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

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

Thus:

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

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.

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  • $\begingroup$ that part of your equation that i deleted was not correct. $\endgroup$ Commented Aug 14, 2017 at 6:53

1 Answer 1

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

means

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

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

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

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

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    $\begingroup$ For completeness one has to take care of the Region of Convergence of the Z-transform $\endgroup$
    – AnVij
    Commented Aug 14, 2017 at 7:04
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    $\begingroup$ 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.) $\endgroup$ Commented Aug 14, 2017 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
    Commented Aug 14, 2017 at 11:39
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    $\begingroup$ 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. $\endgroup$ Commented Aug 14, 2017 at 15:08

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