Question

3. (a) For each of the RC passive filters shown in Fig. 3, sketch the magnitude in dB and phase as a function of frequency (in Hz), with the frequency on a log scale. Indicate the poles and zeros. For the magnitude plots, indicate the slope (in dB/decade) for each region; for regions that exhibit flat magnitude, indicate the value. Likewise, for regions that exhibit flat phase, indicate the value. (b) Indicate the type of each filter (e.g., lowpass, highpass, bandpass, notch). (c) Derive and sketch the step response of each filter.

Answer 1

v_c (s) = v_in (s) 1/sc/R + 1/sc V_0/v_in = 1Rsc + 1 = [1/Rc/s + 1/Rc] Given R = 1 k ohm, c = 10 pf Rc = (1k) (10 p) = 10^- 3 v_0/v_in = 1/s(VRC) + 1 = [1/s/10^8 + 1] Vertical-bar V_0/v_in Vertical-bar = 1/squareroot (w/10^8)^2 + 1 angle w_sin = -tan^- 1 (w/10^8) b) v_0 = v_in (R)/R + 1/sc = R v_in/(RSC + 1)/SC v_0 = RSC V_in/(RSC + 1) = RC (S) V_In/B/(RC) + 1 v_0/v_in = 10^- s.s/(s/10^8 + 1) (v_0/v_th) = 10^- 8(w)/squareroot w^2/10^16 + 1 v_0/v_in = 90 degree - tan^- 1 (w/10^8) \r\n R_2 = 2k ohm R_2 = 1k ohm c_1 = 10 pf c_2 = 20 pf Req = R_2// 1/sc_2 = R_2 1/sc_2/R_2 + 1/sc_2 = R_2/R_s sc_2 + 1 V_0 (s) = v_in (R_2/R_2 sc_2 + 1)/R_2/R_2 sc_2 + 1 + R_1 + 1/sc_1 = v_in (R_2) (s(1))/R_2 s c_1 + R_1 sc_1 (R_2 s c_2 + 1) + R_2 s c_2 + 1) v_0 (s)/v_in