Best clear cut explanation , this video helps to jump from FT to FFT and understand the computational efficiency. Come with some prep study of FT and do an actual calculation in excel then come here and you are done!
At 9:33 you say "x sub zero" and I found that a bit confusing later in the video, when I forgot where it's came for, but now I have realized that it's more like "x sub oh". :) Thanks for the great video! It really helped me!
no, at 11:10 it is just splitting even and odd index terms (splitting N = 8 DFT into 2 DFTs of N = 4). The full bit reversal (x(0) x(4) x(2) x(6) ) occurs after three stages of splitting - see 15:08
@@CanaDan One tricky bit I’ve learned recently. You will want to skew the visualization horizontally towards the right. For audio, the FFT contains a ton of detailed high frequency information so if you just visualize the FFT as is, it is hard to see the bass/miss areas. You want to skew the visualization to emphasize the bass and mid regions for regular viewing as you might see in a spectrum analyzer in your DAW
Yes, good catch. x[8] should not be on slide 4 at 10:04, but it should be x[0] x[2] x[4] x[6] as Colo Ricatti pointed out. Somehow I skipped x[4] when counting.
frankly, i didn't get this 8:09 - N > n, k isn't great value too because of Nyquist limit (actually, k sample rate, the're no any sense in such calculations - it computes damn lot of ghost frequencies, because it increases sample rate.
This is a good overview of FFT. It would be nice to explain how the DFT convolution sum is derived. Also, the de-interlacing of the inputs was glossed over (not explained clearly) but only the reversed binary notation was mentioned (this is just an after-the-fact observation of How, not an explanation of Why). Readers who dive deeper into the splitting of a larger N-point FFT into two smaller N/2-point FFT’s, or understand the relationships between the twiddle factors (and their periodic nature) would understand and retain better the FFT technique (and be able to conquer any arbitrary size of N-point FFT (N being a power of 2, of course).
I highly suggest going back to second-year advanced calculus textbooks. There're many well-rounded explanations of FFT rather than what we have in DSP textbooks.
9:05 I don't think the sum for even terms Σ(r=0 to (N/2)-1) x[2r] * (w_N/2)^kr is the DFT for N/2 samples since it should evaluate both the function x and the complex exponential at 2r instead of x at 2r and the exponential at r to be the DFT, and same for the sum for odd terms. This comment is supposed to be a question ...
To calculate the cost in the 13:42. It should be 2(N/2)^2+N/2, the next steps will be similar. So the final cost will be 3/2N+1/2N logN. It is because that the w of the lower half are simply negative of those of the upper half. So the cost should not be counted. Of course under big O notation, the final answer is still O(NlogN). You are really helpful, but there are some minor mistakes.
The complexity calculation for the FFT as you explained in the video is incorrect. 2 summations of half size of the array will be N(N/2 • 2), which is N^2. The benefit of FFT is from the fact that W_N^rk = -W_N^r(k-N) when k ≥ N. This is due to the symmetric property of the exponential function: e^j2π = -e^jπ Now you only have to compute the FT for half the array and the other half can be constructed by negating those terms, so you end up with complexity N(N/2) for the first split.
Yes. There is an annotation at 10:07 that points out the problem, but that may not be visible on your device. I have also uploaded a corrected version of this video to my channel, called The Fast Fourier Transform (FFT) Algorithm (c)
As an electronics technician of only a 2 year Associate in Science Technology degree in electronics for 37 years, and retired now, who helped engineers design, test breadboard and build / troubleshoot practical low frequency RLC circuits and also at the microwave frequencies for building 'cascaded microwave stripline amplifiers' in the 4 to 30 GHZ range for use in the real world, I had never used Fourier Transforms nor Laplace Transforms, I only used High Tech Oscilloscopes, Spectrum Analyzers, Vector Network Analyzers, SMITH CHARTS, Voltmeters, Ammeters, Frequency counters and computers, and sometimes used ALGEBRA and TRIGONOMETRY to solve electronics circuits problems In the laboratory and in the REAL WORLD, I have great RESPECT for the Geniuses of ELECTROMAGNETICS Pioneers, who had derived the mathematical physics of all the basics the we based our technology from: James Clerk Maxwell, Hertz, Gauss, Faraday, Lenz, Oersted, Henry, Steinmetz, Heaviside, Tesla, Weber, and many more, including STEVE WOZNIAC.
i am gonna implement FFT in verilog and previously i was a moderate knowledgeabout FFT but after watching this video get through from this Thanks saving my life :v
How would the butterfly diagram (@ 15:06) change if there were numerous iterations of inputs? For example if N=128 but the 8 channels structure is maintained? Thanks
You missed one of the main parts: How to compute X[k+N/2]. So you only take k from 0 to N/2. Based on this, you didn't prove that the asymptotics is 2*(N/2)^2.
