Laplace transform heavyside function

1. Jun 20, 2009

elcotufa

1. The problem statement, all variables and given/known data
find the transform of
$$sin (wt) u(t - b)$$

I have to get it into the form for the time shift but I do not know how to deal with the unit function

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

2. Relevant equations

3. The attempt at a solution

2. Jun 20, 2009

xepma

The Laplace transform is defined as:
$$\mathcal{L}\lbrace f(t)\rbrace = \int_0^{\infty} e^{-st}f(t) dt$$
Now, let's put in
$$f(t) = g(t) u(t-b)$$
where u(t-b) is zero for t<b and 1 for t>b. Plugging this in:

$$\int_0^{\infty} e^{-st}f(t) dt = \int_0^{\infty} e^{-st}g(t) u(t-b) dt$$

Now split this integral into two parts:

$$\int_0^{\infty} e^{-st}g(t) u(t-b) dt = \int_0^{b} e^{-st}g(t) u(t-b) dt + \int_b^{\infty} e^{-st}g(t) u(t-b) dt$$

The first integral vanishes, since u(t-b) is zero on the integration domain. In the second the u-function is simply 1. I'll let you solve the remaining steps.

3. Jun 20, 2009

elcotufa

ahhh cool

now lets say I have
$$(t-1)^4 u(t)$$

function is 1 from 0 < t < 1 and $$(t-1)^4$$ for t> 1

I can easily do the transform of t^4 how would i play with it to just do a time shift after the transform or do I have to go the integral route and do parts?

4. Jun 22, 2009

Defennder

If you do not have to work it out from first principles (and I do not think you have to), you could do it by using this L-transform formula:
$$L[f(t-a)u(t-a)] = e^{-as}F(s) \text{ where F(s) is L.T. of } f(t)$$.

Of course in the case of $$(t-1)^4 u(t)$$, it's not in a form which yields itself readily to that formula. In such a case you may have to work out the binomial expansion of (t-1)^4 before applying that formula.