# Thread: Help with unit step function and laplace transform

1. ## Help with unit step function and laplace transform

I could not figure out how to get the LaTex to look right with piece wise functions, so I just made a picture. Here are the 2 problems:

Its sopposed to be a piece wise function but I cannot figure out do it with latex.

For no. 1 How do I set it up? Once I get past that, Im sure I can do the rest. Heres what I have:
$\displaystyle 3U(t) - e^4^t^-^1^1U(t-3)$ Is this correct?

For problem no.2, I just need to get to the undertermined coefficient part. So heres what I do for the problem:
$\displaystyle s^2Y(s) - sy(0) - y'(0) - 3sY(s) - y(0) = L{sin(t-2\pi ) U(t-2\pi )}$

Solving for $\displaystyle Y(s)$

$\displaystyle Y(s) =\frac{ e^-^2^\pi^s}{s(s^2+ 1)(s-3)} + \frac{1}{s(s-3)}$

$\displaystyle Setting up Undertermined Coefficients:$

$\displaystyle e^2^\pi ^s = \frac{As+B}{s^2+1} + \frac{C}{s} + \frac{D}{s-3}$

Does this look correct so far? Thanks guys!

2. 1. For each piece, you can regard the Heaviside step function as a switch that turns that piece on or off. I think you need to have something more like the following:

$\displaystyle f(t)=3(\overbrace{U(t)}^{\text{turns it on}}-\underbrace{U(t-2)}_{\text{turns it off}})+e^{4t-11}U(t-3).$

2. Your Y(s) is correct. I'm not sure what your last line is doing, though. Why not just take the inverse transform of what you have? That's the beauty of the LT approach: it takes care of initial conditions, piece-wise defined functions, and inhomogeneous terms in a fairly straight-forward manner, when it works at all.

3. How would i take the inervse transform of what I have? Dont you have to use partial fractions?

4. Originally Posted by Jeonsah
How would i take the inervse transform of what I have? Dont you have to use partial fractions?

Yes but ignore the e factor for a bit and use

$\displaystyle \frac{1}{s(s-3)(s^2+1)}=\frac{A}{s}+\frac{B}{s-3}+\frac{Cs+D}{s^2+1}$

After you have completed partial fraction multiply every term of the solution by

$\displaystyle e^{-2\pi s}$

You will need to use partial fractions on the 2nd fraction as well

5. sweet thanks. So let me get this straight. The term:
$\displaystyle \frac{Cs+D}{S^2 + 1}$

Will need partial fractions again? These partial fractions are killin me haha. So would I solve it on its on by doing:
$\displaystyle \frac{Cs+D}{S^2+1} = ?$

Sorry for the nub mistakes, I just dont recall learning learning partial fractions very much in high school. Thanks.

6. Originally Posted by Jeonsah
sweet thanks. So let me get this straight. The term:
$\displaystyle \frac{Cs+D}{S^2 + 1}$

Will need partial fractions again? These partial fractions are killin me haha. So would I solve it on its on by doing:
$\displaystyle \frac{Cs+D}{S^2+1} = ?$

Sorry for the nub mistakes, I just dont recall learning learning partial fractions very much in high school. Thanks.
No, I was unclear I meant the other term in the original expression.

$\displaystyle \frac{1}{s(s-3)}=\frac{A}{s}+\frac{B}{s-3}$

They can be combined and done at the same time because the first term has the factor of

$\displaystyle e^{-2\pi s}$

7. Thank you very much man. You are extremel help full. I really appreciate it. I gotta go to work now, but i got a couple more questions ill post up later. But thank you very the help!!!!!!!!!!!!!!!!!!!!!!!