Why is the output of the FFT symmetrical?

[@MarkNewmanEducation]

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That's very kind of you. Please share with students you think this could help.

[@JohnVahedi]

OK so you literally touched on every question, no matter how obscure, with fantastic graphics to boot. Thank you.

[@MarkNewmanEducation]

Thanks. This was a topic that was confusing me and these were all the questions that I wanted to find the answers to. I am happy to have helped others find the answers too.

[@robin.2210]

wow Mark, this is brilliant! After spending hours watching almost all of youtube's FFT education content, in just these 10 minutes so many pieces of the puzzle fall into their places. While there are beautiful visualisations and verbalised intuitions about the algorithm, I've always missed that very pragmatic and concise perspective. Thank you so much!

[@MarkNewmanEducation]

Amazing! I love giving people a Eureka moment. Really glad to have helped.

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[@123string4]

at 5:02 what if the test signal isn't in phase with the signal? Do you sweep the signal from 0 to 2pi to see if it matches? Does that mean any algorithm implementing this would at least be O(n^2)

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[@bijoydas6044]

if frequency is +ve then its angular frequency is also +ve it tells us that it moves in anticlockwise direction, if frequency is -ve then its angular frequency is also -ve it tells us that it moves in clockwise direction,

[@MarkNewmanEducation]

Correct. Negative frequency can also affect certain real-world signals. We'll be investigating this in the next video which is currently in production.

[@bijoydas6044]

@@MarkNewmanEducation your teaching way of fft encourage me to curious about it , now i am making a project on fft in our college. So thank you very much for your priceless effort.🙏🙏👍👍

[@vex18th]

hi if all the frequency scores for frequencys above nyquist frequency (here 8 Hz) are unreliable, why do we test for frequencys above 8 Hz, if we already know the results for frequencys in the range f_n to R ??? love the videos

[@alessandrobertulli425]

I might be wrong, but IIRC this symmetry only arises if the starting signal is real valued. If it was complex (which may occur in certain calculations) then there would be no symmetry anymore. But please fact-check this answer, I'm not so sure

[@현영오의골목식당]

I understand that the result of 2Hz and 14Hz are same. but why 14Hz can change to -2Hz?

[@MarkNewmanEducation]

It's not that the 14Hz can change into -2Hz as such, it's that the frequency response is periodic, it keeps repeating itself. A signal that is discrete in the time-domain will be periodic in the frequency domain. So the real and imaginary values of the frequency domain response at -2 Hz will be the same as those of 14 Hz.

[@acf2802]

It's not true to say that a sampled signal only tells you what's going on at the sample points and "anything" can happen in between. If you are sampling a signal with frequency content only bellow Fs/2 then your samples uniquely describe a single continuous bandlimited signal. You know exactly what is happening between samples.

[@MarkNewmanEducation]

The proviso being that you are sampling at greater than the Nyquist rate of course. The problem is you don't necessarily know what the maximum frequency of a random signal is unless you have some prior knowledge of the signal or you filter it to ensure that no higher frequency is getting through. The point I was trying to make is that in the sampling process, you are effectively throwing away information about the signal and that causes the repeating nature of the frequency spectrum. As far as the FT is concerned, those higher frequencies could be there. The FT has no idea whether you filtered your signal or not.