1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
    [/itex]

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted



Similar Discussions: Convolution Question
  1. Convolution question (Replies: 9)

  2. Convolution Question (Replies: 7)

Loading...