# $f \star (g \,* \,h) = (f \star g) * h$?

In this paper that I'm reading, I see this equation: $$S_0 \star (S_0 * N) = N,$$ where $$\star$$ denotes cross-correlation, $$*$$ denotes convolution and $$S_0 \star S_0 = \delta$$. Is it true in general that $$f \star (g*h) = (f \star g)*h$$ and if not, why is it true here?

• Please explain what $\delta$ means. As to the general result, try to use the frequency domain where convolutions become products and crossscoorelations become dot products to see if the result holds. – Dilip Sarwate Dec 1 '20 at 3:00
• Does this help? – TimWescott Dec 1 '20 at 4:02
• $\delta$ denotes the Dirac delta. – SaltSolutionPromo Dec 1 '20 at 6:21

It's easiest to look at this in the frequency domain. We have $$\mathcal{F} (x*y) = X \cdot Y$$ $$\mathcal{F} (x \star y) = X' \cdot Y$$ where ' is the complex conjugate

So we get $$\mathcal{F} (f \star (g * h)) = F' \cdot G \cdot H$$ $$\mathcal{F} ((f \star g) * h) = (F \cdot G)' \cdot H = F' \cdot G' \cdot H$$

These are only identical if G is real.

In your case you let's simplify the notation to $$f = S_0, g = S_0, h = H$$ $$S_0$$ has a unit impulse as autocorrelation, which means that the spectrum of the autocorrelation is just unity, i.e. $$1$$

$$\mathcal{F} (f \star f) = \mathcal{F} (\delta) = 1 = F' \cdot F$$

We use that property and can show that

$$\mathcal{F} (f \star (f * h)) = F' \cdot F \cdot H = H$$ since $$F' \cdot F = 1$$

The equation requires that pairwise product are well defined. If so, cross-correlation can be rewritten as a convolution with one function conjugated and time-reversed:

$$a(t) \star b(t) = a(t) \ast\overline{b}(-t) = \overline{a}(-t) \ast b(t) \,.$$

Hence, using associativity of the convolution (or $$a\ast b)\ast c = a\ast (b\ast c)$$): $$S_0 \star (S_0 \ast N) = \overline{S_0}(-t) \ast (S_0 \ast N) = (\overline{S_0}(-t) \ast S_0 )\ast N = (S_0 \star S_0 )\ast N\,.$$

This is for instance used in seismic: instead of analyzing the subsurface with something mimicking an impulse (like dynamite), which should be avoided for several reasons, one may use seismic sources called seismic vibrators which:

propagate energy signals into the Earth over an extended period of time as opposed to the near instantaneous energy provided by impulsive sources.

They typically are sweeps (similar to chirps) covering a wide range of frequency. The sweep-shaped source signal was generated by controlled vibrations whose autocorrelation is Dirac-like. The data recorded (seismic trace) has then to be cross-correlated with the pilot to yield an impulse response.