Chegg Solutions for Microelectronic Circuits (Adel S Sedra, Kenneth C Smith) (Z-Library)_parte_1884

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Step of 5 17.055P Refer to Figure .22(a) for the second- order filter circuit and also refer to figure 7.13(a) for the first-order op amp-RC circuit given in the textbook Cascade both the second-order filter circuit and the first-order op amp-RC circuit and design the required Butterworth filter. The 3-dB bandwidth of fifth-order (N=5) Butterworth filter is, The expression for the maximum gain is, Substitute for Simplify further. Step of 5 The resonant frequency is equal to the pass band edge frequency. The expression for the DC gain magnitude of the first-order op amp-RC filter circuit is, Substitute for Choose 10 for and R2 The expression for resonant frequency is, 1 Calculate the capacitance Substitute for and for R2 1 nF Therefore, the value of capacitance 1 nF. Step of 5 Consider the transfer function of the second-order filter T(s) R2 Here, K is the DC gain The resonant frequency is, R2 Substitute R for R, R2 and R, and C for and R R R Consider the value of capacitance 100 Substitute for for R = Therefore, the value of resistances R, R2 and is 100 Ω and the value of capacitances and is 100 nF. Step of The expression for the quality factor of the filter is, Substitute 5 for N. 0.809 2Q Q=0.618 Therefore, the quality factor is 0.618. Step of 5 Consider the relation between resonant frequency and quality factor Substitute for 618 for and for 1 0.618 0.618 Therefore, the value of resistance is Therefore, the component values of the fifth-order Butterworth filter circuit are, for the first-order op amp-RC filter, and for the second-order filter circuit. nF