State and prove final value theorem

Statement

If X(z) = Z[x(k)] and (1 − z⁻¹)X(z) has all its poles inside the unit circle |z| = 1 (so that x(k) tends to a finite limit), then

x(∞) = lim (k → ∞) x(k) = lim (z → 1) (z − 1) X(z) = lim (z → 1) (1 − z⁻¹) X(z)

Proof

Use the shifting property Z[x(k + 1)] = z X(z) − z x(0). Then

Z[x(k + 1) − x(k)] = z X(z) − z x(0) − X(z) = (z − 1) X(z) − z x(0)

By definition of the z-transform,

Σ (k = 0 to ∞) [x(k + 1) − x(k)] z⁻ᵏ = (z − 1) X(z) − z x(0)

Take the limit z → 1 on both sides. The left side becomes a telescoping sum:

lim (n → ∞) Σ (k = 0 to n) [x(k + 1) − x(k)] = lim (n → ∞) [x(n + 1) − x(0)] = x(∞) − x(0)

The right side becomes lim (z → 1) (z − 1) X(z) − x(0).

Hence

x(∞) − x(0) = lim (z → 1) (z − 1) X(z) − x(0)

so

x(∞) = lim (z → 1) (z − 1) X(z)

Condition: the theorem is valid only if x(∞) exists, i.e. all poles of (z − 1)X(z) lie inside the unit circle.