Question 17.2: Find the Chebyshev transfer function that meets the same low...
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.
Substituting Amax = 1 dB into Eq. (17.21) yields \epsilon = 0.5088. By trying various values for N in Eq. (17.22)
\epsilon = \sqrt{10_{A_{max}/ – 1}} (17.21)
A(ω_{s}) = 10 \log \left[1 + \epsilon^{2} cosh^{2} (Ncosh ^{-1}(ω_{s}/ω_{p})) \right] (17.22)
we find that N = 4 yields A(ωs) = 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
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) k = 1, 2, . . . ,N (17.23)
p1,p5 = ωp (−0.0895 ± j0.9901)
p2,p4 = ωp (−0.2342 ± j0.6119)
p5 = ωp (−0.2895)
The transfer function is obtained by substituting these values in Eq. (17.24) as
T(s) = \frac{K ω^{N}_{p}}{\epsilon 2 ^{N – 1}(s − p_{1})(s − p_{2}) · · · (s − p_{N} )} (17.24)
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}} (17.25)
where ωp = 2π × 104 rad/s.