Integrals and Convolution: How to Group Multiple Functions Together?

[divB]

Hi,

I shall show the following:

$$(f*g) \star (f*g) = (f\star f)*(g\star g)$$

where $*$ denotes convolution and $\star$ cross-correlation. Writing this in terms of integral & regrouping:

$$\int_{\phi} \left(\int_{\tau_1} f(t - \tau_1) g(\tau_1) d\tau_1\right) \cdot \left(\int_{\tau_2} f(\tau_2) g(t+\phi-\tau_2) d\tau_2\right) d\phi \\<br /> = \int_{\tau_2} \int_{\tau_1} f(t - \tau_1) g(\tau_1) d\tau_1 \int_{\phi} f(\tau_2) g(t+\phi-\tau_2) d\tau_2 d\phi$$

But now I am stuck. How should I bring both f into one integral? Both are functions of a differerent variable and $\int f(x)dx \cdot \int f(x)dx \neq \int f(x)\cdot f(x) dx$...

Thanks for any pointer...

 

[haruspex]

$$\left(\int_{\tau_2} f(\tau_2) g(t+\phi-\tau_2) d\tau_2\right) d\phi$$

I don't think the argument to g is right.

 

[divB]

Hi, through you very much. But why? t-tau2 is from the convolution and (t+phi) because it is the 2nd operand for the x-correlation...

 

[haruspex]

Actually, the left hand integral in the same line was also wrong. Both must mention phi:
$h(t) = (f*g)(t) = \int_{\tau=0}^t f(\tau)g(t-\tau).d\tau$
$(h*h)(t) = \int_\phi h(\phi)h(t-\phi).d\phi = \int_{\phi=0}^t \left(\int_{\tau=0}^t f(\tau)g(\phi-\tau).d\tau\right)\left(\int_{\tau=0}^{t-\phi} f(\tau)g(t-\phi-\tau).d\tau\right).d\phi$

 

[Mute]

Actually, the left hand integral in the same line was also wrong. Both must mention phi:
$h(t) = (f*g)(t) = \int_{\tau=0}^t f(\tau)g(t-\tau).d\tau$
$(h*h)(t) = \int_\phi h(\phi)h(t-\phi).d\phi = \int_{\phi=0}^t \left(\int_{\tau=0}^t f(\tau)g(\phi-\tau).d\tau\right)\left(\int_{\tau=0}^{t-\phi} f(\tau)g(t-\phi-\tau).d\tau\right).d\phi$

Your second line should be $(h \star h)(t)$, as it's a cross-correlation, so it's a plus $\phi$ in the integral arguments rather than a minus $\phi$, and the limits are the real line:

$$(h \star h)(t) = \int_{-\infty}^\infty d\phi~ h(\phi)h(t+\phi) = \int_{\phi=-\infty}^\infty d\phi~ \left(\int_{\tau_1=0}^t d\tau_1~f(\tau_1)g(\phi-\tau_1)\right)\left(\int_{\tau_2=0}^{t+\phi} d\tau_2 f(\tau_2)g(t+\phi-\tau_2)\right). $$

 

[haruspex]

Your second line should be $(h \star h)(t)$, as it's a cross-correlation, so it's a plus $\phi$ in the integral arguments rather than a minus $\phi$,

OK, I missed the distinction in the signs, thanks. So the error in the OP is that $f\left(t-\tau_1\right)$ should have been $f\left(\phi-\tau_1\right)$

 

[divB]

Hi, thank you both of you! That's indeed a mistake.

However, I just don't see it, maybe I miss the forest for the trees (aehm, integrals ...)

In the end, I need to group both "g" and both "f" together into one integral.
However f has distinct variables (tau1 and tau2).
The g has even 4 (!) different variables: t, phi, tau1, tau2

I tried substitution, but I just can't get rid of them ...