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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?

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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? –  J. M. Sep 23 '11 at 1:22
    
this is for a discrete time signal.... I am sure this is correct... –  rrazd Sep 23 '11 at 1:23
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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. –  J. M. Sep 23 '11 at 1:31

1 Answer 1

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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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. –  Arkamis Jan 9 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. –  Seyhmus Güngören Jan 10 at 1:26

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