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Inner product of dirac delta function

  1. Jan 10, 2014 #1
    1. The problem statement, all variables and given/known data

    Find the inner product of f(x) = σ(x-x0) and g(x) = cos(x)

    2. Relevant equations

    ∫f(x)*g(x)dx

    Limits of integration are -∞ to ∞

    3. The attempt at a solution

    First of all, what is the complex conjugate of σ(x-x0)? Is it just σ(x-x0)?

    And I'm not sure how to multiply f(x) and g(x) together. Do I just pick out a value of x0 that will make the dirac delta disappear and be left with the cos(x0)?
    Thanks
     
  2. jcsd
  3. Jan 10, 2014 #2

    strangerep

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    What is ##\sigma(x-x_0)##? Did you mean ##\delta(x-x_0)##?

    If so, the problem involves nothing more than the definition of the Dirac delta distribution. So perhaps you should write down that definition here, since it seems to be missing from your "relevant equations" section.
     
  4. Jan 11, 2014 #3
    edit: Oh, I was using sigma and I was supposed to be using delta; hence the "delta" function. They look similar.

    The dirac delta function just picks out the value of x that makes x - x0 = 0, and in that case, the function equals 1.

    So basically, this problem is just a normal integral, since there's no complex conjugate of the dirac delta function?

    Thanks.
     
    Last edited: Jan 11, 2014
  5. Jan 11, 2014 #4

    strangerep

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    That's the effect of the dirac delta when integrated against another function.
    I don't know which "function" you're referring to here, but I'm reasonably sure you still have some misconceptions about the Dirac delta (since you're still calling it a "function", or so it seems).

    Take a look at the Dirac delta Wiki page. In particular the section titled "As a distribution". The distinction between "distribution" and "ordinary function" is very important for doing anything nontrivial with the Dirac delta.

    For simple cases, you can get away with thinking of it like that. But for less trivial stuff, you must understand the concept of "distribution" mentioned above.

    It's incorrect to say there's "no complex conjugate". Rather the Dirac delta is its own conjugate (like a real number).
     
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