This is probably a FAQ, but I looked for a while and couldn't find an answer. Apologies in advance. What is the "sweet spot" in a spectrum analysis waterfall display, for generating or decoding a narrower signal that's embedded within it? Let's say I'm sampling at 48 kHz, and there's a signal in there that's 10 kHz wide. Where shall I place this signal? If it's too far left (lower frequency), there will be artifacts as it approaches DC. If it's too far right (higher frequency), there will also be artifacts, as it approaches the Nyquist frequency.of 24 kHz. I'm sure there's a good rule of thumb to follow here... any enlightenment would be appreciated. Thank you! Josh Lehan
Question about spectrum analysis and sampling theory
Started by ●July 2, 2011
Reply by ●July 2, 20112011-07-02
On Sat, 2 Jul 2011 14:04:09 -0700 (PDT), Krellan <krellan@gmail.com> wrote:>This is probably a FAQ, but I looked for a while and couldn't find an >answer. Apologies in advance. > >What is the "sweet spot" in a spectrum analysis waterfall display, for >generating or decoding a narrower signal that's embedded within it? > >Let's say I'm sampling at 48 kHz, and there's a signal in there that's >10 kHz wide. Where shall I place this signal? > >If it's too far left (lower frequency), there will be artifacts as it >approaches DC. > >If it's too far right (higher frequency), there will also be >artifacts, as it approaches the Nyquist frequency.of 24 kHz. > >I'm sure there's a good rule of thumb to follow here... any >enlightenment would be appreciated. > >Thank you! > >Josh LehanWhat artifacts are you talking about? Numerically (i.e., mathematically) all of the frequencies within the supported Nyquist region have essentially the same properties (if the transform is linear, which it typically is for that purpose). If there's a non-rectangular window applied to the transforms that are creating the waterfall display then there may be some attenuation that will be a function of the window shape. Eric Jacobsen http://www.ericjacobsen.org http://www.dsprelated.com/blogs-1//Eric_Jacobsen.php
Reply by ●July 2, 20112011-07-02
On 07/02/2011 05:04 PM, Krellan wrote:> This is probably a FAQ, but I looked for a while and couldn't find an > answer. Apologies in advance. > > What is the "sweet spot" in a spectrum analysis waterfall display, for > generating or decoding a narrower signal that's embedded within it?Whoa. Are you observing the signal (spectrum analysis) or generating it? Those are two totally different things. -- Randy Yates % "Watching all the days go by... Digital Signal Labs % Who are you and who am I?" mailto://yates@ieee.org % 'Mission (A World Record)', http://www.digitalsignallabs.com % *A New World Record*, ELO
Reply by ●July 3, 20112011-07-03
On Jul 2, 5:04=A0pm, Krellan <krel...@gmail.com> wrote:> This is probably a FAQ, but I looked for a while and couldn't find an > answer. =A0Apologies in advance. > > What is the "sweet spot" in a spectrum analysis waterfall display, for > generating or decoding a narrower signal that's embedded within it?i am just totally unfamiliar with this use of the semantics. i know what spectral analysis is (or examples of it) and i know what a waterfall display is, but i have never heard of *the* sweet spot or even *a* "sweet spot" regarding either spectral analysis or waterfall display.> Let's say I'm sampling at 48 kHz, and there's a signal in there that's > 10 kHz wide. =A0Where shall I place this signal?"place" the signal? what do you mean? in time? in frequency? place it wherever it turns out to be.> If it's too far left (lower frequency), there will be artifacts as it > approaches DC.are you inquiring about modulating a baseband signal that is 10 kHz wide (is that the double-sided bandwidth?) along the spectrum between 0 and 24 kHz? if 10 kHz is the single-sided bandwidth, then you could theoretically center it anywhere from 5 kHz to 19 kHz without any foldover or aliasing.> If it's too far right (higher frequency), there will also be > artifacts, as it approaches the Nyquist frequency.of 24 kHz. > > I'm sure there's a good rule of thumb to follow here... any > enlightenment would be appreciated.well, i am not sure what it is that you're talking about. i think it's a semantic issue. r b-j
