# If the cosine function is periodic, why does it have a Fourier Transform? [duplicate]

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As far as I understand Fourier Transforms are for non-periodic signals and Fourier Series for periodic signals.

So why is it we can take the Fourier Transform of a cosine when it is a periodic function, assuming the above paragraph is correct?

## marked as duplicate by Matt L., lennon310, MBaz, Stanley Pawlukiewicz, Peter K.♦Oct 2 at 15:12

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• Note that nothing prevents you from finding the Fourier Series of a non-periodic signal in the interval $T$. The series will converge in $T$. Likewise, as long as the integrals make sense, you can the FT of any signal you want. I recall there's a nice explanation in Paul Nahin's "The Science of Radio" if you can get ahold of it. – MBaz Sep 18 at 2:06

## 1 Answer

Indeed there are two things you have to know.

First, it can be shown that the continuous-time Fourier transform can be obtained from the continuous-time Fourier series by letting the period $$T$$ go to infinity.

Second, formally speaking the Fourier transform integral for periodic signals do not converge, hence do not exist. The solution is a generalisation of the Fourier transform by the use of Dirac impulse functions.

The result is an interpretation that the Fourier transform of periodic functions is a sum of scaled Dirac impulses at the Fourier harmonic frequencies, and the scale being the corresponding Fourier series coefficients.