Find the Chebyshev transfer function that meets the same low-pass filter specifications given in Example 17.1: namely, fp = 10 kHz, Amax = 1 dB, fs = 15 kHz, Amin = 25 dB, dc gain = 1.

Question

Find the Chebyshev transfer function that meets the same low-pass filter specifications given in Example 17.1: namely, f_p = 10 kHz, A_max = 1 dB, f_s = 15 kHz, A_min = 25 dB, dc gain = 1.

Accepted Answer

Substituting A_max = 1 dB into Eq. (17.21) yields [latex]\epsilon [/latex] = 0.5088. By trying various values for N in Eq. (17.22)

[latex]\epsilon = \sqrt{10_{A_{max}/ - 1}}[/latex] (17.21)

[latex]A(ω_{s}) = 10 \log \left[1 + \epsilon^{2} cosh^{2} (Ncosh ^{-1}(ω_{s}/ω_{p})) \right][/latex] (17.22)

we find that N = 4 yields A(ω_s) = 21.6 dB and N = 5 provides 29.9 dB. We thus select N = 5. Recall that we required a ninth-order Butterworth filter to meet the same specifications in Example 17.1. The poles are obtained by substituting in Eq. (17.23) as

[latex]p_{k} = −ω_{p} sin \left(\frac{2 k − 1 }{N}\frac{π}{2}\right) sinh \left(\frac{1}{N} sinh^{-1}\frac{1}{\epsilon}\right) + j ω_{p}cos \left(\frac{2 k − 1 }{N}\frac{π}{2}\right) cosh \left(\frac{1}{N} sinh^{-1}\frac{1}{\epsilon}\right) [/latex] k = 1, 2, . . . ,N (17.23)

p₁,p₅ = ω_p (−0.0895 ± j0.9901)
p₂,p₄ = ω_p (−0.2342 ± j0.6119)
p₅ = ω_p (−0.2895)

The transfer function is obtained by substituting these values in Eq. (17.24) as

[latex]T(s) = \frac{K ω^{N}_{p}}{\epsilon 2 ^{N - 1}(s − p_{1})(s − p_{2}) · · · (s − p_{N} )}[/latex] (17.24)

[latex]T(s) = \frac{ω^{5}_{p}}{ 8.1408 (s + 0.2895 ω_{p}) (s^{2} + s 0.4684 ω_{p} + ω^{2}_{p})} × \frac{1}{s^{2} + s 0.1789 ω_{p} + 0.9883 ω^{2}_{p}}[/latex] (17.25)

where ω_p = 2π × 10⁴ rad/s.

Question 17.2: Find the Chebyshev transfer function that meets the same low...

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