Inverse Laplace transform with unit step function

In summary, the conversation revolves around the process of taking the inverse Laplace transform of a unit step function. The individual is struggling to understand a step in the solution, which involves converting a polynomial fraction to a sum of simpler terms. They are seeking clarification and help on how to properly use the Laplace transform in Latex.
  • #1
shaqywacky
4
0
Hello again.

First off, I wasn't sure how to say this in the title but I'm not taking the inverse Laplace transform of a unit step function. I'm taking the Laplace transform of something that comes out to the unit step function.

I have this question, which is a similar version of the question I am trying to solve. This is also the solution but I have no idea what happened.

laplace_unit.png

(sorry, I don't know why that is so small, click to make it bigger)

So the first part of that image is just the inverse Laplace transform I am trying to solve. The first step I saw was to do partial fractions. When I do that I get:
[itex]\frac{e}{s+1} + \frac{1}{s}[/itex]
I'm omitting the inverse Laplace transform here because I don't know how to do it in latex.

But as you can see, the solution maunual got:
[itex]\frac{e^{-s}}{s} - \frac{e^{-s}}{s+1}[/itex]

Clearly there is something I do not understand about this. I can get the correct answer after this, I just don't understand this step.

Any help would be greatly appreciated.

Thank you.
 
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  • #2
Hi !
You question seems confused.
You must not use the same variable (s) for the function and for the Laplace transform of the function.
Please write clearly :
"The Laplace transform of f(x) is F(s)"
or
"The inverse Laplace transform of F(s) is f(x)"
Then, write clearly the expression of the known function : is it f(x) or F(s) ?
 
  • #3
I don't think I did. I never took any inverse Laplace transforms in my post. I'm talking about inside the Laplace transform. I didn't know how to use the Laplace transform in Latex, so I omitted it and made a note of it but I don't think I was being very clear. I'm just omitting the notation of the Laplace transform. So the equations I posted (besides the ones in the picture) are what's inside of the Laplace transform I'm trying to solve.

To try to clarify, I just don't understand the first step that is in the picture. So I don't see how they went from
[itex]\frac{e^{-s}}{s(s-1)}[/itex]

to

[itex]\frac{e^{-s}}{s} - \frac{e^{-s}}{s-1}[/itex]

Thanks.
 
  • #4
Hi !

Do you know how to write a polynomial fraction as a sum of simple terms ?
Have a look to the attachment :
 

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  • #5


Hi there,

The inverse Laplace transform of a unit step function is a common problem in mathematics and engineering. The unit step function, also known as the Heaviside step function, is defined as:

u(t) = {0, t<0; 1, t>=0}

When taking the Laplace transform of a function that results in a unit step function, the inverse Laplace transform will involve a combination of exponential and step functions. In the example you provided, the solution manual used the method of partial fractions to break down the original function into simpler terms. However, it looks like they made a mistake in their calculations.

The correct partial fractions decomposition for your function is:

\frac{e}{s+1} + \frac{1}{s} = \frac{e}{s} + \frac{e}{s+1}

From here, you can use the table of Laplace transforms to find the inverse Laplace transform of each term. The first term will result in the unit step function u(t), while the second term will result in the exponential function e^{-t}. So the final inverse Laplace transform will be:

L^{-1}\{\frac{e}{s+1} + \frac{1}{s}\} = u(t) + e^{-t}

I hope this helps clarify the steps involved in finding the inverse Laplace transform of a function that results in a unit step function. Let me know if you have any further questions. Best of luck with your studies!
 

What is an Inverse Laplace transform?

An Inverse Laplace transform is a mathematical operation that takes a function in the Laplace domain (s-domain) and converts it back into the time domain. It is used to solve differential equations and understand the behavior of linear systems.

What is a unit step function?

A unit step function, also known as the Heaviside function, is a function that is defined as 0 for negative values and 1 for positive values. It is commonly used in engineering and physics to represent a sudden change or "step" in a system.

How is the Inverse Laplace transform of a unit step function calculated?

The Inverse Laplace transform of a unit step function is calculated using the formula f(t) = 1/s * e^(-as), where s is the Laplace variable and a is the point at which the step occurs. This formula can be derived from the definition of the Laplace transform and the properties of the unit step function.

What is the significance of the Inverse Laplace transform with a unit step function?

The Inverse Laplace transform with a unit step function is commonly used in the analysis of control systems and circuit analysis. It allows us to understand the behavior of a system after a sudden change or "step" has occurred, and can help us design and optimize systems for various applications.

Can the Inverse Laplace transform be used for functions other than the unit step function?

Yes, the Inverse Laplace transform can be used for a wide range of functions, including exponential, trigonometric, and polynomial functions. It is a powerful tool in solving differential equations and understanding the behavior of complex systems.

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