Why is convolution of these two function not commutative? $m2(t)=z(t)*f(t)=\int^{\infty}_{-\infty}z(\tau)f(t-\tau)d\tau$


$\tau$ is a variable in place of $t$, which is now a constant.


For $-1<t<1$, $m2(t)=\int^{0}_{t}2e^{2\tau}d\tau =e^{2t}-1$
But if do $m2(t)=f(t)*z(t)=\int^{\infty}_{-\infty}f(\tau)z(t-\tau)d\tau$

For just the $-1<t<1$ case, $m2(t)=\int^{0}_{t}2e^{-2\tau}d\tau =1-e^{-2t}$
How is the two m2 different for the same t interval?

 A: The point is to keep careful track of where $\tau$ has to be for there to be a contribution to the integral, as a function of $t$. In the first case, $\tau \in [0,2]$ is required for $z$ to be nonzero, and $t-\tau \in [0,1]$ is required for $f$ to be nonzero. The latter is equivalent to $\tau \in [-1+t,t]$. So you need the intersection of $[0,2]$ with $[-1+t,t]$. The lower limit is $\max \{ 0,-1+t \}$, the upper limit is $\min \{ t,2 \}$, and we interpret $[a,b]$ with $b<a$ as $\emptyset$. In particular when $t \in [0,1]$ this will be an integral from $t-1$ to $t$ (which seems to be where you made a mistake).
In the second case, $\tau \in [0,1]$ is required for $f$ to be nonzero and $t-\tau \in [0,2]$ is required for $z$ to be nonzero. The latter is equivalent to $\tau \in [-2+t,t]$. So you need the intersection of $[0,1]$ with $[-2+t,t]$. By similar reasoning, this is $[\max \{ 0,t-2 \},\min \{ 1,t \}]$ with the same convention. In particular when $t \in [0,1]$ this is an integral from $0$ to $t$ which you seem to have gotten right.
