# How do I make sense of the cosine wave having Fourier Transform coefficients which have infinite magnitude?

To illustrate my question better, consider the Fourier Transform of an aperiodic (as a periodic cosine wave has a Fourier Transform not Fourier Series) cosine wave

$$f(x) = \begin{cases} \cos(2\pi f_0x), & \text{ 0 \le x \le 2\pi} \\ 0, & \text{else} \end{cases}$$

$$F[f(x)] = \frac{1}{2}(\delta(f - f_0) + \delta(f + f_0))$$

This above implies an aperiodic cosine wave consists of two Fourier Transform coefficients which have infinite magnitude (as a delta function has infinite magnitude). Therefore, does this imply that a perfect cosine wave can not be produced in real life?

• (1) I don't think your FT is correct. (2) Regardless, a pure cosine wave cannot be produced in real life: you'd have to have turned your source on before the universe even began! Not to mention the need for an infinite amount of energy. – MBaz Sep 17 at 23:57
• @MBaz I often see questions stating "The Fourier Transform of a cosine wave", I thought they couldn't possibly be referring to a periodic cosine wave, so it shouldn't it be an aperiodic wave like I've shown? – vecohah Sep 18 at 0:03
• Please see this question and answers for a hint: dsp.stackexchange.com/q/6038/41790 – Ed V Sep 18 at 0:21
• @vecohah Mathematically it is possible to find the FT of a periodic cosine wave, but you get Dirac deltas. The FT of the truncated cosine doesn't have any deltas. – MBaz Sep 18 at 2:00

Your function is both integrable, and square integrable, because it is continuous on the finite support $$[0\;2\pi]$$. In this context ($$L_1 \cap L_2$$), its Fourier transform is pretty well defined as a classical function (not a distribution like the Dirac), and continuous. So, your formula is incorrect.