Find moving average with filter

This is a class assignment. Teacher is assuming everyone has experience with filters and stuff. I definitely don't althought the rest of the class seems to.

We're supposed to use the "filter" command to find the running average of a vector containing integers from 1 to 10, ie y1 = 0.5*(x1+x2), y2= 0.5*(x2+x3) etc...

My approach would've been:

x = [1:1:10]

for j = 1:10 k = j+1 y(j) = 0.5*(x(j) + k) end

But they want us to use this filter function. I really can't interpret the documentation since I have no idea what a filter is even.

[Matt J]

In my experience, it's a bad idea to take courses you lack prerequisites for.

Answer 1

[Daniel Shub]

The key part of the documentation is:

y(n) = b(1)*x(n) + b(2)*x(n-1) + ... + b(nb+1)*x(n-nb)
               - a(2)*y(n-1) - ... - a(na+1)*y(n-na)

If you let all the a's be zero you get

y(n) = b(1)*x(n) + b(2)*x(n-1) + ... + b(nb+1)*x(n-nb)

If you then let b(1) = 0.5, and b(2) = 0.5, and zero otherwise

y(n) = 0.5*x(n) + 0.5*x(n-1) = 0.5*(x(n)+x(n-1))

Does that help?

Answer 2

[Shashank Prasanna]

Appreciate your honesty. I recommend you go to wikipedia or your favorite signal processing book and take a deep loop at FIR filters.

Pay close attention to the block diagram. Although I will give you the code below, I want you to look at the block diagram and relate it to the "for" loop you wrote. Think about how the coefficients b0 b1 are the 1/2s and z_inverse delays the input in order to sum them. Think hard till it hits you and you will be like ah! and trust me you need this now for the rest of your course. Here is the block diagram from wiki:

http://en.wikipedia.org/wiki/File:FIR_Filter.svg

and the code, go to the documentation of FILTER to see what each argument means.

data = 1:10;
filter([1 1]/2,1,data,1)

http://www.mathworks.com/help/matlab/ref/filter.html

Answer 3

[Matt Kindig]

Or even easier, go to the "Help" in Matlab, and search "moving average filter". The first or second link (the one entitled "Example: Moving Average Filter") will show you exactly how to do this using the 'filter' function.

[Jan]

Searching in the help is an answer for which I cannot vote enough.

Answer 4

[Image Analyst]

Why not simply use conv()?

filteredSignal = conv(originalSignal, [.5, .5], 'valid');

[Daniel Shub]

Presumably because question number 2 requires an IIR filter ...