The frequency response of a typical moving average filter of length $N$ is given by $H(\omega)=\frac{1}{N}\frac{\sin(\omega N/2) e^{-j \omega ((N-1)/2)}}{\sin(\omega/2)}$. Firstly, isn't the cut off frequency of an average filter equal to 0 Hz (an average filter passes the dc value and filters the rest out)? By that notion i will get the value of $H(0)=1/N$ (using L hospitals rule). But i don't think that's right. What am i missing here? Care to explain.


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The cut-off frequency is whatever you define it to be. One standard definition would be the frequency $\omega_c$ at which the attenuation is $3$dB, i.e., $\big|H(\omega_c)\big|=\frac{1}{\sqrt{2}}$.

As far as we know there is no analytical formula for the exact computation of the $3$-dB cut-off frequency of a moving average filter. More than you ever might want to know about approximately computing the $3$-dB cut-off frequency can be found in the answers to this question.

Note that you made a mistake evaluating $H(0)$. The moving average filter satisfies $H(0)=1$ (and not $H(0)=1/N$). The real-valued amplitude function of a length $N$ moving average filter is


Using $\sin(x)\approx x$ for $x\approx 0$ gives

$$A(\omega)\Big|_{\omega\approx 0}\approx\frac{\frac{N\omega}{2}}{N\frac{\omega}{2}}=1\tag{2}$$

A more straightforward way of seeing this is to note that


where $h[n]$ are the filter coefficients. Since for a moving average filter we have $N$ filter coefficients with values $1/N$, the sum in $(3)$ equals $1$.


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