Mathematically based method to increase bandwidth of ADC

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Wel, you could digitize the signal first, do the calculations in the digital domain, convert it back to analog and then digitize it following your scheme ;o)

Seriously, Nyquist states that the sample rate should be at least twice the signal's _bandwidth_ (which is equivalent to saying "highest frequency" if you start at DC). This means that you can undersample bandlimited signals that do not go down too much, i.e. if there is empty space in the spectrum where the alised signal can go without interfering with anything. In many applications where high frequency signals have to be sampled these signals do not go down to DC. If they do, cut up your signal (DC to f_max) in two or more parts with (band-pass) filters and (under)sample their outputs. Apply correction to taste in the digital domain.

--DF

Heres a short write up of the formulas involved

heh, yeah... that would work! ;)

I think this is effectively similar to what I'm talking about but not formulated in a precise mathematical model. (Although it might be exactly equivilent to what I am doing).

I don't know much about undersampling but I don't see how you could recover higher frequencies. If the ADC has a bandwidth of O then you are completely screwed no matter how you sample if your signal has a higher bandwidth(here the bandwidth of the ADC means the largest bandwidth it can sample properly).

In effect though I think your undersampling is basically what I'm doing but in the time domain by multiplying by the shift factor (exp(I*a*t)*s(t) hat(s)(w - a)).

I think what you are talking about is pretty much exactly what I'm doing though except I didn't realize that you could undersample a bandlimited signal yet recover the actual signal. The difference here is that your signal has to have "room" in the bandwidth to do this while my method doesn't rely on it(but basically same principle).

i.e. If, say, we have a signal that has a bandwidth between [a, a + 0] (0 here is omega and not 0... maybe not a good choice) then if a = 0 and we sample the signal with an ADC that has bandwidth [zero, Omega] we can recover the original signal by digitally shifting the frequency domain by omega. I think this is what you are talking about and is basically what i'm doing except I'm shifting the original signal's frequency domain too instead of leaving it fixed.

Jon

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You need room to shift spectra around no matter how you do it. Without room you will get interference and aliasing. If you shift a low-pass spectrum down you will get problems with the low frequencies that fold back, i.e. aliasing.

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Undersampling a band-pass signal effectively shifts its spectrum down. But you don't need fancy math to do it, just slowing down the clock.

--DF

For calculation speed use an fpga. (or dsp?)

National have a 3 Gsps chip (which uses two A/D in parallell).

There's no free lunch. While your method might shift the frequencies, that's a BAD thing, as it will break all the phase relationships. Plus your high frequencies will get added to any low frequencies in the signal. And you'll get massive multi-pass aliasing.

Looks like a good idea. You might be interested in the following notes on Digital Bandwidth Interleaving by Lecroy:

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ WaveExpert Series Oscilloscope SDA 100G, NRO 9000, WE 9000

The World's Fastest Oscilloscope

The WaveExpert and SDA 100G are the first instruments to combine the high bandwidth and accuracy of a sampling oscilloscope with the speed and flexibility of a real time instrument. These are the first products in the new instrument class called Near Real Time Oscilloscopes (NRO) which eliminate most of the constraints of traditional sampling scopes. The WaveExpert family features high acquisition speed, a responsive GUI and a powerful suite of analysis tools. Enabled by another new LeCroy technology - Accelerated Throughput Architecture (ATA), WaveExpert comes to market with up to 100 GHz bandwidth, signal acquisition speeds 100 times faster, and memory depths 125,000 times deeper than conventional sampling scopes. In addition to much higher throughput, LeCroy ATA allows for signal analysis algorithms that rival capabilities found in only real-time oscilloscopes.

Digital Bandwidth Interleaving Peter J. Pupalaikis Principal Technologist March 29, 2005

The author is Principal Technologist at LeCroy and a co-inventor of the Digital Bandwidth Interleave technology. He has held a variety of titles during his ten year career at LeCroy including digital signal processing engineer and product marketing manager for high performance oscilloscopes. He holds a BSEE from Rutgers University and is a member of Tau Beta Pi, Eta Kappa Nu and the IEEE communications and signal processing societies. He holds several patents in the area of the application of digital signal processing to measurement instruments.

New SDA 18000 18 GHz Serial Data Analyzer: 18 GHz, 60 GS/s, 24 Mpts in 1 Ch mode, 11 GHz, 40 GS/s, 16 Mpts + 2 x 6 GHz, 20 GS/s, 8 Mpts on 4 Ch, 6 GHz, 20 GS/s, 8 Mpts / Ch

Another new feature in the SDA 18000 is the Q-Scale analysis and plot view, an innovation that provides much more insight into jitter than any other method. Q-Scale analysis uses an alternative method of breaking down jitter that identifies subcomponents more accurately instead of classifying them as random jitter, which can happen using traditional methods. It also incorporates improvements in the method used in extrapolating the jitter histogram tails, improving the accuracy, stability and convergence time.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Regards,

Mike Monett

Thats not true. There is no theoretical phase change. If there was then I could not get the original function out of the mathematics... but yet I do.

