When taking the Fourier Transform of a sum of two different frequencies, why doesn't a peak appear that corresponds to the beating frequency of the signal?

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If you take the FFT of the sum of two sine waves, you’ll get peaks only at the frequencies of those two sine waves, because that’s where the energy is.

Beat patterns arise when you multiply sine waves. But, multiplication in the time domain is equivalent to convolution in the frequency domain. So, it turns out that the two situations are related.

You can also arrive at this result by busting out your trig identities. In particular:

sin(ω0t)+sin(ω1t)=2sin(ω0+ω12t)cos(ω0−ω12t)

So, if I set ω0=10 and ω1=11, I get the following result when I add the sine waves:

I get the same result if I multiply 2sin(10.5t)cos(0.5t)$2\sin (10.5t)\cos (0.5t)$:

So, you’re asking why the FFT doesn’t show a peak at the 0.5t$0.5t$ frequency. That’s because there’s no energy there. There’s also no energy at 10.5t$10.5t$.

Fourier analysis decomposes a signal into a weighted sum of sine waves at different frequencies. Your original signal, sin(ω0t)+sin(ω1t)$\sin ({\omega }_{0}t)+\sin ({\omega }_{1}t)$ is precisely the sum of two sine waves. Thus, the FFT will have two peaks, one at ω0${\omega }_0$ and ω1${\omega }_1$.

The beat is equivalent to the result you would get if you multiplied two sine waves at two different, related frequencies: ω0+ω12$\frac{{\omega }_0+{\omega }_1}{2}$ and ω0−ω12$\frac{{\omega }_0-{\omega }_1}{2}$. There’s no energy at the beat frequency though, nor is there any energy at the frequency of the sine wave the beat’s superimposed on.