Understanding Heaviside Step Function - Explained with Examples

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SUMMARY

The discussion centers on the Heaviside step function, particularly its application in solving equations involving Laplace transformations. The Heaviside function, defined as h(u) = 0 for t < u and h(u) = 1 for t ≥ u, acts as an "on-switch" for functions. Participants seek clarification on applying the Heaviside function to specific equations, including F(s) = exp[-2s] / (s^2 + s - 2) and a piecewise function f(t) defined for t < 2 and t ≥ 2. The conversation emphasizes the importance of understanding the Heaviside function in the context of differential equations.

PREREQUISITES
  • Understanding of the Heaviside step function
  • Familiarity with Laplace transformations
  • Basic knowledge of differential equations
  • Ability to interpret piecewise functions
NEXT STEPS
  • Study the properties and applications of the Heaviside step function
  • Learn how to perform Laplace transformations on piecewise functions
  • Explore examples of solving differential equations using the Heaviside function
  • Investigate the relationship between the Heaviside function and impulse functions
USEFUL FOR

Students, mathematicians, and engineers who are working with differential equations and Laplace transformations, particularly those looking to deepen their understanding of the Heaviside step function.

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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:
 
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The heaviside function is most easily explained as a on-switch.

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

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.

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

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 ?
 

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