I was tempted to post this in sci.math, but you're all a clever bunch, anyway, aren't you? OK. Is there a theorem or method that enables me to find the zeros of any sampled waveform? Kind regards, Funky
zeros of a sampled waveform?
Started by ●February 16, 2004
Reply by ●February 16, 20042004-02-16
Funky wrote:> I was tempted to post this in sci.math, but you're all a clever bunch, > anyway, aren't you? > OK. Is there a theorem or method that enables me to find the zeros of any > sampled waveform? > Kind regards, > FunkyWhat are the restrictions on the waveform? Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●February 16, 20042004-02-16
Hello Funky, I've used Newton's iter to find zeros of waveforms. You just need a derivative of your waveform via your favorite method and just hunt until a consecutive pair of samples has a nonpositve product and use the nearer to zero (in a magnitude kind of way) sample's location as your seed in the Newton iter. If the system is a fair bit oversampled, then the derivative becomes quicker to find. I was able to find a peak of a sampled function to within 1/1000th of a sample time. Clay -- Clay S. Turner, V.P. Wireless Systems Engineering, Inc. Satellite Beach, Florida 32937 (321) 777-7889 www.wse.biz csturner@wse.biz "Funky" <hello@hello.com> wrote in message news:c0rge4$ph6$1@news8.svr.pol.co.uk...> I was tempted to post this in sci.math, but you're all a clever bunch, > anyway, aren't you? > OK. Is there a theorem or method that enables me to find the zeros of any > sampled waveform? > Kind regards, > Funky > >
Reply by ●February 16, 20042004-02-16
Clay S. Turner wrote:> Hello Funky, > > I've used Newton's iter to find zeros of waveforms. You just need a > derivative of your waveform via your favorite method and just hunt until a > consecutive pair of samples has a nonpositve product and use the nearer to > zero (in a magnitude kind of way) sample's location as your seed in the > Newton iter. If the system is a fair bit oversampled, then the derivative > becomes quicker to find. I was able to find a peak of a sampled function to > within 1/1000th of a sample time. > > ClayWhat's the need for derivatives as such? If, as you say, the waveform is a fair bit oversampled, linear interpolation between samples of opposite sign should give the zero crossings pretty accurately. If more accuracy is needed, second- or third-order interpolation might be useful in the absence of much noise. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●February 17, 20042004-02-17
"Jerry Avins" <jya@ieee.org> wrote in message news:40317e99$0$3081$61fed72c@news.rcn.com...> What's the need for derivatives as such? If, as you say, the waveform is > a fair bit oversampled, linear interpolation between samples of opposite > sign should give the zero crossings pretty accurately. If more accuracy > is needed, second- or third-order interpolation might be useful in the > absence of much noise.Hello Jerry, The advantages of the derivative is the quadratic convergence. Remember both f(n+x) and f'(n+x) are to be generated from just f[n] where x is in (-1/2, 1/2). Depending on how much accuracy he needs in the position of the zero, he can shortcut the process by cruder approximations. I just mentioned the way I did it for an extreme precision case. And since the interpolation was computationally expensive, a quadratic convergent algo was very effective. Linear interpolation sometimes had a hard time with 1/10 of a sample time[1] and I needed 2 orders of magnitude performance better than that. I was finding the TDOA of short duration radio bursts received at different locations. And I needed to be within 100nSec. The method is a success as it is being used by the police to locate their officers when they key their portable radios. They may not be in their car and so the car's having a GPS is helpful but not enough. Currently this is used in Brevard country Florida. And by slight oversampling I was referring to a case of where the sampling rate is only 3 times the bandwidth instead of where someone may wish to oversample at 10 times or more. Here part of the problem is the backhauling of data where minimizing its size is important. Clay [1] Sometimes the signals we were looking at could be down in the dirt at -110 dBm. If they were all at -65dBm, life would be easy, but we were operating vary close to the Cramer-Rao bound. -- Clay S. Turner, V.P. Wireless Systems Engineering, Inc. Satellite Beach, Florida 32937 (321) 777-7889 www.wse.biz csturner@wse.biz
Reply by ●February 17, 20042004-02-17
Clay S. Turner wrote:> "Jerry Avins" <jya@ieee.org> wrote in message > news:40317e99$0$3081$61fed72c@news.rcn.com... > > >>What's the need for derivatives as such? If, as you say, the waveform is >>a fair bit oversampled, linear interpolation between samples of opposite >>sign should give the zero crossings pretty accurately. If more accuracy >>is needed, second- or third-order interpolation might be useful in the >>absence of much noise. > > > Hello Jerry, > > The advantages of the derivative is the quadratic convergence. Remember both > f(n+x) and f'(n+x) are to be generated from just f[n] where x is in (-1/2, > 1/2). Depending on how much accuracy he needs in the position of the zero, > he can shortcut the process by cruder approximations. I just mentioned the > way I did it for an extreme precision case. And since the interpolation was > computationally expensive, a