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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 ?

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  • $\begingroup$ If you scroll down, PSD is clearly defined differently; they may have equivalence operations in common (e.g. autocorrelation), but not all of them. $\endgroup$ May 5, 2023 at 15:56
  • $\begingroup$ @OverLordGoldDragon I don't see what you mean, how do you mean equivalent operations if they can be defined in terms of the autocorrelation ? $\endgroup$ May 5, 2023 at 15:59
  • $\begingroup$ I'm saying your question is unclear relative to the title. If their equations are different then so are the definitions, by definition. Just because they both have computation steps in common, e.g. both take fft, doesn't make them same. $\endgroup$ May 5, 2023 at 16:00
  • $\begingroup$ @OverLordGoldDragon Its not computation steps in common, it is the fact that both quantities appear to defined in terms of the Fourier transform of the autocorrelation function. If their definition is the same, meaning that their computation is the same, how are they different ? $\endgroup$ May 5, 2023 at 16:04
  • $\begingroup$ Eh, I guess if that makes you think they're same then it's valid to ask it that way. Nevermind. $\endgroup$ May 5, 2023 at 16:12

2 Answers 2

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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.

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  • $\begingroup$ Thanks so much Matt ! That's the second that you've provided a very thorough explanation of something that has been nagging at me and I appreciate it. You seem to have a very deep understanding of a lot of these concepts and I was wondering if perhaps you could recommend any books that you found useful during your studies ? I've always found the DSP books to skim the surface on some of the mathematics which can be frustrating. $\endgroup$ May 6, 2023 at 10:11
  • $\begingroup$ Also I'm assuming that the equation reference you give in the fourth paragraph is meant to be equation $(3)$ ? $\endgroup$ May 6, 2023 at 10:12
  • $\begingroup$ @RickarySanchez: Thx, I've corrected the reference. As for books, getting to know and understand things well is of course a matter of many years of study and experience, but if I were to recommend one single book with a lot of wisdom in it, I'd name "The Fourier Integral And Its Applications" by A. Papoulis. $\endgroup$
    – Matt L.
    May 6, 2023 at 10:26
  • $\begingroup$ Thanks I have a long ways to go but its always good to have a solid reference. Thanks again $\endgroup$ May 6, 2023 at 10:29
  • $\begingroup$ Perhaps it is worth mentioning that the most common example of deterministic power signals is periodic signals for which the limit of the autocorrelation function is also a periodic function, and the power spectral density is a sum of impulses. $\endgroup$ May 6, 2023 at 13:35
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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.

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  • $\begingroup$ I’m still confused because the derivation you show leads to $|X(\omega)|^2$ which to my understanding is the definition of ESD not the PSD which is defined as $\lim_{T\to\infty}|X_T(\omega)|^2$ Also, if they are not defined in terms of the autocorrelation function, but both lead to same expression in terms of the autocorrelation function then why are they no equivalent? $\endgroup$ May 6, 2023 at 7:33
  • $\begingroup$ Do you know the difference between an energy signal and a power signal ? For an energy signal the PSD is zero, for a power signal the ESD is infinite:. They are essentially equivalent: which one you use depends on the type of signal you use. It's always one or the other. $\endgroup$
    – Hilmar
    May 6, 2023 at 17:18

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