Bilateral Laplace / Z transform of a constant

Discussion in 'General Electronics Chat' started by narasimhan, Aug 4, 2013.

  1. narasimhan

    Thread Starter Member

    Dec 3, 2009
    72
    6
    Hi. It has been a long time since I posted here. Was kind of out of touch with electronics etc. So have a basic(maybe dumb) question. I have been pondering if there is a bilateral Laplace/Z transform of a constant. Does it exist? Or it exists but without a ROC? Can a transform exist/valid without ROC?

    By constant I mean the signal(either continuous(for laplace)/discrete(for Z)) is constant for all t/n (-infinity to +infinity). That is why I mentioned bilateral transform.

    The reason is Laplace/Z transform can be considered(I assume) as a super-set for Fourier transform(Continuous time/Discrete time) since many signals/systems(unstable) that does not have Fourier Transform have Laplace/Z transform.
    But in this case Fourier transform exists for a constant but Laplace/Z is debatable(at least I am at loss).
    I faintly remember some explanation for this 4 yrs back but now I forgot the exact answer. A friend of mine asked me this question recently and I did look up Oppenheim(Signals and Systems) but could not locate it. Thanks.
     
  2. narasimhan

    Thread Starter Member

    Dec 3, 2009
    72
    6
    I have found out that Laplace/Z transform of a constant does not exist even though the Fourier transform exists.
    Laplace/Z transform cannot exist without a region of convergence(ROC).
    Though Laplace/Z transform in general can be considered as a super-set of Fourier transform(since in most cases Laplace/Z transform exists even though Fourier transform does not exist) this is one of the exceptions.
     
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