Question

Signal Processing

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Find the transfer function H(s) of the Chebyshev filter in Problem 7 and sketch the mag- nitude of the frequency response [H(jo)

Table 2.4:

Problem 7(Answer is Red):

Problem 2(Answer is Red):

Answer 1

H (s) = k/s^4 + a_3 s^3 + a_2 s^2 + a_1 s^1 + a_0 k = a_0/squareroot 1 + epsilon^2 for order even epsilon = 0.508 k = 0.275628/squareroot 1 + (0.508)^2 k = 0.2457 H (s) = 0.2457/s^4 + 0.95 s^3 + 9.45s^2 + 0.74s + 0.275 replace s = s/2 pi times 23444 H (s) = 0.2457/(s/2 pi times 23444)^4 + 0.95 (s/2 pi times 23444)^3 + 1.45 (s/2 pi times 23444)^2 + 0.74 (s/2 pi times 23444) + 0.275 \r\n Given specifications f_p = 20kHz f_s = 48 kHz alpha_p = 1dB alpha_s = 40 ds epsilon = [10^-.1 alpha_p ds - 1]^1/2 epsilon = [10^0.1 times 1 - 1]^1/2 = [0.2589]^1/2 epsilon = 0.508 Normalized frequency by 20 kHz f_p (norm) = 20 times 10^3/20 times 10^3 = 1, f_s (norm) = 48 times 10^3/20 times 10^3 = 2.4Vertical-bar H (jw) Vertical-bar (inchs) = -20 log epsilon - 6 (n - 1) - 20 n log f_s (norm) - 40 = -20 log (0.508) - 6n + 6 - 20 log_2.4 - 40 = 5.88 - 6n + 6 - 7.604n