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Convolution of two functions

  1. Oct 6, 2016 #1
    1. The problem statement, all variables and given/known data
    I have the two functions below and have to find the convolution [itex]\beta * L[/itex]

    2. Relevant equations
    Assume a<1
    [tex]
    \beta(x)=\begin{cases}
    \frac{\pi}{4a}\cos\left(\frac{\pi x}{2a}\right) & \left|x\right|<a\\
    0 & \left|x\right|\geq a
    \end{cases}
    [/tex]
    [tex]
    L(x)=\begin{cases}
    1 & \left|x\right|<1\\
    0 & \left|x\right|\geq 1
    \end{cases}
    [/tex]

    3. The attempt at a solution
    [tex]\beta * L=\int^\infty_{-\infty}\beta(y)L(x-y)dy=\int^\infty_{-\infty}\beta(x-y)L(y)dy[/tex]
    What I am unsure about is the limits and which integral to use. Any hints would be greatly appreciated.
     
  2. jcsd
  3. Oct 6, 2016 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Start by drawing the two-dimensional ##A## region for the second form:
    $$A_a = \{ (x,y) : -1 \leq y \leq 1, |x-y| \leq a \}.$$
    There are three cases: (1) ##a < 1##; (2) ##a = 1##; and (3) ##a > 1##. The region will look slightly different in cases (1) and (3).


    For some values of ##x## the line vertical line through ##x## will miss region ##A_a## completely, so the ##y##integral will equal 0. For other values of ##x## the vertical line through ##x## will intersect region ##A_a## in a vertical line segment, so will have the form ##y_1(x) \leq y \leq y_2(x)##, and your ##y##-integration will be from ##y_1(x)## to ##y_2(x)##.
     
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