Reply by January 31, 20202020-01-31
bmh161=E6=96=BC 2009=E5=B9=B44=E6=9C=889=E6=97=A5=E6=98=9F=E6=9C=9F=E5=9B=
=9B UTC+8=E4=B8=8B=E5=8D=889=E6=99=8222=E5=88=8655=E7=A7=92=E5=AF=AB=E9=81=
=93=EF=BC=9A
> This is probably a stupid question but, well, I'm not all that smart... >=20 > Instead of going through all the trouble of designing complicated digital > filters, why don't we just pick the frequencies that we want to supress a=
nd
> replace those bins in the FFT with zeros?
You may be interested in the simple experiment using matlab. https://poweidsplearningpath.blogspot.com/2019/04/dftidft.html
Reply by Rick Lyons April 21, 20092009-04-21
On Fri, 10 Apr 2009 08:13:48 +0200, "Robert Lacoste"
<see-alciom-dot-com> wrote:

> >"bmh161" <bmh161@yahoo.com> a &#4294967295;crit dans le message de >news: -9CdnboVQ4Mya0DUnZ2dnUVZ_vednZ2d@giganews.com... >> This is probably a stupid question but, well, I'm not all that smart... >> >> Instead of going through all the trouble of designing complicated digital >> filters, why don't we just pick the frequencies that we want to supress >> and >> replace those bins in the FFT with zeros? >> > >Doing what you propose (FFT of the signal, multiplying with a frequency >mask, IFFT) gives EXACTLY the same result mathematically speaking than a FIR >filter, its just another way to calculate it. In a nutshell :
[snipped by Lyons] Hi Robert, are you sure about that? Let's say I have a 1024-sample time sequence on which I perform a 1024-pt FFT. Next, I zero-out the FFT bins (samples) for the freqs I want to attenuate. Finally I perform a 1024-pt inverse FFT. My resultant "filtered" time sequence is 1024 samples in length (just as was the original time sequence). Now ...what kind of time-domain digital filtering can I perform on the original 1024-sample time sequence where the filter output sequence is EXACTLY the same 1024 samples as the above freq-domain filtering output? I believe the answer is: "There *IS* no such time-domain digital filter." As far as I know, the freq-domain filtering I discussed in my first paragraph above is equivalent to time-domain circular convolution, and time-domain filtering does not perform circular convolution. Robert, if I'm mistaken, then I'd sure like to know about it. See Ya', [-Rick-]
Reply by Rick Lyons April 21, 20092009-04-21
On Fri, 10 Apr 2009 02:08:18 -0700 (PDT), Rune Allnor
<allnor@tele.ntnu.no> wrote:

>On 10 Apr, 02:17, Rick Lyons <R.Lyons@_BOGUS_ieee.org> wrote: >> On Thu, 09 Apr 2009 08:22:55 -0500, "bmh161" <bmh...@yahoo.com> wrote: > >> >Instead of going through all the trouble of designing complicated digital >> >filters, why don't we just pick the frequencies that we want to supress and >> >replace those bins in the FFT with zeros? >... >> &#4294967295;I just wanted to say that >> your question is sensible and that we, the DSP guys here, >> should investigate the process of freq-domain filtering >> in detail > >Rick, did we read the same post? The question is not why >frequency domain filtering is a bad idea, but why one doesn't >use the naive, 'obvious' specs for the filters. I commented on >a similar question not too long ago: > >http://groups.google.no/group/comp.dsp/msg/2d4bd5dfb25b5a23 > >Such questions are covered in most DSP texts - I wouldn't be >surprised if you touched on the subject already. If not, the issue >might be worth half a page's comments in your 3rd edition, as >your book is the first entry point into DSP for lots of users. > >Teachning students how to think 'right' is an admirable (and >difficult!) task; warning against how to think 'wrong' can only >help the effort. > >Rune
Hi Rune, Oh shoot. Maybe I misunderstood the poster's question. For whatever reason, I thought he was asking: "Can we perform an N-point FFT on an N-point time sequence, zero-out the spectral components that we wish to attenuate, and then perform an N-point inverse FFT to obtain a "filtered" time-domain sequence. If the original poster was asking about ways to design a time-domain digital filter, then I completely missed the point of his question. (That wouldn't be the first time for me.) [-Rick-] PS. Rune, I wish you "calm seas."
Reply by Vladimir Vassilevsky April 19, 20092009-04-19

glen herrmannsfeldt wrote:

> Well, say you FFT one track on a CD, maybe five minutes. > (So not a Mahler symphony, for example.) > > Five minutes is 5*60*44100 samples, or 13230000 samples. > FFT is O(N logN), so it shouldn't take all that long on most > current machines. The bin spacing is then 1/(5 minutes) or > about 0.00333 Hz.
FFT with the size of up to ~100M points is no problem as it takes ~minutes on PC. However beware of the numeric artifacts.
> In FFT terms, the signal is periodic. Even if you don't believe > that, the quantization noise will be enough that you can't > detect frequencies closer than 0.00333Hz, anyway, in a 5 minute > signal.
This is not so obvious to me. Can you justify that.
> Interesting question for those building 60Hz notch filters, > how accurate is the power line frequency over short time scales? > Is it better than 0.00333Hz?
According to my measurements, the short term accuracy of AC frequency is about 1e-4. But the fluctiation of the waveform sets the limit to the usefulness of the comb filters before the frequency instability comes into play. Vladimir Vassilevsky DSP and Mixed Signal Design Consultant http://www.abvolt.com
Reply by glen herrmannsfeldt April 19, 20092009-04-19
a@b.com wrote:
(snip, I wrote)

