# limit of the integral in the anti-transform

if I have an autocorrelation on a limited domain

r($$\tau$$)=$$\lim_{T \to \infty}\frac{1}{T} \int_{0}^{T} u(t)u(t+\tau)dt$$

The power spectral density, obviously will have an infinitive domain:

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

If I do the anti-transform of the PSD, which will be the limit of the integral?

from -$$\infty$$ to $$\infty$$ or from $$0$$ to $$T$$??

Thank

I think you are making this more complicated than it needs to be.

The Fourier and autocorrelation integrals are defined from $$-\infty$$ to $$+\infty$$. That's always correct and the saftest way to write them.

If you are integrating over a function with limited support on $$[0,T]$$ than you can adjust the integration interval to $$[0,T]$$ too. The areas outside the support interval is all zero so it makes no difference whether you include them in the integration or not. In this case both versions are identical and both versions are correct. You can use whichever one is more practical for your specific problem or application.

In your specific example: if $$u(t)$$ has support on $$[0,T]$$ then it's autocorrelation $$r_{uu}(t)$$ also has support only on $$[0,T]$$. That means you can calculate the PSD by executing the Fourier integral over either $$[-\infty,+\infty]$$ or $$[0,T]$$. Both are correct and both will give you the same result.

$$S(f)$$ has infinite support so you need to use $$[-\infty,+\infty]$$ for the inverse Fourier integral.

$$S_{uu}(\omega)=\int_{-\infty}^{+\infty}r_{uu}(t) \cdot e^{-j \omega t} d \omega = \int_{0}^{T}r_{uu}(t) \cdot e^{-j \omega t} d \omega$$

$$r_{uu}(t)= \frac{1}{2 \pi} \int_{-\infty}^{+\infty} S_{uu}(\omega) \cdot e^{+j \omega t} d \omega$$