How to take irfft of multidimensional array in julia?-CodePudding

I am new to julia, and I am trying to take the irfft of B, which is a 3d array of size (n/2, n, n) where B = rfft(A). However, the irfft in julia reqires an additional input d for the size of the transformed real array, and I'm unsure of what to put. I tried n and n/2, but both did not seem to work as expected when I printed the resulting matrix out.

CodePudding user response：

Check out this discussion. Presumably any triple of numbers will do the trick, but may or may not give you what you want.

CodePudding user response：

This should work:

using FFTW

function test(n = 16)
    a = rand(n ÷ 2, n, n)
    f = rfft(a)
    @show irfft(f, n ÷ 2   1)
end

test()