Skip to main content

Inverse Z-transform of z(z+2)/(z−1)²

Short answer

X(z)z^(k-1) = z^k(z+2)/(z-1)² has a double pole at z=1. The residue is lim_{z→1} d/dz[z^(k+1)+2z^k] = (k+1)+2k = 3k+1. Hence x(k) = 3k + 1.

Obtain the inverse Z-transform of X(z)=z(z+2)/(z-1)² by inversion integral method.

By the inversion integral method,

x(k)=∑residues of X(z)zk−1 at the poles of X(z)zk−1x(k) = \sum \text{residues of } X(z)z^{k-1} \text{ at the poles of } X(z)z^{k-1} X(z)zk−1=z(z+2)zk−1(z−1)2=zk(z+2)(z−1)2X(z)z^{k-1} = \frac{z(z+2)z^{k-1}}{(z-1)^2} = \frac{z^k(z+2)}{(z-1)^2}

For k≥0k \ge 0 the only pole is a double pole at z=1z = 1 (no pole at the origin).

Residue at a double pole

x(k)=1(2−1)!lim⁡z→1ddz[(z−1)2zk(z+2)(z−1)2]=lim⁡z→1ddz[zk+1+2zk]=lim⁡z→1[(k+1)zk+2kzk−1]=(k+1)+2k=3k+1\begin{aligned} x(k) &= \frac{1}{(2-1)!}\lim_{z\to1}\frac{d}{dz}\left[(z-1)^2\frac{z^k(z+2)}{(z-1)^2}\right] \\ &= \lim_{z\to1}\frac{d}{dz}\left[z^{k+1} + 2z^k\right] \\ &= \lim_{z\to1}\left[(k+1)z^k + 2kz^{k-1}\right] \\ &= (k+1) + 2k = 3k + 1 \end{aligned}

Check

k0123
3k+13k+114710

Long division: X(z)=z2+2zz2−2z+1=1+4z−1+7z−2+10z−3+⋯X(z) = \frac{z^2+2z}{z^2-2z+1} = 1 + 4z^{-1} + 7z^{-2} + 10z^{-3} + \cdots, which agrees.

Answer: x(k)=3k+1x(k) = 3k + 1, k=0,1,2,…k = 0, 1, 2, \dots

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Digital Control System chapter 2 questions →