# Infinite impulse response

(Redirected from IIR)

Infinite impulse response (IIR) filters have an impulse response function which is non-zero over an infinite length of time. This is in contrast to finite impulse response filters (FIR) which have fixed-duration impulse responses. The simplest analog IIR filter is an RC filter made up of a single resistor (R) feeding into a node shared with a single capacitor (C). This filter has an exponential impulse response characterized by an RC time constant.

Recursive filters are signal processing filters which re-use one or more output(s) of the filter as inputs. This feedback results in an unending impulse response characterized by either exponentially growing, decaying, or sinusoidal signal output components.

IIR filters may be implemented as either analog or digital filters. In digital IIR filters, the output feedback is immediately apparent in the equations defining the output. Note that unlike with FIR filters, in designing IIR filters it is necessary to carefully consider "time zero" case in which the outputs of the filter have not yet been clearly defined.

In practice electrical engineers find IIR filters to be fast and cheap, but with poorer bandpass filtering and stability characteristics than FIR filters.

Example IIR filters include the Chebyshev filter, Butterworth filter, and the Bessel filter.

 Contents

## Transfer function

In designing IIR filters, engineers often make use of the transfer function:

[itex] H(s) = \frac{Y(s)} {X(s)} [itex]

An IIR filter is typically characterized by its order, which is the number of polynomial terms in the denominator of the transfer function. In digital filter implementations the order is equivalent to the number of time snapshots of the output which are fed back into the input. In analog implementations the order often refers to the number of capacitors or inductors used in the filter circuit.

An IIR filter having [itex]P[itex] feed-forward stages ("b(k)" coefficients) and [itex]Q[itex] feedback ("a(k)" coefficients) stages has the following form:

[itex] y(n) = \left[b(0) x(n) + b(1) x(n-1) + \cdots + b(P) x(n-P)\right] + \left[a(1) y(n-1) + a(2) y(n-2) + \cdots + a(Q) y(n-Q)\right][itex]
[itex] y(n) = \sum_{k=0}^P b(k)x(n-k) + \sum_{k=1}^Q a(k) y(n-k)[itex]

and after taking the z-transform

[itex] H(z) = \frac{\sum_{k=0}^P b(k) z^{-k}} {1 - \sum_{k=1}^Q a(k) z^{-k}} [itex]

See Z-transform#LCCD equation for more details.

## Block diagram

A typical block diagram of an IIR filter looks like the following. The "T" block is a unit delay. The coefficients and number of feedback/feedfoward paths is implementation-dependent.

Missing image
Iir-filter-wiki.png
Image:Iir-filter-wiki.png

## Stability

From the region of convergence of the Z-transform, the stability of an IIR filter can be evaluated. If all poles are within the unit circle (i.e., [itex]|z_p| < 1[itex]) then the filter is stable.

By definition of the region of convergence, then if the system is stable then [itex]\sum_{k=-\infty}^{\infty} |h(k)| < \infty[itex] will hold true. Otherwise, if the system is unstable then the sum will be [itex]\infty[itex].

IIR filters are sometimes preferred over FIR filters because an IIR filter can achieve a much sharper transition region roll-off than an FIR filter of the same order.

## Example

Let the transfer function of a filter H be

[itex]H(z) = \frac{A(z)}{B(z)} = \frac{1}{1 - a z^{-1}}[itex] with ROC [itex]a < |z|[itex] and [itex]0 < a < 1[itex]

which has a pole at a, is stable and causal. The time-domain impulse response is

[itex]h(n) = a^{n} u(n)[itex]

which is non-zero for [itex]n >= 0[itex].

• Art and Cultures
• Countries of the World (http://www.academickids.com/encyclopedia/index.php/Countries)
• Space and Astronomy