Reply by ●July 3, 20112011-07-03
On 7/2/2011 2:04 PM, Krellan wrote:> This is probably a FAQ, but I looked for a while and couldn't find an > answer. Apologies in advance. > > What is the "sweet spot" in a spectrum analysis waterfall display, for > generating or decoding a narrower signal that's embedded within it? > > Let's say I'm sampling at 48 kHz, and there's a signal in there that's > 10 kHz wide. Where shall I place this signal? > > If it's too far left (lower frequency), there will be artifacts as it > approaches DC. > > If it's too far right (higher frequency), there will also be > artifacts, as it approaches the Nyquist frequency.of 24 kHz. > > I'm sure there's a good rule of thumb to follow here... any > enlightenment would be appreciated. > > Thank you! > > Josh LehanThe question doesn't make complete sense to me. But I'll try: "Generating or decoding a narrower signal that's embedded in "it"? Do you mean that's part of the original signal? I will assume so. I think a big part of the issue and the answer lies in: > Let's say I'm sampling at 48 kHz, and there's a signal in there that's > 10 kHz wide. Where shall I place this signal? I don't know what "place" means but there is a big part of the situation unmentioned: What is the size of the temporal window???? Let's assume that you want the 10kHz signal to show up in a single bin of the waterfall. In that case we divide 48kHz by 10 kHz to get 4.8 bins - so let's say 5 bins. So, N=5 will resolve the 10 kHz. That would be a 5 X 1/48,000 = 0.104 msec time window. Somehow I doubt all this and figure there must be a typo ..... like maybe 10Hz instead of 10kHz? That would result in 4800 bins .. maybe a lot but more like it, eh? Then a time window of 0.1 sec. You say that the signal of interest is 10kHz "wide" but don't say what the center frequency is. It's the center frequency that "places" it in the spectrum unless you have an analyzer that grabs a sub-band somehow. I seem to recall things like this that some called called "super vernier" but that's just jargon. Looks like a spectrum analyzer on a ham radio receiver display for a single band or band segment. Dale will know no doubt - but you need to define things better. Fred
Reply by ●July 3, 20112011-07-03
"Eric Jacobsen" <eric.jacobsen@ieee.org> wrote in message news:4e0f8bce.89268966@www.eternal-september.org...> On Sat, 2 Jul 2011 14:04:09 -0700 (PDT), Krellan <krellan@gmail.com> > wrote: > >>This is probably a FAQ, but I looked for a while and couldn't find an >>answer. Apologies in advance. >> >>What is the "sweet spot" in a spectrum analysis waterfall display, for >>generating or decoding a narrower signal that's embedded within it? >> >>Let's say I'm sampling at 48 kHz, and there's a signal in there that's >>10 kHz wide. Where shall I place this signal? >> >>If it's too far left (lower frequency), there will be artifacts as it >>approaches DC. >> >>If it's too far right (higher frequency), there will also be >>artifacts, as it approaches the Nyquist frequency.of 24 kHz. >> >>I'm sure there's a good rule of thumb to follow here... any >>enlightenment would be appreciated.If possible, I would centre the 10 KHz wide signal around 12 KHz i.e. from 7 KHz to 17 KHz.
