An FIR filter with transfer function H(z) = 1 + z⁻² (y[n] = x[n] + x[n − 2]) runs at a sampling frequency of 8 kHz. Which of the following is the frequency of a sine-wave input for which the output is 0?

Answer: B. 2 kHz

The solutions of 1 + z⁻² = 0, that is, the zeros, are z = ±j from z² = −1, which lie on the unit circle at ω = ±π/2. ω = π/2 corresponds to f = ω·fs/(2π) = fs/4 = 2 kHz; a 2 kHz sine wave has exactly the opposite sign two samples earlier, so the sum is 0. A zero on the unit circle creates a notch that completely removes that frequency. At 4 kHz (the Nyquist frequency), z⁻² = 1, so the gain is 2.

• A. At 1 kHz (ω = π/4), z⁻² = e^(−jπ/2) = −j and |1 − j| = √2 ≈ 1.41, so the output is not 0.
• B. Correct. The zeros z = ±j lie at ω = π/2, that is, fs/4 = 2 kHz.
• C. At the Nyquist frequency (z = −1), z⁻² = 1 and the gain is 1 + 1 = 2.