Why the correct plot of a $x(t)$ signal is $2r(t-1) - 2r(t-3) - 4u(t-3)$ and not $4r(t-1) + 4u(t-3)$

Question

I'm new with this kind of mathematics and I would like to know why this graph is represented by

$x(t)=2r(t-1) - 2r(t-3) -4u(t-3)$

if I were to represent the signal above with basic functions (unit step and unit ramp functions) I would say that the signal $x(t)$ is $x(t)=4r(t-1)+4u(t-3)$ but is wrong and I can't understand why.

What am I missing?

Answer 1

Since the unit ramp can also be defined as $r(t)=t \, u(t)$, then $r(t-1)=0$ at $t=0$ and $r(t-1) = 2$ at $t=3$, then the term $2r(t-1)$ correctly reproduce the signal up to $t=3$.

Thereafter, that term would continue to "ramp-up": to stop it and make it constant at the value of $4$, from $t=3$ onwards, you shall deduct $2r(t-3)$.

So your graph is $2r(t-1)-2r(t-3)$.

If you deduct also the term $4u(t-3)$ you are bringing the signal to $0$ (for $3 \le t$) which is not what you draw.