What is the cut-off frequency of a moving average filter?

I need to design a moving average filter that has a cut-off frequency of 7.8 Hz. I have used moving average filters before, but as far as I'm aware, the only parameter that can be fed in is the number of points to be averaged... How can this relate to a cut-off frequency?

The inverse of 7.8 Hz is ~130 ms, and I'm working with data that are sampled at 1000 Hz. Does this imply that I ought to be using a moving average filter window size of 130 samples, or is there something else that I'm missing here?

Answer

The moving average filter (sometimes known colloquially as a boxcar filter) has a rectangular impulse response:

$$ h[n] = \frac{1}{N}\sum_{k=0}^{N-1} \delta[n-k] $$

Or, stated differently:

$$ h[n] = \begin{cases} \frac{1}{N}, && 0 \le n < N \\ 0, && \text{otherwise} \end{cases} $$

Remembering that a discrete-time system's frequency response is equal to the discrete-time Fourier transform of its impulse response, we can calculate it as follows:

$$ \begin{align} H(\omega) &= \sum_{n=-\infty}^{\infty} x[n] e^{-j\omega n} \\ &= \frac{1}{N}\sum_{n=0}^{N-1} e^{-j\omega n} \end{align} $$

To simplify this, we can use the known formula for the sum of the first $N$ terms of a geometric series:

$$ \sum_{n=0}^{N-1} e^{-j\omega n} = \frac{1-e^{-j \omega N}}{1 - e^{-j\omega}} $$

What we're most interested in for your case is the magnitude response of the filter, $|H(\omega)|$. Using a couple simple manipulations, we can get that in an easier-to-comprehend form:

$$ \begin{align} H(\omega) &= \frac{1}{N}\sum_{n=0}^{N-1} e^{-j\omega n} \\ &= \frac{1}{N} \frac{1-e^{-j \omega N}}{1 - e^{-j\omega}} \\ &= \frac{1}{N} \frac{e^{-j \omega N/2}}{e^{-j \omega/2}} \frac{e^{j\omega N/2} - e^{-j\omega N/2}}{e^{j\omega /2} - e^{-j\omega /2}} \end{align} $$

This may not look any easier to understand. However, due to Euler's identity, recall that:

$$ \sin(\omega) = \frac{e^{j\omega} - e^{-j\omega}}{j2} $$

Therefore, we can write the above as:

$$ \begin{align} H(\omega) &= \frac{1}{N} \frac{e^{-j \omega N/2}}{e^{-j \omega/2}} \frac{j2 \sin\left(\frac{\omega N}{2}\right)}{j2 \sin\left(\frac{\omega}{2}\right)} \\ &= \frac{1}{N} \frac{e^{-j \omega N/2}}{e^{-j \omega/2}} \frac{\sin\left(\frac{\omega N}{2}\right)}{\sin\left(\frac{\omega}{2}\right)} \end{align} $$

As I stated before, what you're really concerned about is the magnitude of the frequency response. So, we can take the magnitude of the above to simplify it further:

$$ |H(\omega)| = \frac{1}{N} \left|\frac{\sin\left(\frac{\omega N}{2}\right)}{\sin\left(\frac{\omega}{2}\right)}\right| $$

Note: We are able to drop the exponential terms out because they don't affect the magnitude of the result; $|e^{j\omega}| = 1$ for all values of $\omega$. Since $|xy| = |x||y|$ for any two finite complex numbers $x$ and $y$, we can conclude that the presence of the exponential terms don't affect the overall magnitude response (instead, they affect the system's phase response).

The resulting function inside the magnitude brackets is a form of a Dirichlet kernel. It is sometimes called a periodic sinc function, because it resembles the sinc function somewhat in appearance, but is periodic instead.

Anyway, since the definition of cutoff frequency is somewhat underspecified (-3 dB point? -6 dB point? first sidelobe null?), you can use the above equation to solve for whatever you need. Specifically, you can do the following:

1. 
   

   Set $|H(\omega)|$ to the value corresponding to the filter response that you want at the cutoff frequency.

   
   


2. 
   

   Set $\omega$ equal to the cutoff frequency. To map a continuous-time frequency to the discrete-time domain, remember that $\omega = 2\pi \frac{f}{f_s}$, where $f_s$ is your sample rate.

   
   
   


3. 
   

   Find the value of $N$ that gives you the best agreement between the left and right hand sides of the equation. That should be the length of your moving average.