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I Frequency of a constant function

  1. Apr 11, 2016 #1
    Just a quick question that I feel should be simple, but I'm unable to come up with a satisfactory answer.

    A constant signal has an arbitrarily small period (rather, it has no fundamental period), and so it seems to me that this means the frequency of a constant signal grows without bound. However, in Fourier analysis, for instance, we treat constant signals as having a frequency component only at ##f = 0##. Why, mathematically, can we not say that a constant signal has (approaching) infinite frequency since ##f = 1/T## and a constant function has arbitrarily small ##T##? I mean, certainly from an experimental basis (i.e. Designing a high pass filter to get rid of a constant component of a signal), ##f=0## corresponds to a constant signal, but I'd like a mathematical reason for this.
     
  2. jcsd
  3. Apr 11, 2016 #2

    BvU

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    You get a constant signal from ##\cos (\omega t)## if you let ##\omega \downarrow 0##, definitely not if you let ##\omega \rightarrow \infty##
     
  4. Apr 11, 2016 #3
    Ah yes, of course. Then what's the issue with looking at it in terms of ##f=1/T##?
     
  5. Apr 11, 2016 #4

    BvU

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    You can look at it from the perspective of ##T\rightarrow\infty##: the signal never crosses the zero...
     
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