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Convolutions (differential Equations)

  1. May 22, 2016 #1
    1. The problem statement, all variables and given/known data
    Compute the convolution f*g for the given function f and g.
    f(t)=cost g(t)=U2(t)


    2. Relevant equations
    f*g=∫f(t-u)g(u)du

    3. The attempt at a solution
    So I pretty much only know how to plug in the functions into the integral for convolutions. Not really sure how to evaluate it with the U2(t)

    f*g=∫cos(t-u)U2(u)du

    so do I evaluate thi integral directly or are there other methods to evaluate this.?
     
  2. jcsd
  3. May 22, 2016 #2

    Mark44

    Staff: Mentor

    What does U2(t) mean in this context? I think I know, but I want to be sure that you do, as well.
     
  4. May 22, 2016 #3
    well it is a heaviside function. The U2(t) has a discontinuity at 2. 2 is when the function gets "turned on" so to speak.
     
  5. May 22, 2016 #4
    sorry for the late reply
     
  6. May 22, 2016 #5

    Mark44

    Staff: Mentor

    What effect does this have on your integral as well as on the limits of integration?
     
  7. May 22, 2016 #6
    Well it causes for the integral to be evaluated from 2 to t. Because the function would be 0 when t is less than 2
     
  8. May 22, 2016 #7

    Mark44

    Staff: Mentor

    OK, so what does your convolution integral look like then?
     
  9. May 22, 2016 #8
    t2cos(t-u)U2(u)du
     
  10. May 23, 2016 #9

    Mark44

    Staff: Mentor

    But what is U2(u) on that interval? What's the value of any Heaviside function?
     
    Last edited: May 23, 2016
  11. May 23, 2016 #10
  12. May 23, 2016 #11

    Mark44

    Staff: Mentor

    You're not sure?
     
  13. May 23, 2016 #12
    no I'm sure
     
  14. May 23, 2016 #13

    Mark44

    Staff: Mentor

    So, if U2(t) = 1 for t ≥ 2, what does the integral now look like? This is what I was asking you in post #7.

    U2 should NOT appear in the integral.
     
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