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Analytic solution of a Convolution Integral

  1. Feb 13, 2015 #1
    1. The problem statement, all variables and given/known data
    The question is in the attached image . My problem starts when dealing with the limits of integration . I need an analytic procedure of solving such problems without involving graphical method . The equations of the graphs of h(t) and x(t) are easily derived . image.jpg

    2. Relevant equations
    See attached image:
    The convolution integral of 2 functions is defined as :
    [itеx] y(t) = \int_{-infinity}^{+infinity}h(\tau}*x(t-\tau)d\tau [\itеx]

    3. The attempt at a solution
    See attached image

    Attached Files:

    Last edited: Feb 13, 2015
  2. jcsd
  3. Feb 13, 2015 #2


    Staff: Mentor

    The integral in your second attachment (both attachments have image.jpg as their filename) is not difficult to write using LaTeX.
    $$y(t) = \int_{t - 4}^{t - 2} h(\tau) d\tau $$
    The LaTeX script that produces this integral is y(t) = \int_{t - 4}^{t - 2}h(\tau}d\tau. Use two $ symbols at each end. More info here: https://www.physicsforums.com/help/latexhelp/
  4. Feb 13, 2015 #3


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    ##h(\tau) ## is a linear function, and should be easy to integrate if you break it into two pieces.
  5. Feb 13, 2015 #4
    The problem , as i mentioned , is in the intervals of integration . We should consider to which interval t belongs . The overlapping confuses me
  6. Feb 13, 2015 #5


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    h(t) is non-zero for 1<t<3, so your integral will be zero for t<3, it will grow on 3<t<5, then shrink down again on 5<t<7 and be zero from that point on.
    So from the outset, restrict t to between 3 and 7 and see where that gets you.
  7. Feb 13, 2015 #6
    I seriously didn't understand . How did you know that the integral will be zero for t<3 and grow in....?
    How am i supposed to know the different intervals of t for integration ?
  8. Feb 13, 2015 #7


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    ##f(t) = \int_{t-4}^{t-2} h(\tau) d\tau = \int_{2}^{4} h(t-\tau) d\tau##
    If the intersection: ##[t-4, t-2] \cap [1,3] ## is empty, then the function is zero, since the integrand is zero over the entire integral.
    This intersection is maximized when the two ranges match exactly, i.e. t=5.
    Other than that, you are evaluating a portion of the area under the tent of h.
  9. Feb 13, 2015 #8


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    Are you able to write the integral of h as a function of x?
    something like ## f(x) = \left\{ \begin{array} {l l } \displaystyle \int_1^x g(t) dt & 1<x\leq 2 \\ \displaystyle \int_1^2 g(t) dt + \int_2^x k(t) dt & 2< x\leq 3 \end{array} \right. ##
  10. Feb 13, 2015 #9

    Ray Vickson

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    Always draw a picture! Look at the feasible region in ##(\tau,t)##-space.
  11. Feb 13, 2015 #10


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    I have to admit I'm always puzzled by students when they insist on a non-graphical method of analyzing a problem. Analyzing the problem graphically is probably one of the most powerful tools you have and much less error-prone than trying to apply a non-intuitive, non-graphical method.
  12. Feb 13, 2015 #11
    Unfortunately , i'm not . It seems I need to do further readings :(
  13. Feb 13, 2015 #12


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    In your attached photo, you already have the piecewise definition of h as:
    ## \left\{ \begin{array} {l l} t-1 & 1<t\leq 2 \\ 3-t & 2<t\leq 3 \\ 0 & \text{Otherwise} \end{array} \right. ##
    What happens if you take the integral of this for t from 0 to 2? From 1 to 3? From -2 to 0? From 2 to 4?
    If you can answer these questions, you should know enough to complete this problem.
  14. Feb 14, 2015 #13
    But the integral is from t-4 to t-2 . The intervals of t that we should consider are not the same as h(tau) 's intervals of definition . For now , i obtained 7 cases of t
  15. Feb 14, 2015 #14


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    What if you made a shift and said x= t-4 and x+2 = t-2?
    Either way, you will get an integral that depends on some variable which can be evaluated as a function that has maximum of 1 and is zero for all but a small interval of length 4.
    7 cases? Could you show what you have so far?
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