# Homework Help: Laplace transform of unit step function

1. Aug 22, 2013

### Icetray

1. The problem statement, all variables and given/known data

3. 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! :)

2. Aug 23, 2013

### 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.

3. Aug 23, 2013

### Icetray

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! :)

4. Aug 23, 2013

### Ray Vickson

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

5. Aug 23, 2013

### LCKurtz

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?

6. Aug 23, 2013

### Icetray

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?