Inverse Laplace involving heaviside function

In summary, Wolfram Alpha's answer for the inverse Laplace transform of the given function is -3 + u(x-4)x -1u(x-4).
  • #1
reed2100
49
1

Homework Statement


Give the inverse Laplace transform of F(s) = (-3/s) + (e^-4s)/(s^2) + (3e^-4s)/s

Homework Equations


Inverse Laplace [e^(-cs) F(s)] = f(x-c)u(x-c)

The Attempt at a Solution



I'll break this into 3 parts.

Part 1 - (-3/s)
-3/s = -3(1/s) -> inverse laplace of -3(1/s) = -3

Part 2 - (e^(-4s))/(s^2)
By looking at the e component, I see that in this case "c" = 4.

Rearrange to (e^(-4s)) (1/s^2)
So the inverse laplace = u(x-4)*x [ u(x-4) being for the heaviside component and x being the inverse laplace of 1/s^2]

Part 3 - (3e^-4s / s)

e component still says that c = 4.
Rearrange into 3(e^-4s) (1/s)
Inverse laplace becomes 3 * u(x-4) * 1 = 3u(x-4)

So MY answer would be that the inverse laplace of the complete F(s) is

-3 + u(x-4)x + 3u(x-4)

The CORRECT answer that wolfram spat out and that is an answer choice on the online practice test is
f(x) = u(x-4)x - 3 - u(x-4)

I see the similarities but that makes it even more confusing to me how they got
-u(x-4) instead of +3u(x-4). I don't see how that works out.
Any and all help is greatly appreciated.
 
Physics news on Phys.org
  • #2
reed2100 said:

Homework Statement


Give the inverse Laplace transform of F(s) = (-3/s) + (e^-4s)/(s^2) + (3e^-4s)/s

Homework Equations


Inverse Laplace [e^(-cs) F(s)] = f(x-c)u(x-c)

The Attempt at a Solution



I'll break this into 3 parts.

Part 1 - (-3/s)
-3/s = -3(1/s) -> inverse laplace of -3(1/s) = -3

Part 2 - (e^(-4s))/(s^2)
By looking at the e component, I see that in this case "c" = 4.

Rearrange to (e^(-4s)) (1/s^2)
So the inverse laplace = u(x-4)*x [ u(x-4) being for the heaviside component and x being the inverse laplace of 1/s^2]

Part 3 - (3e^-4s / s)

e component still says that c = 4.
Rearrange into 3(e^-4s) (1/s)
Inverse laplace becomes 3 * u(x-4) * 1 = 3u(x-4)

So MY answer would be that the inverse laplace of the complete F(s) is

-3 + u(x-4)x + 3u(x-4)

The CORRECT answer that wolfram spat out and that is an answer choice on the online practice test is
f(x) = u(x-4)x - 3 - u(x-4)

I see the similarities but that makes it even more confusing to me how they got
-u(x-4) instead of +3u(x-4). I don't see how that works out.
Any and all help is greatly appreciated.

You should have -3 + u(x-4)(x-4) + 3u(x-4)
 
  • #3
Ray Vickson said:
You should have -3 + u(x-4)(x-4) + 3u(x-4)
Oh ok, I see it now. That's a careless mistake. So when fixing that , the u(x-4)(x-4) becomes u(x-4)x - u(x-4)4.
Combining it all together you get -3 + u(x-4)x -1u(x-4).

Thank you for your help!
 

Related to Inverse Laplace involving heaviside function

1. What is the inverse Laplace transform of a Heaviside function?

The inverse Laplace transform of a Heaviside function u(t-a) is e^(as)/s, where a is the time delay.

2. How do you handle a Heaviside function in a Laplace transform?

To handle a Heaviside function in a Laplace transform, you need to use the time delay property, which states that the Laplace transform of u(t-a) is e^(-as)/s.

3. Can a Heaviside function be used in a Laplace transform for a piecewise function?

Yes, a Heaviside function can be used in a Laplace transform for a piecewise function. It is commonly used to represent a sudden change in the function at a specific time, such as a step function.

4. How do I find the inverse Laplace transform of a function involving multiple Heaviside functions?

If the function involves multiple Heaviside functions, you can use the properties of the Laplace transform to break down the function into smaller parts. Then, use the inverse Laplace transform for each part and combine the results to find the overall inverse Laplace transform.

5. Can a Heaviside function be used in a Laplace transform for a differential equation?

Yes, a Heaviside function can be used in a Laplace transform for a differential equation. It can be used to represent a sudden change or discontinuity in the solution of the differential equation at a specific time.

Similar threads

  • Calculus and Beyond Homework Help
Replies
10
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
147
  • Calculus and Beyond Homework Help
Replies
2
Views
252
  • Calculus and Beyond Homework Help
Replies
2
Views
335
  • Calculus and Beyond Homework Help
Replies
8
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
926
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
852
  • Calculus and Beyond Homework Help
Replies
5
Views
338
Back
Top