I'm having a hard time understanding how the convolution integral works (for Laplace transforms of two functions multiplied together) and was hoping someone could clear the topic up or link to sources that easily explain it. Note: I don't understand math notation very well, so a watered-down explanation would do just fine.
$$(f * g)(t) = \int_0^t f(\tau)g(t-\tau)\ \mathrm{d}\tau$$
This is what my textbook has written. What do those lowercase t-like symbols represent (I haven't seen them before).
Best Answer
Intuitively speaking when you are given two signals/or functions $f$ and $g$. You time reverese one of the signals, it doesnt matter which one, and shift it by a value of $t$ then you simply multiply and then sum the area under the intersection.
If you consider a function say function of $x$, then time reversal means inserting $-x$ wherever you see $x$ in this function.
Example:
Question1: Assume you have a function $f(x)$ that is $1$ if $x\in[0,1]$, and $0$ elsewhere then how should you plot $f(-x)$?
Question2: Assume you have $g(x)=f(x)$ is there any intersecting area between $f(x)$ and $g(x)$?
Question3: Now shift $g(-x)$ by $0.5$, that is to find $g(-x+0.5)$. How does it look like when you plot it?
Question4: Where does the intersecting region lie in this case $x\in?$ what is the are of the intersecting region? Answer: below white area$=0.5$ at $x=0.5$ shift.
Question5: If you select the shifting parameter not $0.5$ but all reals in $[0,2]$ what function should you get at the output? check $x=0.5$ and see $f*g(x)$ is $0.5$ as found at step 4
EDIT: You define convolution integral in $[0,t]$ for bounded signals. The integral limits depend on where your signal is non-zero.
If you have two signals as you suggested $f(t)=e^{at}$ and $g(t)=e^{bt}$ then the first question: what is the relation between $a$ and $b$? are they positive? where is the function defined? For example when $a$ and $b$ are some positive terms then we have the following integral
$$h(t)=\int f(\tau)g(t-\tau)d\tau= \int e^{a\tau}e^{b(t-\tau)}d\tau=e^{bt}\int e^{(a-b)\tau}d\tau=\Bigg]_{\tau\in\Omega}\frac{e^{(a-b)\tau}}{a-b}$$
clearly $\Omega=\mathbb{R}$ is not possible because the integral does not converge.