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Convolution of a linear and rectangular function

  1. Dec 12, 2013 #1
    Hi everyone

    1. The problem statement, all variables and given/known data
    I want to to calculate the convolution of the following two functions
    [tex] h(t)=\left\{\begin{array}{ll} t, & 0 \leq t \leq 10 \\
    0, & otherwise\end{array}\right. [/tex] and the function
    [tex] x(t)=\left\{\begin{array}{ll} A, & 0 \leq t \leq 10 \\
    0, & otherwise\end{array}\right. [/tex]


    2. Relevant equations
    Convolution theorem



    3. The attempt at a solution
    I tried to solve it, but I have problems with the constant A.

    I wrote
    [tex] h(t) * x(t) = \int_{0}^{10} h(\tau) *x(t-\tau) d\tau =\int_{0}^{10} \tau \cdot A d\tau[/tex]

    I'm not sure about the last step, can anyone help me ?

    Thanks for your help in advance

    edit: actually I think the limits are wrong aswell? Because If I'd do it this way I get a number as a result and not a function. But I don't know what the limits should be then?!
     
  2. jcsd
  3. Dec 12, 2013 #2

    jbunniii

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    Please note that the above is correct only for ##t=10##. To deal with ##A##, recognize that it simply stands for some number which does not vary with ##t## or ##\tau##, so you can slide it out of the integral just as if it were a number:
    $$\int_{0}^{10} \tau \cdot A d\tau = A \int_{0}^{10} \tau d\tau$$
    Yes, you're right. The integral you wrote gives you ##h*x(10)##. In general, you should expect one or both of the integration limits to depend on ##t##. You won't be able to write down a single integral which is correct for all values of ##t##. Instead you will have to consider several cases.

    I highly recommend sketching plots of ##h(\tau)## and ##x(t - \tau)## as functions of ##\tau##. This will be very helpful to determine the endpoints of the integral. You should find that it makes sense to consider the following four cases:

    (1) ##t < 0##
    (2) ##0 \leq t \leq 10##
    (3) ##10 \leq t \leq 20##
    (4) ##t > 20##
     
  4. Dec 12, 2013 #3
    Thank you for your quick reply. But I don't understand why you differ the fours different cases. I'd take t<0 0<t<10 and t>10 but why do you mention the one between 10 and 20 ?
     
  5. Dec 12, 2013 #4

    jbunniii

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    I'll assume you have sketched the graphs of ##h(\tau)## and ##x(t - \tau)## as functions of ##\tau##. I'm too lazy to scan and attach an image, but your graphs should have the following characteristics:

    • ##h(\tau)## is nonzero between ##\tau = 0## and ##\tau = 10##
    • ##x(t - \tau)## is nonzero between ##\tau = t-10## and ##\tau = t##
    Now, case (1) is when the graph of ##x(t-\tau)## is entirely to the left of ##h(\tau)##, and the nonzero portions do not overlap. This happens when the right edge of ##x(t-\tau)## comes before the left edge of ##h(\tau)##, i.e., when ##t < 0##.

    Case (2) is when the left edge of ##h(\tau)## is between the left and right edges of ##x(t - \tau)##. This corresponds to the case ##0 \leq \tau \leq 10##.

    Case (3) is when the left edge of ##x(t - \tau)## is between the left and right edges of ##h(\tau)##. This corresponds to the case ##10 \leq \tau \leq 20##.

    Case (4) is when the left edge of ##x(t - \tau)## comes after the right edge of ##h(\tau)##. This corresponds to the case ##\tau > 20##.
     
  6. Dec 12, 2013 #5
    Thank you very much again. I will try it tomorrow and if I have further problems I post again in this thread. Much appreciated your help.:smile:
     
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