How can I use conv() in a function handle and keep original vector size, but not use 'same'/'valid'?

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Kevin Jansen
Kevin Jansen 2022-1-24
编辑： Kevin Jansen 2022-1-24
Hi! I am trying to use conv() inside a function handle, while keeping the original vector size of one of the vectors:
So in the example code I convolve a and b, which returns c as [1,4,10,16,22,28,27,18].
I want this to be [1,4,10], as it is the size of a.
This is my code, which does not work:
a = [1,2,3]
b = [1,2,3,4,5,6]
c = @(x) conv(a,b(x(1)))(1:numel(a))
The following does work, but requires multiple lines (which unfortunately is not possible with function handles, so not an option for me):
c = conv(a,b)
c = c(1:numel(a))
If I use 'same' or 'valid', the output is not the same as using conv() and keeping the first columns of the original vector.
using 'same' returns: [16,22,28], using 'valid' returns: [ ], both not [1,4,10].
All help would be appreciated! :)

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Walter Roberson
Walter Roberson 2022-1-24
Ran in:
a = [1,2,3]
a = 1×3
1 2 3
b = [1,2,3,4,5,6]
b = 1×6
1 2 3 4 5 6
FirstN = @(V,N) V(1:N)
FirstN = function_handle with value:
@(V,N)V(1:N)
c = @(x) FirstN(conv(a,b(x(1))), numel(a))
c = function_handle with value:
@(x)FirstN(conv(a,b(x(1))),numel(a))
However... you are using x(1) as an index into b, and x(1) is a scalar, so b(x(1)) would be a scalar, and you would be doing conv() of a vector and a scalar. The result of convolution of a vector and a scalar is the same as multiplying the vector by the scalar. With a = [1 2 3], you cannot get out [1 4 10] with any scalar b(x(1)) as that is not linear scaling.
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Kevin Jansen
Kevin Jansen 2022-1-24
编辑：Kevin Jansen 2022-1-24
Thanks! I will try this out. You are right, I wrote my example wrong. For the convolution in the example I actually did conv(a,b). In my code b is actually equal to poisspdf(1:188,x(1)), that's why I put x(1) there, Thanks again!

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