Reply by ●July 4, 20112011-07-04
On Jul 2, 11:04=A0pm, Krellan <krel...@gmail.com> wrote:> This is probably a FAQ, but I looked for a while and couldn't find an > answer. =A0Apologies in advance. > > What is the "sweet spot" in a spectrum analysis waterfall display, for > generating or decoding a narrower signal that's embedded within it?...> I'm sure there's a good rule of thumb to follow here... any > enlightenment would be appreciated.I think you might want to contemplate the task: 'Analysis' is about extracting whatever information is contained within the data as given; *not* about manipulating the data to give a desired answer, image or result. [ And yes, do contemplate this. This is one of those cases that might sound sarcastic up front, but where the nuances really are that fine and essential. ] If you are given a signal and want to produce a waterfall display, then you produce the waterfall display. There is, at the outset, little or nothing you can do to modify the appearance of the display, given the data. If you objective is technical analysis. If, on the other hand, you are doing some esthetical visualization of a music waveform, where details don't matter and you 'prettify' the data anyway, you are free to do whatever you want. Rune
Reply by ●July 4, 20112011-07-04
On Saturday, July 2, 2011 2:24:29 PM UTC-7, Eric Jacobsen wrote:> What artifacts are you talking about? Numerically (i.e., > mathematically) all of the frequencies within the supported Nyquist > region have essentially the same properties (if the transform is > linear, which it typically is for that purpose). If there's a > non-rectangular window applied to the transforms that are creating the > waterfall display then there may be some attenuation that will be a > function of the window shape.Thanks. I probably used the wrong terminology. Digital sampling loses pre= cision as you approach the Nyquist frequency, degenerating completely to ju= st square waves once you reach the Nyquist frequency. Let's say I'm sampli= ng a signal that is 1 kHz wide, and I'm sampling at 48 kHz, giving me a tot= al range of 0 to 24 kHz to work in (Nyquist frequency of 24 kHz). If the s= ignal of interest is from 23 kHz to 24 kHz within my sampling, then it won'= t be as precise as if it is from somewhere in the middle, say 12 kHz to 13 = kHz. That's what I meant by "artifacts". Josh
Reply by ●July 4, 20112011-07-04
On Sunday, July 3, 2011 1:18:18 PM UTC-7, Andrew Holme wrote:> If possible, I would centre the 10 KHz wide signal around 12 KHz i.e. from 7 > KHz to 17 KHz.Thanks! 12 kHz is half of the Nyquist frequency in this case (I'm sampling at 48 kHz, so my useful range is from 0 to 24 kHz). In other words, if I'm using a waterfall display, this puts my signal of interest at dead center within the waterfall. I wonder what the derivation for this is, or is it just a good "rule of thumb"? Josh
Reply by ●July 4, 20112011-07-04
On 07/04/2011 04:05 PM, JoSH Lehan wrote:> On Saturday, July 2, 2011 2:24:29 PM UTC-7, Eric Jacobsen wrote: >> What artifacts are you talking about? Numerically (i.e., >> mathematically) all of the frequencies within the supported Nyquist >> region have essentially the same properties (if the transform is >> linear, which it typically is for that purpose). If there's a >> non-rectangular window applied to the transforms that are creating the >> waterfall display then there may be some attenuation that will be a >> function of the window shape.> Thanks. I probably used the wrong terminology. Digital sampling > loses precision as you approach the Nyquist frequency, degenerating > completely to just square waves once you reach the Nyquist > frequency. Let's say I'm sampling a signal that is 1 kHz wide, and > I'm sampling at 48 kHz, giving me a total range of 0 to 24 kHz to > work in (Nyquist frequency of 24 kHz). If the signal of interest is > from 23 kHz to 24 kHz within my sampling, then it won't be as > precise as if it is from somewhere in the middle, say 12 kHz to 13 > kHz. That's what I meant by "artifacts".Josh, Theoretically speaking, you're incorrect. There is absolutely no degradation all the way up to (but not including) 24 kHz when sampling at 48 kHz. Now the practical side of things is a different matter... Perhaps it would be more fruitful if you told us what you need to accomplish? Then we could suggest some approaches. -- Randy Yates % "Watching all the days go by... Digital Signal Labs % Who are you and who am I?" mailto://yates@ieee.org % 'Mission (A World Record)', http://www.digitalsignallabs.com % *A New World Record*, ELO