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Definition of convolution question

  1. Jun 11, 2009 #1
    [tex]
    (f*g)=\int_{-\infty}^{+\infty}f(\tau)g(t-\tau)
    [/tex]

    i was told that because of the definition of the function
    we can substitute the itervals to
    [tex]
    (f*g)=\int_{0}^{t}f(\tau)g(t-\tau)
    [/tex]


    why??
     
  2. jcsd
  3. Jun 11, 2009 #2

    Cyosis

    User Avatar
    Homework Helper

    What is the definition of the function you were provided?
     
  4. Jun 11, 2009 #3
    a convolution function is 0 at t<0
    and f(t)>0 at t>=0

    i am not sure if this is a correct definition of a convolution function
    ??
     
  5. Jun 11, 2009 #4

    Mark44

    Staff: Mentor

    The two integrals are equal if f([itex]\tau[/itex]) = 0 for [itex]\tau[/itex] <= 0 and for [itex]\tau[/itex] > t.

    I don't believe there is such a thing as a "convolution function." Convolution is an operation involving two functions, and defined as you show in your first integral. You can get to the second integral if function f is nonzero on the interval above, and zero everywhere else.
     
  6. Jun 12, 2009 #5
    i am using the type of function you talked about.
    i was told that if we put a value in tau which is bigger then t
    then we get a negative time for which it defined as 0.

    so what the problem in that??

    why we cant put values bigger then "t" into tau??
     
  7. Jun 12, 2009 #6

    Mark44

    Staff: Mentor

    Let's cut to the chase here. What is the function f you are using? Be sure to include the domain for this function.

    Instead of saying stuff like this, and reporting things that some mysterious person whispered to you
    just tell us what the function is.
     
  8. Jun 12, 2009 #7
    i am not having any function
    i am using a function of type you were talking about
     
  9. Jun 12, 2009 #8
    If you look at the physical motivation it makes sense. Think of f(t) as your driving function e.g. an electric voltage, and g(t) as your impulse response e.g. the current through a capacitor. Then if f(t) begins at t=0, then the response only needs to be tracked on the interval from[0,t]. The current at time t can be calculated by adding up the impulse responses on the interval [0,t].
     
  10. Jun 12, 2009 #9
    i am not having any function
    i am using a function of type you were talking about
     
  11. Jun 13, 2009 #10
    but what happens after time of t>tau
    ?
     
  12. Jun 13, 2009 #11

    Mark44

    Staff: Mentor

    f(t) = 0 for those values, so
    [tex]\int_t^{\infty} f(\tau) d\tau = 0[/tex]
    If a function is identically zero over some interval, its integral is zero over that interval as well. It's really not very complicated.
     
  13. Jun 13, 2009 #12
    ye but lets take a step function or shock delta function
    they are defined at t<0 as 0
    but after they dont have any 0 value
    like in the step function (which is 1 till infinity)
     
  14. Jun 13, 2009 #13
    the solution is simpler than that.
    its just when t>tau the function gets a negative time
    which is by the original definition gives us 0.
     
  15. Jun 13, 2009 #14

    Mark44

    Staff: Mentor

    That doesn't make any sense. Time can't increase through positive values, and then all of a sudden become negative.
     
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