I'm trying to understand the principle of correlaction functions.
I have two rectangular signals:

$s(\tau) = a_1 \cdot (\sigma (\tau) - \sigma (\tau - T_1))$
$g(\tau) = a_2 \cdot (\sigma (\tau) - \sigma (\tau - T_2))$

To find the correlation function, I need to take a look at five intervals:
1. $ t < -T_1 $
2. $ -T_1 \leq t < - T_1 + T_2 $
3. $ -T_1 + T_2 \leq t < 0 $
4. $ 0 \leq t < T_2 $
5. $ t > T_2 $

I know, that the correlation for 1. and 5. is 0. The question is: how do I find the correlation function for 2, 3 and 4?

$\varphi_{sg}^E = \int_{-\infty}^{\infty} s(\tau) \cdot g(t + \tau) d\tau$

I know that for 2. it should be $ \varphi_{sg}^E (t) = (t + T_1) \cdot a_1 a_2$. But as far as I know, you can't integrate heaviside functions. So how do I find the solution for this without "really" integrating?

Despite the question, how to integrate the heaviside function, I have a second problem. I should have written it more clearly.
I want to find a function (or functions) which look like the one in the screenshot, when you combine them. The question is, from where to where I need to integrate to achieve this.
The 5 intervals given at the beginning should help, but unfortunately they don't help me.
enter image description here


You only need to integrate over the region where the two square signals have overlap. Both signals start at $t=0$, lasting $T_1$ and $T_2$, respectively, and without lag between them.

Since you seem to assume $T_2>T_1$, then their correlation is non-zero over $(0,T_1)$,

$$\int_0^{T_1} s(\tau)g(\tau)d\tau = a_1 a_2 T_1$$

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    $\begingroup$ The key idea is that you can eliminate the Heaviside functions from the definite integral of the product of Heaviside functions and other functions by integrating the product of the other functions over the interval where all of the Heaviside functions are "on." $\endgroup$ – Brian Borchers Aug 23 at 1:24
  • $\begingroup$ Correct, only when they are simultaneously active. $\endgroup$ – Quanto Aug 23 at 1:48
  • $\begingroup$ Thanks Quanto and @BrianBorchers for the answer and the clarifying comment. In combination, they helped a lot. I forgot to clarify one thing in my question, so I updated it. Maybe you can help again. $\endgroup$ – TimSch Aug 25 at 16:19

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