Filtering code- unknown filter

4 次查看（过去 30 天）
Coo Boo
Coo Boo 2012-8-17
Hi to all,
I have a code which filters the signal 'x' with length of N, but I don't know what kind of filtering that is? The code is:
%Filtering.
fL=5; % Setting fL
if fL>1
for i=fL+1:N
x(i-fL)=0;
for j=1:fL
x(i-fL) = x(i-fL)+(1/fL)*x(i-j+1);
end
end
end
Can you help me to know that? thanks,

请先登录，再回答此问题。

采纳的回答

Matt Fig
Matt Fig 2012-8-17
编辑：Matt Fig 2012-8-17
It is a forward moving average. That is, the value at index I is the average of the fL values I+1:I+fL. Like other moving averages, it can be considered a low-pass filter.
Here is a simplification of the code that makes this more clear:
y = x; % Compare final x with final y after both loops.
for jj = 1:length(y)-fL
y(jj) = sum(y(jj+1:jj+fL)/fL)
end
What is interesting is that it stops short of carrying to the end of the vector. Most moving averages carry through with less terms as necessary, like this:
for jj = 1:length(y)-1
y(jj) = sum(y(jj+1:min(jj+fL,length(y)))/fL) ;
end
Also y above to compare:
MA = fliplr(filter(ones(1,5)/5,1,fliplr(y)));
only the first and the last elements differ because of the way MAs work.

更多回答（1 个）

Azzi Abdelmalek
Azzi Abdelmalek 2012-8-17
编辑：Azzi Abdelmalek 2012-8-17
x=rand(104,1);y=x
N=size(x,1);
fL=5; % Setting fL
if fL>1
for i=fL+1:N
x(i-fL)=0;
for j=1:fL
x(i-fL) = x(i-fL)+(1/fL)*x(i-j+1);
end
end
end
cla;plot(abs(fft(x)));hold on ;plot(abs(fft(y)),'r')
it seems to be a low pass filter, comparing the Fourier transform of x and y
your original: red
your filtred signal: blue
  3 个评论
显示 1更早的评论
Azzi Abdelmalek
Azzi Abdelmalek 2012-8-17
just look at the half of the plot (from 0 to 52), because froù 53 to 104 it's just a symetric
Coo Boo
Coo Boo 2012-8-17
编辑：Coo Boo 2012-8-17
I think, this will show better for bigger value of fL,e.g. fL=20
plot(abs(fft(x(1:N-fL))));hold on ;plot(abs(fft(y)),'r')

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Single-Rate Filters 的更多信息

标签

产品

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by