X[k] = Xe[k] + Wn,k*Xo[k]. Index for X runs from 0 - 7, but that of Xe and Xo runs from 0 - 3. Please reply how to deal with X's when the index runs above 3. (I'm assuming N=8)
I don't know if you ever check comments for this particulair video since it's been a while since you've uploaded it, but I have a question: how do you get O(((N^2)/2)+N)? I know you explained it in the video (at around 12:00), but I don't understand it. Thanks in advance!
where can i learn more about what "k" and "w"(omega) and all these greek symbols mean? i'm studying too many things and it is just hard to remember what they mean.
How it is n = 2r for even and n = 2r+1 for odd.. Can you explain please?.. Lets say if n = 8, for even -> n=2r -> 2*4 = 8. whereas, for odd -> n=2*r+1 => 9
Let us take the case that N is 8, the samples are numbered X[0], X[1], X[2], ... till X[7]. By dividing them into even and odd samples, X[0], X[2], X[4], and X[6] are grouped together, and X[1], X[3], X[5], and X[7] are grouped together. The value of r ranges from 0 to N/2 - 1. Thus, We can say that r is in the range of 0 to 3 (N/2 - 1 = 8/2 - 1 = 4 -1 = 3), making 2r and 2r+1 fall in the same range as well. Hope this makes things clear :)
This video is 10 years old and is still the most complete and concise video on the subject 💯
5:30 to skip review of big-O and DFT stuff, good video thanks
Thank you so much!! So clear!! I read the book, resulting in feeling hopeless, and your explanation just lightens up my world. ^_^ Thanks again!
your photo is so cute
I wish I could say the same...
Awesome video thank you! Explained in ten minutes what I couldn't understand my professor was trying to say in three hours.
Best clear cut explanation , this video helps to jump from FT to FFT and understand the computational efficiency.
Come with some prep study of FT and do an actual calculation in excel then come here and you are done!
At 9:33 you say "x sub zero" and I found that a bit confusing later in the video, when I forgot where it's came for, but now I have realized that it's more like "x sub oh". :)
Thanks for the great video! It really helped me!
no, at 11:10 it is just splitting even and odd index terms (splitting N = 8 DFT into 2 DFTs of N = 4). The full bit reversal (x(0) x(4) x(2) x(6) ) occurs after three stages of splitting - see 15:08
Super helpful! I'm trying to make an audio spectrum analyzer. This has gotten me closer to understanding the concept, thanks!
did you build it ? can you share the circuit diagram?
me too. iv been looking around kinda everwhere for good information on how this all works and how to calculate audio into something viewable
@@CanaDan One tricky bit I’ve learned recently. You will want to skew the visualization horizontally towards the right. For audio, the FFT contains a ton of detailed high frequency information so if you just visualize the FFT as is, it is hard to see the bass/miss areas. You want to skew the visualization to emphasize the bass and mid regions for regular viewing as you might see in a spectrum analyzer in your DAW
Yes, good catch. x[8] should not be on slide 4 at 10:04, but it should be x[0] x[2] x[4] x[6] as Colo Ricatti pointed out. Somehow I skipped x[4] when counting.
Spent 20 mins trying to figure if there was something I'd missed lol. Will come back to the comments more often in the future
the best i have watched so far..
Somebody please get this guy a raise
Finally someone explains FFT that people can understand!!!!
哪来的8啊
Nan Wu 诶。。。 你怎么也看这个了
Nan Wu 什么8？
Steven Z 就扫了一眼封面就看到了x[8]
真的是眼力很好啊！
This video had so much potential but glossed over so so much.
Feel bad for Mr. Gauss
Thank you, this was super helpful in understanding the transition from DFT to DFFT.
At 12:00, shouldn't the first left numbers be 0, 2, 4 and 6?
