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Z-transform of X(s) = 1/(s(s+1))

Short answer

Using partial fractions, X(z)= (1-e^{-T})z/((z-1)(z-e^{-T})).

Obtain the z-transform of X(s)=1/s(s+1).

Expand X(s)X(s) in partial fractions, take the z-transform of each sampled time function, and combine.

Step 1: Partial fractions

X(s)=1s(s+1)=1s−1s+1X(s) = \frac{1}{s(s+1)} = \frac{1}{s} - \frac{1}{s+1}

Step 2: Time function and its samples

x(t)=1−e−t,t≥0  ⇒  x(kT)=1−e−kTx(t) = 1 - e^{-t}, \quad t \ge 0 \;\Rightarrow\; x(kT) = 1 - e^{-kT}

Step 3: Z-transform of each term

X(s)X(s)x(kT)x(kT)X(z)X(z)
1s\dfrac{1}{s}1111−z−1\dfrac{1}{1-z^{-1}}
1s+1\dfrac{1}{s+1}e−kTe^{-kT}11−e−Tz−1\dfrac{1}{1-e^{-T}z^{-1}}
X(z)=11−z−1−11−e−Tz−1=(1−e−Tz−1)−(1−z−1)(1−z−1)(1−e−Tz−1)=(1−e−T)z−1(1−z−1)(1−e−Tz−1)\begin{aligned} X(z) &= \frac{1}{1 - z^{-1}} - \frac{1}{1 - e^{-T}z^{-1}} \\ &= \frac{(1 - e^{-T}z^{-1}) - (1 - z^{-1})}{(1-z^{-1})(1-e^{-T}z^{-1})} \\ &= \frac{(1 - e^{-T})z^{-1}}{(1-z^{-1})(1-e^{-T}z^{-1})} \end{aligned}

In positive powers of zz:

X(z)=(1−e−T)z(z−1)(z−e−T)X(z) = \frac{(1 - e^{-T})z}{(z-1)(z - e^{-T})}

Numerical form for T=1T = 1 s

e−1=0.3679e^{-1} = 0.3679, so

X(z)=0.6321z(z−1)(z−0.3679)=0.6321zz2−1.3679z+0.3679X(z) = \frac{0.6321z}{(z-1)(z-0.3679)} = \frac{0.6321z}{z^2 - 1.3679z + 0.3679}

Check

  • Initial value: lim⁡z→∞X(z)=0=x(0)\lim_{z\to\infty}X(z) = 0 = x(0).
  • Final value: lim⁡z→1(1−z−1)X(z)=1−e−T1−e−T=1=x(∞)\lim_{z\to1}(1-z^{-1})X(z) = \dfrac{1-e^{-T}}{1-e^{-T}} = 1 = x(\infty).

Answer: X(z)=(1−e−T)z−1(1−z−1)(1−e−Tz−1)=(1−e−T)z(z−1)(z−e−T)X(z) = \dfrac{(1-e^{-T})z^{-1}}{(1-z^{-1})(1-e^{-T}z^{-1})} = \dfrac{(1-e^{-T})z}{(z-1)(z-e^{-T})}.

✓ Answer checked by Claude Opus 5.5 (high effort)

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