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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

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    $\begingroup$ A Fourier transform is invertible. This means dimensionality can not shrink, because if it did, then we would have lost information on the way and inversion would be impossible. $\endgroup$ Jun 24 '19 at 6:28
  • $\begingroup$ Thank u mathreadler for your answer. Of course I will add this when I will explain this to my supervisor. But I would like to give him a very rigorous demonstration with some mathemathical passages that he can follow. I would like to demonstrate it doing a dimensional analysis of the Fourier Transform formula, but I am not sure about it, because it seems to me that becouse of the differential the units do change. $\endgroup$ Jun 24 '19 at 7:01
  • $\begingroup$ Do you mean units when you say dimension? If you have a function $f(t)$ and perform a DFT to get $\hat{f}(k)$ then if $t$ has unit $T$ then $k$ will have unit $1/T$. $f$ and $\hat{f}$ will have the same dimension. All this follows directly from the definition of a DFT. $\endgroup$
    – Winther
    Jun 26 '19 at 1:59
  • $\begingroup$ Ok thank you Winther. Yes I meant units! Has the fourier transform instead a unit of EU/s (where EU is engineering unit)? $\endgroup$ Jun 26 '19 at 12:05
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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.

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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)

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