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Need help with Impulse function and unit step function at singularity

  1. May 14, 2005 #1
    I have some trouble wrapping my head around singularity

    One of assignment question is to show that the unit function is not defined at 0. To do that, I need to show

    [tex]\lim_{\Delta\to0}[u_{\Delta}(t)\delta(t)]=0[/tex]
    [tex]\lim_{\Delta\to0}[u_{\Delta}(t)\delta_{\Delta}(t)]=\frac{1}{2}\delta(t)[/tex]


    Also, I need to show that the following is identical to u(t)
    [tex]g(t)=\int u(t)\delta(t-\tau)d\tau[/tex]
    integrating from negative infinity to positive infinity


    One more question, what's the derivative of the impulse function?

    PS: what's the tex code for integration from infinity to infinity? I tried \int_-\infty^+\infty, but the tex output is messed up
     
  2. jcsd
  3. May 14, 2005 #2

    dextercioby

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    Of course.Click on this

    [tex] \int_{-\infty}^{+\infty} [/tex]

    and use {} wherever necessary.

    Daniel.
     
  4. May 16, 2005 #3
    thanks dex,


    still need help with the signularity functions ...
     
  5. Sep 27, 2009 #4
    Hello very body
    i got confused understanding how we can get the derivative of V(t) equal to Hv(t) :


    V= (1/R) e^(-t/RC) U(t) its derivative V' = Hv(t) = -(1/R²C)e^(-T/RC) U(t) +(1/R)&(t)
    where : U(t) is the unite step fonction and &(t) is the delta fonction
    please some help
    thanks for your time
     
  6. Sep 29, 2009 #5
    Second part:

    The Kronecker delta is non zero only when the parameter in its brackets is equal to zero. So for the integral, its only when tao is equal to t that delta(t) is non zero. Set tau=t, and you have the result.

    The derivative of the step function is the Kronecker delta function. check this out
    http://mathworld.wolfram.com/HeavisideStepFunction.html
     
  7. Sep 29, 2009 #6
    It is Dirac Delta function that is mentioned here. Kronecker Delta is different.
     
  8. Sep 30, 2009 #7
    Oh sorry, I meant the Dirac delta function, not the Kronecker, my bad.
     
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