Frequency of a constant function

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  • Thread starter axmls
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  • #1
axmls
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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.
 

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  • #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##
 
  • #3
axmls
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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##

Ah yes, of course. Then what's the issue with looking at it in terms of ##f=1/T##?
 
  • #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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