>>If you take the FFT on a signal that is at least as long as >>the source, there are no in between bins. However, FFT
> Could you give a practical numerical example, or more detail about > what you mean here? I guess that's different then when all > frequencies of the sampled signal are an integer multiple of the bin > size (so fall exactly on an FFT bin).
Well, say you FFT one track on a CD, maybe five minutes. (So not a Mahler symphony, for example.) Five minutes is 5*60*44100 samples, or 13230000 samples. FFT is O(N logN), so it shouldn't take all that long on most current machines. The bin spacing is then 1/(5 minutes) or about 0.00333 Hz. In FFT terms, the signal is periodic. Even if you don't believe that, the quantization noise will be enough that you can't detect frequencies closer than 0.00333Hz, anyway, in a 5 minute signal. Interesting question for those building 60Hz notch filters, how accurate is the power line frequency over short time scales? Is it better than 0.00333Hz? -- glen
Reply by April 18, 20092009-04-18
On Thu, 9 Apr 2009 18:49:35 +0000 (UTC), glen herrmannsfeldt
<gah@ugcs.caltech.edu> wrote:

>Greg Berchin <gberchin@comicast.net.invalid> wrote: >(snip on filtering using FFT) > >> 1. You're falling into the classic "if I just zero the appropriate FFT >> bins, I'll have a perfect lowpass/bandpass/highpass/whateverpass >> filter" trap. Consider what happens in-between FFT bins -- you have >> no control over frequencies that do not correspond to FFT bins, and >> they will not automatically be zero just because the adjacent FFT bins >> are zero. If you FFT longer and longer time signals in order to be >> able to explicitly zero more and more of those in-between frequencies, >> eventually you will realize that you have to FFT an infinitely long >> time signal in order to actually accomplish what you desire. > >If you take the FFT on a signal that is at least as long as >the source, there are no in between bins. However, FFT
Could you give a practical numerical example, or more detail about what you mean here? I guess that's different then when all frequencies of the sampled signal are an integer multiple of the bin size (so fall exactly on an FFT bin).
Reply by zzbu...@netscape.net April 17, 20092009-04-17
On Apr 9, 9:22&#4294967295;am, "bmh161" <bmh...@yahoo.com> wrote:
> This is probably a stupid question but, well, I'm not all that smart... > > Instead of going through all the trouble of designing complicated digital > filters, why don't we just pick the frequencies that we want to supress and > replace those bins in the FFT with zeros?
Well, that's just a particular type of filter called a notch. It's not that it doesn't work, it just has so many overused strange side effects, it's one of the reasons Optical Computers, Fiber Optics Signalling, Holograms, HDTV, CD, and DVD were invented.
Reply by Randy Yates April 14, 20092009-04-14
"hariharan.gk" <hariharan1983@yahoo.com> writes:

> I believe MIPS is also a concern when you do FFT/IFFT just for
> filtering. MIPS are less than time-domain convolution - that's really the reason we use it. -- % Randy Yates % "With time with what you've learned, %% Fuquay-Varina, NC % they'll kiss the ground you walk %%% 919-577-9882 % upon." %%%% <yates@ieee.org> % '21st Century Man', *Time*, ELO http://www.digitalsignallabs.com
Reply by hariharan.gk April 14, 20092009-04-14
I believe MIPS is also a concern when you do FFT/IFFT just for filtering.


Reply by foxcob April 13, 20092009-04-13
>>This is probably a stupid question but, well, I'm not all that smart... >> >>Instead of going through all the trouble of designing complicated >digital >>filters, why don't we just pick the frequencies that we want to supress >and >>replace those bins in the FFT with zeros? >> >> >> > >Then, with the undesirable frequencies zeroed out, take the inverse FFT
to
>get back to the time domain. >
Others have already mentioned some of the problems with the method you describe. One approach to implementing a filter this way is to use the window method of FIR design without ever working in the time domain. With the window method, the desired frequency response is plugged into the appropriate FFT bins. A brick wall filter would be typical (zeroing many FFT bins). The iFFT is then taken to create the impulse response of the desired length and the result is then multiplied by a window function (see Kaiser, Von Hann, Hamming, etc). This impulse response is the coefficients for your FIR filter. However, since multiplication in the time domain is synonymous with convolution in the frequency domain, you can take your original FFT bin values, and convolve them with the frequency response of the desired window function. You should be able to adapt this to a overlap-add implementation . I'm pretty sure this works correctly, but you should investigate and work it out for yourself. This is just coming off the top of my head and may have oversights or just plain gross errors... Jacob