I was reading this post and found your comment on cyclostationary. My understanding of cyclostationary are signals that there statistical measurements are periodic in time. From my understanding also, this cyclostationary feature appears on "modulated" signals due to the built in periodicity in the carrier. However, I am not sure if a pure binary signaly after demodulation is considered as cyclostationary, most digital communication books assume the binary source as a random normal distribution (stationary process). Do you agree with me on this point or you think I am missing a point?
Can a binary signal be cyclostationary?
Started by ●October 31, 2008
Reply by ●October 31, 20082008-10-31
>I was reading this post and found your comment on cyclostationary. My >understanding of cyclostationary are signals that there statistical >measurements are periodic in time. From my understanding also, this >cyclostationary feature appears on "modulated" signals due to the builtin>periodicity in the carrier. However, I am not sure if a pure binarysignaly>after demodulation is considered as cyclostationary, most digital >communication books assume the binary source as a random normal >distribution (stationary process). Do you agree with me on this point or >you think I am missing a point?Let me clarify more here, this post is placed mistakenly as new post rather than a follow up to ongoing thread! But at any-case, I had this wonder about the cyclostationary processes that I would like to discuss it with you probably in a separate thread since this feature is used by some signal detection technique to distinguish "weak" signals from noise. Noise are not cyclostationary process while modulated signals are. Any thoughts about this subject :)
Reply by ●October 31, 20082008-10-31
On Oct 31, 8:54�pm, "ytach" <ytachw...@ou.edu> wrote:> >I was reading this post and found your comment on cyclostationary. My > >understanding of cyclostationary are signals that there statistical > >measurements are periodic in time. From my understanding also, this > >cyclostationary feature appears on "modulated" signals due to the built > in > >periodicity in the carrier. However, I am not sure if a pure binary > signaly > >after demodulation is considered as cyclostationary, most digital > >communication books assume the binary source as a random normal > >distribution (stationary process). Do you agree with me on this point or > >you think I am missing a point? > > Let me clarify more here, this post is placed mistakenly as new post > rather than a follow up to ongoing thread! But at any-case, I had this > wonder about the cyclostationary processes that I would like to discuss it > with you probably in a separate thread since this feature is used by some > signal detection technique to distinguish "weak" signals from noise. Noise > are not cyclostationary process while modulated signals are. Any thoughts > about this subject :)i would say no..pure binary signal cannot be a cyclostationary signal..and regarding detecting a weak signal..i guess Cyclo spectral density (csd) and other cyclostationary related methods (cyclic autocorrelation) will be applied to find out weak signal in the noise..where (as per what you told) modulated signals will peak out..leaving out noise..
Reply by ●October 31, 20082008-10-31
PARTICLEREDDY (STRAYDOG) wrote:> i would say no..pure binary signal cannot be a cyclostationary > signal..???? If a binary signal can change its value only at nT, then the signal is cyclostationary. VLV
Reply by ●October 31, 20082008-10-31
> > >PARTICLEREDDY (STRAYDOG) wrote: > > >> i would say no..pure binary signal cannot be a cyclostationary >> signal.. > >???? > >If a binary signal can change its value only at nT, then the signal is >cyclostationary. > >VLV > >Mmmm.... I see your point here. I would agree with you based on mu basic understanding to cyclostationary (process with periodic statistics) but how can you prove that? any reference to understand that better? Why digital communication text boox assume a normally distributed data source, or even more than that, the basic approximation of the quantization error is a normal distributed noise (found to be not accurate but still alot of people accept that). The point about detection using scd is absolutely a good point since I have read that for signal detection. More clarification will be appreciated VLV :)
Reply by ●November 1, 20082008-11-01
>> >> >>PARTICLEREDDY (STRAYDOG) wrote: >> >> >>> i would say no..pure binary signal cannot be a cyclostationary >>> signal.. >> >>???? >> >>If a binary signal can change its value only at nT, then the signal is >>cyclostationary. >> >>VLV >> >> > >Mmmm.... I see your point here. I would agree with you based on mu basic >understanding to cyclostationary (process with periodic statistics) buthow>can you prove that? any reference to understand that better? Why digital >communication text boox assume a normally distributed data source, oreven>more than that, the basic approximation of the quantization error is a >normal distributed noise (found to be not accurate but still alot ofpeople>accept that). > >The point about detection using scd is absolutely a good point since I >have read that for signal detection. > >More clarification will be appreciated VLV :) > >Let me also elaborate on my comment on modeling the quantization error as normally distributed noise (Stationary process.) The quantization noise can be seen as a binary signal as well (in away except that it has multi-level rather than binary level but I do not feel that the level issue will affect the modeling as a stationary or cyclostationary signal). If that is true, then this will be the reason I thought that binary signals are stationary process. VLV, I am very interested to know from you your clarification on this point and thanks in advance sharing this knowedge with the group.
Reply by ●November 1, 20082008-11-01
>If a binary signal can change its value only at nT, then the signal is >cyclostationary. > >VLVWrong. Counterexample: unit step function, i.e., u(t) = 0 for t<0, and u(t)=1 for t>=0. Emre
Reply by ●November 1, 20082008-11-01
On Nov 1, 10:50 am, "emre" <egu...@ece.neu.edu> wrote:> >If a binary signal can change its value only at nT, then the signal is > >cyclostationary. > > >VLV > > Wrong. Counterexample: unit step function, i.e., u(t) = 0 for t<0, and > u(t)=1 for t>=0. > > EmreBut your example is not a stochastic process :-). How can it be stationary or cyclostationary or non-stationary? I think that VLV means nT for n \in \Integers. To the original poster, this is a well-answered question in just about any digital communication book.
Reply by ●November 1, 20082008-11-01
>> >If a binary signal can change its value only at nT, then the signal is >> >cyclostationary. >> >> >VLV >> >> Wrong. Counterexample: unit step function, i.e., u(t) = 0 for t<0,and>> u(t)=1 for t>=0. >> >> Emre > >But your example is not a stochastic process :-). How can it be >stationary or cyclostationary or non-stationary?Hey Julius, Cyclostationarity does not imply randomness, unlike stationarity. (I just learned about this, too. You can find this with a quick search on the web.) Vladimir's statement only says "a binary signal".>I think that VLV means nT for n \in \Integers.The unit step function, u(t) above, satisfies this. It changes value only for n = 0, which is an integer. I just don't believe in the correctness of the statement. I am not trying to nit-pick. :-) Emre
Reply by ●November 2, 20082008-11-02
On Nov 2, 1:02�am, "emre" <egu...@ece.neu.edu> wrote:> >> >If a binary signal can change its value only at nT, then the signal is > >> >cyclostationary. > > >> >VLV > > >> Wrong. �Counterexample: �unit step function, i.e., u(t) = 0 for t<0, > and > >> u(t)=1 for t>=0. > > >> Emre > > >But your example is not a stochastic process :-). �How can it be > >stationary or cyclostationary or non-stationary? > > Hey Julius, > > Cyclostationarity does not imply randomness, unlike stationarity. (I just > learned about this, too. �You can find this with a quick search on the > web.) �Vladimir's statement only says "a binary signal". > > >I think that VLV means nT for n \in \Integers. > > The unit step function, u(t) above, satisfies this. �It changes value only > for n = 0, which is an integer. > > �I just don't believe in the correctness of the statement. I am not trying > to nit-pick. �:-) > > Emreplz refer to these files on matlab implementationand you can quickly check the binary signal also. http://www.mathworks.fr/matlabcentral/fileexchange/16742
