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Prove shifting theorem of z-transform

Short answer

By definition, Z[x(t+nT)] = sum_{k=0}^∞ x((k+n)T) z^{-k}. Substituting m=k+n gives z^n sum_{m=n}^∞ x(mT) z^{-m} = z^n [X(z) - sum_{k=0}^{n-1} x(kT) z^{-k}], assuming x(t)=0 for t<0.

If x(t) = 0 for t < 0 and Z[x(t)] = X(z) for t ≥ 0, then prove that Z[x(t + nT)] = zⁿ [X(z) - Σ_{k=0}^{n-1} x(kT) z^{-k}].

Proof (shift to the left / advance)

By the definition of the z-transform, with x(t) = 0 for t < 0,

Z[x(t + nT)] = Σ (k = 0 to ∞) x((k + n)T) z⁻ᵏ

Put m = k + n, so k = m − n and m runs from n to ∞:

Z[x(t + nT)] = Σ (m = n to ∞) x(mT) z^−(m − n) = zⁿ Σ (m = n to ∞) x(mT) z⁻ᵐ

The sum from n to ∞ equals the full sum minus the first n terms:

Σ (m = n to ∞) x(mT) z⁻ᵐ = Σ (m = 0 to ∞) x(mT) z⁻ᵐ − Σ (m = 0 to n − 1) x(mT) z⁻ᵐ = X(z) − Σ (m = 0 to n − 1) x(mT) z⁻ᵐ

Therefore

Z[x(t + nT)] = zⁿ [ X(z) − Σ (k = 0 to n − 1) x(kT) z⁻ᵏ ]

Special cases

  • n = 1: Z[x(t + T)] = z X(z) − z x(0)
  • n = 2: Z[x(t + 2T)] = z² X(z) − z² x(0) − z x(T)

These are the forms used to solve difference equations.

✓ Answer checked by Claude Opus 5.5 (high effort)

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