dimension after fast fourier transform

Question

I am doing a frequency analysis. Reading some literature I have seen that when you perform an fft on a time history of a variable, the dimension of this variable remains the same also after the application of the fft.

In particular, in my case study I have a variable with dimension [L^2/T^2]. Once I perform a fft on this variable the dimensions remain [L^2/T^2].

Now I have two question: 1. Can you confirm that the dimensions does not change after a fft? 2. Can you help me to demonstrate this (the fact that the dimensions does not change after fft) very very rigorously? Unfortunately I have to demonstrate to someone that does not believe it. Thanks

Luca

Answer 1

The Discrete Fourier Transform (DFT) can be written as a matrix-vector multiplication.

$$\text{dft}({\bf v}) = \bf Dv$$

The Fast Fourier Transform (FFT) can be viewed as simply a factorization of the DFT matrix:

$$\text{fft}({\bf v}) = \underset{\bf D}{\underbrace{\bf F_k F_{k-1} \cdots F_2 F_1}} \cdot \bf v$$

Where we first multiply $\bf v$ with $\bf F_1$, then resulting vector by $\bf F_2$ and so on.

If we can prove that this $\bf D$ matrix is invertible then we are done.

It is a quite simple proof showing that $\bf D$ is invertible. Quite easy to show that it is unitary, even. It should be present in any elementary course in Transform Theory. It might even be left as exercise for the student.

For inspiration you can maybe look at this question on Vandermonde matrix invertibility.

Answer 2

Thank you. What I cannot understand now is the fact that , on the contrary, if I perform a Fourier transform, and we look at the formula, the unit changes.

For example if we do the fourier transform of a unit [L^2/T^2], we obtain [L^2/T^2]*[T]=[L^2/T], but if we do the fft the unit does not change (as you demonstrated).

Sorry for all these questions (I am studying it but unfortunately I cannot find easily the answers for these questions)