# IFFT function in julia

**URL:** <https://discourse.julialang.org/t/ifft-function-in-julia/63454>\
**Category:** Signal and Image Processing\
**Tags:** fftw, matlab, matrices\
**Created:** [June 23, 2021, 4:09pm UTC](https://discourse.julialang.org/t/ifft-function-in-julia/63454 "2021-06-23T16:09:37Z")\
**Posts on this page:** 1\
**Showing post:** 8

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**Author:** ![apo383](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/apo383/32/11272_2.png) [@apo383](https://discourse.julialang.org/u/apo383)\
**Post date:** [June 23, 2021, 7:56pm UTC](https://discourse.julialang.org/t/ifft-function-in-julia/63454/8 "2021-06-23T19:56:33Z")

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@Sukera already answered almost the same question for `fft`, which boiled down to the fact that Matlab by default computes the `fft` for each column, whereas Julia computes a multi-dimensional one.

> [@Getting different output for fft function in julia and matlab](https://discourse.julialang.org/t/getting-different-output-for-fft-function-in-julia-and-matlab/63349/4):
>
> You can just give fft the dimensions: help?\> fft fft(A [, dims]) Performs a multidimensional FFT of the array A. The optional dims argument specifies an iterable subset of dimensions (e.g. an integer, range, tuple, or array) to transform along. [...] julia\> m …

The same applies to `ifft`. I’d like to gently suggest looking at the documentation, `?ifft` in Julia and `help ifft` in Matlab.

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