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
