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I was looking at a table of Fourier transform pairs, and one entry is really confusing me. There's one on the second page that states $$ \mathcal{F}(\cos(\omega_0t))(\omega) = \pi(\delta(\omega - \omega_0) + \delta(\omega + \omega_0)) $$ I know that a Fourier transform is supposed to map a function in the time domain to an equivalent function in the frequency domain, so lets see what that means in this case. The period of $\cos(\omega_0t)$ is $2\pi/\omega_0$, meaning the frequency in rad/s is $\omega_0$. Thus at $\omega = \pm\omega_0$, the frequency domain should be $1$, because the amplitude of $\cos$ is $1$. When we plug it in, however, we get $\pi$: $$ \pi(\delta(0) + \delta(2\omega_0)) = \pi(1 + 0) = \pi $$ Why is this true? Isn't the amplitude of the cosine $1$, not $\pi$?

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  • $\begingroup$ When you plug $\pm\omega_0$, you get "$\pi\infty$", if that makes sense. And with the usual convention, the coefficient is $1/2$, not $\pi$. $\endgroup$ – Yves Daoust Jul 10 '15 at 13:42
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The Dirac distribution needs an integration to lead to finite values.

And that might also explain the factor $\pi$, because that integration is the inverse Fourier transformation and that might have a compensating factor in your case.

Your text lists $$ f(t) = \frac{1}{2\pi} \int\limits_{-\infty}^{\infty} \!\! F(\omega)\,e^{j\omega t}\, d\omega $$

so this gives $$ f(t) = \frac{1}{2\pi} \int\limits_{-\infty}^{\infty} \!\! \pi \left[ \delta(\omega - \omega_0) + \delta(\omega + \omega_0) \right] \, e^{j\omega t}\, d\omega = \frac{1}{2}\left[ e^{j\omega_0 t} + e^{-j\omega_0 t}\right] = \cos(\omega_0 t) $$ so we could read the transform as $$ F(\omega) = 2\pi \, \frac{1}{2} \left[ \delta(\omega - \omega_0) + \delta(\omega + \omega_0) \right] $$ where the $2\pi$ is due to the specific constants choosen for the Fourier transformation pair.

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Evaluating the Dirac delta function at zero would give you $\infty$. Remember that the Dirac delta is not, strictly speaking, a function, but rather a distribution. The Fourier transform gives the distribution of the function in the frequency domain, so the cosine function is made of of only two frequencies, with no spreading.

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  • $\begingroup$ Yes that's weird, it only takes a value if you integrate it together with someone else. Like a function chameleon, a funchameleon. $\endgroup$ – mathreadler Jul 10 '15 at 13:47
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    $\begingroup$ I remember I asked my analysis professor if the set of integrable functions could be viewed as a commutative ring with identity under addition and convolution. He asked me what the identity was, I said the delta function. Then he told in no uncertain terms that the Dirac delta was NOT a function, and not to treat it like one. $\endgroup$ – Alex S Jul 10 '15 at 13:50

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