Convolution theory Or Laplace Transform

Discussion in 'Homework Help' started by hitmen, Apr 21, 2009.

  1. hitmen

    Thread Starter Active Member

    Sep 21, 2008
    159
    0
    This question came out for my exam.
    Impulse response h(t) = u(t) - u(t-1)
    Find output when input is x(t) = (t^2 + t)u(t)

    Do I use the equation ∫x(τ)h(t-τ)dτ from 0 to infinity or do I use laplace transform?

    Since u(t) - u(t-1) is one from 0 to 1, I reduced the integral to the limits 0 to 1 and continued on from there. Am I right or wrong?

    However, when I use laplace transform I get a totally diff answer. Which is right?
     
  2. SawabyPlus

    Member

    Feb 27, 2009
    14
    0
    The convolution is not properly carried out in the time domain. more specifically, limits of the integral are wrong.
    convolution is made by shifting the flipped version of one signal by t, then integrate the product of this signal with the other signal for all values of t.
    as u've chosen to shift h, the output is divided into two intervals:
    Partial overlapping: 0<=t<1, limits are (0) to (t).
    complete overlapping : 1<= t, limits are (t-1) to (t)
    otherwise, output is zero.
    u should then get similar values.
     
  3. guitarguy12387

    Active Member

    Apr 10, 2008
    359
    12
    Yup. Just to add... convolution is often much more easily done in the frequency domain... if you can easily find the transforms (i.e. step function has a well known transform) all you have to do is transform x(t) and h(t) to get X(s) and H(S) (or X(jw) and H(jw) depending on which transform you use) and then multiply H(S) and X(S). Then take the inverse transform. And there's your answer. Its a property of convolution.
     
  4. hitmen

    Thread Starter Active Member

    Sep 21, 2008
    159
    0
    let start off with the basic:

    how do you solve ∫(τ^2 + τ)u(τ)u(t-τ)dτ from -infinity to infinity?

    I am stuck because there are 2 unit function
     
  5. guitarguy12387

    Active Member

    Apr 10, 2008
    359
    12
    The easiest way to actually evaluate this integral (i.e. setting up the bounds) is by a graphical approach. Are you familiar with this at all?
     
  6. guitarguy12387

    Active Member

    Apr 10, 2008
    359
    12
    Also, it may be helpful to think of h(t) as a shifted rect() function.
     
  7. hitmen

    Thread Starter Active Member

    Sep 21, 2008
    159
    0
    I am familiar with it. That is to change the bounds from the limits 0 to infinity.
     
  8. etuzuner

    Member

    Mar 21, 2009
    23
    0
    Actually, graphical way of thinking is more easier than that.

    As you know the unit step function makes lowest limit of integral 0. If you think the other positions, like SawabyPlus said, you can see that it is very easy to calculate.
    If you're allowed to do it, then I suggest you to try that.

    Best regards
     
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