# Convolution theorem for cross-correlation

Forgive me is this is an ill-posed question.

Is there any such thing as a 'convolution theorem' for the cross-correlation.

Namely, the convolution theorem states that: $$x[n] * h[n] = \sum_{k=0}^{\infty}h[k] x[n-k] = X(f)H(f)$$ Where $X(f)$ is the Fourier transform of $x(t)$.

Is there any equivalent theorem for cross-correlation, i.e.: $$\textrm{corr}(x[n],h[n])=r_{xh}[k]= \sum_{k=0}^{\infty}h[k] x[n+k] = F_?[X(f),H(f)]$$

In other words, is there a way to express $r_{xh}[k]$ in terms of $X(f)$ and $H(f)$, or even in terms of the autocorrelation $r_x[k]$ and $r_h[k]$.

I tried self-deriving this but couldn't come up with anything.

Thanks!

• Your cross-correlation is a function of $n$ (not $k$). – Gilles Mar 29 '16 at 22:00

Yes. Cross correlation is the same as convolution with a time reversed signal. That corresponds to conjugate komplex in the frequency domain.

$R(\omega) = X(\omega ) \cdot H(\omega )^*$

To stick to proper notations, if $\mathcal F \left\{x[n]\right\}=X(\omega)$ and $\mathcal F \left\{h[n]\right\}=H(\omega)$, what you mean is the Fourier transform of a convolution gives the product of the transforms of each term:

$$\mathcal F \left\{x[n] * h[n]\right\} = \mathcal F\left\{\sum_{k=0}^{\infty}h[k] x[n-k]\right\} =X(\omega)\cdot H(\omega)$$

As already mentioned, there is a similar formulation for correlation, the correlation theorem. And again, the Fourier transform of the correlation: $$\mathcal F\left\{\mathrm{corr}\left(x[n],h[n]\right)\right\}=\mathcal F\left\{r_{xh}[n]\right\}= \mathcal F\left\{\sum_{k=0}^{\infty}h[k] x[n+k] \right\}= X(\omega)\cdot H^*(\omega)$$

The product $X(\omega) H^*(\omega)$ is called cross-energy density spectrum and noted $S_{xh}(\omega)$, and this is just $\mathcal F\left\{r_{xh}[n]\right\}$. The $^*$ sign is for conjugate.

Note that your convolution/correlation are on discrete-time signals $x[n]$ and $h[n]$. So $X(\omega)$ and $H(\omega)$ are transforms of $x[n]$ and $h[n]$ respectively, and not of continuous-time signals $x(t)$ and $h(t)$.