1. ## Step function?

I have a laplace transform of F(P) = [1-exp(-p)]/[p(1+exp(-2p)] and it tells me to find the laplace transform of y(t) and it gives me a graph where:

y(0) = 0
y(1) = 1
y(2) = 1
y(3) = 0
y(4) = 0
y(5) = 1
y(6) = 1
and the pattern continues

I think this is come kind of heaviside step function but how do i find out the equation to it?

I have a laplace transform of F(P) = [1-exp(-p)]/[p(1+exp(-2p)] and it tells me to find the laplace transform of y(t) and it gives me a graph where:

y(0) = 0
y(1) = 1
y(2) = 1
y(3) = 0
y(4) = 0
y(5) = 1
y(6) = 1
and the pattern continues

I think this is come kind of heaviside step function but how do i find out the equation to it?
You need to post your questions more clearly.

Are you trying to find f(t) given that its laplace transform is F(P) = [1-exp(-p)]/[p(1+exp(-2p)]?

3. yes I am and the graph I was given gives the answer of f(t) but I just don't know how to find the equation of that graph

yes I am and the graph I was given gives the answer of f(t) but I just don't know how to find the equation of that graph
Here are two things (the first you should already have been taught):

Theorem: If f(t) is a periodic function of period T then $\displaystyle LT[f(t)] = \frac{1}{1 - e^{-pt}} \int_0^T f(t) e^{-pt} \, dt$.

In light of the above theorem, you need to find $\displaystyle LT^{-1} \left[ \frac{1 - e^{-p}}{p}\right]$ and then define f(t) as an appropriate periodic function.

Now you need to think hard about these two things and then make a strong attempt at the question before asking for more help.

5. Is it still included as a periodic function if the lines are straight lines? as in from 0 to 1 it is a straight line with y = x then a flat line from 1 to 2 then from 3 to 4 is is a line from one to 0 thats downward sloping..not sure if that made sense

6. Originally Posted by mr fantastic
Here are two things (the first you should already have been taught):

Theorem: If f(t) is a periodic function of period T then $\displaystyle LT[f(t)] = \frac{1}{1 - e^{-pt}} \int_0^T f(t) e^{-pt} \, dt$.

In light of the above theorem, you need to find $\displaystyle LT^{-1} \left[ \frac{1 - e^{-p}}{p}\right]$ and then define f(t) as an appropriate periodic function.

Now you need to think hard about these two things and then make a strong attempt at the question before asking for more help.
My mistake. The above is not directly relevant to the question. I misread the denominator of the expression in post #1 as having a minus instead of a plus. If you re-arrange the form of the expression by multiplying numerator and denominator by $\displaystyle 1 - e^{-2p}$ then the ideas I have suggested will work I think.