# What is a moving average system?

Can someone elaborate on what a moving average system is?

I know that the system is defined as: $$y[n] = \frac{x[n] + x[n-1] + x[n-2]}{3}$$ How would we draw $y[n]$ given that we have a graph with discrete values for $x[n]$? Can someone actually draw a sample discrete time $x[n]$ graph and show how the corresponding $y[n]$ graph is generated?

• Huh, wait. The form I'm accustomed to goes like $\dfrac{x_{n-1}+x_n+x_{n+1}}{3}$. The form you gave is the one usually used for endpoints... could you check your source again? Commented Sep 23, 2011 at 1:22
• this is for a discrete time signal.... I am sure this is correct... Commented Sep 23, 2011 at 1:23
• Anyway... you're already aware of the "sliding window" analogy, right? Say you have the sequence $(x_1,x_2,x_3,x_4,x_5,x_6,x_7)$. The first pass replaces $x_3$ with the mean of $x_1,x_2,x_3$. Then, you shift by one place, replacing $x_4$ with the mean of $x_2,x_3,x_4$, and so on, up until you're replacing $x_7$ with the mean of $x_5,x_6,x_7$. Note that the group of points being taken always has overlap with the previous one. Commented Sep 23, 2011 at 1:31
• @Guesswhoitis. the OP is a causal filter, yours is not. the way to see how this works is to consider a sliding window of 3 samples and at each point, the output $y[n]$ is the mean of the previous 3 input samples $x[n],x[n-1],x[n-2]$ in the window Commented Sep 9, 2015 at 20:02
• @oldrinb, that's why I provided the second comment; I know that there are a number of moving averages; I just mentioned the one I'm used to. Commented Oct 6, 2015 at 12:53

Let's consider a few terms of the sequence $y$:

$$y[0] = \frac{x[0]+x[-1]+x[-2]}{3}$$

$$y[1] = \frac{x[1]+x[0]+x[-1]}{3}$$

$$y[2] = \frac{x[2]+x[1]+x[0]}{3}$$

Notice how the values of $y$ are always an average of three values. Also, notice how the indices of $x$ "shift" to the right in the expressions, and the next value gets shifted in. This is as if we have a longer sequence, $\{x[-2],x[-1],x[0],x[1],x[2]\}$, and we have a window of three consecutive values of the sequence, and the window shifts over one place to the right for each term in the $y$ sequence. $$\{\color{red}{x[-2],x[-1],x[0]},x[1],x[2]\}$$ $$\{x[-2],\color{red}{x[-1],x[0],x[1]},x[2]\}$$ $$\{x[-2],x[-1],\color{red}{x[0],x[1],x[2]}\}$$

This is sometimes called a sliding-window average as well because of this property.

As for a concrete example, let's consider the sequence $$x[n] = n^2/10, \;\; n\geq 0 \\ x[n] = 0, \;\; n<0$$ This is plotted below.

Now, if you do the calculations, you average the first three points, then the second three points, then the third three points, etc. I leave the calculations out, but the result it as follows:

The corresponding numerical values are given in the table below.

n   x       y
-2  0       0
-1  0       0.0333
0   0       0.1667
1   0.1     0.4667
2   0.4     0.9667
3   0.9     1.6667
4   1.6     2.5667
5   2.5     3.6667
6   3.6     4.9667
7   4.9     6.4667
8   6.4     8.1667
9   8.1     10.0667
10  10      12.1667

• I know this answer is quite old, but if you don't mind, can you list down the points being averaged in the graph? I am blind and I cannot read the graph. Can you list down the values in the points? Thanks! Commented Mar 24, 2019 at 15:40
• I've never thought about the accessibility of my posts to blind people; thank you for expanding my horizons! I'll answer you in a comment here and try to merge the comment into the answer soon. The first un-averaged list of numbers, beginning at n=-2, is 0, 0, 0, 0.1, 0.4, 0.9, 1.6, 2.5, 3.6, 4.9, 6.4, 8.1, 10. The three-point averaged numbers, starting at the same n=-2, are 0, 0.033333333, 0.166666667, 0.466666667, 0.966666667, 1.666666667, 2.566666667, 3.666666667, 4.966666667, 6.466666667, 8.166666667, 10.06666667, 12.16666667. Commented Apr 25, 2019 at 18:35

There is an efficient way to compute these moving averages.

Consider the $k$-long moving average $y(n) =\frac1{k}\sum_{j=0}^{k-1} x(n-j)$. Once you have a particular $y(n)$, then

$\begin{array}\\ y(n+1)-y(n) &=\frac1{k}\sum_{j=0}^{k-1} x(n+1-j)-\frac1{k}\sum_{j=0}^{k-1} x(n-j)\\ &=\frac1{k}\left(\sum_{j=0}^{k-1} x(n+1-j)-\sum_{j=0}^{k-1} x(n-j)\right)\\ &=\frac1{k}\left(\sum_{j=-1}^{k-2} x(n-j)-\sum_{j=0}^{k-1} x(n-j)\right)\\ &=\frac1{k}\left(x(n+1)+\sum_{j=0}^{k-2} x(n-j) -(\sum_{j=0}^{k-2} x(n-j)+x(n-k+1)\right)\\ &=\frac1{k}\left(x(n+1)-x(n-k+1)\right)\\ \end{array}$

You need to convolve your data $\mathbf{x}$ with the impluse response of the corresponding FIR filter $\mathbf{h}$. You can learn the details from here.

• This answer would be better if the relevant source material were summarized, rather than using a non-permanent link to the top level of a 1700-word Wiki page. Alternatively, this is well-suited as a comment. Commented Jan 9, 2015 at 19:55
• @Arkamis I would write definitely more if the question would not be asked 3 years ago. I can write alot and it is a loss of time if the OP is already away. It is therefore better to spend less time at the beginning and if the OP is still alive then he/she will definitely contact you for details. Commented Jan 10, 2015 at 1:26
• While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. Commented Jun 29, 2015 at 21:25
• @AlonsoDelfín yes more explanation can be given if the OP has interest in it.. No interest basically no answer but still I gave an answer.. Next step will be to delete it, if more requests will arise. Commented Jun 29, 2015 at 21:55
• The review was divided. It's more than a link-only answer, but it could profit from being expanded. Commented Jun 29, 2015 at 22:12