- 2081 Chaitra (new course) · 2 marks
Z-transform property proof
Short answer
From y(k)-y(k-1)=x(k), Z-transform gives Y(z)-z^{-1}Y(z)=X(z). Hence Y(z)= (z/(z-1))X(z).
If Z[x(k)] = X(z) and y(k) = Σ_{m=0}^{k} x(m), then show that Y(z) = (z/(z − 1)) X(z).
We use y(k) = Σ_{m=0}^{k} x(m) (the running sum of x).
Proof. Write the difference of consecutive terms, with y(−1) = 0:
y(k) − y(k − 1) = x(k)
Take the Z-transform of both sides. By the right-shift property, Z[y(k − 1)] = z⁻¹ Y(z) (since y(−1) = 0):
Y(z) − z⁻¹ Y(z) = X(z)
Y(z) (1 − 1/z) = X(z)
Y(z) = X(z) / (1 − z⁻¹) = (z/(z − 1)) X(z)
Alternative (direct) proof: Y(z) = Σₖ z^(−k) Σ_{m≤k} x(m). Interchanging the sums,
Y(z) = Σ_m x(m) Σ_{k≥m} z^(−k) = Σ_m x(m) z^(−m) · 1/(1 − z⁻¹) = X(z) · z/(z − 1)
for |z| > 1. ∎
✓ Answer checked by Claude Opus 5.5 (high effort)
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