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 28, 2016 #1

    RJLiberator

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    Screen Shot 2016-04-28 at 6.57.38 PM.png

    2. Relevant equations
    f*f = integral from -inf to inf of f(t)f(x-t) dt = integral from -inf to inf of f(x-t)f(t)dt

    3. The attempt at a solution

    My question concerns some issues with part a.

    I break the problem up into two cases.

    Case 1: When rectangle is sliding into the rectangle, since f(x-t) is moving.
    I expect it to be from -2 ≤ x ≤ 0

    In setting up the integral, I'm unsure how to analyze it based off my notes.

    What I currently have:
    Integral from -1 to 0 of 1*(1+t) dt where 1 represents f(x-t) and (1+t) represents f(t) part.
    Unfortunately, this results in the wrong solution, as it then equals 1/2

    Clearly, "x" needs to be part of the bounds based on the answer. My problem is, I don't understand how to get the bounds. I have drawn the structure with the rectangle being stationary and then the other rectangle sliding through it. I am thinking that I need to get the bounds from this drawing.
    Any insight here is what I need help with and likely will guide me to understanding this problem.

    Case 2: When rectangle is sliding out of the rectangle.
    I expect it to be from 0 ≤ x ≤ 2


    The answer is
    f*f(x) = x+2 if -2≤x≤0, = 2-x if 0≤x≤2, = 0 otherwise
     
  2. jcsd
  3. Apr 28, 2016 #2

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    In ##f*f = \int_{-\infty}^\infty f(x-x')f(x') \, dx'##, ##f(x')## is centered at the origin and ##f(x'-x)## is obtained after horizontal-flipping ##f(x')## and then translate it by the amount ##x##. For ##x>0##, ##f(x-x')## is on the left of ##f(x')## while for ##x<0## it's on the right of ##f(x')##.

    Now, when ##x<-2##, there is no intersection between the two functions and the convolution integral returns zero. When ##-2<x<0##, there is nonzero intersection which takes the form of a rectangle. Find the left and right bounds of this intersection rectangle and compute the area of this intersection.

    Then move to the case when ##0<x<2## where ##f(x-x')## is on the left of ##f(x')##. Find the left and right bounds of the intersection and same as before compute the area of this intersection.
     
    Last edited: Apr 29, 2016
  4. Apr 28, 2016 #3

    RJLiberator

    User Avatar
    Gold Member

    I'm drawing this part out and I find that from x = -1 to x = 0 we just have a square.

    1 unit on the x direction and 1 unit on the y direction.
    Meaning the area is 1*1 = 1.

    So then that would explain the part of the integral that goes to 1, aka f(x-1) = 1.

    would the integral simply be
    integral from -1 to x+1 of 1*dt
    therefore the answer then becomes (x+2) which solves the problem?

    If so, the bounds then make some sense to me.
     
  5. Apr 28, 2016 #4

    RJLiberator

    User Avatar
    Gold Member

    And then for case 2 we have
    integral from (x-1) to 1 of 1*dt ==> 2-x

    Or perhaps I have them confused case 1 vs case 2.
     
  6. Apr 28, 2016 #5

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    From ##f(t)## you need ##-1 \leq t \leq 1## and from ##f(x-t)## you need ##-1 \leq x-t \leq 1##, or ##x-1 \leq t \leq x+1##. So, when ##-2 < x < 0## your integration goes from ##t = -1## to ##t = x+1##. When ##0 < x < 2## the integration goes from ##t = x-1## to ##t = 1##.
     
  7. Apr 28, 2016 #6

    RJLiberator

    User Avatar
    Gold Member

    Excellent. Then part f*f is solved and 80% understood.

    Next, we have f*f*f

    So, this separates into three cases. Since we use our solution in the above part to make it f*(solution)

    Case 1: from -3≤ x ≤ -1

    We get integration from 0 to x+3 of t dt which results in (1/2)(x+3)^2

    Case 3 we go 1≤x≤3

    We get integration from 0 to 3-x of t*dt which results in (1/2) (3-x)^2

    Do any of these cases (Bounds) make sense. The answers are correct, but I'm struggling to form the bounds.

    I have (what I think) is the right drawing of the situation. (f*f) creates a triangle from -2 to 2 on x. Then we have a box going through it in three cases.
     
  8. Apr 28, 2016 #7

    RJLiberator

    User Avatar
    Gold Member

    Edit: Nevermind, I understand case 1 and 3 now, I had a similar problem in my notes.

    So the integration for case 1 goes from t = -1 to t = x+2 on (1+t)dt
    similarly, integration for case 3 goes from -2+x to 1 on (1-t)dt
     
  9. Apr 28, 2016 #8

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    This gets increasingly messy as you take more convolutions. Instead, I prefer to use transform methods. First, lets translate from ##f(x)= 1\{-1 \leq x \leq 1\}## to ##g(x) = 1\{ 0 \leq x \leq 2 \}##. In other words, ##f(x) = g(x+1)##. So, if we can find ##g_n = g*g* \cdots *g## (the ##n##-fold convolution of ##g##) we can get the ##n##-fold convolution ##f_n## of ##f## as ##f_n(x) = g_n(x+n)##.

    So, the Laplace transform of ##g## is
    [tex] \hat{g}(s) = \int_0^{\infty} g(x) e^{-sx} \, dx = \frac{1-e^{-2s}}{s} [/tex]
    The nice thing about transforms is (among others) the fact that the transform of a convolution is the product of the transforms. Thus, the transform ##\hat{g_n}(s)## of ##g_n## is
    [tex] \hat{g_n}(s) = \left(\hat{g}(s) \right)^n = \frac{(1-e^{-2s})^n}{s^n} [/tex]
    We can expand this as
    [tex] \hat{g_n}(s) = \sum_{k=0}^n (-1)^k {n \choose k} \frac{e^{-2ks}}{s^n} [/tex]
    Now we can find the inverse transform of each term separately (using various properties of Laplace transforms) and so obtain an explicit, closed-form formula for ##g_n(x)## on ##0 \leq x \leq 2n##.

    However, that is just how I would do it; you might not be permitted to use those methods, in which case good luck dealing with increasingly messy integration regions.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



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

  2. Convolution Question (Replies: 0)

Loading...