How can get the difference equation of a $\mathcal{Z}$-transform transfer function with time delay? How does a time delay influence the difference equation?

For example:

$$H(z) = \frac{8z^{94}}{z-0.9}$$ where $H(z)=\frac{Y(z)}{U(z)}$

I want to get the difference equation in terms of $y[k+1]$

  • 1
    $\begingroup$ The output of that filter, as presented, responds to inputs 93 steps into the future. Did you mean the numerator to be $8x^{-94}$, or the denominator to be $z^{93}(z - 0.9)$, perhaps? Please edit your question if so. $\endgroup$
    – TimWescott
    Commented Dec 1, 2022 at 22:39

1 Answer 1


The z transform for a time difference is given as:

$$y[n-k] \Leftrightarrow Y(z)z^{-k}$$

The transfer function $H(z)$ is related to the input $X(z)$ and output $Y(z)$ as:

$$H(z) = \frac{Y(z)}{X(z)}$$

Putting the transfer function in terms of negative powers of z (by dividing numerator and denominator by the same powers of z) makes it very easy to then get the difference equation in terms of delayed copies of the output and the input. This is likely a homework problem so rather than solving it directly I will show the solution for a different example (leaving the OP to solve the specific case given as an exercise, and to see how the result for that will also be non-causal):

$$H(z)= \frac{z^5}{z^2-2z+1}$$

Multiply numerator and denominator by $1/z^2$:

$$H(z)= \frac{z^3}{1-2z^{-1}+z^{-2}}$$

$$\frac{Y(z)}{X(z)}= \frac{z^3}{1-2z^{-1}+z^{-2}}$$

$$Y(z)(1-2z^{-1}+z^{-2}) = X(z)z^3$$

$$Y(z)-2Y(z)z^{-1}+Y(z)z^{-2} = X(z)z^3$$

Now from the first relationship given we easily get the difference equation:


Or in terms of what is output $y[n]$ given in this case a future input (non causal!) and past outputs:



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