Numerical 3

Find the Inverse Chebyshev filter for the following steps.

p = 1000 rad/sec αp = αmax = 0.25db

s = 1400 rad/sec αs = αmin = 18db

\[
n = \frac{\cosh^{-1}\!\left(\sqrt{\frac{10^{\alpha_{\min}/10} - 1}{10^{\alpha_{\max}/10} - 1}}\right)}{\cosh^{-1}\!\left(\tfrac{1}{\omega_p}\right)} = 4.51 \approx 5
\]

\[
\varepsilon = \sqrt{10^{\alpha_{\min}/10} - 1} = 0.1269
\]

Pole location is given by

\[
S_k = \sin(u_k)\sinh(v) + j\cos(u_k)\cosh(v)
\]

\[
u_k = \tfrac{1}{n}\cdot \tfrac{(2k+1)\pi}{2}, \quad k=1,2,\dots
\]

\[
v = \pm \tfrac{1}{n}\sinh^{-1}\!\left(\tfrac{1}{\varepsilon}\right)
\]

\[
u_0 = \tfrac{1}{5}\cdot \tfrac{(2\cdot 0+1)\pi}{2} = \tfrac{\pi}{10}, \quad
u_1 = \tfrac{3\pi}{10}, \quad
u_2 = \tfrac{5\pi}{10}, \quad
u_3 = \tfrac{7\pi}{10}, \quad
u_4 = \tfrac{9\pi}{10}
\]

\[
u_5 = \tfrac{11\pi}{10}, \quad
u_6 = \tfrac{13\pi}{10}, \quad
u_7 = \tfrac{15\pi}{10}, \quad
u_8 = \tfrac{17\pi}{10}, \quad
u_9 = \tfrac{19\pi}{10}
\]

So,

\[
v = \pm \tfrac{1}{n}\sinh^{-1}\!\left(\tfrac{1}{\varepsilon}\right) = 0.5252
\]

\[
S_k = \sin(u_k)\sinh(v) + j\cos(u_k)\cosh(v)
\]

\[
S_0 = 0.1698 + j1.085
\]
\[
S_1 = 0.447 + j0.67
\]
\[
S_2 = 0.5496
\]
\[
S_3 = 0.447 - j0.67
\]
\[
S_4 = 0.1698 - j1.085
\]
\[
S_5 = -0.1698 + j1.085
\]
\[
S_6 = -0.447 - j0.67
\]
\[
S_7 = -0.5496
\]
\[
S_8 = -0.447 - j0.67
\]
\[
S_9 = -0.1698 + j1.085
\]

Then the poles of Inverse Chebyshev are given by

\[
|S_0| = \tfrac{1}{S_0} = 0.1367 - j0.889
\]
\[
|S_1| = \tfrac{1}{S_1} = 0.674 - j1.033
\]
\[
|S_2| = \tfrac{1}{S_2} = 1.828
\]
\[
|S_3| = \tfrac{1}{S_3} = 0.674 + j1.033
\]
\[
|S_4| = \tfrac{1}{S_4} = 0.1367 + j0.889
\]
\[
|S_5| = \tfrac{1}{S_5} = -0.1367 + j0.889
\]
\[
|S_6| = \tfrac{1}{S_6} = -0.674 - j1.033
\]
\[
|S_7| = \tfrac{1}{S_7} = -1.828
\]
\[
|S_8| = \tfrac{1}{S_8} = -0.674 - j1.033
\]
\[
|S_9| = \tfrac{1}{S_9} = -0.1367 - j0.889
\]

Similarly for zeros locations

\[
\omega(k) = \sec\!\left(\tfrac{k\pi}{2n}\right), \quad k=1,3,5,\dots \; \text{odd values}
\]

\[
\omega(1) = \sec\!\left(\tfrac{\pi}{10}\right) = 1.051
\]
\[
\omega(3) = \sec\!\left(\tfrac{3\pi}{10}\right) = 1.701
\]
\[
\omega(5) = \sec\!\left(\tfrac{5\pi}{10}\right) = \infty
\]

In the above equations, poles and zeros are calculated assuming \(\omega_p = 1\) but \(\omega_s = 1400 \, \text{rad/sec}\), so we need to frequency scale it by factor of

\[
K_f = 1400
\]