Neu wrote:>> Recapping my conjecture: >> For unknown (deterministic) bandlimited signals, the necessary and >> sufficient condition for Oli's property to hold is that the signal be >> sampled at twice the Nyquist rate. > > > I dont think thats 100% true. If the signal is narrowband and periodic > for instance, we > can devise a scheme that samples *below* the Nyquist based on the > signal's periodicity.If you already know the signal, why bother to sample it? (I was using a sampling scope in 1963.) Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������

# When does sampled autocorrelation = autocorrelation of sampled sequence?

Started by ●February 28, 2009

Reply by ●March 2, 20092009-03-02

Reply by ●March 2, 20092009-03-02

>> Recapping my conjecture: >> For unknown (deterministic) bandlimited signals, the necessary and >> sufficient condition for Oli's property to hold is that the signal be >> sampled at twice the Nyquist rate. > > >I dont think thats 100% true. If the signal is narrowband and periodic >for instance, we >can devise a scheme that samples *below* the Nyquist based on the >signal's periodicity.For a periodic signal the integral / summation for the autocorrelation may diverge. There is a cyclic version, but it is different from Oli's definition. Still, I don't see why you assumed periodicity to start with. If you are referring to "sub-Nyquist" sampling for narrowband signals, that requires passband filtering to recover the signal. See the property in Oli's original post, there is no such thing there. The desired property is that r(t) = int h(tau) h(tau-t) dt and r[n] = sum_k h[k] h[k-n] such that r[n] = r(n Ts). I'd be very interested to see how this can hold at a lower sampling rate than (2 x Nyquist) without specific knowledge of the signal other than that it is narrowband. Emre

Reply by ●March 2, 20092009-03-02

emre wrote:>>> Recapping my conjecture: >>> For unknown (deterministic) bandlimited signals, the necessary and >>> sufficient condition for Oli's property to hold is that the signal be >>> sampled at twice the Nyquist rate. >> >> I dont think thats 100% true. If the signal is narrowband and periodic >> for instance, we >> can devise a scheme that samples *below* the Nyquist based on the >> signal's periodicity. > > For a periodic signal the integral / summation for the autocorrelation may > diverge. There is a cyclic version, but it is different from Oli's > definition. Still, I don't see why you assumed periodicity to start with. > > If you are referring to "sub-Nyquist" sampling for narrowband signals, > that requires passband filtering to recover the signal. See the property > in Oli's original post, there is no such thing there. The desired property > is that > r(t) = int h(tau) h(tau-t) dt > and > r[n] = sum_k h[k] h[k-n] > such that r[n] = r(n Ts). > > I'd be very interested to see how this can hold at a lower sampling rate > than (2 x Nyquist) without specific knowledge of the signal other than that > it is narrowband.I've just been thinking about this again, and I think I've found a family of counter-examples (although it's late, so please double-check my logic!): Let h(t) be a raised-cosine filter with zero-crossings at kT/2 (for integer k), so its total support in the frequency domain is 4/T, i.e. 4.fs. Clearly, all sampling points (other than n=0) lie on zero-crossings, so r[n] = delta[n] (Kronecker delta). The continuous-time autocorrelation will give another filter that has zero-crossings at kT/2, and so r(nT) = r[n]. If this logic holds, then it should also hold for any constant integer divisor (i.e. zero-crossings at kT/M), therefore a bandlimited signal of arbitrary bandwidth satisfies the requirement... -- Oli

Reply by ●March 2, 20092009-03-02

>I've just been thinking about this again, and I think I've found a >family of counter-examples (although it's late, so please double-check >my logic!): > >Let h(t) be a raised-cosine filter with zero-crossings at kT/2 (for >integer k), so its total support in the frequency domain is 4/T, i.e. >4.fs. Clearly, all sampling points (other than n=0) lie on >zero-crossings, so r[n] = delta[n] (Kronecker delta). The >continuous-time autocorrelation will give another filter that has >zero-crossings at kT/2, and so r(nT) = r[n]. > >If this logic holds, then it should also hold for any constant integer >divisor (i.e. zero-crossings at kT/M), therefore a bandlimited signal of >arbitrary bandwidth satisfies the requirement... > >-- >OliOli, I agree that you can come up with many examples that satisfy your property. I'll give you a trivial example: h(t) = 0. My point, however, is that the necessary condition for an unknown bandlimited signal h(t) to satisfy your property is that h(t) be sampled at rate (2 x Nyquist). Therefore, a counterexample to this claim can not contain a fully specified signal, since the assumption here is that the signal is *unknown*. That may sound squirrelly, but it makes sense if you don't know exactly what signal you are going to sample ahead of time (other than its bandwidth), but want to make sure that you calculate r[n] correctly from samples of h(t). If, however, your interest is limited to raised-cosine or another family of signals, then of course you can say more specific things about that family. Emre

