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X(z) via convolution integral, repeated pole

Short answer

Sum residues of X(s)z/(z-e^{sT}) at poles s=-1 (double), s=-2, s=-3. For T=1: X(z)=0.1839z/(z-0.3679)^2 - 0.75z/(z-0.3679) + z/(z-0.1353) - 0.25z/(z-0.0498).

Given X(s)=1/((s+1)²(s+3)(s+2)) and T=1. Obtain X(z) by using convolution integral in left half plane.

Use the convolution integral closed in the left half-plane. The z-transform is the sum of residues of X(s)zz−esTX(s)\dfrac{z}{z-e^{sT}} at the poles of X(s)X(s). This is valid because X(s)X(s) has 4 poles and no zeros.

X(z)=∑poles of X(s)Res[1(s+1)2(s+2)(s+3)⋅zz−esT]X(z) = \sum_{\text{poles of } X(s)}\text{Res}\left[\frac{1}{(s+1)^2(s+2)(s+3)}\cdot\frac{z}{z - e^{sT}}\right]

Poles: a double pole at s=−1s = -1, and simple poles at s=−2s = -2 and s=−3s = -3.

Residue at s=−2s = -2

R2=[1(s+1)2(s+3)⋅zz−esT]s=−2=1(1)(1)⋅zz−e−2T=zz−e−2TR_2 = \left[\frac{1}{(s+1)^2(s+3)}\cdot\frac{z}{z-e^{sT}}\right]_{s=-2} = \frac{1}{(1)(1)}\cdot\frac{z}{z - e^{-2T}} = \frac{z}{z - e^{-2T}}

Residue at s=−3s = -3

R3=[1(s+1)2(s+2)⋅zz−esT]s=−3=1(4)(−1)⋅zz−e−3T=−14⋅zz−e−3TR_3 = \left[\frac{1}{(s+1)^2(s+2)}\cdot\frac{z}{z-e^{sT}}\right]_{s=-3} = \frac{1}{(4)(-1)}\cdot\frac{z}{z - e^{-3T}} = -\frac{1}{4}\cdot\frac{z}{z - e^{-3T}}

Residue at the double pole s=−1s = -1

Let F(s)=1(s+2)(s+3)F(s) = \dfrac{1}{(s+2)(s+3)}, so F(−1)=12F(-1) = \dfrac12 and

F′(s)=−2s+5(s+2)2(s+3)2  ⇒  F′(−1)=−34F'(s) = -\frac{2s+5}{(s+2)^2(s+3)^2} \;\Rightarrow\; F'(-1) = -\frac{3}{4} R1=dds[F(s)zz−esT]s=−1=F′(−1)zz−e−T+F(−1)Tze−T(z−e−T)2=−34⋅zz−e−T+Te−T2⋅z(z−e−T)2\begin{aligned} R_1 &= \frac{d}{ds}\left[F(s)\frac{z}{z-e^{sT}}\right]_{s=-1} = F'(-1)\frac{z}{z-e^{-T}} + F(-1)\frac{Tze^{-T}}{(z-e^{-T})^2} \\ &= -\frac{3}{4}\cdot\frac{z}{z-e^{-T}} + \frac{T e^{-T}}{2}\cdot\frac{z}{(z-e^{-T})^2} \end{aligned}

General result

X(z)=Te−T2⋅z(z−e−T)2−34⋅zz−e−T+zz−e−2T−14⋅zz−e−3TX(z) = \frac{Te^{-T}}{2}\cdot\frac{z}{(z-e^{-T})^2} - \frac{3}{4}\cdot\frac{z}{z-e^{-T}} + \frac{z}{z-e^{-2T}} - \frac{1}{4}\cdot\frac{z}{z-e^{-3T}}

Substitute T=1T = 1

e−1=0.36788e^{-1} = 0.36788, e−2=0.13534e^{-2} = 0.13534, e−3=0.04979e^{-3} = 0.04979, and Te−T2=0.18394\frac{Te^{-T}}{2} = 0.18394:

X(z)=0.18394z(z−0.36788)2−0.75zz−0.36788+zz−0.13534−0.25zz−0.04979X(z) = \frac{0.18394z}{(z-0.36788)^2} - \frac{0.75z}{z-0.36788} + \frac{z}{z-0.13534} - \frac{0.25z}{z-0.04979}

Combined into a single ratio:

X(z)=0.030919z3+0.023057z2+0.000939zz4−0.92088z3+0.27828z2−0.030011z+0.000912X(z) = \frac{0.030919z^3 + 0.023057z^2 + 0.000939z}{z^4 - 0.92088z^3 + 0.27828z^2 - 0.030011z + 0.000912}

The denominator is (z−0.36788)2(z−0.13534)(z−0.04979)(z-0.36788)^2(z-0.13534)(z-0.04979).

Check

The inverse Laplace transform is x(t)=14[(2t−3)e−t+4e−2t−e−3t]x(t) = \tfrac14\left[(2t-3)e^{-t} + 4e^{-2t} - e^{-3t}\right]. Its samples are:

kk0123
x(k)x(k)00.030920.051530.03979

Long division of X(z)X(z) gives 0.03092z−1+0.05153z−2+0.03979z−3+…0.03092z^{-1} + 0.05153z^{-2} + 0.03979z^{-3} + \dots, which matches.

Answer: X(z)=0.18394z(z−0.3679)2−0.75zz−0.3679+zz−0.1353−0.25zz−0.0498X(z) = \dfrac{0.18394z}{(z-0.3679)^2} - \dfrac{0.75z}{z-0.3679} + \dfrac{z}{z-0.1353} - \dfrac{0.25z}{z-0.0498}.

✓ Answer checked by Claude Opus 5.5 (high effort)

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