Pole-zero, magnitude response of y=x+x(n−2)

System function and pole-zero plot

y(n) = x(n) + x(n − 2). Taking the z-transform:

Y(z) = X(z)(1 + z⁻²), so H(z) = 1 + z⁻² = (z² + 1)/z²

• Zeros: z² + 1 = 0, so z = ±j (on the unit circle, at angles ±90°).
• Poles: z² = 0, a double pole at z = 0.
• Impulse response: h(n) = {1, 0, 1}, n = 0, 1, 2 (FIR).

 Im
  1 |       o (0,+j)
    |
 ---+---(2)x---------- Re      x (double) = pole at origin
    |                          o = zeros at +j and −j
 -1 |       o (0,−j)

(the dashed unit circle passes through both zeros)

Magnitude response

H(e^(jω)) = 1 + e^(−j2ω) = e^(−jω)(e^(jω) + e^(−jω)) = 2cos(ω) e^(−jω)

|H(e^(jω))| = 2|cos ω|, phase = −ω (linear, with jumps of π where cos ω changes sign).

 |H|
  2 |*                          *
    |  *                     *
  1 |     *               *
    |         *       *
  0 +-------------*------------> w
    0   pi/4   pi/2   3pi/4   pi

It is a notch (band-stop) filter with a null at ω = π/2 (the zeros ±j), maximum gain 2 at ω = 0 and ω = π.

Causality and stability

• Causal: h(n) = 0 for n < 0, and the output depends only on x(n) and x(n−2) (past and present). ROC: |z| > 0, which is the exterior of a circle (including ∞).
• Stable: Σ|h(n)| = 2 < ∞; equivalently the only poles are at z = 0, inside the unit circle, and the ROC includes |z| = 1. Every FIR system is stable.