Positive and negative frequencies

Comments · 24

• SA

  @satanaz · 2 years ago

  Amazing content! So clear and straight to the point.<br>Must have taken a ton of work to digest it down to this, but still make it understandable, without oversimplifying it, thank you so much for making this publicly available! <br>I&apos;m a mechanical engineer, and needed some refreshing of the theory on DSP and signal processing. Even after I refreshed what I needed, by watching 2 of your videos in this series, I went back and binge-watched the whole 17 videos, just because your way of teaching is so clear, concise and captivating.

  1

• EX

  @exoticcoder5365 · 2 years ago

  <a href="https://www.youtube.com/watch?v=Nupda1rm01Y&amp;t=555">9:15</a> Maths is very good 👍🏻

• NA

  @nathanielrindlaub6186 · 4 years ago

  Thank you so much for this video series; it has been truly invaluable! At <a href="https://www.youtube.com/watch?v=Nupda1rm01Y&amp;t=100">1:40</a> you mention that the FT of complex values performs a little differently and you allude to possibly touching on that in more depth in a later video. Did you ever make a video discussing FTs on complex signals? I am struggling to understand how complex sampling works in software defined radio (which typically uses IQ data) and I imagine such a video would help a great deal. I have scoured YouTube and elsewhere and can not for the life of me understand why bandwidth = sample rate with IQ data.

  1

• JA

  @jamiele9250 · 5 years ago

  Thank you^^

• DE

  @devrimturker · 4 years ago

  When I watch rolling wheel of car, at high speed, it seems like it is rolling backward. Is it due to aliasing, can we call negative frequency?

  3

• CH

  @charliechang2897 · 5 years ago (edited)

  Thanks for the great video. But I think it should be hz = linspace(0, srate, N+1) at <a href="https://www.youtube.com/watch?v=Nupda1rm01Y&amp;t=705">11:45</a>, because the first point of dft corresponds to 0 hz and the (N+1) th point corresponds to the sampling rate. Note that in Matlab, &quot;y = linspace(x1,x2,n) generates n points. The spacing between the points is (x2-x1)/(n-1).&quot; (x1 and x2 are included in y)

• MU

  @mugdhagarg5100 · 3 years ago

  Per what you&apos;ve explained, I can see why there would be aliasing beyond the Nyq freq., but not why the frequencies would keep decreasing as the signal frequencies increase. Please explain such things better? People like me get stuck forever if things are said without their full explanations.

• AL

  @aleksinuutila2315 · 5 years ago

  This is completely unrelated, but the spectrum at <a href="https://www.youtube.com/watch?v=Nupda1rm01Y&amp;t=42">0:42</a> is identical to the UV-vis spectrum of tetrangulol.

• AN

  @AnonymousePhx · 4 years ago

  Thank you for making such complex (no pun intended) subjects accessible to people like me who thought they missed the boat on advanced mathematics and were forever screwed in understanding LFP analyses! :) &nbsp;<br><br>I&apos;m still a bit hung up on the necessity of using complex sine waves to compute the Fourier transform - do you have any suggested readings, video, or code that steps through the alternative strategies of how it would look to compute Fourier using real sine waves? Thank you so much for what you do.

• CH

  @chansonjoy · 6 years ago

  I found it&apos;s a bit too difficult to follow at <a href="https://www.youtube.com/watch?v=Nupda1rm01Y&amp;t=575">9:35</a> as the transit is too fast and miss some link. I tried to explain to my self as following:<br>so Fourier coefficient (FC) is nothing more than a dot product FC = |A||B|cos(θ)<br>cos(θ) = 1/2e(-jk) + 1/2e(jk)<br>therefor FC = |A||B|(1/2e(-jk) + 1/2e(jk))<br>FC = 1/2|A||B|e(-jk) + 1/2|A||B|e(jk))<br>From above we can see FC is composed by 2 parts: positive and negative frequency and the magnitude of these 2 parts is half of |A||B|

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• AM

  @amrelsherbiny4811 · 4 years ago

  Thank you so much for this amazing series, it really helped me a lot. I have a question regarding the negative frequencies please. I&apos;m sampling a signal at a rate of 1000hz, so my Nyquist frequency is 500hz. After that, I filter the signal with a bandpass filter between 20-450hz. When I plot the Fourier transform results, I&apos;m still seeing positive and negative frequencies, which are mirror images centered around 0. What I don&apos;t understand is, if the negative frequencies are the ones after Nyquist frequency, and I am already filtering the signal below Nyquist frequency of the signal, how do they still exist?

  1

• LO

  @lonebot2003 · 1 month ago (edited)

  This is not accurate it has same content like all other videos trying to explain negative frequencies, If you think you are suddenly an expert after this video, try to design a image reject radio by using negative frequencies! Yup, you can&apos;t do it since this video just has the standard explanation you find everywhere