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Convolution Calculation (piecewise function)

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  1. Mar 14, 2016 #1
    1. The problem statement, all variables and given/known data
    Compute the convolution ##(f*h)(t)## where

    $$f(t) = \left\{\begin{matrix}1, \ \ for \ \ |t|<1 \\ 0, \ \ \ \ otherwise \end{matrix}\right.$$

    and

    $$h(t) = \left\{\begin{matrix}2|t|-1, \ \ for \ \ |t|<1/2 \\ 0, \ \ \ \ otherwise \end{matrix}\right.$$

    2. Relevant equations

    Convolution integral: ##(f*h)(t)= \int^\infty_{-\infty} f(\tau) h(t-\tau) d \tau##

    3. The attempt at a solution

    Here I first made a sketch of the two functions:

    plot.jpg

    f(t) is non-zero on [-1, 1] and g(t) is non-zero on [-0.5, 0.5], so adding these two we find that (f ∗ g)(t) should be nonzero on [-1.5, 1.5]. The list of change points of (f ∗ g)(t) is {-1.5, -0.5, 0.5, 1.5}. For values t that are outside this range the convolution is zero.

    So, do I need to calculate the convolution for 3 separate cases?

    I mean ##-1.5 \leq t \leq -0.5##, ##-0.5 \leq t \leq 0.5##, and ##0.5 \leq t \leq 1.5##?

    Here is my attempt:

    First case: we want to integrate from -1.5 to -0.5 using the formula given above (##(f*g)(t)=\int^{0.5}_{-1.5} f(\tau) h(t-\tau) d \tau##). ##h(t)## is zero. But for ##f(t)## the function is zero on the interval ##(-1.5, -1)##, whereas it is equal to 1 from ##(-1, -0.5)##, so what value do I need to be using for the function? :confused:

    Any help is greatly appreciated.
     
  2. jcsd
  3. Mar 14, 2016 #2

    Samy_A

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    If ##f## is zero on an interval, the integral simply vanishes there. Then you get:
    ##(f*h)(t)=\displaystyle \int^{+ \infty}_{-\infty} f(\tau) h(t-\tau) d \tau = \int^{1}_{-1} h(t-\tau) d \tau##.

    (EDIT: I had written ##(f*g)(t)## instead of ##(f*h)(t)##, typo now corrected.)
     
    Last edited: Mar 15, 2016
  4. Mar 15, 2016 #3

    Samy_A

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    To add to my previous post:

    I would consider a substitution ##\sigma =t- \tau##.
    Also notice that both ##f## and ##h## are even functions, so their convolution will also be an even function.
    Finally, when you distinguish between the different relevant cases for ##t## (ie ##t\leq -1.5##, ##-1.5 \leq t \leq -1##, ##-0.5 \leq t \leq 0##), the results you obtain for each of the segments must be equal on their mutual endpoints.

    EDIT: the relevant cases were meant to be ##t\leq -1.5##, ##-1.5 \leq t \leq -1##, ##-1 \leq t \leq -0.5##, ##-0.5 \leq t \leq 0##
     
    Last edited: Mar 15, 2016
  5. Mar 15, 2016 #4
    Thank you Samy for your response.

    For the case ##-1.5 \leq t \leq -0.5##, could you please explain how you got the integral limits to be ##(-1,1)## instead of ##(-1.5, -0.5)##?

    I think alternatively it is also possible to rewrite the integral limits like:

    $$t+0.5 \leq \tau \leq t+1.5$$

    In the interval ##[-1.5, -0.5]##, ##x+0.5## is between -1 and 0. Whereas ##x+1.5## is between 0 and 1. So the integral can be rewritten:

    $$(f*h)(t)=\int^{t+1.5}_{t+0.5} h(t- \tau) \ d \tau$$

    I am not sure how to proceed from here. How exactly does a substitution help? (this is my first convolution problem of this type)

    But shouldn't the three cases be the following?

    (i) ##-1.5 \leq t \leq -0.5##

    (ii) ##-0.5 \leq t \leq 0.5##

    (iii) ##0.5 \leq t \leq 1.5##

    For values of t less than -1.5 or greater than 1.5, (f∗g)(t) is zero.
     
  6. Mar 15, 2016 #5

    Samy_A

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    I wasn't considering cases yet.

    I simply started from the definition: ##\displaystyle (f*h)(t)=\displaystyle \int^{+ \infty}_{-\infty} f(\tau) h(t-\tau) d \tau##.
    Since ##f## is 0 outside of ##[-1,1]##, the integration limits can be replaced by ##-1## and ##+1##.
    Since ##f## is 1 in ##[-1,1]##, the ##f(\tau)## just becomes ##1##, giving the result: ##\displaystyle (f*h)(t)=\displaystyle \int^{+1}_{-1} h(t-\tau) d \tau##.
    I made a mistake in listing the cases, sorry for that.

    It really depends on how you want to tackle the integral. (There are probably a number of ways to do it, I'm not sure the one I suggest is the shortest.)