Yes
Yes
Yes
Yes
yes
frankly, i didn't get this 8:09 - N > n, k isn't great value too because of Nyquist limit (actually, k sample rate, the're no any sense in such calculations - it computes damn lot of ghost frequencies, because it increases sample rate.
This is a good overview of FFT. It would be nice to explain how the DFT convolution sum is derived. Also, the de-interlacing of the inputs was glossed over (not explained clearly) but only the reversed binary notation was mentioned (this is just an after-the-fact observation of How, not an explanation of Why). Readers who dive deeper into the splitting of a larger N-point FFT into two smaller N/2-point FFT’s, or understand the relationships between the twiddle factors (and their periodic nature) would understand and retain better the FFT technique (and be able to conquer any arbitrary size of N-point FFT (N being a power of 2, of course).
I highly suggest going back to second-year advanced calculus textbooks. There're many well-rounded explanations of FFT rather than what we have in DSP textbooks.
i just want to say that i love you man
9:05
I don't think the sum for even terms Σ(r=0 to (N/2)-1) x[2r] * (w_N/2)^kr is the DFT for N/2 samples since it should evaluate both the function x and the complex exponential at 2r instead of x at 2r and the exponential at r to be the DFT, and same for the sum for odd terms.
This comment is supposed to be a question ...
GREAT THANK YOU , HELPED A LOT
Thank you very very much ,Barry, very nice and deep explanation. Your way of explanation is very good. Thank you so much.
To calculate the cost in the 13:42. It should be 2(N/2)^2+N/2, the next steps will be similar. So the final cost will be 3/2N+1/2N logN. It is because that the w of the lower half are simply negative of those of the upper half. So the cost should not be counted. Of course under big O notation, the final answer is still O(NlogN). You are really helpful, but there are some minor mistakes.
FANTASTIC
there is a lot of interesting points at 13:41. please check that, if it is wrong.
Yes, those end up being the twiddle factors - see them in the example diagram for N=8 at 15:14
all important things at a place!!!
Not sure why you left out the recursive relation between the odd and even functions and the DFT. I was so confused where the speed gain was from.
The complexity calculation for the FFT as you explained in the video is incorrect.
2 summations of half size of the array will be N(N/2 • 2), which is N^2.
The benefit of FFT is from the fact that
W_N^rk = -W_N^r(k-N) when k ≥ N.
This is due to the symmetric property of the exponential function:
e^j2π = -e^jπ
Now you only have to compute the FT for half the array and the other half can be constructed by negating those terms, so you end up with complexity
N(N/2) for the first split.
this is beautiful, thanks man!
In butterfly diagram it should be x[4] instead of x[8] ,right?
Yes. There is an annotation at 10:07 that points out the problem, but that may not be visible on your device. I have also uploaded a corrected version of this video to my channel, called The Fast Fourier Transform (FFT) Algorithm (c)
@1:21 it should be N complex multiplies and N complex adds (not N-1) ... there are as many multiplies as there are adds
As an electronics technician of only a 2 year Associate in Science Technology degree in electronics for 37 years, and retired now, who helped engineers design, test breadboard and build / troubleshoot practical low frequency RLC circuits and also at the microwave frequencies for building 'cascaded microwave stripline amplifiers' in the 4 to 30 GHZ range for use in the real world, I had never used Fourier Transforms nor Laplace Transforms, I only used High Tech Oscilloscopes, Spectrum Analyzers, Vector Network Analyzers, SMITH CHARTS, Voltmeters, Ammeters, Frequency counters and computers, and sometimes used ALGEBRA and TRIGONOMETRY to solve electronics circuits problems In the laboratory and in the REAL WORLD,
I have great RESPECT for the Geniuses of ELECTROMAGNETICS Pioneers, who had derived the mathematical physics of all the basics the we based our technology from: James Clerk Maxwell, Hertz, Gauss, Faraday, Lenz, Oersted, Henry, Steinmetz, Heaviside, Tesla, Weber, and many more, including STEVE WOZNIAC.
Excellent explanation. Helped a lot. Thanks!
i am gonna implement FFT in verilog and previously i was a moderate knowledgeabout FFT but after watching this video get through from this Thanks saving my life :v
How would the butterfly diagram (@ 15:06) change if there were numerous iterations of inputs? For example if N=128 but the 8 channels structure is maintained? Thanks
Wan , you're the best .
Great explanation on Butterfly algorithm!
