Laplace transform of unit step function

[Icetray]

Homework Statement

The Attempt at a Solution

I know that u(t) is a unit step function and holds a value of either 0 or 1. In laplace transform, when we integrate f(t) from 0 to infinity, we take u(t) to be 1.

In this case, since u(t) is u(-t), does this mean it holds a value of 0? Does not make sense to me that both answers are 0.

Thanks in advance for the help. :) Any links to notes on similar laplace transforms would be very much appreciated as well! :)

 

[HallsofIvy]

I don't understand what you want to do. The Laplace transform is defined only for functions on the positive real numbers. u(t) is 1 for 0\le t\le 1, 0 otherwise. So u(-t) is 1 for -1\le t\le 0. Since both of your x(t) is 0 for all positive t, their Laplace transform is 0.

 

[Icetray]

I don't understand what you want to do. The Laplace transform is defined only for functions on the positive real numbers. u(t) is 1 for 0\le t\le 1, 0 otherwise. So u(-t) is 1 for -1\le t\le 0. Since both of your x(t) is 0 for all positive t, their Laplace transform is 0.

Thank you so much for your reply Halls of Ivy! It's much appreciated.

The exact question is actually just "Find the Laplace transform of:". I apologize for leaving it out.

So just to confirm, the answers for both questions are actually 0? Seems a little weird that they've give us two questions on the same "concept".

Could it be possible that instead of a unit step function, it could be some sort of shift?

Once again, thank you so much! :)

 

[Ray Vickson]

Thank you so much for your reply Halls of Ivy! It's much appreciated.

The exact question is actually just "Find the Laplace transform of:". I apologize for leaving it out.

So just to confirm, the answers for both questions are actually 0? Seems a little weird that they've give us two questions on the same "concept".

Could it be possible that instead of a unit step function, it could be some sort of shift?

Once again, thank you so much! :)

Careful: the usual definition of the unit step function u(t) is
u(t) = \left\{ \begin{array}{cl}1 &amp; \text{ if } t \geq 0\\<br /> 0 &amp; \text{ if } t &lt; 0.<br /> \end{array} \right.
So, as said already, your functions are multiples of u(-t), so are 0 for t >= 0. Thus, their Laplace transforms are zero.

 

[LCKurtz]

Thank you so much for your reply Halls of Ivy! It's much appreciated.

The exact question is actually just "Find the Laplace transform of:". I apologize for leaving it out.

So just to confirm, the answers for both questions are actually 0? Seems a little weird that they've give us two questions on the same "concept".

I agree. I would almost bet there is something missing or misunderstood about notation or transcription here. Have you copied the complete problem exactly, word for word?

 

[Icetray]

I agree. I would almost bet there is something missing or misunderstood about notation or transcription here. Have you copied the complete problem exactly, word for word?

Yups, that's all the question says. Nothing before that and no hints after that. :( I guess I'll just stick to 0 then.

Is it possible though that it could be a shifting problem instead of a unit step function?