Power Spectral Density and Energy Spectral Density are both Fourier transform of autocorrelation?

Question

I've been looking into both of these quantities and feel like I have a good understanding intuitively of what they each represent but according to Wikipedia both the PSD and the ESD can be computed as the Fourier Transform of the Autocorrelation function of the time-domain signal. $$ PSD = \lim _{T\to \infty }{\frac {1}{T}}|{\hat {x}}_{T}(f)|^{2}\, = \int_{-\infty}^{\infty}R_{xx}(\tau)e^{-i2\pi f\tau }\,d\tau ={\hat {R}}_{xx}(f) $$ $$ ESD \triangleq \left|{\hat {x}}(f)\right|^{2} = \int_{-\infty}^{\infty}R_{xx}(\tau)e^{-i2\pi f\tau }\,d\tau $$

Perhaps I have misunderstood but if I have not could someone help me make sense of that ?

Answer 1

The apparent contradiction is caused by the fact that the autocorrelation function is defined differently for different signal classes. For deterministic signals with finite energy, the autocorrelation is defined as

$$r_x(\tau)=\int_{-\infty}^{\infty}x(t)x(t+\tau)dt\tag{1}$$

Its Fourier transform is the squared magnitude of the Fourier transform of the signal, which equals the energy density spectrum:

$$\mathcal{F}\big\{r_x(\tau)\big\}=\big|X(\omega)\big|^2\tag{2}$$

For deterministic signals with finite (but non-zero) power, the autocorrelation is defined as a limit:

$$r_x(\tau)=\lim_{T\to\infty}\frac{1}{T}\int_{-T/2}^{T/2}x(t)x(t+\tau)dt\tag{3}$$

The Fourier transform of the autocorrelation given by $(3)$ is the power spectrum of the deterministic signal $x(t)$. It can be shown that the power spectrum can also be directly expressed in terms of $x(t)$:

$$S_x(\omega)=\lim_{T\to\infty}\frac{1}{T}\left|\int_{-T/2}^{T/2}x(t)e^{-j\omega t}dt\right|^2\tag{4}$$

Finally, if a signal is modeled as a wide-sense stationary process, its autocorrelation is defined in terms of the following expectation:

$$R_X(\tau)=E\big\{X(t)X(t+\tau)\big\}\tag{5}$$

As to be expected, the corresponding power spectrum is given by the Fourier transform of $(5)$.

Hence, even though in all cases we compute the Fourier transform of an autocorrelation function, the results are different due to the different definitions of the autocorrelation function for various signal types.

Answer 2

The difference between PSD and ESD relates the difference between power signals and energy signals.

They are NOT defined through the Fourier Transform (FT) of the autocorrelation but as the squared magnitude of the signal spectrum. You CAN calculate them as the FT of the autocorrelation due the following mathematical relationship

$$r_{xx}(t) = x(t)*x(-t) \leftrightarrow R_xx(\omega) = X(\omega)\cdot X^{'}(\omega) = |X(\omega)|^2 = \text{PSD}(\omega)$$

In other words the autocorrelation can be calculated as the convolution between the signal and it's time inverse. In the frequency domain this is equivalent the product of the spectrum with it's conjugate and hence you get the PSD.