I think the worst case digital signal, in terms of resulting maximum amplitude,
for sinc reconstruction is the sequence:
...-1,+1,-1,+1,-1,+1,+1,-1,+1,-1,+1,-1...
The question I can't answer is what is the magnitude of the reconstruction
exactly between the two +1 samples. I think it comes down to:
+inf
(2/pi) * sum{ 1/(.5 + n) }
n=0
but forget how to evaluate that. Is that correct and if so can anyone help me
with its evaluation?
This isn't homework, it's incipient Alzheimers. :-)
Thanks,
Bob
--
"Things should be described as simply as possible, but no simpler."
A. Einstein
Worst Case Signal for Reconstruction
Started by ●June 23, 2003
Reply by ●June 23, 20032003-06-23
On Sun, 22 Jun 2003, Bob Cain wrote:> I think the worst case digital signal, in terms of resulting maximum > amplitude, for sinc reconstruction is the sequence: > > ...-1,+1,-1,+1,-1,+1,+1,-1,+1,-1,+1,-1...You are right and we have had this before: http://groups.google.com/groups?threadm=3D55D545.5FF8C854%40ieee.org The answer is, you get infinite peak amplitude from sinc reconstruction of that signal. -olli
Reply by ●June 23, 20032003-06-23
Olli Niemitalo wrote:> > On Sun, 22 Jun 2003, Bob Cain wrote: > > > I think the worst case digital signal, in terms of resulting maximum > > amplitude, for sinc reconstruction is the sequence: > > > > ...-1,+1,-1,+1,-1,+1,+1,-1,+1,-1,+1,-1... > > You are right and we have had this before: > > http://groups.google.com/groups?threadm=3D55D545.5FF8C854%40ieee.org > > The answer is, you get infinite peak amplitude from sinc reconstruction of > that signal. >Thanks, Olli. Not sure how I missed that one. Bizzare result. Bob -- "Things should be described as simply as possible, but no simpler." A. Einstein
Reply by ●June 23, 20032003-06-23
Bob Cain wrote:> > I think the worst case digital signal, in terms of resulting maximum amplitude, > for sinc reconstruction is the sequence: > > ...-1,+1,-1,+1,-1,+1,+1,-1,+1,-1,+1,-1... > > The question I can't answer is what is the magnitude of the reconstruction > exactly between the two +1 samples.Are you looking for amplitude of fundamental frequency component of meander?> I think it comes down to: > > +inf > (2/pi) * sum{ 1/(.5 + n) } > n=0 > > but forget how to evaluate that. Is that correct and if so can anyone help me > with its evaluation?That can't be correct because the sum diverges. Vladimir Vassilevsky, Ph.D. DSP and Mixed Signal Design Consultant http://www.abvolt.com
Reply by ●June 23, 20032003-06-23
Hello Bob, The problem you are talking about, critical sampling, has the neat feature that the perceived amplitude is dependent on the phase between the sampling clock and the extrema of the sinusoid. If they are in phase, then the following link shows how to reconsitute the cosine function. http://personal.atl.bellsouth.net/p/h/physics/nyquist.pdf [1] If the sampling occured at the zeroes of the function, then the amplitude info is lost. For other phase shifts, there will be a cosine(phase shift) factor that has to be accounted for. Clay [1] This was writtern in response to a question posed by R B-J last year, and I posted a simple outline of a proof. Together we worked up this detailed version, and would like to give him credit here for that. "Bob Cain" <arcane@arcanemethods.com> wrote in message news:3EF68911.FEB41BAD@arcanemethods.com...> I think the worst case digital signal, in terms of resulting maximumamplitude,> for sinc reconstruction is the sequence: > > ...-1,+1,-1,+1,-1,+1,+1,-1,+1,-1,+1,-1... > > The question I can't answer is what is the magnitude of the reconstruction > exactly between the two +1 samples. I think it comes down to: > > +inf > (2/pi) * sum{ 1/(.5 + n) } > n=0 > > but forget how to evaluate that. Is that correct and if so can anyonehelp me> with its evaluation? > > This isn't homework, it's incipient Alzheimers. :-) > > > Thanks, > > Bob > -- > > "Things should be described as simply as possible, but no simpler." > > A. Einstein
Reply by ●June 23, 20032003-06-23
Vladimir Vassilevsky wrote:> > Bob Cain wrote: > >...> > > I think it comes down to: > > > > +inf > > (2/pi) * sum{ 1/(.5 + n) } > > n=0 > > > > but forget how to evaluate that. Is that correct and if so can anyone help me > > with its evaluation? > > That can't be correct because the sum diverges. >I don't think it diverges. +inf sum(1/n) n=0 diverges, but barely. Anything even marginally smaller does not. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●June 23, 20032003-06-23
