# why is the z transform transfer function 1/(z-1) called an integrator?

I am reading up on delta sigma modulators and there this term $\frac{1}{z-1}$ that appears repeatedly and is referred to as an "integrator". Why is this so ?

## 2 Answers

There are a couple reasons. One is that $(1-z^{-1})$ represents $x[n]-x[n-1]$ which is a finite difference over a very small period of time. and that is an approximation to a differentiator. The reciprocal is

$$\frac{1}{1-z^{-1}} = \frac{z}{z-1}$$

which is the inverse operator. We normally call the inverse operation of differentiation, we call that "integration".

Another reason is simply to implement that term as a transfer function of a tiny little LTI system:

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

or

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

that translates to

$$y[n] - y[n-1] = x[n-1]$$

or

$$y[n] = y[n-1] + x[n-1]$$

so the current output sample is the previous output added to the (slightly delayed) input. the output is an accumulation of the input. similarly an integral of an input $x(t)$ is an accumulation of that input until the present time $t$.

i would say a better representation of a discrete-time integrator is $\frac{z}{z-1}$. that would correspond to

$$y[n] = y[n-1] + x[n]$$

Robert's explanation is nice. Although, for finite difference, 𝑥[𝑛]−𝑥[𝑛−1] translates to 𝑥[𝑛+1]−𝑥[𝑛], which translates to X(z).z^(-1) - X(z) = X(z).{(1-z)/z}. And for the case of actual differentiation, the finite difference will generally be divided by sampling period T. Those other expressions from text books eg. z/(z-1) tend to get lost in translation, which is a life story of lots of teachings in many subjects.