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

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  • čas přidán 22. 02. 2020
  • Explains how the Discrete Fourier Transform (DFT) / Fast Fourier Transform (FFT) vector relates to real frequencies and real signals.
    Related videos: (see: iaincollings.com)
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    For a full list of Videos and accompanying Summary Sheets, see the associated website: www.iaincollings.com

Komentáře • 19

  • @kristinefrancesbaldas6306

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

  • @anamikasharma6540
    @anamikasharma6540 Před 3 lety +1

    Thank you very much. Your videos are very helpful.

  • @spencerduncan7006
    @spencerduncan7006 Před 3 lety

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

  • @shankaraenagandula1561

    Thanks, prof. It is very useful for me.

  • @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

  • @paulooctavioaraujo3087

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

    • @iain_explains
      @iain_explains  Před 3 lety

      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
      @elijahsmith7552 Před 10 měsíci

      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
      @iain_explains  Před rokem +1

      Short answer: No. Hopefully this video helps: "What is Negative Frequency?" czcams.com/video/gz6AKW-R69s/video.html

    • @wojciechzajaczkowski9080
      @wojciechzajaczkowski9080 Před rokem

      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
      @iain_explains  Před rokem

      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?" czcams.com/video/TLWE388JWGs/video.html and "Visualising Complex Numbers with an Example" czcams.com/video/hXl5uX6Ysh0/video.html

  • @amansaxena1479
    @amansaxena1479 Před rokem

    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
      @iain_explains  Před rokem

      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.