from "Signals and Systems Demystified", 2006, page 142, Example 6-3: (http://www.gatestudymaterial.com/study-material/signals%20and%20systems/text%20books/SIGNALS%20AND%20SYSTEMS%20BY%20DAVID%20MCMOHAN.pdf)

Given signal:


has autocorrelation function:


Find the energy content of the signal:

So I take the Fourier transform of $R_{xx}(\tau)$ to obtain energy spectral density $S_{xx}(\omega)$:

$$F\{R_{xx}(\tau)\}=(1/(2a))\text{ } F\{e^{-a\tau}\}$$ $$S_{xx}(\omega) = (1/(2a))\text{ }(2a/(a+j\omega))$$ $$S_{xx}(\omega) = 1 / (a+j\omega)$$

Now according to the book, equation 6.9, "if x(t) is real valued, then the energy spectral density (S) is:

$$ S_{xx}(\omega) = | X(\omega)|^{2}$$

However, when I look at my Fourier transform I find a complex number:

$$S_{xx}(\omega) = 1 / (a+j\omega)$$

my question is: what is going on? why is it complex when the book says it should be a real magnitude?

(I suppose I could look inside a real "signals and system" textbook and obtain the answer. But, I've already went through half the "Demystified" book to turn around. I'm just curious, what they are doing here.)

  • 1
    $\begingroup$ The "realness" constraint is unnecessary. What they are forgetting is the weak-sense stationarity that is required for the Wiener-Kinchin theorem to be true. I.e. your signal must be w.s.s. (but can be complex, no problem), so that $\mathcal F\{R_{xx}\} = S_{xx}(\omega) = \left| \mathcal F\{x\}\right|^2$. That's the only paragraph I've been reading from that book, but right now, I'm not a big fan. $\endgroup$ Commented Dec 2, 2018 at 0:01
  • $\begingroup$ ah, they only consider deterministic signals, not random processes. Then, the Wiener-Kinchin is irrelevant. $\endgroup$ Commented Dec 2, 2018 at 0:31
  • $\begingroup$ AutoCorreation function must be Hermitian symmetric. Yours is not? so it be $R_{xx}(\tau) = \frac{ e^{-a |\tau|} }{ 2 a}$ have you tried this also ? $\endgroup$
    – Fat32
    Commented Dec 2, 2018 at 0:45
  • $\begingroup$ that's a good point... its not an even function without the absolute value on the tau. $\endgroup$
    – Bill Moore
    Commented Dec 2, 2018 at 0:53

1 Answer 1


The so called auto-correlation function that's used in the example (which was found in a previous example) is not correct.

First note that it's the deterministic auto-correlation computation, compared to its theoretical counterpart which requires that the signals be random for their ACF to be defined at all.

Anyway, so the following integral signifies one definition of deterministic ACF of a given deterministic signal $x(t)$ as :

$$R_{xx}(\tau) = \int_{-\infty}^{\infty} x(t)x(t-\tau) dt$$

applying this definition to the signal $x(t) = e^{-a t} u(t)$ with $a >0$ we have:

$$ \begin{align} R_{xx}(\tau) &= \int_{-\infty}^{\infty} x(t)x(t-\tau) dt \\ \\ &= \int_{-\infty}^{\infty} e^{-a t} u(t) e^{-a (t-\tau)} u(t-\tau) dt \\ \\ &= e^{a \tau} \int_{-\infty}^{\infty} e^{-2 a t} u(t) u(t-\tau) dt \\ \\ &= e^{a \tau} \int_{\max\{0,\tau\}}^{\infty} e^{-2 a t} dt \\ \\ &= \begin{cases}{ e^{a\tau} \int_{\tau}^{\infty} e^{-2at} dt ~~~,~~~\tau \geq 0 \\ \\ \\ e^{a\tau} \int_{0}^{\infty} e^{-2at} dt ~~~,~~~\tau < 0 }\end{cases}\\ \\ &= \begin{cases}{ e^{a\tau} \frac{e^{-2a\tau}}{2a} ~~~,~~~\tau \geq 0 \\ \\ \\ e^{a\tau} \frac{1}{2a} dt ~~~,~~~\tau < 0 }\end{cases}\\ \\ &= \begin{cases}{ \frac{e^{-a\tau}}{2a} ~~~,~~~\tau \geq 0 \\ \\ \\ \frac{e^{a\tau}}{2a} dt ~~~,~~~\tau < 0 }\end{cases}\\ \\ R_{xx}(\tau) &= \frac{e^{-a |\tau|}}{2a} \\ \\ \end{align} $$

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    $\begingroup$ Thanks. Explains the mystery of the absolute value... $\endgroup$
    – Bill Moore
    Commented Dec 2, 2018 at 1:36
  • 1
    $\begingroup$ that change fixes the book... adding the absolute value on the tau changes the Fourier transform to $1 / (a^{2} + w^{2})$ which checks with $S_{xx}(\omega) = | X(\omega)|^{2}$ being a real value. $\endgroup$
    – Bill Moore
    Commented Dec 2, 2018 at 2:10
  • $\begingroup$ @BillMoore that's nice! you can accept the answer if it helped ;-) $\endgroup$
    – Fat32
    Commented Dec 2, 2018 at 8:48

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