Editing Finite impulse response (section)

==Moving average example==
{{multiple image   
| direction = vertical   
| width     = 300   
| footer    =    
| image1    = FIR Filter (Moving Average).svg
| alt1      = Block diagram of a simple FIR filter (second-order/3-tap filter in this case, implementing a moving average smoothing filter)
| caption1  = Block diagram of a simple FIR filter (second-order/3-tap filter in this case, implementing a moving average smoothing filter)

| image2    = MA2PoleZero C.svg   
| alt2      = Pole–zero diagram   
| caption2  = [[Pole–zero diagram]] of the example second-order FIR smoothing filter

| image3    = Frequency_response_of_3-term_boxcar_filter.svg   
| alt3      = Magnitude and phase responses of the example second-order FIR smoothing filter
| caption3  = Magnitude and phase responses of the example second-order FIR smoothing filter

| image4    = Amplitude & phase vs frequency for 3-term boxcar filter.svg   
| alt4      = Amplitude and phase responses of the example second-order FIR smoothing filter
| caption4  = Amplitude and phase responses of the example second-order FIR smoothing filter
}}

A [[moving average]] filter is a very simple FIR filter. It is sometimes called a [[Boxcar function|boxcar]] filter, especially when followed by [[Decimation (signal processing)|decimation]], or a [[Sinc filter#Frequency-domain sinc|sinc-in-frequency]]. The filter coefficients, <math display="inline"> b_0, \ldots, b_N</math>, are found via the following equation:
:<math>b_i=\frac{1}{N+1}</math>

To provide a more specific example, we select the filter order: 
:<math>N = 2</math>

The impulse response of the resulting filter is''':'''
:<math>h[n] = \frac{1}{3}\delta[n] + \frac{1}{3}\delta[n-1] + \frac{1}{3}\delta[n-2]</math>

The block diagram on the right shows the second-order moving-average filter discussed below. The transfer function is''':'''

:<math>H(z) = \frac{1}{3} + \frac{1}{3}z^{-1} + \frac{1}{3}z^{-2} = \frac{1}{3}\frac{z^{2} + z + 1}{z^{2}}.</math>

The next figure shows the corresponding [[pole–zero diagram]].  Zero frequency (DC) corresponds to (1, 0), positive frequencies advancing counterclockwise around the circle to the Nyquist frequency at (−1, 0).  Two poles are located at the origin, and two zeros are located at <math display="inline"> z_1 = -\frac{1}{2} + j\frac{\sqrt{3}}{2}</math>, <math display="inline"> z_2 = -\frac{1}{2} - j \frac{\sqrt{3}}{2}</math>.

The frequency response, in terms of [[Normalized frequency (digital signal processing)|normalized frequency]] ''ω'', is''':'''
:<math>
\begin{align}
H\left(e^{j\omega}\right) &= \frac{1}{3} + \frac{1}{3}e^{-j\omega} + \frac{1}{3}e^{-j2\omega}\\
&= \frac{1}{3}e^{-j\omega}\left(1+2\cos(\omega)\right).
\end{align}
</math>

The magnitude and phase components of <math display="inline"> H\left(e^{j\omega}\right)</math> are plotted in the figure. But plots like these can also be generated by doing a [[discrete Fourier transform]] (DFT) of the impulse response.{{
efn-ua|See {{slink|Discrete-time Fourier transform|Sampling the DTFT|nopage=y}}.
}} 
And because of symmetry, filter design or viewing software often displays only the [0, π] region. The magnitude plot indicates that the moving-average filter passes low frequencies with a gain near 1 and attenuates high frequencies, and is thus a crude [[low-pass filter]]. The phase plot is linear except for discontinuities at the two frequencies where the magnitude goes to zero.  The size of the discontinuities is π, representing a sign reversal. They do not affect the property of linear phase, as illustrated in the final figure.