Reply by ●March 9, 20092009-03-09

On Mar 2, 4:13�pm, Jerry Avins <j...@ieee.org> wrote:> Neu wrote: > >> Recapping my conjecture: > >> � � For unknown (deterministic) bandlimited signals, the necessary and > >> sufficient condition for Oli's property to hold is that the signal be > >> sampled at twice the Nyquist rate. > > > I dont think thats 100% true. If the signal is narrowband and periodic > > for instance, we > > can devise a scheme that samples *below* the Nyquist based on the > > signal's periodicity. > > If you already know the signal, why bother to sample it? (I was using a > sampling scope in 1963.) > > JerryI didn't say we new the complete signal but only some of its properties. It could be possible that one would wish to estimate its amplitude.... By the way, regarding the original post, it appears that from section 10-5 from"Probability, Random Variables and Stochastic Process" (Papoulis & Pillai) as long as R(t) is bandlimited and sampled above the nyquist rate than it should yield the original autocorrelation (note that the constraint is on the autocorrelation sequence and not in the process and the original post suggests).

Reply by ●March 9, 20092009-03-09

> For a periodic signal the integral / summation for the autocorrelation may > diverge. �There is a cyclic version, but it is different from Oli's > definition. �Still, I don't see why you assumed periodicity to start with. > > If you are referring to "sub-Nyquist" sampling for narrowband signals, > that requires passband filtering to recover the signal. �See the property > in Oli's original post, there is no such thing there. �The desired property > is that > � �r(t) = int �h(tau) h(tau-t) dt > and > � �r[n] = sum_k �h[k] h[k-n] > such that �r[n] = r(n Ts). > > I'd be very interested to see how this can hold at a lower sampling rate > than (2 x Nyquist) without specific knowledge of the signal other than that > it is narrowband. > > EmreHi Emre, I referring to the determinitic signals you mentioned in the previous post. An example of subsampling can be on Shaum's Outlines on Digital Signal Processing pg 121. That offcourse is due to the fact already stated that Nyquist is sufficient but not necessary condition...

Reply by ●March 9, 20092009-03-09

On Mar 2, 8:42�pm, "emre" <egu...@ece.neu.edu> wrote:> >> Recapping my conjecture: > >> � � For unknown (deterministic) bandlimited signals, the necessary and > >> sufficient condition for Oli's property to hold is that the signal be > >> sampled at twice the Nyquist rate. > > >I dont think thats 100% true. If the signal is narrowband and periodic > >for instance, we > >can devise a scheme that samples *below* the Nyquist based on the > >signal's periodicity. > > For a periodic signal the integral / summation for the autocorrelation may > diverge. �There is a cyclic version, but it is different from Oli's > definition. �Still, I don't see why you assumed periodicity to start with. > > If you are referring to "sub-Nyquist" sampling for narrowband signals, > that requires passband filtering to recover the signal. �See the property > in Oli's original post, there is no such thing there. �The desired property > is that > � �r(t) = int �h(tau) h(tau-t) dt > and > � �r[n] = sum_k �h[k] h[k-n] > such that �r[n] = r(n Ts). > > I'd be very interested to see how this can hold at a lower sampling rate > than (2 x Nyquist) without specific knowledge of the signal other than that > it is narrowband. > > Emre> For a periodic signal the integral / summation for the autocorrelation may > diverge. There is a cyclic version, but it is different from Oli's > definition. Still, I don't see why you assumed periodicity to start with.> If you are referring to "sub-Nyquist" sampling for narrowband signals, > that requires passband filtering to recover the signal. See the property > in Oli's original post, there is no such thing there. The desired property > is that > r(t) = int h(tau) h(tau-t) dt > and > r[n] = sum_k h[k] h[k-n] > such that r[n] = r(n Ts).> I'd be very interested to see how this can hold at a lower sampling rate > than (2 x Nyquist) without specific knowledge of the signal other than that > it is narrowband.> EmreHi Emre, I referring to the determinitic signals you mentioned in the previous post. I am not sure I understand. Why does requirement of pass-band filtering r[n] from sampling reconstruction would violate the property? In other words, where exactly is the conflict here?

Reply by ●March 9, 20092009-03-09

>Hi Emre, > >I referring to the determinitic signals you mentioned in the previous >post. I am not sure I understand. Why does requirement of pass-band >filtering r[n] from sampling reconstruction would violate the >property? > In other words, where exactly is the conflict here? >Hi, Oli's property, the way I understand it, is that r(nT) = sum_k h[n] h[n-k]. My original claim was regarding the condition under which this equality holds (nothing more, or less.) Otherwise, I certainly agree that you can design a scheme to reconstruct r(t) as long as you have sufficiently many samples of h[n]. So I guess we are on the same page as far as that goes. :-) Emre