Question

Part A Review The input signal to a filter contains components that range in frequency from 100 IIz to 50 kIIz. We wish to reduce the amplitude of the 50-kHz component by a factor of 200 by passing the signal through a first-order lowpass filter What half-power frequency is required for the filter? Express your answer to three significant figures and include the appropriate units. Correct Part B By what factor is à component at 6 kHx changed in amplitude in passing through this filter? Express your answer using three significant figures H(F)1-10. I 24 Provious Answors Request Answer Submit X Incorrect Try Again; 5 attempts remaining

Answer 1

Part A: To achieve a reduction of 50khz component by a factor of 200 we must have Vertical-bar H(50k) Vertical-bar = 1/squareroot 1 + (f/f_B)^2 f = 50khz 1/200 = 1/squareroot 1 + (50k/f_B)^2 squaring on both sides (1/200)^2 = 1/1 + (50k/f_B)^2 4 + (50k/f_B)^2 = (1/1/200)^2 1 + (50k/f_B)^2 = 40000 50k/f_B = squareroot 40000 - 1 = 199.9975 50k/199.9975 = f_B f_B = 250Hz f_B = 250hz part B: at 5khz the resulting value of the transfer function H(F) = 1/1 + j(F/F_B) = 1/1 + j(5k/250) = 1/1 + 20j = 2.493 times 10^- 3 - 0.049870 = 0.04993 Vertical-bar H(e) Vertical-bar = 0.4993 thus the 5khz component is changed in amplitude by a factor of 0.04993