On Sat, 15 Dec 2012 23:51:54 -0800, Mac Decman <dearman.mark@gmail.com> wrote:><snip>PPS. Same as above picture. Sweep modulation changed so same IF will require 30kHz of bandwidth. http://foxy.cascadeacoustic.com/worksheet1.jpg Mark DeArman
Nyquist For Chirps
Started by ●December 15, 2012
Reply by ●December 16, 20122012-12-16
Reply by ●December 16, 20122012-12-16
Phase or frequency modulation is nonlinear. Nyquist stops here... For low modulation indices (=> slow chirp), looking at only the first Bessel zone is a good approximation. But if you wobble the frequency too "aggressively", higher order terms with a wider bandwidth will pop up in the spectrum.
Reply by ●December 16, 20122012-12-16
On Saturday, December 15, 2012 4:30:47 AM UTC-8, John Harkes wrote:> So when sampling a signal Nyquist (assuming first Nyquist zone) says you get half your clock rate. I have always followed the rule that you get closer to 40% of your clock rate. My question is what about chirps or swept FM. > > > > Lets assume you chirp from 100MHz to 500MHz then I would want to run my A/D at 1.25GHz to swallow this signal. If I chirp from 100MHz to 500MHz in 100ns that is a very aggressive sweep rate, but everything works OK. If I change my sweep period from 100ns to 10ns then my demodulator starts to break down and everything becomes very distorted. I can fix everything by increasing the A/D clock rate. That makes me think this is an issue where I am not meeting some sampling criteria for my sweep rate even though I am able to sample a 500MHz signal with a 1.25GHz A/D. This is all done in Matlab so it is not a case of hardware that cannot keep up. If there is a rule of thumb for sampling rate and sweep rate I would love to know what it is. Any help would be greatly appreciated. > > > > Thanks.First I would like to say thanks for all the respones. Let me state my question another way and see what is said. Lets assume the 5GHz sample rate and a 100ns sweep and I sweep from 0 to 100MHz. That gives me 500 samples to represent 100MHz of bandwidth or 200KHz per sample. Now keep the clock and sweep period the same but now lets sweep from 0 to 2GHz. I still have 500 samples but now must represent 2GHz of spectrum with those points or 4MHz per sample. If I change the sweep from 100ns to 10ns then I get only 50 points to represent 2GHz of spectrum or 40MHz per point. At some point the signal has to break down and I would think there is a rule to explain it.
Reply by ●December 16, 20122012-12-16
On Sun, 16 Dec 2012 08:42:12 -0800 (PST), John Harkes <jharkes20653@gmail.com> wrote:>On Saturday, December 15, 2012 4:30:47 AM UTC-8, John Harkes wrote: >> So when sampling a signal Nyquist (assuming first Nyquist zone) says you = >get half your clock rate. I have always followed the rule that you get clo= >ser to 40% of your clock rate. My question is what about chirps or swept F= >M. >>=20 >>=20 >>=20 >> Lets assume you chirp from 100MHz to 500MHz then I would want to run my A= >/D at 1.25GHz to swallow this signal. If I chirp from 100MHz to 500MHz in = >100ns that is a very aggressive sweep rate, but everything works OK. If I = >change my sweep period from 100ns to 10ns then my demodulator starts to bre= >ak down and everything becomes very distorted. I can fix everything by inc= >reasing the A/D clock rate. That makes me think this is an issue where I a= >m not meeting some sampling criteria for my sweep rate even though I am abl= >e to sample a 500MHz signal with a 1.25GHz A/D. This is all done in Matlab= > so it is not a case of hardware that cannot keep up. If there is a rule o= >f thumb for sampling rate and sweep rate I would love to know what it is. = >Any help would be greatly appreciated. >>=20 >>=20 >>=20 >> Thanks. > >First I would like to say thanks for all the respones. Let me state my que= >stion another way and see what is said. Lets assume the 5GHz sample rate a= >nd a 100ns sweep and I sweep from 0 to 100MHz. That gives me 500 samples t= >o represent 100MHz of bandwidth or 200KHz per sample. Now keep the clock a= >nd sweep period the same but now lets sweep from 0 to 2GHz. I still have 5= >00 samples but now must represent 2GHz of spectrum with those points or 4MH= >z per sample. If I change the sweep from 100ns to 10ns then I get only 50 = >points to represent 2GHz of spectrum or 40MHz per point. At some point the= > signal has to break down and I would think there is a rule to explain it.A relevant topic to study is the Time-Bandwidth product of linear FM pulses in the context of radar systems, or Pulse Compression Ratio for linear FM. There are tradeoffs with respect to bandwidth and pulse length, and they're usually discussed in the radar contexts mentioned above. As has been mentioned by other posters, sidelobe control comes into play among other things. Eric Jacobsen Anchor Hill Communications http://www.anchorhill.com
