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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 ?

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3 Answers 3

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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

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

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  • $\begingroup$ 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. $\endgroup$ Jul 30, 2020 at 4:47
  • $\begingroup$ 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 $$ Y(z) = z^{-1} Y(z) + X(z) $$ $\endgroup$ Jul 30, 2020 at 13:00
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Robert's explanation is nice. Although, for finite difference, π‘₯[𝑛]βˆ’π‘₯[π‘›βˆ’1] translates to π‘₯[𝑛+1]βˆ’π‘₯[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.

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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.

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