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

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

enter image description here

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?

Phill Alexakis
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    What is your definition of $r(t)$? I would expect $r(t)=\max(0,t)$ or $r(t)=\min(1,\max(0,t))$, but that does not fit either variant. The ramp function in that case is continuous, while the step function has a jump. Their non-trivial combination should have a jump at $t=3$, which the graph does not show. – Lutz Lehmann Sep 07 '19 at 17:24
  • $r(t)=t$ when $t>0$ and $r(t)=0$ when $t<0$ – Phill Alexakis Sep 07 '19 at 17:34
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    This is $r(t)=\max(0,t)$, as said, a continuous function. – Lutz Lehmann Sep 07 '19 at 17:39

1 Answers1

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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.

G Cab
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  • So $-4u(t-3)$ should bring the signal down to 0? for $3 \le t$? – Phill Alexakis Sep 07 '19 at 17:45
  • why is it not necessary to mention the $u(t)$? – Phill Alexakis Sep 07 '19 at 17:48
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    @PhillAlexakis: with the first term you establish the ramp; deducting the second ramp, you block the signal at the value($4$) it has reached at $t=3$;deducting a constant signal of $4$ from $t=3$ you clearly bring the whole signal to $0$. If you make a simple sketch of the ramps and Heaviside functions involved you can easily see that. – G Cab Sep 07 '19 at 18:37