Z-transform of a running sum

$y(k)$ is the running sum (accumulation) of $x(k)$. Its z-transform is $Y(z) = \dfrac{X(z)}{1-z^{-1}} = \dfrac{z}{z-1}X(z)$.

Step 1: Write a difference equation

$$\begin{aligned} y(k) &= x(0) + x(1) + \dots + x(k-1) + x(k) \\ y(k-1) &= x(0) + x(1) + \dots + x(k-1) \end{aligned}$$

Subtracting:

$$y(k) - y(k-1) = x(k), \qquad k = 0, 1, 2, \dots$$

This also holds at $k = 0$ because $y(-1) = 0$ (given $y(k) = 0$ for $k < 0$), so $y(0) = x(0)$.

Step 2: Take the z-transform

Use the right-shift (delay) theorem $\mathcal{Z}[y(k-1)] = z^{-1}Y(z)$, which needs $y(k) = 0$ for $k < 0$ (given):

$$Y(z) - z^{-1}Y(z) = X(z)$$ $$Y(z) = \frac{1}{1-z^{-1}}X(z) = \frac{z}{z-1}X(z)$$

Alternative view (convolution)

$y(k) = \sum_{h=0}^{k} x(h)\cdot 1(k-h)$ is the convolution of $x(k)$ with the unit step $1(k)$. Convolution in time is multiplication in $z$:

$$Y(z) = X(z)\cdot \mathcal{Z}[1(k)] = X(z)\cdot\frac{1}{1-z^{-1}}$$

Example

Let $x(k) = 1$ for all $k \ge 0$ (unit step), so $X(z) = \dfrac{z}{z-1}$. Then $y(k) = k+1$ and

$$Y(z) = \frac{z}{z-1}\cdot\frac{z}{z-1} = \frac{z^2}{(z-1)^2}$$

Check: $\mathcal{Z}[k+1] = \dfrac{z}{(z-1)^2} + \dfrac{z}{z-1} = \dfrac{z + z(z-1)}{(z-1)^2} = \dfrac{z^2}{(z-1)^2}$, which agrees.

Remarks

• $\dfrac{1}{1-z^{-1}}$ is the pulse transfer function of a digital accumulator (summer). It has a pole at $z = 1$.
• This result is the discrete version of integration ($1/s$) and is used in the integral term of a digital PID controller.

Answer: $Y(z) = \dfrac{X(z)}{1 - z^{-1}} = \dfrac{z\,X(z)}{z-1}$.