quadratic convergent algo was very effective. > Linear interpolation sometimes had a hard time with 1/10 of a sample time[1] > and I needed 2 orders of magnitude performance better than that. I was > finding the TDOA of short duration radio bursts received at different > locations. And I needed to be within 100nSec. The method is a success as it > is being used by the police to locate their officers when they key their > portable radios. They may not be in their car and so the car's having a GPS > is helpful but not enough. Currently this is used in Brevard country > Florida. And by slight oversampling I was referring to a case of where the > sampling rate is only 3 times the bandwidth instead of where someone may > wish to oversample at 10 times or more. Here part of the problem is the > backhauling of data where minimizing its size is important. > > Clay > > [1] Sometimes the signals we were looking at could be down in the dirt > at -110 dBm. If they were all at -65dBm, life would be easy, but we were > operating vary close to the Cramer-Rao bound.Clay, That's an impressive display of technology, to be bland about it. Given that there are only discrete points, I don't see off hand how to iterate to improve the estimate, unless one first does local band-limited interpolation. That's the most interesting abstract point I see. Note that I wrote "derivatives as such". A few points, say three, are used to calculate the derivative, then it and one of the points together locate the zero crossing. In the end, one has a function of three points that can be applied directly. With three points, I can also do quadratic interpolation without thinking about derivatives, and my hunch is that the two analytical approaches will yield the same function of three points. Thanks for the enlightenment. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●February 17, 20042004-02-17
"Jerry Avins" <jya@ieee.org> wrote in message news:4031512c$0$3081$61fed72c@news.rcn.com...> Funky wrote: > > > I was tempted to post this in sci.math, but you're all a clever bunch, > > anyway, aren't you? > > OK. Is there a theorem or method that enables me to find the zeros ofany> > sampled waveform? > > Kind regards, > > Funky > > What are the restrictions on the waveform?None, except the obvious that it's band limited and sampled at the Nyquist frequency. Funky.> Jerry > -- > Engineering is the art of making what you want from things you can get. > ����������������������������������������������������������������������� >
Reply by ●February 17, 20042004-02-17
"Clay S. Turner" <CSTurner@WSE.Biz> wrote in message news:mjcYb.8852$fE4.6151@bignews5.bellsouth.net...> Hello Funky, > > I've used Newton's iter to find zeros of waveforms. You just need a > derivative of your waveform via your favorite method and just hunt until a > consecutive pair of samples has a nonpositve product and use the nearer to > zero (in a magnitude kind of way) sample's location as your seed in the > Newton iter. If the system is a fair bit oversampled, then the derivative > becomes quicker to find. I was able to find a peak of a sampled functionto> within 1/1000th of a sample time. > > ClayI was thinking more on the lines of an unknown, bandlimited, Nyquist sampled waveform. Funky> Clay S. Turner, V.P. > Wireless Systems Engineering, Inc. > Satellite Beach, Florida 32937 > (321) 777-7889 > www.wse.biz > csturner@wse.biz > > > > "Funky" <hello@hello.com> wrote in message > news:c0rge4$ph6$1@news8.svr.pol.co.uk... > > I was tempted to post this in sci.math, but you're all a clever bunch, > > anyway, aren't you? > > OK. Is there a theorem or method that enables me to find the zeros ofany> > sampled waveform? > > Kind regards, > > Funky > > > > > >
Reply by ●February 18, 20042004-02-18
"Funky" <hello@hello.com> wrote in message news:c0u29n$kpq$1@newsg3.svr.pol.co.uk...> I was thinking more on the lines of an unknown, bandlimited, Nyquistsampled> waveform. > Funky > >Hello Funky, That's the type of waveform I did this on. Even when I found the peak of the autocorrelation function, it is done similarly, except Newton's iter now uses the 1st and 2nd derivatives instead of just the original function and its 1st derivative. How accurately do you need to find the positions of the zeroes? Clay -- Clay S. Turner, V.P. Wireless Systems Engineering, Inc. Satellite Beach, Florida 32937 (321) 777-7889 www.wse.biz csturner@wse.biz
Reply by ●February 18, 20042004-02-18
"Clay S. Turner" <CSTurner@WSE.Biz> wrote in message news:%hOYb.100$ld7.37@bignews3.bellsouth.net...> "Funky" <hello@hello.com> wrote in message > news:c0u29n$kpq$1@newsg3.svr.pol.co.uk... > > > I was thinking more on the lines of an unknown, bandlimited, Nyquist > sampled > > waveform. > > Funky > > > > > > Hello Funky, > > That's the type of waveform I did this on. Even when I found the peak ofthe> autocorrelation function, it is done similarly, except Newton's iter now > uses the 1st and 2nd derivatives instead of just the original function and > its 1st derivative. How accurately do you need to find the positions ofthe> zeroes?I want it to be as accurate as I please. I'm interested in alternative methods to interpolation. Never came across the Newton iter method before, but I'll be googling for the next day or so... Funky> Clay > > -- > Clay S. Turner, V.P. > Wireless Systems Engineering, Inc. > Satellite Beach, Florida 32937 > (321) 777-7889 > www.wse.biz > csturner@wse.biz > > > >
