# Unit step functions to find Laplace Transform

1. Feb 5, 2013

### izen

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

f(t) = cos (pi*t) if 1$\leq$ t <4 and 0 elsewhere

using unit step functions to find Laplace Transform

2. Relevant equations

3. The attempt at a solution

I came up with the unit step function f(t) = cos(pi*t) u(t-1) - cos(pi*t) u(t-4)
in order to use the second shifting theorem f(t) must in the format of f(t-a) in this case a = 1 and 4.

I add 1 and subtract 1 , add 4 and subtract 4 and I get

f(t) = cos(pi(t-1+1) u(t-1) - cos(pi(t-4+4) u(t-4)
= cos (pi(t-1)) u(t-1) +u(t-1) - cos(pi*(t-4) u(t-4)+u(t-4) << i think this step is not correct
It should get -cos (pi(t-1)) u(t-1) - cos(pi*(t-4) u(t-4) but i dont know how to get there

2. Feb 5, 2013

### jbunniii

It is not correct. Instead, let's go back to the previous step:
$$f(t) = \cos(\pi(t-1+1)) u(t-1) - \cos(\pi(t-4+4)) u(t-4)$$
I can write this as follows:
$$f(t) = \cos(\pi(t-1)+\pi) u(t-1) - \cos(\pi(t-4)+4\pi) u(t-4)$$
Now what can you say in general about $\cos(x + \pi)$? What about $\cos(x + 4\pi)$?

3. Feb 5, 2013

### Ray Vickson

Does the problem statement say you MUST use the unit step function? If not, just using the original form is as easy a way as any:
$$\cal{L}(f)(s) = \int_0^4 e^{-st} \cos(\pi t) \, dt.$$

4. Feb 5, 2013

### izen

Thank you jbunniii I got it After that we use the trig identity of cos(x+y)=cos(x)-cos(y)-sin(x)sin(y)
Thank you Ray Vickson for the tip