1. ## Heaviside Function

The Heaviside Function, H(t) is defined as

H(t) =
1 if t>=0
0 if t<0

It is used in the study of electric circuits to represent the sudden surge of electric current, or voltage, when a switch is instantaneously turned on

My student asked me the following question,
"since time cannot be negative, then why the function is defined for negative values of t?"

What is the best explanation? is it because usually we want to have a function that is defined for all real numbers?

Thanks!

2. The Heaviside Function, H(t) is defined as

H(t) =
1 if t>=0
0 if t<0

It is used in the study of electric circuits to represent the sudden surge of electric current, or voltage, when a switch is instantaneously turned on

My student asked me the following question,
"since time cannot be negative, then why the function is defined for negative values of t?"

What is the best explanation? is it because usually we want to have a function that is defined for all real numbers?

Thanks!

--------------------------------------------------------------------

analogy
the graph of the function basically turns "on" ((H(t)=1) => turns on signal) if the switch is turned on hence you get time for a surge, so if the switch is off, for the duration it is off it could be said to be negative up until your turn it on, at that moment the instaneous time=0 hence H(0)=1,((H(t<0)=0) => turns off signal). Now, we know that time cannot be taken away, i.e. go back in time.but rather the time(t) is actually in respect to the moment you turn it on t=0, so for generality, now t isused for the function to be defined for all real numbers.

another analogy-
so the unprocessing of the signal i.e. reversing time (going from right to left on a graph of H(t)) in the function H(t) is basically undoing w/e we did to it. that is it was a value of one when t>=0, but when we go back in time, "we undo what happened" and bring it back to 0 for t<0
and it is easier to have a function for all real numbers.

3. No, there must be a better explanation, because if you remove negative values of $\displaystyle t$, then your function becomes $\displaystyle H(t) = 1$, which is a bit useless (although in my opinion it was useless before anyway, no offense).
Maybe you could also tell him that the input is not necessarily time (I'm not sure about that), and hence the necessity of defining the value for negative inputs ...
You could also tell him that it looks nicer and is easier when defined on all reals
Maybe some research could be interesting here ?

4. Originally Posted by Bacterius
No, there must be a better explanation, because if you remove negative values of $\displaystyle t$, then your function becomes $\displaystyle H(t) = 1$, which is a bit useless (although in my opinion it was useless before anyway, no offense).
Maybe you could also tell him that the input is not necessarily time (I'm not sure about that), and hence the necessity of defining the value for negative inputs ...
You could also tell him that it looks nicer and is easier when defined on all reals
Maybe some research could be interesting here ?

yeah i was wrong the first time but after some research im positive im right now.
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}