Understanding Heaviside Step Function - Explained with Examples

Click For Summary

Discussion Overview

The discussion focuses on understanding the Heaviside step function, its properties, and its application in solving specific equations, particularly in the context of Laplace transforms. Participants seek clarification on how to apply the Heaviside function to given mathematical expressions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant expresses confusion about the Heaviside step function and requests detailed explanations on its application to specific equations.
  • Another participant describes the Heaviside function as an "on-switch," explaining that it outputs zero until a certain point, after which the function behaves as it would without the Heaviside multiplication.
  • There is uncertainty regarding the notation used in the equations presented, with a participant questioning whether it relates to Laplace transformations.
  • A participant reiterates their request for clarification on the equations involving the Heaviside function, indicating they are puzzled by the notation.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the application of the Heaviside function or the interpretation of the equations, as confusion and requests for clarification persist.

Contextual Notes

There are unresolved questions about the notation used in the equations and the specific context of the Laplace transformations, which may affect the understanding of the Heaviside function's application.

slider527
Messages
9
Reaction score
0
Heaviside? HELP

hey there peeps,

could u guys help me out with heaviside step function..cu zi don't understand anything bout how it works and how to apply to a particular problem...

could u guys explain in detail how to solve the next equations using heaviside:
exp[-2s]
F(s)= --------
s^2+s-2
and
__
|
| 0 for t<2
f(t) <
| t^2-2t+2 for t (greater or equal to ) 2
|__


ThnQ ...it means a lot to me :smile:
 
Physics news on Phys.org
The heaviside function is most easily explained as a on-switch.

[tex]h(u) = \left\{ \begin{array}{rcl}<br /> \ 0 & \mbox{for} & t < u\right \\ 1 & \mbox{for} & t \geqq u<br /> \end{array}\right[/tex]

Thus if you multiply a function with the Heaviside function, the output are zero until the t = u, from that point, the function is similar to what it would look like without multiplying with Heaviside.

I didn't understand your notation so didn't get the equation, was it laplace?
 
fannemel said:
The heaviside function is most easily explained as a on-switch.

[tex]h(u) = \left\{ \begin{array}{rcl}<br /> \ 0 & \mbox{for} & t < u\right \\ 1 & \mbox{for} & t \geqq u<br /> \end{array}\right[/tex]

Thus if you multiply a function with the Heaviside function, the output are zero until the t = u, from that point, the function is similar to what it would look like without multiplying with Heaviside.

I didn't understand your notation so didn't get the equation, was it laplace?

well Laplace transformations are used to solve differential equations...
thnqs anyway,...mayb i can figure out woth u just said :confused:
 
could u guys explain in detail how to solve the next equations using heaviside:
exp[-2s]
F(s)= --------
s^2+s-2
and
__
|
| 0 for t<2
f(t) <
| t^2-2t+2 for t (greater or equal to ) 2
|__

I was kind of left somewhat puzzled what actually comes out of this notation ... could you clarify a bit ?
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 2 ·
Replies
2
Views
6K
  • · Replies 1 ·
Replies
1
Views
19K