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Chegg Solutions for Microelectronic Circuits (Adel S Sedra, Kenneth C Smith) (Z-Library)_parte_1817

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Step of 5 17.012P The transmission zero is, The poles of the transfer functions are, Step of 5 The expression for the transfer function of a second-order low-pass filter is, Substitute 2 for -0.25+j for and -0.25-j for kx(s-2) = k(s-2) = (s+0.25-j)(s+0.25+j) k(s-2) = k(s-2) = Simplify further. T(s) = k(s-2) k(s-2) = Step of 5 Consider the expression (1) Substitute 0 for S and calculate DC gain -2k = 1.0625 Substitute 1 for 1.0625 -1.0625 2 k=0.53125 Step of 5 Recall equation (1). Substitute 000000000for k. -0.53125(s-2) = = -17(s-2) = = -17(s-2) Therefore, the transfer function T(s) is -17(s-2) Step of 5 The expression for the gain as approaches to infinity is, DC gain Substitute -17(s-2) for -17(s-2) DC gain = lim = lim -17 = lim =0 Therefore, the DC gain as approaches to infinity is 0

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