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Let $F(t)$ be a function of $t$, given by $$F(t) = t U(t)-(t-1) U(t-1) + (t-2) U(t-2) - (t-3) U(t-3)$$ where $U$ is the Heaviside step function.

I am getting confused regarding its graphical representation. Any help would be highly appreciated.

Fakemistake
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Jasmine
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  • What is $F(t)$ for $t<0$. – hamam_Abdallah Nov 18 '18 at 18:50
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    Draw each term on the RHS alone. For example draw $tU(t)$, $(t-1)U(t-1)$, and so on and add/subtract them graphically – Fakemistake Nov 18 '18 at 18:55
  • I added the solution in my answer below. I think now it's better not to think about how each term looks like, it's better to think piecewise from $0$ to $1$, $1$ to $2$ and so on because $U(x)$ is zero for $x<0$. – Fakemistake Nov 25 '18 at 09:18
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    Notice $tU(t) = \max(0,t)$. throwing the command Plot[Max[t,0] - Max[t-1,0] + Max[t-2,0] - Max[t-3,0],{t,-2,5}] to WA, you will get this graph. – achille hui Nov 25 '18 at 09:24

1 Answers1

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Hint: The first term is

$$tU(t)= \begin{cases} 0 & t<0\\ t & t\geq 0 \end{cases}$$

the second

$$(t-1)U(t-1)=\begin{cases} 0 &t<1\\ t-1 & t\geq 1 \end{cases}$$

and so on... Can you take it from here?

Edit:

$$F(t)= \begin{cases} 0, & t<0\\ t, & 0\leq t<1\\ 1, & 1\leq t<2\\ t-1,& 2\leq t<3\\ 2, &t\geq 3 \end{cases}$$

Fakemistake
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  • I'm afraid I cannot. I am able to draw functions like U(t)+2U(t-1)+4U(t-2) - 3U(t-3), whose graph is a combination of step-like figures. But my professor said that the first term of the given function would give a straight line with 45 degree slope, which will continue till t=1 on the t axis, and so on. I, on the other hand, always thought that these functions yield some horizontal lines at different steps. Perhaps the graph could clarify everything. – Jasmine Nov 18 '18 at 19:43
  • Maybe you can do the next step when you subtract these two functions. What value holds for negative $t$, for $t\geq 1$ and for $0\leq t<1$? – Fakemistake Nov 18 '18 at 21:20