If I have sampled a signal at say fs=1MHz sampling rate and I capture N=100 samples of that signal, my resolution bandwidth in frequency domain will be fs/N = 10kHz. 100 samples correspond to 100us of capture. If sampling frequency is doubled and if the capture duration is also doubled, I would expect the resolution bandwidth also to increase by a factor of x. But it is not the case. Resolution bandwidth remains the same. Why is that? What is the intuition behind this?
FFT Resolution bandwidth
Started by ●March 3, 2008
Reply by ●March 3, 20082008-03-03
Sumeer,
The resolution should change. If it is not happening
then we have to see the reason why?
case 1 : Fs = 1mhz
duration = 100 us (N = 100)
resolution = 10khz
case 2 : Fs = 2mhz
duration = 200 us (N = 400)
resolution = 2mhz/400 = 5khz
Hence the resolution in case 2 will be 5khz.
Regards
Bharat Pathak
Arithos Designs
www.Arithos.com
>If I have sampled a signal at say fs=1MHz sampling rate and I capture
N=100
>samples of that signal, my resolution bandwidth in frequency domain will
be
>fs/N = 10kHz.
>
>100 samples correspond to 100us of capture.
>
>If sampling frequency is doubled and if the capture duration is also
>doubled, I would expect the resolution bandwidth also to increase by a
>factor of x.
>
>But it is not the case. Resolution bandwidth remains the same. Why is
>that?
>What is the intuition behind this?
>
>
>
>
>
Reply by ●March 4, 20082008-03-04
"sumeer" <sumeerbhatara@yahoo.com> wrote in message news:TamdnZlrkr3aM1HanZ2dnUVZ_i2dnZ2d@giganews.com...> If I have sampled a signal at say fs=1MHz sampling rate and I capture > N=100 > samples of that signal, my resolution bandwidth in frequency domain will > be > fs/N = 10kHz. > > 100 samples correspond to 100us of capture. > > If sampling frequency is doubled and if the capture duration is also > doubled, I would expect the resolution bandwidth also to increase by a > factor of x. > > But it is not the case. Resolution bandwidth remains the same. Why is > that? > What is the intuition behind this?My intuition is that you made a mistake. Check these numbers out with relation to your application: Increase the sampling frequency by 2. Increase the capture duration by 2. This means you will have 2x2=4 times the samples. When transformed you will have 4 times the samples over 2 times the frequency so the resultion *must be* 2 times greater. Fred
Reply by ●March 4, 20082008-03-04
> >"sumeer" <sumeerbhatara@yahoo.com> wrote in message >news:TamdnZlrkr3aM1HanZ2dnUVZ_i2dnZ2d@giganews.com... >> If I have sampled a signal at say fs=1MHz sampling rate and I capture >> N=100 >> samples of that signal, my resolution bandwidth in frequency domainwill>> be >> fs/N = 10kHz. >> >> 100 samples correspond to 100us of capture. >> >> If sampling frequency is doubled and if the capture duration is also >> doubled, I would expect the resolution bandwidth also to increase by a >> factor of x. >> >> But it is not the case. Resolution bandwidth remains the same. Why is >> that? >> What is the intuition behind this? > >My intuition is that you made a mistake. Check these numbers out with >relation to your application: > >Increase the sampling frequency by 2. >Increase the capture duration by 2. >This means you will have 2x2=4 times the samples. >When transformed you will have 4 times the samples over 2 times the >frequency so the resultion *must be* 2 times greater. > >FredThanks Fred, Sorry for the mistake in my question. If I double fs and also double N, the resolution bandwidth remains the same. i.e. if I sample twice as fast and als capture double the number of samples, the resoltion bandwidth does not increase.
