Inverse Laplace involving heaviside function

  • Thread starter reed2100
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  • #1
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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.
 

Answers and Replies

  • #2
Ray Vickson
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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
49
1
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!
 

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