How does the Discrete Fourier Transform DFT relate to Real Frequencies?

Explains how the Discrete Fourier Transform (DFT) / Fast Fourier Transform (FFT) vector relates to real frequencies and real signals.
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For a full list of Videos and accompanying Summary Sheets, see the associated website: www.iaincolling...

[@spencerduncan7006]

You're videos are saving my grade!!! thank you thank you thank you

[@iain_explains]

I'm so glad!

[@shankaraenagandula1561]

Thanks, prof. It is very useful for me.

[@iain_explains]

Great. I'm glad it was helpful.

[@sivasaikrishnamarthy178]

Great explanation sir, few quick queries: (1) Isn't it the case that (0-pi) plot for DFT is a mirror image to (pi - 2pi) ? Please let me know. (2) From your video, M is the discrete vector length of DFT. N is the sampling frequency of signal. Does M should always be same as N ? M cannot be greater or less than N ? (3) DFT is called to be sampled version of DTFT. so we cannot construct a complete signal (highly resoluted signal )in time domain unless M or N tends to --> infinity right? Thanks again

[@iain_explains]

Answers: (1) Yes, (2) M=N, (3) Yes

[@amansaxena1479]

06:15 FOR REAL SIGNALS THE NYQUIST SAMPLING RATE IS TWICE THE maximum FREQUENCY OF THE SIGNAL. BUT FOR 10ms PERIOD SIGNAL, frequency is 100hz hence nyquist rate=200hz. Am i missing something professor?

[@iain_explains]

It's not a 10ms period signal. It's a 10ms sampling period. Which means the sampling rate is 1/10ms = 100 Hz, which means the highest frequency that can be fully recovered from the samples is 50 Hz.

[@paulooctavioaraujo3087]

The sampling period is the time between two samples? or is it the whole sampling time?

[@iain_explains]

It is the time between every sample in the time domain vector x. (so the answers to your questions are "yes" to the first question, "no" to the second question.)

[@elijahsmith7552]

I had the same question. Thanks

[@wojciechzajaczkowski9080]

Is this about two sided spectrum, right ? I wonder if the negative frequencies could be separately modulated and "bounded" together with the positive, building the single frequency at the end. Then only if we multiply such a signal with some fc carrier frequency, than the negative and positive will start to distinguishable ? Is this really that the single real sin function may carry the four signals (negative sin+cos and positive sin+cos), making it possible to modulate 2 separate symbols at the same time ?

[@iain_explains]

Short answer: No. Hopefully this video helps: "What is Negative Frequency?" kzbin.info/www/bejne/nauZcn6NYrdpb9U

[@wojciechzajaczkowski9080]

Thank You for that link. Actually I was wondering if positive and negative frequencies could cancel each other making pure cosine (horizontal complex movement) or pure sin (vertical complex movement) or any other combination (diagonal complex movement)? Is that possible assuming that we address positive and negative frequency with a different symbol in OFDM ?

[@iain_explains]

Perhaps you didn't quite understand the "Negative Frequency" video. Negative frequency does not exist. It is just a representation of a phasor that is rotating in the negative direction. These videos might help: "How do Complex Numbers relate to Real Signals?" kzbin.info/www/bejne/in26dmZuba-KfdU and "Visualising Complex Numbers with an Example" kzbin.info/www/bejne/nonPZqiOa76mnpI

[@anamikasharma6540]

Thank you very much. Your videos are very helpful.

[@iain_explains]

Glad you like them!

[@kristinefrancesbaldas6306]

this video of yours is really helpful. Thanks a bunch!

[@iain_explains]

I'm so glad!