Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Find $L\{F(t)\}$ if

$$F(t) = \begin{cases} \sin t & \text{between }0 < t < \pi \\ 0 & \text{between } \pi < t < 2\pi \end{cases}$$

Really stumped by this one. Please can you work this out?

share|improve this question
    
I'm stumped too... –  David Mitra Jan 20 '13 at 2:10
1  
Do you wish to find the Laplace transform of $F$? –  David Mitra Jan 20 '13 at 2:11
    
yes, I do! The laplace transform of F(t) –  user58944 Jan 20 '13 at 2:17
    
You need to specify $F$ for all $t\ge 0$, or say it is periodic. As stated problem is incomplete. –  Maesumi Jan 20 '13 at 6:04

1 Answer 1

I presume the function is periodic. Have you tried considering

$$\mathcal{L}[f(t)] = \int_{0}^{\infty} e^{-pt}f(t) dt = \sum_{n=0}^{\infty}\int_{n\pi}^{(n+1)\pi} e^{-pt}f(t) = \sum_{n=0}^{\infty}\int_{2n\pi}^{(2n+1)\pi} e^{-pt}\sin(t)$$

But this is just

$$\sum_{n=0}^{\infty} \frac{(e^{p\pi} + 1)(e^{\pi(-(2n+1))p})(p\sin(2\pi n) + \cos (2 \pi n))}{p^2 + 1}$$

Since $\sin (2\pi n) = 0$ and $\cos (2 \pi n) = 1$, this becomes

$$\sum_{n=0}^{\infty} \frac{(e^{p\pi} + 1)(e^{\pi(-(2n+1))p})}{p^2 + 1}$$

And in fact we can pull out terms involving $p$ to obtain

$$ \frac{(e^{\pi} + 1)e^{-p\pi}}{p^2 +1}\sum_{n=0}^{\infty} e^{-2n p \pi}$$

And we evaluate the sum to obtain

$$ \frac{(e^{\pi} + 1)e^{-p\pi}}{p^2 +1}\left(\frac{e^{2\pi p}}{e^{2 \pi p} - 1} \right) = \frac{(e^{\pi p} +1)\mbox{csch} (\pi p)}{2(p^2 + 1)}$$


If this function is not periodic, but $0$ for $t> 2\pi$, then

$$\mathcal{L}[f(t)] = \int_{0}^{\infty} e^{-pt}f(t) dt = \int_{0}^{\pi} e^{-pt}\sin(t) dt = \frac{e^{-p\pi} + 1}{p^2 + 1}$$

share|improve this answer
    
Thank you for your help, I think the answer is expected in the s domain –  user58944 Jan 20 '13 at 2:13
    
wow! Thank you for taking the time to do all of this. You're a great help! What is the csch though? –  user58944 Jan 20 '13 at 2:32
    
The function $\mbox{csch}$ is the hyperbolic cosecant. It's defined by $\mbox{csch}z = \frac{2}{e^{z} - e^{-z}}$ –  Isaac Solomon Jan 20 '13 at 2:33
    
OK, I think this might be a more complex solution than I should have! Thanks for guiding me through this anyway. I'll try and get it into the form of s –  user58944 Jan 20 '13 at 2:39
    
This is in the form of $s$, I've just written $p$ for $s$, because it's what I'm used to. –  Isaac Solomon Jan 20 '13 at 2:40

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.