Which of the following gives the gains at DC and at the Nyquist frequency of the first-order IIR low-pass filter given by the exponential smoothing y[n] = 0.2x[n] + 0.8y[n − 1]?

Answer: C. 1 at DC and about 0.11 at the Nyquist frequency

The transfer function is H(z) = 0.2/(1 − 0.8z⁻¹); at DC (z = 1) the gain is 0.2/(1 − 0.8) = 1, and at the Nyquist frequency (z = −1) it is 0.2/(1 + 0.8) = 1/9 ≈ 0.11 (about −19 dB). The new input is weighted by 0.2 and the previous output by 0.8; the weights sum to 1, so a constant input passes unchanged. Remembering only the previous output gives an average of past inputs with exponentially decreasing weights. The pole at z = 0.8 lies inside the unit circle, so the filter is stable, and the closer the feedback coefficient is to 1, the stronger the smoothing and the slower the response.

• A. The numerator of this filter is a constant and it has no zero on the unit circle, so there is no frequency where the gain is 0.
• B. These are the gains of y[n] = x[n] + 0.8y[n − 1], with the input coefficient 0.2 replaced by 1 (1/0.2 = 5, 1/1.8 ≈ 0.56).
• C. Correct. At z = 1, 0.2/0.2 = 1; at z = −1, 0.2/1.8 ≈ 0.11.