Reply by ●December 16, 20122012-12-16
On Sun, 16 Dec 2012 08:42:12 -0800, John Harkes wrote:> On Saturday, December 15, 2012 4:30:47 AM UTC-8, John Harkes wrote: >> So when sampling a signal Nyquist (assuming first Nyquist zone) says >> you get half your clock rate. I have always followed the rule that you >> get closer to 40% of your clock rate. My question is what about chirps >> or swept FM. >> >> >> >> Lets assume you chirp from 100MHz to 500MHz then I would want to run my >> A/D at 1.25GHz to swallow this signal. If I chirp from 100MHz to >> 500MHz in 100ns that is a very aggressive sweep rate, but everything >> works OK. If I change my sweep period from 100ns to 10ns then my >> demodulator starts to break down and everything becomes very distorted. >> I can fix everything by increasing the A/D clock rate. That makes me >> think this is an issue where I am not meeting some sampling criteria >> for my sweep rate even though I am able to sample a 500MHz signal with >> a 1.25GHz A/D. This is all done in Matlab so it is not a case of >> hardware that cannot keep up. If there is a rule of thumb for sampling >> rate and sweep rate I would love to know what it is. Any help would be >> greatly appreciated. >> >> >> >> Thanks. > > First I would like to say thanks for all the respones. Let me state my > question another way and see what is said. Lets assume the 5GHz sample > rate and a 100ns sweep and I sweep from 0 to 100MHz. That gives me 500 > samples to represent 100MHz of bandwidth or 200KHz per sample. Now keep > the clock and sweep period the same but now lets sweep from 0 to 2GHz. > I still have 500 samples but now must represent 2GHz of spectrum with > those points or 4MHz per sample. If I change the sweep from 100ns to > 10ns then I get only 50 points to represent 2GHz of spectrum or 40MHz > per point. At some point the signal has to break down and I would think > there is a rule to explain it.Well, not so much just one rule as a bunch of useful methods. What you are missing is that the spectrum of a signal that sweeps from DC to 100MHz is _not_ 100MHz. That's part of the "Nyquist didn't say that" thing: because the frequency is sweeping, the signal spectrum is wider than just the difference between the highest and lowest frequency. The rule of thumb for FM radio is that the total bandwidth of a signal is twice the frequency deviation plus twice the bandwidth of the modulating signal. So if you have 100kHz deviation and a 20kHz signal, the resulting signal out there in radio-land is approximately 240kHz wide (given that it's FM it pretty much needs to be approximate, because FM is a nonlinear modulation technique). For baseband signals you halve the bandwidth (because you're only considering from 0Hz to whatever), so in your frequency ramp case you can probably do a first-order cut of the spectral content by considering the spectral content of a 10ns-long ramp, figuring a bandwidth, then adding that to your 100MHz. The most useful method is probably still the Nyquist rate, but it is the Nyquist rate given the useful bandwidth of the _whole_ signal, not the presumed (but incorrect) "bandwidth" found by taking the highest instantaneous frequency. -- My liberal friends think I'm a conservative kook. My conservative friends think I'm a liberal kook. Why am I not happy that they have found common ground? Tim Wescott, Communications, Control, Circuits & Software http://www.wescottdesign.com
Reply by ●December 17, 20122012-12-17
On Saturday, December 15, 2012 4:30:47 AM UTC-8, John Harkes wrote: ...> My question is what about chirps or swept FM....> If there is a rule of thumb for sampling rate and sweep rate I would love to know what it is. Any help would be greatly appreciated. > > Thanks.On Sunday, December 16, 2012 8:42:12 AM UTC-8, John Harkes wrote:> On Saturday, December 15, 2012 4:30:47 AM UTC-8, John Harkes wrote:...