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Z-transform of t² e^(−at)

Short answer

For x(t)=t²e^(-at), sampled as T²k²(e^(-aT))^k. Using Z[k²] = z(z+1)/(z-1)³ and scaling property, X(z) = T²e^(-aT)z(z+e^(-aT))/(z-e^(-aT))³.

Find the Z-transform of x(t)=t²e⁻ᵃᵗ (for k≥ 0).

Sampled form: x(kT)=(kT)2e−akT=T2k2(e−aT)kx(kT) = (kT)^2 e^{-akT} = T^2 k^2 (e^{-aT})^k. Let r=e−aTr = e^{-aT}.

Step 1: Z[k2]\mathcal{Z}[k^2]

From Z[1]=zz−1\mathcal{Z}[1] = \frac{z}{z-1} and Z[k x(k)]=−zddzX(z)\mathcal{Z}[k\,x(k)] = -z\frac{d}{dz}X(z):

Z[k]=z(z−1)2,Z[k2]=−zddzz(z−1)2=z(z+1)(z−1)3\mathcal{Z}[k] = \frac{z}{(z-1)^2}, \qquad \mathcal{Z}[k^2] = -z\frac{d}{dz}\frac{z}{(z-1)^2} = \frac{z(z+1)}{(z-1)^3}

Step 2: multiply by rkr^k (scaling: replace zz by z/rz/r)

Z[k2rk]=(z/r)(z/r+1)(z/r−1)3=rz(z+r)(z−r)3\mathcal{Z}[k^2 r^k] = \frac{(z/r)(z/r+1)}{(z/r-1)^3} = \frac{rz(z+r)}{(z-r)^3}

Result

X(z)=T2e−aTz (z+e−aT)(z−e−aT)3X(z) = \frac{T^2 e^{-aT} z\,(z + e^{-aT})}{(z - e^{-aT})^3}

Answer: Z[t2e−at]=T2e−aTz(z+e−aT)(z−e−aT)3\mathcal{Z}[t^2e^{-at}] = \dfrac{T^2e^{-aT}z(z+e^{-aT})}{(z-e^{-aT})^3} (equivalently T2e−aTz−1(1+e−aTz−1)(1−e−aTz−1)3\dfrac{T^2e^{-aT}z^{-1}(1+e^{-aT}z^{-1})}{(1-e^{-aT}z^{-1})^3}).

✓ Answer checked by Claude Opus 5.5 (high effort)

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