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Laplace and ROC of function(- [e^(-at)]u(-t))

  1. Nov 15, 2016 #1
    1. The problem statement, all variables and given/known data
    If laplace of [e^-(at)] u(t) is 1/(s+a) and ROC is s > -a
    Find laplace and ROC of -e^(-at) u(-t)

    2. Relevant equations
    Laplace is integral over minus infinity to plus infinity of f(t) e^(-st) dt

    3. The attempt at a solution
    Well i integrated f(t) over the limits with e(-st)u(-t) dt
    the u(-t) changes limits to from (-inf to +inf) to (-inf to zero.)
    So i integrated it and got Laplace as 1/(s+a)
    Now ROC is region for which laplace goes positive that is more than zero
    So in case of (1/(s+a)) s+a must be > 0 for F(s) to be positive.
    For that s > (-a)

    The book says that laplace is right but ROC is s < (-a)
    Book didnt integrate. It went through other formula.
    like L(f(t)) is F(s)
    then laplace of f(-t) is F(-s)
    so Laplace of e^(at)u(-t) becomes 1/(-s + a) ROC is -s + a > 0 so -s > -a
    Now a becomes -a
    so Laplace of e ^ (-at) u(-t) becomes 1/(-s -a) = -1 /(s + a) ROC is s + a > 0 so s > -a
    Multiply both side f(t) and F(s) by -1
    Laplace of -1 * e ^ (at) u(-t) becomes 1 / (s + a) Here it said that ROC remains same as power remains same. And ROC is -s > a
    But shouldnt ROC for (1/(s+a) ) be s + a > 0, so s > -a
    I'm confused as to why is ROC different. What does underlined part mean?
     
  2. jcsd
  3. Nov 15, 2016 #2

    I like Serena

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    Hi jaus tail! :oldsmile:

    Simply calculating the Laplacian gives us:
    $$\mathcal L[-e^{-at} u(-t)] = \int_{-\infty}^\infty e^{-st}\cdot-e^{-at} u(-t)dt
    = \int_{+\infty}^{-\infty} e^{s\tau}\cdot e^{a\tau} u(\tau)d\tau
    =\int_{+\infty}^{0} e^{(s+a)\tau}d\tau
    =\frac{1}{s+a}e^{(s+a)\tau} \Big|_{+\infty}^0
    =\frac{1}{s+a}
    $$
    with ROC ##(s+a)<0\Rightarrow s<-a##.
     
  4. Nov 16, 2016 #3
    Thanks for the reply.
    But isn't ROC values for which denominator of laplace is positive. Like in the below exam from screen show of book.
    upload_2016-11-16_21-17-43.png


    Here ROC is s > -a, means s + a is more than 0. Laplace of function is 1 / (s + a), so it means denominator of laplace must be positive for ROC, right?

    But then same book as other example as:
    upload_2016-11-16_21-20-45.png
    Why is -a to right of Y axis? here ROC is s < -a, means s + a < 0, whereas above it was s + a > 0.
     

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  5. Nov 16, 2016 #4

    I like Serena

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    Not quite.
    ROC is the Region Of Convergence.
    That's where the Laplacian converges.
    In our case it converges iff:
    $$\lim_{\tau\to+\infty} e^{(s+a)\tau} < \infty$$
    And that's only if ##(s+a)<0##.
    All other rules about ROC derive from that and in case of doubt we have to go back to when the integral converges exactly.

    In both your examples we have ##e^{-(s+a)t}## inside the integral.
    After integration that will turn out unchanged due to the nature of the exponential function.
    In the first example the corresponding infinite boundary is ##+\infty## meaning that we require that ##(s+a) > 0## to ensure the integral converges.
    And in the second example the infinite boundary is ##-\infty## meaning that we require that ##(s+a) < 0##.
     
  6. Nov 16, 2016 #5
    Oh...so by converges it means the value must never go infinite. Like an energy signal? The value must always be bounded. Right?

    Another doubt: In second ROC picture why have they put -a, in right side of Y axis? Shouldn't -a be on left side?
     
  7. Nov 16, 2016 #6

    I like Serena

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    Right!
    ##a## could be either positive or negative - we don't know. So it's an arbitrary choice to put ##-a## left or right of the y-axis
     
    Last edited: Nov 16, 2016
  8. Nov 18, 2016 #7
    Thanks for the help.
     
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