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

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! :)

Answer 1

[Walter Roberson]

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.

[Kevin Jansen]

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!

[Kevin Jansen]

It worked! Thanks a lot!