> First I would like to say thanks for all the respones. Let me state my question another way and see what is said. Lets assume the 5GHz sample rate and a 100ns sweep and I sweep from 0 to 100MHz. That gives me 500 samples to represent 100MHz of bandwidth or 200KHz per sample. Now keep the clock and sweep period the same but now lets sweep from 0 to 2GHz. I still have 500 samples but now must represent 2GHz of spectrum with those points or 4MHz per sample. If I change the sweep from 100ns to 10ns then I get only 50 points to represent 2GHz of spectrum or 40MHz per point. At some point the signal has to break down and I would think there is a rule to explain it.John If you have failed to find a coherent answer it is because you have failed to ask a coherent question. If you generate a signal following some rules, you get a signal that follows the rules. What does "break down" mean? Are you generating a signal to be detected? to measure time delay? to modulate with information to be transmitted, received and demodulated? Those are things that might break down as you change signal characteristics. So, what work are you trying to do so that we know what "works" and "breaks down" mean? In your second post you suggest Fs = 5GHs for 3 sweeps: 1) 100nSec, 0-100Mhz gives 5 cycles in 500 samples 2) 100nSec, 0-2GHz gives 100 cycles in 500 samples 3) 10nSec, 0-2Ghz gives 10 cycles in 50samples What are you trying to do with these that gives meaning to something "breaks down"? Dale B. Dalrymple
Reply by ●December 17, 20122012-12-17
On Sun, 16 Dec 2012 06:26:01 -0600, "mnentwig" <24789@dsprelated> wrote:>Phase or frequency modulation is nonlinear. Nyquist stops here... > >For low modulation indices (=> slow chirp), looking at only the first >Bessel zone is a good approximation. But if you wobble the frequency too >"aggressively", higher order terms with a wider bandwidth will pop up in >the spectrum.Thanks for the reply, that was my understanding. I would love to find some resources, I'll have to check IEEE, for proofs for different types of modulation. Mark DeArman
Reply by ●December 17, 20122012-12-17
This might be of interest, even though it's the other way round (design a frequency-modulated waveform for a given power spectrum, that is, to suppress "sidelobes"). http://www.ll.mit.edu/mission/aviation/publications/publication-files/atc-reports/Labitt_1994_ATC-223_WW-15318.pdf One key word to look out for is "Key Fowle Haggarty" (KFH) "A Method of Sidelobe Suppression in Phase-Coded Pulse Compression Systems"
Reply by ●December 17, 20122012-12-17
On Dec 15, 2:27�pm, Tim Wescott <t...@seemywebsite.please> wrote:> > I'm fairly sure that a chirp with a nice linear frequency ramp has a > closed-form Fourier transform. �I'm sure that if it does, it involves > Bessel functions. �The older I get (and the better computers get at > numerical calculations), the more I translate "has Bessel functions" to > mean, in a practical sense "run away, now".Hi Tim, Actual an LFM pulse does not have a closed form Fourier Transform. For pulses which have a high enough time-bandwidth product - you can use the Principle of Stationary Phase to get an approximation for the Fourier Transform. In those cases it turns out to be an LFM chirp in the Frequency domain. The POSP basically says that the fast oscillations cancel themselves out and the main contribution for the Fourier Integral comes from the stationary point. The other assumption you'll see with the POSP is that the envelope is slowly varying i.e. s(t)=A(t)exp(jw0t +jBt^2) , where A(t) is real then A(t)) is the envelope. For an LFM pulse with a low time bandwidth product (small chirp rates) I don't believe anyone has come up with a good approximation - at least not that I'm aware of. Details of this type of analysis appear in Papoulis' "Signal Analysis", or "The Fourier Integral and Its Applications" Cheers, Dave
Reply by ●December 17, 20122012-12-17
On Saturday, December 15, 2012 5:30:47 AM UTC-7, John Harkes wrote: <snip> Turn off your Matlab for a while and use your brain. 8^) You are mistaken in thinking that the band limits of a LFM chirp is the same as the starting and stopping frequencies. (Don't feel bad--the guy who invented FM thought the same thing.) You want to know the Fourier transform of a chirp. You haven't mentioned any window on the chirp so I'll assume rectangular. If you know how, write an equation for the Fourier transform of the chirp. You'll say, "I don't know how to evaluate that." So then look it up in a table of integrals. You will find that the FT of the chirp is a Fresnel integral. (Yes, the optics guy Fresnel--the math is the same as near field lens optics.) You will see that the time-bandwidth product (duration of the pulse times the difference between start and stop frequencies) is a parameter of the Fresnel integral. Now, turn on Matlab again and plot Fresnel integrals as a function of frequency with time-bandwidth product as a parameter and you will see the true spectrum of your chirp. From this you can pick a point "far enough" down on the spectrum to set your half-sampling frequency. Hint--for slow chirps the spectrum approximates to a rectangular pulse in frequency with the edges at the start and stop frequencies, but you are talking about very fast chirps thus your question. Jerry
