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Using the FFT to mimic the Fourier Transform I have this question.

The definition that I use for the PSD $S(f)$ is:

$$S(f)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\exp\left(-i\tau2\pi f\right)r(\tau)d\tau$$

where $r(t)$ is the correlation function.

The FFT in MATLAB use this formula

$$X(k) = \sum_{n=1}^N x(n)\exp\left(-j2\pi\frac{(k-1)(n-1)}{N}\right), \qquad 1 \leq k \leq N.$$

So in order to replace the PSD I did:

S=abs(fft(r))*delta_t

where delta_t is the time step. In other words, translating the code in formula I did:

$$S(k)=\sum_{n=1}^{N}\exp\left(-i2\pi \frac{(k-1)(n-1)}{N}\right)r(n)\Delta T$$

After using this formula I applied this, checking the result $$\int_{-\infty}^{\infty}S(f)df=r.m.s. \big\{x(t)\big\}$$

where $x(t)$ is the original signal. This formula is perfectly verified.

Now the problem is that I noticed that in the MATLAB's fft there is no coefficient $\frac{1}{2\pi}$.

Can you give me an explanation of why this happens? I know that there are a lot of different coefficient that I can use for the Fourier transform pair, but I don't know if this is related to my problem.

Thanks

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I would simply say that it doesn't matter so much. The Fourier Transformation and its inverse are a pair and the two formulas are entangled and require a normalization factor, which is up to conventions. See eg enter link description here

If your theory/model does not require a special convention, it is basically a degree of freedom that you can choose to fit your needs.

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