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CT Convolutions and their bounds

  1. Sep 22, 2006 #1
    This question deals with an example found in Signals and Systems, 2ed, page 99, ex 2.7. But I can summarize it here:

    EDIT - my latex doesn't seem to work ...

    [tex]

    x(t) = (1, 0 < t < T) or (0, otherwise)

    [/tex]

    [tex]

    h(t) = (t, 0 < t < 2T) or (0, otherwise)
    [/tex]

    I realize there are five bounds to consider: (t<0, 0 <t <T, T < t < 2T, 2T < t <3T, 3T < t)

    However, for instance,

    [tex]

    \\int_T^0 x(\tau) h(t-\tau) d\tau = \\int_T^0 h(t-\tau) d\tau = t\tau - \tau^2/2 (from 0 to T)
    [/tex]

    does not equal the example's answer of
    [tex]
    t^2/2
    [/tex]

    nor does

    (integrate from 2T to 3T)
    [tex]

    \\int_2*T^3*T x(\tau) h(t-\tau) d\tau = \\int_T^0 h(t-\tau) d\tau = t\tau - \tau^2/2 (from 2T to 3T)
    [/tex]

    does not equal the example's answer of
    [tex]
    -t^2/2 + Tt + 3/2*T^2
    [/tex]

    So my question is this:

    Once I realize the bounds that the convolution must be evaluated for, how do i convert the bounds so that I can integrate them and derive the correct answer?

    edit- x(tau) is a square pulse of height 1, from 0 to T (not tau but T)

    and

    h(t-tau) is a downward sloping right trianlge (90degree corner on the left) with a base from t-2T to t.

    Thank you for your time.
     
    Last edited: Sep 22, 2006
  2. jcsd
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