=2000 \pi \frac{\left[(2.414)^{1 / 2}+(2.414)^{-1 / 2}\right]}{2}=2197 \pi\\ \end{aligned} $

$ \begin{aligned} s_k & =a \cos \phi_k+j b \sin \phi_k, \quad k=1,2 \\\\ \phi_k & =\frac{\pi}{2}+\frac{(2 k-1) \pi}{2 N} \quad k=1,2 \\\\ \phi_1 & =\frac{\pi}{2}+\frac{\pi}{4}=135^{\circ} \\\\ \phi_2 & =\frac{\pi}{2}+\frac{3 \pi}{4}=225^{\circ} \\\\ s_1 & =a \cos \phi_1+j b \sin \phi_1=-643.46 \pi+j 1554 \pi \\\\ s_2 & =a \cos \phi_2+j b \sin \phi_2=-643.46 \pi-j 1554 \pi\\ \end{aligned}\\ $

Step 5:

The denominator of $H(s)=(s+643.46 \pi)^2+(1554 \pi)^2$

Step 6:

The numerator of $H(s)=\frac{(643.46 \pi)^2+(1554 \pi)^2}{\sqrt{1+\varepsilon^2}}=(1414.38)^2 \pi^2$

The transfer function $H(s)=\frac{(1414.38)^2 \pi^2}{s^2+1287 \pi s+(1682)^2 \pi^2}$.