Now there might be phase distortion due to the implementation though... but thats why I asked. If you cared to look at the pdf you would see the equations and would see how simple it is. BTW shifting frequencies do not change the phase relationship as it is constant. Also note that all filter produces some type of phase distortion so even though it might be a bad thing it happens all the time.

i.e.

given a function f(t), it has a fourier transform F(w) =

Now F(w) is, in general, a complex valued function... so

F(w) = |F(w)|*exp(I*arg(F(w))

Now you are saying there is a phase change so one would have

G(w) = |F(w)|*exp(I*(arg(F(w)) + h(w)) = F(w)*exp(I*h(w))

but taking the inverse fourier transform of gives

g(t) = 1/sqrt(2*Pi)*int(F(w)*exp(I*(w*t + h(w))),w=-infinity..infinity)

Surely if h(w) is not 0 then one cannot get g(t) = f(t).. and if there is such an h(w) then it doesn't matter because you get the exact same time dependent function... (which pretty much means that they must have the same phase relationships to ==> h(w) = 0).

So, the point being is that there is no phase change or it would have poped out because the methods are basically what is happenin above.

Actually, if you care to look at the pdf you will see that there is a phase change when going into the ADC... just like you said, yet it gets cancelled end the end.

So, sure, there is a phase change but it is only a temporary one. Now this might have some practical significance in reducing the quality but mathematically it is not important.

i.e., since the process is an approximation there might be significant degredation of the signal to make it impractical.

Thanks. I was interested in this at first as that was the most logical thing to do. I asked about it here before but a lot of people said it wasn't that good of an idea because its hard to implement.... I'll check it out though. Might learn something ;)

(Although it would be nice to know how practical my method is too ;)

Thanks again, Jon

I'm not sure what you mean but if you are saying to use something like that to do the circuitry I'm talking about then I think it probably won't work. These devices are bandlimited too and are digital? So you might as well just try and make a faster ADC?

i.e., it would be useless to compute the convolution digitally because you have to go through and ADC in the first place... (although maybe a mixture of analog and digital can work)

wow. thats a lot. But I bet its hard to use?

What my method is suppose to let you do, theoretically(if no flaws), is let you use any ADC's for *any* signal by creating a standarized front end to condition the signal. Theres a trade off of speed vs bandwidth though yet these could be solved by using more ADC's.

Oh, and I guess basically my method would require running ADC's in parallel which is somewhat similar to the interleaving. It might not be as effective as that though... because instead of dealing with the clock issue you have to worry about the delay of the frequency shifting circuitry. But my method would require much more complicated circuitry along with many more potential things that would introduce inaccuracies.

Okay, my phase intuition was wrong.

But I still think you're getting something for nothing, which is usually wrong.

How can a signal of 1/10th the bandwidth carry the same information?

You're giving up something, perhaps signal to noise ratio.

Does the slower ADC need X times the amplitude resolution? If so, you havent gained anything. Adding resolution to an A/D is harder than adding speed. Speed you can easily get with multiple A/D's and delay lines, that's a no-brainer.

And of course there's the chicken-egg problem-- to do the math requires the high speed signal in digital format, which is what you're trying to avoid. Ouch.

Of course it's hard to implement. That's why the LeCroy scope costs $150K or something like that.

John

On 13 Jul 2006 06:39:22 -0700, Ancient_Hacker wrote in Msg.

I think he wanted to use several slow ADCs.

robert

Right. But what I don't get is why:

Lecroy has a 100GHz bandwidth scope that only samples at 1GHz. And then they have a 1GHz bandwidth scope that samples at 10 GSa/s.

Why the disparity? Why not have the ultra high sample rates on the wide analog bandwidth unit?

I need to direct sample an X band radar signal (ie 10 GHz)-- but find that ADCs with analog bandwidths are only available to 3GHz. How to overcome the analog bandwidth limitations of the ADC? You can overcome the sampling rates by parallel conversions on phase-shifted clocks. But what is the work-around for _analog_ bandwidth limitations???

Bo

The first is an equivalent-time sampler [1] that takes only one sample each repetition of the signal, ie one sample per trigger, and reconstructs the waveform. The second has a really fast ADC and memory, and can acquire a full waveform from a single trigger.

An equivalent-time sampler has a fast diode-based sample-and-hold ahead of a very slow ADC.

John

When Tek designed the 7104 1 GHz analog scope, they considered splitting up the input into frequency bands, amplifying each, and recombinimg at the CRT. They opted for brute-force silicon plus other tricks, in the end. The story is in one of Jim Williams' books.

John

Splitting up and recombining bands sounds like something of a nightmare... I suppose that with the rather limited resolution of a 'scope it might work OK, but at least personally brute-force approach seems like a *much* better solution in terms of lowering technical risk.

No, it doesn't. You are splitting up one signal into several... each "subsignal" carries part of the bandwidth. The first carries the low frequencies, the second carries higher frequencies, etc... It would be just like filtering a signal into "bands"... but then all those signals are "mapped" into the same frequency range that can be handled by the ADC(s).

Nope. Ideally you are not giving up anything. Its very simple. For all practical purposes though it migh not be worth it because you have to approximate some things(such as using a finite convolution and finite kernel).

No, its resolution independent.

huh? The math doesn't require any thing from electronics. Mathematically it works perfectly and is based on a very simple concept(and is similar to the basics of wavelet theory using haar wavelets).