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Homework Help: Signals - Integration of Heavyside Step & Dirac Delta Functions

  1. Sep 6, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex]
    \int_{-\infty}^{\infty}{u(t)e^{-t}(\delta(t+1)+\delta(t-1))dt
    [/tex]


    2. Relevant equations

    [tex]
    \int_{-\infty}^{t}{u(t)dt = \left\{\begin{array}{cc}0,&\mbox{ if }
    t< 0\\t, & \mbox{ if } t>0\end{array}\right.
    [/tex]


    [tex]
    \int_{-\infty}^{\infty}{f(t)\delta(t-a)dt} = f(a)
    [/tex]


    3. The attempt at a solution

    [tex]
    \int_{-\infty}^{\infty}{u(t)e^{-t}(\delta(t+1)+\delta(t-1))dt
    [/tex]

    [tex]
    = \int_{-\infty}^{\infty}{u(t)e^{-t}{\delta(t+1)dt}
    + \int_{-\infty}^{\infty}{u(t)e^{-t}{\delta(t-1)dt}
    [/tex]

    I obviously could use the second relevant equation if the u(t) term was not in these integrals.

    I am stuck. Could someone point me in the right direction?
     
    Last edited: Sep 6, 2010
  2. jcsd
  3. Sep 6, 2010 #2
    [tex]
    = \int_{-\infty}^{\infty}{u(t)e^{-t}{\delta(t+1)dt}
    + \int_{-\infty}^{\infty}{u(t)e^{-t}{\delta(t-1)dt}
    [/tex]

    [tex]
    = \int_{0}^{\infty}{e^{-t}{\delta(t+1)dt}
    + \int_{0}^{\infty}{e^{-t}{\delta(t-1)dt}
    [/tex]

    Is that correct?
     
  4. Sep 6, 2010 #3

    vela

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    The upper limit of the integral should be t, not ∞.
    Why does the presence of u(t) stop you? What's the definition of the Heaviside step function?
     
  5. Sep 6, 2010 #4

    vela

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    Yes.
     
  6. Sep 6, 2010 #5
    [tex]

    = \int_{0}^{\infty}{e^{-t}{\delta(t+1)dt}
    + \int_{0}^{\infty}{e^{-t}{\delta(t-1)dt}

    [/tex]

    [tex]

    = [e^{-t}]_{t=-1}
    + [e^{-t}]_{t=1}

    [/tex]



    [tex] [e^{-t}]_{t=-1} = 0 [/tex] because [tex] t=-1 [/tex] is out of the limits of integration.



    [tex] [e^{-t}]_{t=-1}
    + [e^{-t}]_{t=1}
    = \frac{1}{e}
    [/tex]


    Does that look right?
     
  7. Sep 6, 2010 #6

    vela

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    Yes and no. You figured out the answer correctly, but what you wrote isn't correct. For one thing, the first integral isn't equal to [itex][e^{-t}]_{t=-1}[/itex] since [itex][e^{-t}]_{t=-1}=e[/itex]. Second, [itex][e^{-t}]_{t=-1}[/itex] is always equal to e; it's never equal to 0. What you should have written is simply that the first integral is equal to 0 because the delta function is zero over the interval of integration.
     
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