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If I have two sets of samples $a$ and $b$ and I have already calculated $A = FFT(a)$ and $B = FFT(b)$ is it possible to calculate $FFT(a + b)$ without performing another FFT?

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    $\begingroup$ Yes, it is possible by Einstein's famous formula $FFT(a+b)=FFT(a)+FFT(b)$. $\endgroup$
    – Matt L.
    Feb 1 '18 at 10:51
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    $\begingroup$ I can honestly say that, compared to the complexity of the other problems you tackle in your questions, I'm a bit disappointed that you didn't do enough research (or, just write down the DFT formula) to figure this one out. So, with a heavy heart, I downvoted for lack of own research. $\endgroup$ Feb 1 '18 at 12:47
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    $\begingroup$ @Marcus Müller, I just had a complete brain meltdown on what to search for, the subject of FFTs is just crammed with web pages that didn't help. I did do a reasonable amount of searching, but was struggling with what search term to use. It's just a hobby for me. $\endgroup$
    – keith
    Feb 1 '18 at 12:54
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Fourier Transform is linear, hence if $\mathcal F[x(t)]=X(f)$ and $a$, $b$ are complex numbers, then:

$$\mathcal{F}[ax(t)+by(t)] = a X(f) + bY(f) $$

So in your case, simply sum the results together.

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