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Inverse Z-transform by inversion integral

Short answer

For X(z) = z(1-e^(-aT))/((z-1)(z-e^(-aT))), compute residues of X(z)z^(k-1) at z=1 and z=e^(-aT). Residue at z=1 is 1; at z=e^(-aT) is –e^(-akT). Summing residues gives x(kT) = 1 – e^(-akT).

Obtain the inverse z-transform of X(z)=z(1-e⁻ᵃᵀ)/((z-1)(z-e⁻ᵃᵀ)) using inversion integral method.

The inversion integral gives

x(k)=12πj∮CX(z)zk−1 dz=∑residues of X(z)zk−1 at its polesx(k) = \frac{1}{2\pi j}\oint_C X(z)z^{k-1}\,dz = \sum \text{residues of } X(z)z^{k-1} \text{ at its poles}

Form X(z)zk−1X(z)z^{k-1}

X(z)zk−1=(1−e−aT) zk(z−1)(z−e−aT)X(z)z^{k-1} = \frac{(1-e^{-aT})\,z^k}{(z-1)(z-e^{-aT})}

For k=0,1,2,…k = 0, 1, 2, \dots there is no pole at z=0z=0; the poles are z=1z = 1 and z=e−aTz = e^{-aT} (both simple, assuming a>0a > 0).

Residues

At z=1z = 1:

K1=(z−1)(1−e−aT)zk(z−1)(z−e−aT)∣z=1=1−e−aT1−e−aT=1K_1 = \left.(z-1)\frac{(1-e^{-aT})z^k}{(z-1)(z-e^{-aT})}\right|_{z=1} = \frac{1-e^{-aT}}{1-e^{-aT}} = 1

At z=e−aTz = e^{-aT}:

K2=(1−e−aT)zkz−1∣z=e−aT=(1−e−aT)e−akTe−aT−1=−e−akTK_2 = \left.\frac{(1-e^{-aT})z^k}{z-1}\right|_{z=e^{-aT}} = \frac{(1-e^{-aT})e^{-akT}}{e^{-aT}-1} = -e^{-akT}

Result

x(k)=K1+K2=1−e−akT,k=0,1,2,…x(k) = K_1 + K_2 = 1 - e^{-akT}, \quad k = 0, 1, 2, \dots

Check: x(0)=0x(0) = 0 and lim⁡z→∞X(z)=0\lim_{z\to\infty}X(z) = 0 (initial value theorem) agree; x(∞)=1x(\infty) = 1 and lim⁡z→1(z−1)X(z)=1\lim_{z\to1}(z-1)X(z) = 1 (final value theorem) agree. This is the sampled form of x(t)=1−e−atx(t) = 1 - e^{-at}.

Answer: x(kT)=1−e−akTx(kT) = 1 - e^{-akT}.

✓ Answer checked by Claude Opus 5.5 (high effort)

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