Laplace Transforms: Explaining Unit Step Function

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In summary, the Laplace transform of the unit step function u(t) is 1/s and multiplying it by e^{as} results in the Laplace transform of u(t-a).
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EXPLAIN LAPLACE TRANSFORM OF UNIT STEP FUNTION?

i.e L{u(t)} = 1/s
 
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  • #2
The unit step function is defined as::

[tex]
u(t)=\begin{cases} 0, & t < 0 \\ 1, & t \ge 0 \end{cases}
[/tex]

Now take the Laplace transform.

[tex]
L[u(t)]=\int_0^\infty u(t) e^{-st} dt=\int_0^\infty 1*e^{-st} dt
[/tex]

Because on the interval [itex]0 \leq x < \infty, u(t)=1[/itex].

You should be able to work it out now.

The same holds for the two-sided Laplace transform, because on the interval [itex]-\infty<x<0[/itex] the unit step function is 0.
 
  • #3
Perhaps what you really want is L(u(t-a)).

[tex]u(t-a)=\begin{cases} 0, & t < a \\ 1, & t \ge a \end{cases}[/tex]

Then
[tex]L(u(t-a))= \int_0^\infty u(t-a)e^{-st} dt= \int_a^\infty e^{-st}dt[/tex]
Let v= t- a. Then t= v+ a, dv= dt, when t= a, v= 0, and when t= [itex]\infty[/itex], v= [itex]\infty[/itex]. The integral becomes
[tex]L(u(t-a)= \int_0^\infty e^{s(v+a)}dv= \int_0^\infty e^{sv}e^{sa}dv[/tex]
[tex]= e^{sa}\int_0^\infty e^{sv}dv= e^{sa}e^{-st}dt[/tex]
and that last integral is the Laplace transform of 1.

That u(t-a) step function multiplying a function basically multiplies the Laplace transform of the function by [itex]e^{as}[/itex].
 

1. What is a Laplace Transform?

A Laplace Transform is a mathematical tool used to convert a function in the time domain into a function in the complex frequency domain. It is often used in engineering and physics to solve differential equations and analyze systems.

2. What is a Unit Step Function?

A Unit Step Function is a piecewise function that is equal to 0 for all values less than a specific value, and equal to 1 for all values greater than or equal to that value. It is often used to model a sudden change or jump in a system.

3. How are Laplace Transforms and Unit Step Functions related?

Laplace Transforms can be used to solve differential equations involving Unit Step Functions. By taking the Laplace Transform of a system with a Unit Step Function, the function can be transformed into a simple algebraic expression, making it easier to solve.

4. What is the inverse Laplace Transform?

The inverse Laplace Transform is the opposite of the Laplace Transform. It takes a function in the frequency domain and converts it back into the time domain. This is useful for obtaining the original function after solving a differential equation using Laplace Transforms.

5. Why are Laplace Transforms important in science and engineering?

Laplace Transforms are important because they provide a powerful tool for solving complex differential equations and analyzing systems. They allow for the transformation of a problem from the time domain to the frequency domain, where it can be easily solved using algebraic methods. This makes them useful in a wide range of fields, including physics, engineering, and mathematics.

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