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Convolution help.

  1. Mar 28, 2006 #1
    I'm having trouble understanding convolution. In particular, in convolving h(t) and x(t), I have no idea what to do when x(t) = u(t-1). So for example, if h(t) = exp(-at)u(t) an x(t) = u(t-1). Is this even the right set up? I don't think it is.

    [tex]\int_o^t e^{-a(t-T)}u(t-1)dT[/tex]

    the other idea I had was

    [tex]\int_1^t e^{-a(t-T)}u(t)dT[/tex]

    but I doubt this too.
    Help?
     
    Last edited: Mar 28, 2006
  2. jcsd
  3. Mar 28, 2006 #2
    lemme just be clear about your notation. Is this right?

    [tex]x(t) = u(t-1)[/tex]

    where [itex]u(t)[/itex] is the unit step function, ie:

    [tex]
    u(x) = \left \lbracket
    \begin{array}{cc}
    0 & \mbox{if } x<0 \\
    1 & \mbox{if } x \geq 0 \\
    \end{array}
    \right.
    [/tex]

    and [tex]h(t) = e^{-at}u(t)[/tex].

    the convolution of x and h, denoted [itex]x \ast h[/itex], is given by,
    [tex] (x \ast h)(t) = \int_{-\infty}^{\infty}x(T)h(t - T)dT[/tex]

    Also, convolution is commutative so [itex] x \ast h = h \ast x[/itex]. Therefore choose the one that makes your calculation easier. From your post it looks like you chose [itex] x \ast h[/itex]

    your first goal is to come up with the correct expressions for

    [tex] h(t - T) = \ ?? [/itex]
    [tex]x(T) = \ ?? [/tex]
     
    Last edited: Mar 28, 2006
  4. Mar 28, 2006 #3
    [tex] h(t - T) = e^{-a(t-T)} [/tex]
    [tex]x(T) = u(T-1) [/tex]

    Maybe?
     
  5. Mar 28, 2006 #4
    In your first post you said [itex] h(t) = e^{-at}u(t)[/itex]. What happened to [itex]u[/itex] in [itex]h(t-T)[/itex]?

    Thats the correct expression for x(T).

    What's [itex]u(t)[/itex]? Is it the unit step function?
     
    Last edited: Mar 28, 2006
  6. Mar 28, 2006 #5
    yeah it is, and that's what's catching me a little i think; is the u(t) in the h(t) related to the u(t) which x(t) is equal to?
     
  7. Mar 28, 2006 #6
    yes. recall from algebra that the graph of [itex]u(t-1)[/itex] is the graph of [itex]u(t)[/itex] shifted to the right 1 unit.

    If [itex]f,g,[/itex] and [itex]h[/itex] are functions such that [itex]f(t) = g(t)h(t)[/itex], and [itex]a[/itex] is a real number, then [itex]f(t-a) = g(t-a)h(t-a)[/itex]. Use this to come up with an expression for [itex]h(t-T)[/itex].
     
    Last edited: Mar 28, 2006
  8. Mar 28, 2006 #7
    I feel like a n00bie.

    So
    [tex] h(t - T) = e^{-a(t-1)} [/tex]?
     
  9. Mar 28, 2006 #8
    [tex]h(t) = e^{-at}u(t)[/tex]

    to get [itex]h(t-T)[/itex] replace all instances of [itex]t[/itex] with [itex]t-T[/itex]
     
  10. Mar 28, 2006 #9
    so h(t-T):

    [tex]h(t-T) = e^{-a(t-T)}u(t-T)}[/tex]

    forgive me its late and I'm having a really hard time understanding this.
     
  11. Mar 28, 2006 #10
    yes :smile:

    now you can substitute the expresions for [itex]x(T)[/itex] and [itex]h(t-T)[/itex] in the convolution integral.
     
    Last edited: Mar 28, 2006
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