z-transform of t·e^{-at}

Let the sampling period be T, so x(kT) = kT e^(−akT) and Z[x(t)] = Σ x(kT) z⁻ᵏ.

Start with the standard result (geometric series, |z| > e^(−aT)):

Z[e^(−at)] = Σ (e^(−aT)/z)ᵏ = z/(z − e^(−aT))

Use the multiplication-by-t property, Z[t x(t)] = −T z d/dz X(z):

Z[t e^(−at)] = −T z d/dz [ z/(z − e^(−aT)) ]

d/dz [ z/(z − q) ] = [ (z − q) − z ]/(z − q)² = −q/(z − q)², where q = e^(−aT).

So Z[t e^(−at)] = −T z · (−q)/(z − q)² = T q z/(z − q)².

Answer:

Z[t e^(−at)] = T e^(−aT) z / (z − e^(−aT))²

For T = 1 this is e^(−a) z/(z − e^(−a))². Region of convergence |z| > e^(−aT).