# I Frequency of a constant function

1. Apr 11, 2016

### axmls

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. Apr 11, 2016

### BvU

You get a constant signal from $\cos (\omega t)$ if you let $\omega \downarrow 0$, definitely not if you let $\omega \rightarrow \infty$

3. Apr 11, 2016

### axmls

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

4. Apr 11, 2016

### BvU

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