Convolution without any Built-in Commands
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Hey guys, I'm trying to learn how convolution works without any built in fft or
conv commands, and I'm not quite sure how to write it. I'm starting at index 1 in
this case, so y(n) =(x*h)(n)= sum from 1 to infinity of (x(k)h(n-k+1)), and
trying to use just arrays and no loops, while also creating a stem plot. The lack
of a for loop is giving me the most trouble, any suggestions?
3 件のコメント
Matt Fig
2012 年 11 月 24 日
Sounds like homework. Please show the code you have written so far.
SB
2012 年 11 月 24 日
編集済み: Image Analyst
2012 年 11 月 25 日
SM60
2019 年 3 月 7 日
please do deconv() function using for loops
回答 (2 件)
Matt J
2012 年 11 月 24 日
TOEPLITZ isn't a built-in command. You could use that to implement convolution, e.g.,
>> A=toeplitz([2 3, zeros(1,8)],[2 1 0 0 0 0 0 0 0 0 ]), b=rand(10,1);
A =
2 1 0 0 0 0 0 0 0 0
3 2 1 0 0 0 0 0 0 0
0 3 2 1 0 0 0 0 0 0
0 0 3 2 1 0 0 0 0 0
0 0 0 3 2 1 0 0 0 0
0 0 0 0 3 2 1 0 0 0
0 0 0 0 0 3 2 1 0 0
0 0 0 0 0 0 3 2 1 0
0 0 0 0 0 0 0 3 2 1
0 0 0 0 0 0 0 0 3 2
>> A*b-conv(b,1:3,'same')
ans =
1.0e-15 *
0
0.4441
0.8882
0
0
0.8882
-0.4441
0
0
0
3 件のコメント
Matt J
2012 年 11 月 25 日
Why don't you test it against the output of conv() to see if you get the same results?
Image Analyst
2012 年 11 月 25 日
I think you mean x(n) ** h(n), which is the usual textbook notation for convolution, rather than (x*h)(n). The code it's not exactly the way I'd do it (padding with zeros, etc.) but it's easy enough to test, like Matt suggested. I would use the double for loop though.
Jan
2012 年 11 月 25 日
Even CONV is not a built-in function. But it calls conv2(), which is built-in now.
conv(a,b) can be calculated using:
c = filter(a, 1, b);
when the shorter of the inputs is padded by zeros. And filter() can by run as M-file also, see http://www.mathworks.com/matlabcentral/answers/9900#answer_13623. And considering the special inputs, this code can be improved.
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