# i need help with applying Laplace transform

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• Nov 9th 2010, 06:19 AM
maliZeleni
i need help with applying Laplace transform
hello,

I need to apply Laplace transform on this signal

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

and i'm just stuck with this one :D so can anybody please help me with this :D

any help will be most appreciated :D:D:D
• Nov 9th 2010, 06:21 AM
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?
• Nov 9th 2010, 06:29 AM
maliZeleni
Quote:

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 ?
• Nov 9th 2010, 06:33 AM
Ackbeet
Technically, you'd have

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

Now what?
• Nov 9th 2010, 06:43 AM
maliZeleni
Quote:

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 ?

$\displaystyle \mathcal{L}[u(t)] = \frac {1}{S}$
• Nov 9th 2010, 06:44 AM
Ackbeet
No, no. You're done with the LT. You should plug these results back into your original equation. What do you get?
• Nov 9th 2010, 06:50 AM
maliZeleni
Quote:

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 :D
• Nov 9th 2010, 06:55 AM
Ackbeet
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.
• Nov 9th 2010, 07:05 AM
maliZeleni
Quote:

Originally Posted by Ackbeet

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

Does that suggest anything to you?

Unfortunately no :D

yes T is positive number...
• Nov 9th 2010, 07:08 AM
Ackbeet
• Nov 9th 2010, 07:21 AM
maliZeleni
Quote:

Originally Posted by Ackbeet

so this should be :

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

? :D
• Nov 9th 2010, 07:23 AM
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?
• Nov 9th 2010, 07:27 AM
maliZeleni
Quote:

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 :D:D:D

as you realise i'm terrible with English , and i don't understand what's LHS ?
• Nov 9th 2010, 07:29 AM
Ackbeet
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.