How can I derive the general response of an IIR filter from its transfer function? I know that:

$$H(z)=\frac{1}{1+\sum\limits_{m=1}^N{a_m z^{-m}}}$$


$$Y(z)=X(z)H(z)=\frac{X(z)}{1+\sum\limits_{m=1}^N{a_m z^{-m}}}$$

The general response is:

$$y[n]=-\sum\limits_{m=1}^N{a_m y[n-m]}+x[n]$$

Where the above expression comes from?

Thank you for your time.

  • $\begingroup$ that part of your equation that i deleted was not correct. $\endgroup$ Aug 14 '17 at 6:53

$$Y(z)=X(z)H(z)=\frac{X(z)}{1+\sum\limits_{m=1}^N{a_m z^{-m}}}$$


$$\left(1+\sum\limits_{m=1}^N{a_m z^{-m}} \right)Y(z) = X(z)$$

$$Y(z) + \sum\limits_{m=1}^N{a_m Y(z) z^{-m}} = X(z)$$

$$y[n] + \sum\limits_{m=1}^N{a_m y[n-m]} = x[m]$$

$$y[n] = - \sum\limits_{m=1}^N{a_m y[n-m]} + x[m]$$

  • 1
    $\begingroup$ For completeness one has to take care of the Region of Convergence of the Z-transform $\endgroup$
    – AnVij
    Aug 14 '17 at 7:04
  • 2
    $\begingroup$ i don't see why. (no one is saying it's stable. but it's simply how this all-pole filter is implemented from the given transfer function.) $\endgroup$ Aug 14 '17 at 7:19
  • $\begingroup$ It's not about stability. The map from signal to z-domain is only invertible if you include the region of convergence with the z-domain representation. The region of convergence is implied if you want a causal impulse response, but it should at least be mentioned. $\endgroup$
    – Jazzmaniac
    Aug 14 '17 at 11:39
  • 1
    $\begingroup$ it's bullshit. there is no other mapping from $Y(z)$ than to $y[n]$ and there is no other mapping from $Y(z) z^{-m}$ than to $y[n-m]$ and there is no other mapping from $X(z)$ than to $x[n]$. you don't need ROC to apply the fundamentals of linearity (additivity and homogeneity). you don't need ROC in any manner to answer this question. $\endgroup$ Aug 14 '17 at 15:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.