# 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 ?

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

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

• Can you please elaborate how the term x(n-1) lead to x(n) in the final output equation ? Note: I am not able to comment on your post since I dont have enough reputation. Jul 30, 2020 at 4:47
• The final equation is $$y[n] = y[n-1] + x[n]$$ and that comes from a transfer function of $$H(z) = \frac{Y(z)}{X(z)} = \frac{z}{z-1} = \frac{1}{1-z^{-1}}$$ that results in $$Y(z)(1-z^{-1}) = X(z)$$ which is the same as $$Y(z) = z^{-1} Y(z) + X(z)$$ Jul 30, 2020 at 13:00

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.

Note $$\frac{z}{z - 1}$$ does not have constant phase of $$90$$ degrees, like a continuous integrator has.

The bilinear integrator $$\frac{z + 1}{z - 1}$$ has $$90$$ degree phase across the whole frequency range. This is used in mapping continuous $$s$$-transform filters to discrete $$z$$-transform filters. It can be extended in an infinite series that converges on the continuous integrator.