1d Convolution using Matlab's conv() function

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Ba Ba Black Sheep!
Ba Ba Black Sheep! 2018-10-27
编辑： Bruno Luong 2019-9-16
According to the documentation(https://www.mathworks.com/help/matlab/ref/conv.html),
len(output) = len(input) + len(kernel) - 1
So, I figured out
  • In case of conv(u,v,"full"):
len(pad) = len(kernel) - 1
For instance, according to Matlab commandline:
u = [1 2 1 3]
v = [2 0 1]
w = [2 4 3 8 1 3]
Coz,
len(w) = len(u) + len(v) - 1
= 4 + 3 - 1
= 6
len(pad) = len(v) - 1
= 3 - 1
= 2
So, according to calculation:
0 0 1 2 1 3 0 0
1 0 2
---------------
0 0 2 = 2
. . . . . .
. . . . . .
0 0 1 2 1 3 0 0
1 0 2
-------------------
3 0 0 = 3
  • In case of conv(u,v,"same"):
u = [1 2 1 3]
v = [2 0 1]
w = [4 3 8 1]
Coz,
len(w) = len(u)
= 4
len(pad) = floor(len(v) / 2)
= floor(3 / 2)
= 1
So, according to calculation:
0 1 2 1 3 0
1 0 2
-----------
0 0 4 = 4
. . . . . .
. . . . . .
0 1 2 1 3 0
1 0 2
---------------
1 0 0 = 1
But, the problem arises in case of the following example:
u = [1 2 1 3 1]
v = [2 0 1 0]
The following one is okay:
  • In case of conv(u,v,"full"):
w = [2 4 3 8 3 3 1 0]
len(w) = len(u) + len(v) - 1
= 5 + 4 - 1
= 8
len(pad) = len(v) - 1
= 4 - 1
= 3
So,
0 0 0 1 2 1 3 1 0 0 0
0 1 0 2
---------------------
0 0 0 2 = 2
. . . . .
. . . . .
0 0 0 1 2 1 3 1 0 0 0
0 1 0 2
---------------------------
0 0 0 0 = 0
But, the following one has issues:
  • In case of conv(u,v,"same"):
w = [3 8 3 3 1]
Coz,
len(w) = len(u)
= 5
len(pad) = floor(len(v) / 2)
= floor(4 / 2)
= 2
So, according to calculation:
0 0 1 2 1 3 0 0
0 1 0 2
---------------
0 0 0 4 = 4
0 0 1 2 1 3 0 0
0 1 0 2
---------------
0 1 0 2 = 3
0 0 1 2 1 3 0 0
0 1 0 2
---------------
0 2 0 6 = 8
0 0 1 2 1 3 0 0
0 1 0 2
-----------------
0 1 0 0 = 1
0 0 1 2 1 3 0 0
0 1 0 2
-------------------
0 3 0 0 = 3
I.e. output = [4 3 8 1 3] which doesn't match the Matlab output.
What is going on here?

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采纳的回答

Bruno Luong
Bruno Luong 2018-10-28
编辑：Bruno Luong 2019-9-16
Ba Ba Black Sheep wrote
len(pad) = floor(len(v) / 2)
Actually the above is wrong,
In case CONV using with of 'same' option, the 0-pad on the head (left) side of u is:
floor((length(v)-1)/2)
and on the tail (right) side of u is:
ceil((length(v)-1)/2) = floor(length(v)/2)
For v that has odd-length, both are equal. The the result is as if extracted from the FULL result with equal chopping both sides in order to have the same length than the first argument u.
For v that has even-length, the zero-pad has 1 less element on the left than on the right. Therefore the chop size (from FULL result) has 1-more element on the left than on the right.

更多回答（1 个）

David Goodmanson
David Goodmanson 2018-10-28
Hi B^3S,
Your last example, the one you are having problems with, is incorrect. You should be using [1 2 1 3 1] but you are using [1 2 1 3] instead. With the right u,
conv(u,v)
ans = 2 4 3 8 3 3 1 0
For the 'same' option, conv picks the centermost 5 (in this case) elements. It's not documented very well if at all, but when there are an odd number of extra elements on the ends, conv seems to cut out one more unused element on the left hand side than the right hand side. Hence
conv(u,v,'same')
ans = 3 8 3 3 1

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