Reply by ●March 4, 20082008-03-04
"sumeer" <sumeerbhatara@yahoo.com> wrote in message news:zISdnWrK-Z_deFDanZ2dnUVZ_iydnZ2d@giganews.com...> > >>"sumeer" <sumeerbhatara@yahoo.com> wrote in message >>news:TamdnZlrkr3aM1HanZ2dnUVZ_i2dnZ2d@giganews.com... >>> If I have sampled a signal at say fs=1MHz sampling rate and I capture >>> N=100 >>> samples of that signal, my resolution bandwidth in frequency domain > will >>> be >>> fs/N = 10kHz. >>> >>> 100 samples correspond to 100us of capture. >>> >>> If sampling frequency is doubled and if the capture duration is also >>> doubled, I would expect the resolution bandwidth also to increase by a >>> factor of x. >>> >>> But it is not the case. Resolution bandwidth remains the same. Why is >>> that? >>> What is the intuition behind this? >> >>My intuition is that you made a mistake. Check these numbers out with >>relation to your application: >> >>Increase the sampling frequency by 2. >>Increase the capture duration by 2. >>This means you will have 2x2=4 times the samples. >>When transformed you will have 4 times the samples over 2 times the >>frequency so the resultion *must be* 2 times greater. >> >>Fred > > Thanks Fred, > > Sorry for the mistake in my question. > > If I double fs and also double N, the resolution bandwidth remains the > same. > > i.e. if I sample twice as fast and als capture double the number of > samples, the resoltion bandwidth does not increase. >Right. If you sample twice as fast and capture double the number of samples then the time span remains the same - and the frequency resolution remains the same accordingly. It is the time span NT that determines the frequency resolution, which is its reciprocal, 1/NT. Fred
Reply by ●March 4, 20082008-03-04
sumeer wrote:> ... if I sample twice as fast and als capture double the number of > samples, the resoltion bandwidth does not increase.Right. You're sampling for the same length of time. f = 1/T Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●March 4, 20082008-03-04
On Mar 4, 4:55 pm, Jerry Avins <j...@ieee.org> wrote:> sumeer wrote: > > ... if I sample twice as fast and als capture double the number of > > samples, the resoltion bandwidth does not increase.Only if you equate DFT bin separation with the term resolution bandwidth. For some signal models this might be appropriate.> Right. You're sampling for the same length of time. f = 1/TThe information in the extra samples represents potential higher frequency spectra now available in unaliased form due to the higher sample rate. IMHO. YMMV. -- rhn A.T nicholson d.0.t C-o-M
Reply by ●March 4, 20082008-03-04
Ron N. wrote:> On Mar 4, 4:55 pm, Jerry Avins <j...@ieee.org> wrote: >> sumeer wrote: >>> ... if I sample twice as fast and als capture double the number of >>> samples, the resoltion bandwidth does not increase. > > Only if you equate DFT bin separation with the term > resolution bandwidth. For some signal models this > might be appropriate. > >> Right. You're sampling for the same length of time. f = 1/T > > The information in the extra samples represents potential > higher frequency spectra now available in unaliased form > due to the higher sample rate.If the lower rate was adequate for sampling and the analog signal is unchanged, there is no higher frequency spectrum. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●March 4, 20082008-03-04
"Jerry Avins" <jya@ieee.org> wrote in message news:hIudnQde-fTIYVDanZ2dnUVZ_hudnZ2d@rcn.net...> Ron N. wrote: >> On Mar 4, 4:55 pm, Jerry Avins <j...@ieee.org> wrote: >>> sumeer wrote: >>>> ... if I sample twice as fast and als capture double the number of >>>> samples, the resoltion bandwidth does not increase. >> >> Only if you equate DFT bin separation with the term >> resolution bandwidth. For some signal models this >> might be appropriate. >> >>> Right. You're sampling for the same length of time. f = 1/T >> >> The information in the extra samples represents potential >> higher frequency spectra now available in unaliased form >> due to the higher sample rate. > > If the lower rate was adequate for sampling and the analog signal is > unchanged, there is no higher frequency spectrum. > > Jerry > --...er no *important* higher frequency spectral components. After all, this is real world it seems. I know of no other definition of "resolution".... so I bend to Jerry's treatment. I think the other term would be "bandwidth" or, in a narrower sense, "frequency extent". Fred
Reply by ●March 5, 20082008-03-05
On Mar 4, 5:50 pm, Jerry Avins <j...@ieee.org> wrote:> Ron N. wrote: > > On Mar 4, 4:55 pm, Jerry Avins <j...@ieee.org> wrote: > >> sumeer wrote: > >>> ... if I sample twice as fast and als capture double the number of > >>> samples, the resoltion bandwidth does not increase. > > > Only if you equate DFT bin separation with the term > > resolution bandwidth. For some signal models this > > might be appropriate. > > >> Right. You're sampling for the same length of time. f = 1/T > > > The information in the extra samples represents potential > > higher frequency spectra now available in unaliased form > > due to the higher sample rate. > > If the lower rate was adequate for sampling and the analog signal is > unchanged, there is no higher frequency spectrum.1) The OP did not specify such. 2) There are no perfectly bandlimited signals in nature, since they would be of infinite extent, so there is never no higher frequency spectra, only the issue of whether it is far enough below the desired noise floor.
