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Convolution & Systems

  1. Oct 13, 2015 #1

    Zondrina

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    Hi everyone, I'm having some trouble understanding how convolutions are applied to systems.

    Suppose I'm given the impulse response of a system as ##g(t) = u(t)## and I'm also given the system input ##x(t) = u(t) - u(t-2)##.

    The system output should then be given by:

    $$y(t) = g(t) * x(t) = \int_0^t g(t - x) x(x) \space dx = \int_0^t u(t-x)u(x) \space dx - \int_0^t u(t-x)u(x-2) \space dx$$

    The integrals should reduce to this I think:

    $$\int_0^t u(t-x) \space dx - \int_2^t u(t-x) \space dx$$

    How do I finish this integration? I thought about setting ##v = t - x##, but I am unsure.
     
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  3. Oct 13, 2015 #2

    Svein

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    Your convolution formula is not according to the standard (https://en.wikipedia.org/wiki/Convolution). It should be [itex]y(t)=g(t)*x(t)=\int_{-\infty}^{\infty}g(z)x(t-z)dz=\int_{-\infty}^{\infty}u(z)(u(t-z)-u(t-z-2))dz [/itex].
     
  4. Oct 13, 2015 #3

    Zondrina

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    Why would this matter? The convolution is a commutative operation: ##f*g = g*f##.

    Also, in the context of systems it is usually taken for granted that ##0 \leq t < \infty## and the convolution limits are taken between ##0## and ##t##.

    I believe I have figured out the next logical step in the integration:

    $$\int_0^t u(t-x) \space dx - \int_2^t u(t-x) \space dx = \int_0^2 u(t-x) \space dx$$

    Okay wait then the area under this function for ##t \geq 2## would simply be equal to ##2##. It would be equal to ##t## for ##0 \leq t \leq 2##.

    Therefore:

    $$\int_0^2 u(t-x) \space dx = t[u(t) - u(t - 2)] + 2u(t - 2) = tu(t) + [2 - t]u(t - 2)$$

    I guess the answer was to think about the area under the function.
     
    Last edited: Oct 13, 2015
  5. Oct 14, 2015 #4

    Svein

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    I do not question that. What I did was to use the definition of convolution on your problem.
    Maybe - but you need to show that this is applicable to the problem.
     
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