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?
$\endgroup$ 53 Answers
$\begingroup$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 12.1667There 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} $
$\endgroup$ $\begingroup$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.
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