How is a^t handled at t = kT?

Put b = a^T so a^t = b^k and Z[a^t] = z/(z - b)

Put b = a/T so Z[a^t] = z/(z - b)

Z[a^t] = z/(z - a) regardless of T

How is Z[t^2 a^t] obtained from Z[a^t]?

Apply the operator -T z d/dz twice

Apply -T z d/dz once and square the result

Differentiate once and multiply by t^2

Integrate twice with respect to z

Short answer

Let b=a^T. Using Z[t x(t)]=-Tz dX/dz twice gives Z[t^2 a^t]=T^2 b z(z+b)/(z-b)^3. With T=1, it reduces to a z(z+a)/(z-a)^3.

Obtain the z-transform of t² aᵗ.

Let the sampling period be T and put b = aᵀ, so that aᵗ at t = kT equals bᵏ.

Step 1. Z[aᵗ] = Σ (b/z)ᵏ = z/(z − b).

Step 2. Multiplication by t: Z[t x(t)] = −T z d/dz X(z). Apply it twice.

Z[t aᵗ] = −T z d/dz [ z/(z − b) ] = −T z · (−b)/(z − b)² = T b z/(z − b)²

Z[t² aᵗ] = −T z d/dz [ T b z/(z − b)² ] = −T² b z · [ (z − b)² − 2z(z − b) ]/(z − b)⁴

= −T² b z · [ (z − b) − 2z ]/(z − b)³ = −T² b z (−z − b)/(z − b)³

Answer:

Z[t² aᵗ] = T² b z (z + b)/(z − b)³, with b = aᵀ

i.e. Z[t² aᵗ] = T² aᵀ z (z + aᵀ) / (z − aᵀ)³, |z| > aᵀ.

For T = 1: a z (z + a)/(z − a)³.

✓ Answer checked by Claude Opus 5.5 (high effort)

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