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Homework Help: Integration Problem

  1. Jan 29, 2005 #1
    I have to find the convolution of two identical square functions.
    [tex]
    F(t)=\left\{\begin{array}{cc}1,&\mbox{ if } 0\leq t < 1\\0, & \mbox otherwise\end{array}\right.
    [/tex]

    I need to calculate F * F
    [tex](F * F)(t) = \int_{\infty}^{-\infty} F(u)F(t-u) du[/tex]

    The first thing I did was change integration limit to [0,1] and F(u) to 1. My main problem is F(t-u).

    F(t-u) = 1 when
    0 < t-u < 1
    -1 < u-t < 0
    t-1 < u < t

    I think thus far I haven't made any mistakes. Now what do I do to substitue F(t-u)? I think my options are to either redefine the integration limit (but it should be [0,1]) or somehow write F(t-u) as a algebraic function. I don't see how I can do the latter. If someone can point me in the right direction I would appreciate it.

    oh and the resulting convolution is a hat function.
     
    Last edited: Jan 29, 2005
  2. jcsd
  3. Jan 29, 2005 #2

    AKG

    User Avatar
    Science Advisor
    Homework Helper

    Would this not just be

    (F * F)(t) = 1 - |t - 1| for 0 < t < 2, 0 otherwise?

    The integrand will have a value of 1 where both F(u) and F(t - u) take on the value of 1, which is when their arguments both take on values in [0, 1); the integrand is zero elsewhere. If you can characterize the interval on which both F(u) and F(t - u) are 1, then the length of that interval will be the value of the integral. That interval is then where both t-1 < u < t as you found and where 0 < u < 1. It is the intersection of (t-1, t] and [0, 1). You can see that there will be a non-empty intersection only if t > 0 and t-1 < 1, i.e. t < 2, so the integrand will only take on non-zero values ever if 0 < t < 2. Otherwise, it will be 0 everywhere, and the integral will be zero everywhere. In fact, when t = 0, we get:

    [tex]\int _{-\infty} ^{\infty}F(u)F(t - u)du = \int _{-\infty} ^{\infty}F(u)F(0 - u)du = \int _0 ^01du = 0[/tex]

    So we can even say that F*F(t) = 0 unless 0 < t < 2, as I did near the top of my post. To figure out what F*F will do on that interval, you need to figure out the length of the intersection of [0,1) and (t-1,t] for the values of t in (0,2). This should be pretty simple, if not, pick a few values of t and do it yourself to find out how to define F*F for these values of t. It will look like a triangle with values of 0 at 0 and 2, coming up to a point in the middle with a value of 1 at t=1 (I suppose this shape is the "hat" you're talking about).
     
  4. Jan 30, 2005 #3
    thanks for help

    I didn't think about defining the intersection of the intervals that way but it really makes sense. However to define what F*F looks like in general for arbitrary functions of F (not given in algebraic form), I just need to calculate a few values (or graph) and find an equation that fits?

    edit:
    I decided to define it piecewise since I need to calculate the convolution of F with (F*F). It sure gets messy but I think I got it.

    F*F is nonzero on interval 0<t<2
    so

    0<t<1
    [tex](F * F)(t) = \int_{t}^{0} du = t[/tex]

    1<t<2
    [tex](F * F)(t) = \int_{1}^{t-1} du = 2-t[/tex]


    If anyone is wondering what all this is for, this is an assignment from my intro to wavelets and splines class. The original F (box function) is a local averaging linear functional (operates on functions). F*F produces a hat function and F*(F*F) produces a bell curve like function (called quadratic B-spline). So more applications of this averaging function F will produce even smoother functions I believe. I would say this is due to the fact that the "time" scale of F is rather large compare to the functions it operates on. It's pretty neat I think. Once again thanks AKG for your help.
     
    Last edited: Jan 30, 2005
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