    I suggested the substitution ##\sigma =t-\tau##.
    The integral then becomes ##\displaystyle (f*h)(t)=\displaystyle -\int^{t+1}_{t-1} h(\sigma) d \sigma \ \ \ \ (1)##.
    (There is no deep reason for this substitution. I think that dealing with ##h(\sigma)## is less error prone than dealing with ##h(t-\tau)## when treating the different cases.)
    Notice that ##h## vanishes outside of ##[-0.5,+0.5]##, and that the expression for ##h## is different for positive and negative ##\sigma##. This will be important when treating the various cases.

    Now for the cases.
    a) For ##t\leq -1.5##, the convolution is 0, as you already wrote.

    b) For ##-1.5 \leq t \leq -1##, the integration limits will be ##-0.5 \to t+1 (\leq 0)##, so you can use the expression for ##h## as valid for negative ##\sigma##.

    c) For ##-1 \leq t -0.5##, the integration limits will be ##-0.5 \to t+1##, where ##0 \leq t+1 \leq 0.5##. In this case the integral (1) will consist of two parts: the integral from ##-0.5## to ##0## and the integral from ##0## to ##t+1## (with different expressions for ##h(\sigma)##).

    d) For ##-0.5 \leq t \leq 0##, the integration limits will be ##-0.5 \to 0.5##, and the integral will also consist of two parts (##-0.5 \to 0, 0 \to 0.5##).
    Since ##h## is even and the integration domain is symmetric around the y-axis, you can compute one of the integrals and multiply the result by 2.

    e) Finally, for ##t \geq 0##, make use of the fact that ##f*h## is an even function, and use (with the needed precautions) the results obtained for negative ##t##.
     
    Last edited: Mar 15, 2016
  7. Mar 15, 2016 #6
    Thank you for the explanation. Yes ##h## is ##-2t-1## for negative σ. So, for instance for the case ##-1.5 \leq t \leq -1## this is what I found:

    $$(f*h)(t) = -\int^{t+1}_{-0.5} h(\sigma) d \sigma = -\int^{t+1}_{-0.5} (-2t-1) \ dt = - \left( (t+1)^2 -(t+1)-0.25 \right)$$

    Is this correct?

    This function looks like an inverted parabola. This is the region of interest:

    graph.jpg

    I just want to make sure it is done correctly so I can calculate the other integrals, and sketch ##(f*h)(t)## to make sure it is an even symmetric function.
     
  8. Mar 15, 2016 #7

    Samy_A

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    I find something different.
    If I'm not mistaken, your function is equal to -0.5 in -1.5. That should be 0.

    Your integral is the same as mine for that case, but the result is different.
    When I computed the convolution this morning, I had a number of sign errors. It looks like that's what you have here too.
     
    Last edited: Mar 15, 2016
  9. Mar 15, 2016 #8
    Oops, I did the integral again and there was a sign error as you suggested:

    $$- \int^{t+1}_{-0.5} (-2t-1).dt = -\Bigg[ \frac{-2 t^2}{2} -t \Bigg]^{t+1}_{-0.5}$$

    $$= - \left( \Big[ -(t+1)^2 - (t+1) \Big] - \Big[ -(-0.5)^2 - (-0.5) \Big] \right) = (t+1)^2 + (t+1) + 0.25$$

    This time it's a right way up parabola:

    graph.jpg
    Is that what you have also?
     
  10. Mar 15, 2016 #9

    Samy_A

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    Yes, I have the same result. ##(f*h)(t)=t²+3t+\frac{9}{4}## for ##t \in [-1.5, -1]##.
    Notice that now the value in -1.5 is 0. The value of the first derivative in -1.5 is also 0. That is as expected, as ##f*h## is 0 below -1.5, and is a differentiable function.
     
  11. Mar 16, 2016 #10
    Thank you so much for the explanation.

    For case (c) I found:

    $$(f*h)(t)= - \Big[ \int^0_{-0.5} f(t) h(\sigma) d \sigma + \int^{t+1}_{0} f(t) h(\sigma) d \sigma \Big] =- \Big[ \int^0_{-0.5} (-2t-1) d t + \int^{t+1}_{0} (2t-1) d t \Big]$$

    $$= -(t+1)^2 + (t+1) + 0.25 = -t^2-t+\frac{1}{4}$$

    This function is the inverted version of y=x2+3x+(9/4) we found for part (b).

    For case (d) I found:

    $$(f*h)(t)= - 2 \int^{0.5}_0 (2t-1)(f(t)=1) dt = 0.5$$

    The graph in Matlab looks like this so far for ##t \leq 0## (parts (b), (c), and (d) are shown in magenta, blue and green respectively):

    plot1.jpg

    Since the ##(f ∗ h)(t)## is continuous and even I think the other half (in blue) should look like this:

    graph1.jpg

    Is this correct?
     
  12. Mar 16, 2016 #11

    Samy_A

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    Yes, I found the same results for cases c) and d).

    As ##f*h## is even, you indeed can just take the mirror image of the graph for positive ##t##.
    To be clear, from ##-1.5 \to -0.5## the graph segments are parabolic, not straight lines (and likewise on the positive side).
     
  13. Mar 17, 2016 #12
    I get it now. Thank you so much for your time and help.
     
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