You missed one of the main parts: How to compute X[k+N/2]. So you only take k from 0 to N/2. Based on this, you didn't prove that the asymptotics is 2*(N/2)^2.
I seen your channel mate, really love the content. Subscribed straight away, We should connect!
X[k] = Xe[k] + Wn,k*Xo[k]. Index for X runs from 0 - 7, but that of Xe and Xo runs from 0 - 3. Please reply how to deal with X's when the index runs above 3. (I'm assuming N=8)
my question too
fuck me this is hard
Like si crees que Sergio y compañía no se han visto el vídeo, pero la recomendación ahí queda.
you are my hero .. I think I will use the website too .. thanks
Really good video!!! Clearly explained!!! thankyou!
I just understood FFT for the first time!
14:53 he cancels out the N^2/N (=N) saying "for capital N large enough, this term dominates". Someone, please help me, why do you drop the +N?
Awesome explanation. Made it very clear. Thanks
This video saved my ass. Thank you Barry
Thank you for sharing your knowledge.
I don't know if you ever check comments for this particulair video since it's been a while since you've uploaded it, but I have a question: how do you get O(((N^2)/2)+N)? I know you explained it in the video (at around 12:00), but I don't understand it. Thanks in advance!
perfect explanation, thank you !
there might be sign errors: -1 should be used for X(k+N/2)
I'm having a hard time finding the equation you are referring to. Could you give me a time stamp in the video and be a bit more specific? Thanks.
Hi, sorry for the unclear description.There might be a sign error in the example graphs.
CAI Jianping why dont you just give the time in the video when the example graph appears
where can i learn more about what "k" and "w"(omega) and all these greek symbols mean? i'm studying too many things and it is just hard to remember what they mean.
Very cool, but maybe it lacks some schemas to express more the ideas of the simplification.
Very good thank you! Nice and clear and well explained.
When you pull the WN^k term out at 7:55, is that term known as the twiddle coefficient?
where is the input X[4] in the block diagram?
Thanks a lot Sir.. You are awesome! ;-)
Thankyou for the video, its great
just what i wanted..thanx a lot :)
There shouldn't be an x[8] term, right? Because then N would be 9?
well done brother
Great video! Thanks
How it is n = 2r for even and n = 2r+1 for odd.. Can you explain please?.. Lets say if n = 8, for even -> n=2r -> 2*4 = 8. whereas,
for odd -> n=2*r+1 => 9
Let us take the case that N is 8, the samples are numbered X[0], X[1], X[2], ... till X[7]. By dividing them into even and odd samples, X[0], X[2], X[4], and X[6] are grouped together, and X[1], X[3], X[5], and X[7] are grouped together. The value of r ranges from 0 to N/2 - 1. Thus, We can say that r is in the range of 0 to 3 (N/2 - 1 = 8/2 - 1 = 4 -1 = 3), making 2r and 2r+1 fall in the same range as well. Hope this makes things clear :)
10:08 its x[0] x[2] x[4] x[6]
where's all the cosines and sines in the equations?
Thanks a lot that was really helpful. Great Job :)
ist there an advantage of bit reversed positioning ?
Great video. Thank you.
How do you choose appropriate W values ?
what will it be if N = 30 ?
Could somebody explains what if the signal is 10 points sampled(which is not 2^N), please?
pad it with 0, or loop it
Great video
sorry i dont get it... first example N = 8.... second example N = 8 again but more stages... help pls
While solving decimation in time and frequency algorithm of fft we used only DFT why not IDFT sir.
beautifully explained (y)
Linear filtering methods based on dft problems
Is that the power of j or i?
If you are electrical engineer like me, we use j
Thanks a lot...
Thanks a ton.!!!
I just read the fast and fourious transform.
Its awesome :)) !..
really good n useful
very nice
what is "j" ????
это i комплексная)
спасибо)
Good one
awesome
15:07 muck fe
Good video but the ads are very distracting. if your intention to help, minimize or remove the ads. thumb down.
Can anyone make a subtitle :目?
Sorry but gauss was not, Fourier was
15:33 ♡♡
interesting！
LMAO imagine waiting 31 years for your image to be processed xD
well mama what about how well we are doing in skewl... the Kids
هاذه صورت بابا اني هبه احبك هواي انته وعالتك اني حبكم هواي باي
Crap audio.
Ffg