Quick question about my understanding of cross correlations and cross spectral densities.

Let $C_{ab}$ be the cross correlation between the signals $a$ and $b$, i.e., $C_{ab}(\tau) = \langle a(t)b(t+\tau)\rangle$. Let $S_{ab}$ be the cross spectral density of the signals $a$ and $b$, i.e., $S_{ab}=\mathcal{F}(a)^* \mathcal{F}(b)$ and by the convolution theorem/cross correlation theorem $S_{ab} = \mathcal{F}(C_{ab}(\tau))$.

Say $x(t)$ is some signal, and $\nu(t)$ is some uncorrelated noise signal. Let $s(t)=x(t)+\nu(t)$ Thus, $C_{xs} = C_{xx}+C_{x\nu}$ from the linearity of the expectation value being calculated in the cross correlation. By initial assumption that $x$ and $\nu$ are uncorrelated, we get $C_{xs}=C_{xx}$. (This calculation so far also appears here on page 5, for example so I think it is correct).

Now, if we take the fourier transform of both sides of this, we get $S_{xs}=S_{xx}$. Is this correct so far? But does this not mean that $\mathcal{F}(x)^* \mathcal{F}(s) = \mathcal{F}(x)^*\mathcal{F}(x)$? and hence $\mathcal{F}(s)=\mathcal{F}(x)$, and hence even $s(t)=x(t)$? Clearly there is something wrong here. What mistake am I making?


EDIT: On a similar note, if $C_{x\nu}=0$, does that not mean that $\mathcal{F}(x)^*\mathcal{F}(\nu)=$ and hence (by taking the absolute value of both sides and squaring) that $S_{xx} S_{\nu\nu}=0$? This is patently false. At each frequency this would require either the power in my signal to be zero, or the power in the noise to be zero. Even if my noise is white noise then this is clearly false. I'm doing something fundamentally wrong here. What? Any help?


2 Answers 2


$S_{xy}=S_{xx}$ is true but this does not mean $F_x*F_y = F_x*F_x$ when $y$ is random.

The actual definition is $$S_{xy}(f)=\lim_{T\rightarrow \infty}\textbf{E}\{F^*_xF_y\}$$ for random signals. Since $x$ and $v$ are uncorrelated, this would be equal to $$\lim_{T\rightarrow \infty}\textbf{E}\{F^*_xF_x\} = S_{xx}.$$ But your next assumption is not correct.

$F_x$, $F_y$ are only based on single realization of $x$ and $y$ respectively. For random signal, $S_{xy} \ne F_x^*F_y$

  • 2
    $\begingroup$ Apologies for the edit. I prefer the ${\tt displaystyle}$ way of showing maths, and prefer this way rather than the inline version. $\endgroup$
    – Peter K.
    Apr 30, 2020 at 12:51
  • $\begingroup$ Please, no need to apologize!! Such feedbacks are good. Really improved the readability and visibility of the answer. $\endgroup$
    – jithin
    Apr 30, 2020 at 12:56

Previous answers have already discussed some of the mathematical issues that arise after the Fourier transforms in the question, so I will try to impart some more easily earned intuition.

Even if we accept $C_{xs} = C_{xx}$ for all times$^{\dagger}$, we must remember that the quantity at each time arises from computing a mean over an ensemble (a single number for each time). Agreement of such functions does not imply agreement of the individual realizations of the process.

The Fourier transforms are not transforms of the products of the signals themselves. In particular, the Fourier transform of $C_{xx}$ is not equal to a convolution of $\mathcal{F}(x)$ with itself, and the transform of $C_{xs}$ is not equal to the convolution of $\mathcal{F}(x)$ and $\mathcal{F}(s)$.$^{\dagger\dagger}$ To see what the Fourier transform of $C_{xx}$ is, see the Wiener-Khinchin theorem.

Rather, the transforms of $C_{xx}$ and $C_{xs}$ are transforms of functions that merely summarize one statistical quantity at each time. You cannot infer equality of the products of the signals from agreement of these functions.

For more intuition, imagine all the random variables that have equal means but have different probability laws. Knowing just one parameter of a probability distribution does not tell you very much about the random variables that behave according to that distribution. The same ambiguity is found here when comparing $x(t)x(t + \tau)$ and $x(t)s(t + \tau)$, for each $t$ and each $\tau$.

$\dagger$ In truth, each of these cross-correlations should be a function of two time variables until we specify some other assumptions on the signal and the noise, such as stationarity.

$\dagger\dagger$ In fact, if $x$ and $s$ have nonzero power spectral densities, then for most realizations the traditional Fourier transform does not exist.


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