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Unilateral and Bilateral Laplace Transform in Solving Differential Equations

  1. Nov 21, 2012 #1
    Why is it that the unilateral lateral Laplace transform is used when given initial conditions that are non-zero. Is there a reason that explains why it would be wrong to use the bilateral Laplace transform instead?

    I know bilateral does not have any input of initial conditions but that does not explain why as it should still give a valid result. Is there some conditions that conflict with the bilateral transform being used when there is no initial rest?

  2. jcsd
  3. Nov 21, 2012 #2


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    There are functions which have unilateral Laplace transforms but not bilateral Laplace transforms. In the bilateral transform, the transformed function ##f(t)## basically has to tend to zero (no slower than exponentially) as ##t \rightarrow \pm \infty##. Otherwise, the exponential term ##\exp(-st)## will blow up in one of those limits and the transform doesn't exist. However, the unilateral Laplace transformed function only has to grow slower than an exponential function in order for the transform to exist.

    Also, for physical applications, the bilateral laplace transform apparently does not respect causality, so if you need a causal function, as is often the case in time series applications, a bilateral laplace transform is not suitable.
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