Question

Problem 2 Consider an FIR filter with the following impulse response: h [n] [1 -2 3] (a) What is the gain at 2 0.67 rads/sample? (b) What is the filter output if the input is x(n] - [1 2 3 2 1?

Answer 1

Using DTFT analysis equation. To get output we use linear convolution.

h[n] = [1 -2 3] a) h[n] = [1 -2 3] H(e^jw) = sigma^infinity_n = -infinity h[n] e^-jwn = sigma^2_n = 0 h[n] e^-jwn = h[0] + h[1] e^-jw + h[2] e^-j2w = 1 - 2 e^-jw + 3 e^-j2w = 1 - 2 [cos w - j sin w] + 3[cos 2w - j sin 2 w] H(e^jw) = [1 - 2 cos w + 3 cos 2 w] + j [2 sin w - 3 sin 2w] K(e^j Ohm) = [1 - 2 cos Ohm + 3 cos (2 Ohm)] + j[2 sin Ohm - 3 sin 2 Ohm] gain Vertical-bar H(e^j Ohm) Vertical-bar = squareroot (1 - 2 cos Ohm + 3 cos 2 Ohm)^2 + (2 sin Ohm - 3 sin 2 Ohm)^2Vertical-bar H(e^j Ohm) Vertical-bar _Ohm = 0.6pi = squareroot (0.6545 + 13.435 = 3.753 \r\n b) output if x[n] = [1 2 3 2 1] y[n] = x[n] times y[n] h[n] = [1 -2 3] linear convolution y[0] = 1, y[1] = -2 + 2 = 0 y[2] = 3 - 4 + 3 = 2 y[3] = 6 - 6 + 2 = 2 y[4] = 9 - 4 + 1 = 6 y[5] = 6 - 2 = 4 y[6] = 3 y[0] = 1, y[1] = 0, y[2] = 2, y[3] = 2, y[4] = 6, y[5] = 4, y[6] = 3 output y[n] = {1, 0, 2, 2, 6, 4, 3}