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Math Help - i need help with applying Laplace transform

  1. #1
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    Smile 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 so can anybody please help me with this

    any help will be most appreciated
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  2. #2
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    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?
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  3. #3
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    Quote Originally Posted by Ackbeet View Post
    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 ?
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  4. #4
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    Technically, you'd have

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

    Now what?
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  5. #5
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    Quote Originally Posted by Ackbeet View Post
    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}
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  6. #6
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    No, no. You're done with the LT. You should plug these results back into your original equation. What do you get?
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  7. #7
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    Quote Originally Posted by Ackbeet View Post
    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
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  8. #8
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    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.
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  9. #9
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    Quote Originally Posted by Ackbeet View Post

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

    Does that suggest anything to you?
    Unfortunately no

    yes T is positive number...
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  10. #10
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  11. #11
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    Quote Originally Posted by Ackbeet View Post
    so this should be :

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

    ?
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  12. #12
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    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?
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  13. #13
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    Quote Originally Posted by Ackbeet View Post
    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 ?
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  14. #14
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    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.
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