1
$\begingroup$

Question

Given that $H(x)$ is a Heaviside function, how would I graph $H(x)+2H(x-3)-3H(x-5)$?

I was thinking about shifting the function, but then realized that there are multiple $H$'s

Or do I need to plot points?

  • 0
    I hate to break it on you, but Heaviside function is named after the dude who used it first. – 2017-01-19
  • 1
    @Ivan Neretin The use of "heavy side" by the OP is easily explained : the left hand side of the graphical representation has collapsed onto the $x$ axis due to the overweight. – 2017-01-19
  • 0
    @JeanMarie Yeah, I guess it sounds kinda natural this way, so maybe we all should switch to this spelling. – 2017-01-19
  • 0
    @John Rawls Draw separately the graphical representations of $y=H(x)$, $y=2H(x-3)$ (with a jump "up" of heigth 2 in point (3,0)), $y=-3H(x-5)$ (with a jump down of height 3 in point $(5,0)$), then add them. – 2017-01-19
  • 0
    may start making sense if you try graphing a related function $\mid x\mid +2\mid x-3 \mid-3\mid x-5 \mid$, then come back to this problem. – 2017-01-19

2 Answers 2

3

@JeanMarie provided an excellent comment on how to do it. Also, @JohnJoy provided a very useful interpretation that is worth exploring.

You would just break up the Heaviside function into three distinct pieces and add the results of them over their respective ranges.

The plot should end up being

enter image description here

Here is the same plot using a different tool

enter image description here

2

It helps to write the function like this:

$$f(x)=\begin{align}&\;\;\;\;\;\color{#eb6235}{0}\\&+(\color{#8778b3}{1}-\color{#eb6235}{0})H(x-\color{#e19c24}{0})\\&+(\color{#5d9ec7}{3}-\color{#8778b3}{1})H(x-\color{#e19c24}{3})\\&+(\color{#996633}0-\color{#5d9ec7}{3})H(x-\color{#e19c24}{5})\end{align}$$

where the numbers being subtracted from $x$ in the unit step function represent the breakpoints, and the matching constants correspond to the pieces that make up your piecewise function. As an example, note that $\color{#5d9ec7}{3}$ shows up in the $\color{#5d9ec7}{3}-\color{#8778b3}{1}$ and $\color{#996633}0-\color{#5d9ec7}{3}$ factors corresponding to $H(x-\color{#e19c24}{3})$ and $H(x-\color{#e19c24}{5})$, so your function takes the value $\color{#5d9ec7}{3}$ in the interval $[\color{#e19c24}{3},\color{#e19c24}{5})$.

In this form, you can then actually make a direct conversion:

$$f(x)=\begin{cases}\color{#eb6235}0&\text{if }x<\color{#e19c24}0\\\color{#8778b3}1&\text{if }\color{#e19c24}0\le x<\color{#e19c24}3\\\color{#5d9ec7}3&\text{if }\color{#e19c24}3\le x<\color{#e19c24}5\\\color{#996633}0&\text{if }\color{#e19c24}5\le x\end{cases}$$

  • 0
    I don't really understand why this was downvoted. – 2017-08-28
  • 0
    I don't either, very nice answer (+1). – 2017-08-28