# Thread: i need help with applying Laplace transform

1. ## i need help with applying Laplace transform

hello,

I need to apply Laplace transform on this signal

$\displaystyle x(t)= \sum _{k=0} ^{+\infty} \delta (t-kT)$

any help will be most appreciated

2. Well, the LT is linear, correct? So you should be able to move the Laplace Transform operator inside the sum. What's the LT of the Dirac Delta function you have in there?

3. Originally Posted by Ackbeet
Well, the LT is linear, correct? So you should be able to move the Laplace Transform operator inside the sum. What's the LT of the Dirac Delta function you have in there?

Laplace transform of the function $\displaystyle \delta (t -kT )$ should be $\displaystyle \frac {1}{e^{kTS}}$ correct ?

4. Technically, you'd have

$\mathcal{L}[\delta(t-kT)]=e^{-kTs}u(kT),$ where $u(t)$ is the Heaviside step function.

Now what?

5. Originally Posted by Ackbeet
Technically, you'd have

$\mathcal{L}[\delta(t-kT)]=e^{-kTs}u(kT),$ where $u(t)$ is the Heaviside step function.

Now what?
should i now do LT of Heaviside function ?

$\displaystyle \mathcal{L}[u(t)] = \frac {1}{S}$

6. No, no. You're done with the LT. You should plug these results back into your original equation. What do you get?

7. Originally Posted by Ackbeet
No, no. You're done with the LT. You should plug these results back into your original equation. What do you get?
i get

$\displaystyle \sum _{k=0} ^{+\infty} e^{-kTS} u(kT)$

now i'm confused, because since k goes to infinity this sum converge to zero ?! i think

8. What is T? Is is positive? If so, you can lose the Heaviside step function (it'll just be 1 everywhere). What do you have on the LHS of this equation? Finally, assuming you can ignore the step function, you can rewrite the sum this way:

$\displaystyle\sum_{k=0}^{\infty}(e^{-Ts})^{k}.$

Does that suggest anything to you?

It's true that the sequence $e^{-kTs}$ probably converges to zero. But that doesn't mean the series

$\displaystyle\sum_{k=0}^{\infty}e^{-kTs}$

converges to zero.

9. Originally Posted by Ackbeet

$\displaystyle \sum_{k=0}^{\infty}(e^{-Ts})^{k}.$

Does that suggest anything to you?
Unfortunately no

yes T is positive number...

10. Originally Posted by Ackbeet
so this should be :

$\displaystyle \mathcal{L} [ \sum _{k=0} ^{+ \infty} \delta (t-kT) ] = \frac {1}{1- e^{-ST}}$

?

11. Well, you should probably have a lower-case s in there, but yes. However, you also need to do the LT of the LHS of the original equation in the OP. That's so hard it's easy:

$\mathcal{L}[x(t)]=X(s)=\dfrac{1}{1-e^{-sT}}.$

Make sense?

12. Originally Posted by Ackbeet
Well, you should probably have a lower-case s in there, but yes. However, you also need to do the LT of the LHS of the original equation in the OP. That's so hard it's easy:

$\mathcal{L}[x(t)]=X(s)=\dfrac{1}{1-e^{-sT}}.$

Make sense?
thank you very very much

as you realise i'm terrible with English , and i don't understand what's LHS ?

13. Oh, sorry. I didn't know. Your English is better than some native speakers I've seen.

LHS = Left Hand Side
RHS = Right Hand Side
LT = Laplace Transform

Cheers.