Convolution Integral - Shift property I am trying to prove $f_1(t-T_1)*f_2(t-T_2)=c(t-T_1-T_2)$ given that $f_1(t)*f_2(t)=c(t)$. Here $f_1(t)*f_2(t)=\int_{-\infty}^{\infty}f_1(x)f_2(t-x)dx$.
I can only deduce that $f_1(t-T_1)*f_2(t)=c(t-T_1)$ and $f_1(t)*f_2(t-T_2)=c(t-T_2)$ by directly evaluating the integral and using the commutative property. According to my textbook, the conclusion should follows directly from these two relationships but I am not able to prove it.
If the 'left' function is delayed by $T_1$ seconds, so is $c(t)$. The same thing happens when the 'right' function is delayed by $T_2$ seconds. So it is quite 'logical' for $c(t)$ to be delayed by the sum of the delayed periods of time in $f_1$ and $f_2$. But this verbal explanation does not satisfy me.
I am also having trouble to express $f_1(t-T_1)*f_2(t-T_2)$ in integral form. I cannot only express it because the definition of the convolution integral seems limited. Thanks in advance.
 A: Define a new function via $g(t) = f_2(t - T_2)$. From what you have proven, the following holds:
$f_1(t - T_1) * f_2(t - T_2) = f_1(t - T_1) * g(t) = c_2(t - T_1)$, where $c_2(t) = f_1(t) * g(t) = f_1(t) * f_2(t - T_2) = c(t - T_2)$.
Combining both equations now yields the desired result
$$f_1(t - T_1) * f_2(t - T_2) = c_2(t - T_1) = c((t - T_1) - T_2) = c(t - T_1 - T_2).$$
A: The primary problem is the notation: One should write (f*g)(t) (= the c(t) above) instead of f(t)*g(t).
To shift a function, we define the shift operator S as follows:
S[a,f](t)=f(t-a)
S[a,f] means: the function f with the argument shifted everywhere by a constant -a. That means f shifted everywhere by a. For example, if f(0)=0, then
S[a,f](a)=f(a-a)=0.
Obviously, as shifting by b=2 and then by a=3 is shifting by a+b=5 altogether:
S[a,S[b,f]](t)=S[b,f](t-a)=f(t-a-b)=S[a+b,f](t)
What has already been proved is:
S[a,f*g]=S[a,f]*g
i.e., the shift can be pulled into one of the factors. Now, it's easy to see:
S[a+b, f*g]=S[b+a,f*g]=S[b,S[a,f*g]]=S[b,S[a,f]*g]=S[b,g*S[a,f]]=S[b,g]*S[a,f]=S[a,f]*S[b,g]  QED
