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Homework Help: Convolution Question

  1. Apr 25, 2012 #1
    1. The problem statement, all variables and given/known data
    This is a question from a sample exam, rather than a homework problem.

    [itex]\text{Let }f \text{ be the periodic function with period } 1 \text{ defined for }-1/2\leq t<1/2 \text{ by }f(t)=t^2\\ \text{ and let }g \text{ be the periodic function with period } 1 \text{ defined for }-1/2\leq t<1/2 \text{ by } g(t)=1.
    \\ \text{(a) What is the period of }f\ast g?
    \\ \text{(b) Find the expression of }f \text{ for }1/2\leq t<3/2, \text{ and }-2\leq t\leq -1.
    \\ \text{(c) Show that }f\ast g\text{ is given for }0\leq x\leq 1 \text{ by }
    \\ (f\ast g)(x)= \frac{5}{4}-\frac{1}{3}\left(x-\frac{1}{2}\right)^3+\frac{1}{3}\left(x+\frac{1}{2}\right)^3-\left(x+\frac{1}{2}\right)^2+x.

    2. Relevant equations
    The convolution of two functions [itex] f_1 [/itex] and [itex] f_2 [/itex] both with period [itex] L [/itex] is given by
    [itex]\displaystyle (f_1\ast f_2)(x)=\frac{1}{L}\int_{-L/2}^{L/2}f_1(x-t)f_2(t)dt= \frac{1}{L}\int_{-L/2}^{L/2}f_1(t)f_2(x-t)dt[/itex]

    3. The attempt at a solution
    (a) Is it so that the convolution will have the same period as the two functions which you convolve?
    (b) Clearly the answers are [itex] (t-1)^2\text{ and }(t+3/2)^2[/itex].

    (c)There seems to be something wrong with the question. The first thing that looks wrong is that the last expression is,
    when you expand it, nothing more than the constant [itex]\displaystyle\frac{13}{12}[/itex].

    I calculated the convolution as follows
    [itex]\displaystyle \int_{-1/2}^{1/2}f(t)g(x-t)dt=\int_{-1/2}^{1/2}f(t)dt=\int_{-1/2}^{1/2}t^2dt=\frac{1}{12} [/itex]

    It seems that the lecturer wants us to compute it as follows
    [itex]\displaystyle \int_{-1/2}^{1/2}f(x-t)g(t)dt[/itex]
    I think that he wants us to use answers from part (b) to figure out what values [itex]f [/itex] should take. I don't see why we should do it this way instead of my way or why the answers would be different between his way and my way. I also can't see exactly what to do for his way. I don't see how I would always know where [itex]x-t[/itex] was and hence I don't see how I would always know the value of [itex]f[/itex] at [itex]x-t[/itex]. I know that's what part (d) is for but I can't see what to do.
  2. jcsd
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