Jerry Avins wrote:> > Vladimir Vassilevsky wrote: > > > > Bob Cain wrote: > > > > ... > > > > > I think it comes down to: > > > > > > +inf > > > (2/pi) * sum{ 1/(.5 + n) } > > > n=0 > > > > > > but forget how to evaluate that. Is that correct and if so can anyone help me > > > with its evaluation? > > > > That can't be correct because the sum diverges. > > > I don't think it diverges. > > +inf > sum(1/n) > n=0 > > diverges, but barely. Anything even marginally smaller does not. >The finite sum of 1/n starting at 0 does not make sense. Also it is obvious that the integral of 1/x diverges in the limits from 0 to x as well as from x to infinity. Vladimir Vassilevsky, Ph.D. DSP and Mixed Signal Design Consultant http://www.abvolt.com
Reply by ●June 23, 20032003-06-23
In article 3EF6A4D7.CAEF51E4@arcanemethods.com, Bob Cain at arcane@arcanemethods.com wrote on 06/23/2003 02:57:> > > Olli Niemitalo wrote: >> >> On Sun, 22 Jun 2003, Bob Cain wrote: >> >>> I think the worst case digital signal, in terms of resulting maximum >>> amplitude, for sinc reconstruction is the sequence: >>> >>> ...-1,+1,-1,+1,-1,+1,+1,-1,+1,-1,+1,-1... >> >> You are right and we have had this before: >> >> http://groups.google.com/groups?threadm=3D55D545.5FF8C854%40ieee.org >> >> The answer is, you get infinite peak amplitude from sinc reconstruction of >> that signal. >> > > Thanks, Olli. Not sure how I missed that one. Bizzare result.it doesn't really satisfy the Nyquist criterion anyway. r b-j
Reply by ●June 23, 20032003-06-23
"robert bristow-johnson" <rbj@surfglobal.net> wrote in message news:BB1C9302.1962%rbj@surfglobal.net...> In article 3EF6A4D7.CAEF51E4@arcanemethods.com, Bob Cain at > arcane@arcanemethods.com wrote on 06/23/2003 02:57: > > > > > > > Olli Niemitalo wrote: > >> > >> On Sun, 22 Jun 2003, Bob Cain wrote: > >> > >>> I think the worst case digital signal, in terms of resulting maximum > >>> amplitude, for sinc reconstruction is the sequence: > >>> > >>> ...-1,+1,-1,+1,-1,+1,+1,-1,+1,-1,+1,-1... > >> > >> You are right and we have had this before: > >> > >> http://groups.google.com/groups?threadm=3D55D545.5FF8C854%40ieee.org > >> > >> The answer is, you get infinite peak amplitude from sinc reconstructionof> >> that signal. > >> > > > > Thanks, Olli. Not sure how I missed that one. Bizzare result. > > it doesn't really satisfy the Nyquist criterion anyway.Robert, whoa dude! (It's not meaningful to say that a sequence of numbers either does or does not satisfy the Nyquist criterion. It's also arguable that a "waveform" satisfies the Nyquist criterion.) - *Any* sequence of finite samples is just a sequence of samples. No Nyquist there.... - Any sequence of samples can be analytically reconstructed into an infinitely long continuous waveform given that a sample rate is assigned. If the resulting waveform is sampled at that same assigned sample rate or higher, it (the sampling frequency, not the waveform) can be said to satisfy the Nyquist sampling criterion. No? Fred
Reply by ●June 23, 20032003-06-23
Vladimir Vassilevsky wrote:> > Jerry Avins wrote: > > > > Vladimir Vassilevsky wrote: > > > > > > Bob Cain wrote: > > > > > > ... > > > > > > > I think it comes down to: > > > > > > > > +inf > > > > (2/pi) * sum{ 1/(.5 + n) } > > > > n=0 > > > > > > > > but forget how to evaluate that. Is that correct and if so can anyone help me > > > > with its evaluation? > > > > > > That can't be correct because the sum diverges. > > > > > I don't think it diverges. > > > > +inf > > sum(1/n) > > n=0 > > > > diverges, but barely. Anything even marginally smaller does not. > > > > The finite sum of 1/n starting at 0 does not make sense.Quite so! Nevertheless, +inf sum(1/n) diverges without that term, and n=1 +inf S = sum{ 1/(.5 + n) } does not. It's also clear that n=1 +inf S' = sum{ 1/(.5 + n) } = S + 2 n=0> Also it is obvious that the integral of 1/x diverges in the limits from > 0 to x as well as from x to infinity.+inf There's even no question that integral[dx/(.5 + x)] is unbounded, but 1 +inf that doesn't mean that sum{ 1/(.5 + n) } is. n=0 The harmonic series is marginally divergent. Anything at all that systematically makes the sum smaller, such as removing any term with a repeated digit, or any term that contains the digit 9, arrests the divergence. Reducing each term by .5 makes it convergent by inspection. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
