Reply by Randy Yates August 21, 20092009-08-21
Dave <dspguy2@netscape.net> writes:
> [...] > In the Papoulis book he later states from that point on he assumes the > white noise signals have zero mean. This suggests that his definition > of white noise allows it to have a non-zero mean.
Really? I didn't know that. Nice to know - thanks Dave. --Randy -- Randy Yates % "I met someone who looks alot like you, Digital Signal Labs % she does the things you do, mailto://yates@ieee.org % but she is an IBM." http://www.digitalsignallabs.com % 'Yours Truly, 2095', *Time*, ELO
Reply by Dave August 18, 20092009-08-18
On Aug 15, 9:49=A0pm, Randy Yates <ya...@ieee.org> wrote:
> Randy Yates <ya...@ieee.org> writes: > > karl bezzoto <karl.bezz...@googlemail.com> writes: > > >> Hello, > >> Per definition a white noise signal has a zero mean value. > > > Really? By whos definition? Neither of two definitions I have > > found require the mean to be zero. I examine this question in > > some detail here: > > > =A0http://www.digitalsignallabs.com/white.pdf > > OK, I'm back-pedaling on the Brown conclusion in this paper and > have "published" revision PA2. > > The Brown definition (constant PSD) DOES imply zero-mean, as > several of you here have already noted. However, I still > maintain that the Papoulis definition does not. > > PS: You may have to press "refresh" in your browsers or > clear the cache to get the new version (PA2). > -- > Randy Yates =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0% "She's sweet on =
Wagner-I think she'd die for Beethoven.
> Digital Signal Labs =A0 =A0 =A0 =A0 =A0 =A0 =A0% =A0She love the way Pucc=
ini lays down a tune, and
> mailto://ya...@ieee.org =A0 =A0 =A0 =A0 =A0% =A0Verdi's always creepin' f=
rom her room."http://www.digitalsignallabs.com% "Rockaria", *A New World Re= cord*, ELO =A0 In the Papoulis book he later states from that point on he assumes the white noise signals have zero mean. This suggests that his definition of white noise allows it to have a non-zero mean. Cheers, Dave
Reply by Les Cargill August 16, 20092009-08-16
Randy Yates wrote:
> Les Cargill <lcargill@cfl.rr.com> writes: > >> Randy Yates wrote: >>> karl bezzoto <karl.bezzoto@googlemail.com> writes: >>> >>>> Hello, >>>> Per definition a white noise signal has a zero mean value. >>> Really? By whos definition? Neither of two definitions I have found >>> require the mean to be zero. I examine this question in >>> some detail here: >>> >>> http://www.digitalsignallabs.com/white.pdf >>> >> Can't you say : " If white noise has a nonzero mean, then >> white noise is a signal" ( has nonzero information-content ) , >> leading to a contradiction? > > Hi Les, > > First of all, note that I reversed my conclusion in the Brown definition > (constant PSD) of this paper to conclude such a definition does imply a > zero mean. >
I saw that. *Kewl*. I always wondered that.
> What I think you're hinting at is the concept of randomness. How random > is "random"? Obviously a plain DC value isn't random at all. But what > about a DC value with some noise on it? Is that a little random? A lot > random? COMPLETELY (purely) random? >
Well, think CDMA. If there exists a transform T such that a .. "variable" DC offset may be decoded from something that is otherwise indistinguishable from noise ( we're assuming the transform is secret ), then it's not purely random. Obviously, the sequences in CDMA are *pseudo*random.
> I believe the answer lies in the autocorrelation function (which of > course is uniquely defined by the PSD as well). If the process is > _purely_ random, then there is absolutely no correlation > sample-to-sample. That's what makes it white. >
Yes, that was very nice.
> Papoulis' definition is somewhat odd since he decided to utilize > not the autocorrelation function but the autocovariance function, > in which case the mean doesn't matter. >
His is generative, and may or may not be useful as a description. :)
> Perhaps I'm saying the same thing over and over... or hopefully this > helps.
-- Les Cargill
Reply by Randy Yates August 15, 20092009-08-15
Les Cargill <lcargill@cfl.rr.com> writes:

> Randy Yates wrote: >> karl bezzoto <karl.bezzoto@googlemail.com> writes: >> >>> Hello, >>> Per definition a white noise signal has a zero mean value. >> >> Really? By whos definition? Neither of two definitions I have found >> require the mean to be zero. I examine this question in >> some detail here: >> >> http://www.digitalsignallabs.com/white.pdf >> > > Can't you say : " If white noise has a nonzero mean, then > white noise is a signal" ( has nonzero information-content ) , > leading to a contradiction?
Hi Les, First of all, note that I reversed my conclusion in the Brown definition (constant PSD) of this paper to conclude such a definition does imply a zero mean. What I think you're hinting at is the concept of randomness. How random is "random"? Obviously a plain DC value isn't random at all. But what about a DC value with some noise on it? Is that a little random? A lot random? COMPLETELY (purely) random? I believe the answer lies in the autocorrelation function (which of course is uniquely defined by the PSD as well). If the process is _purely_ random, then there is absolutely no correlation sample-to-sample. That's what makes it white. Papoulis' definition is somewhat odd since he decided to utilize not the autocorrelation function but the autocovariance function, in which case the mean doesn't matter. Perhaps I'm saying the same thing over and over... or hopefully this helps. -- Randy Yates % "She's sweet on Wagner-I think she'd die for Beethoven. Digital Signal Labs % She love the way Puccini lays down a tune, and mailto://yates@ieee.org % Verdi's always creepin' from her room." http://www.digitalsignallabs.com % "Rockaria", *A New World Record*, ELO
Reply by Randy Yates August 15, 20092009-08-15
Randy Yates <yates@ieee.org> writes:

> karl bezzoto <karl.bezzoto@googlemail.com> writes: > >> Hello, >> Per definition a white noise signal has a zero mean value. > > Really? By whos definition? Neither of two definitions I have > found require the mean to be zero. I examine this question in > some detail here: > > http://www.digitalsignallabs.com/white.pdf
OK, I'm back-pedaling on the Brown conclusion in this paper and have "published" revision PA2. The Brown definition (constant PSD) DOES imply zero-mean, as several of you here have already noted. However, I still maintain that the Papoulis definition does not. PS: You may have to press "refresh" in your browsers or clear the cache to get the new version (PA2). -- Randy Yates % "She's sweet on Wagner-I think she'd die for Beethoven. Digital Signal Labs % She love the way Puccini lays down a tune, and mailto://yates@ieee.org % Verdi's always creepin' from her room." http://www.digitalsignallabs.com % "Rockaria", *A New World Record*, ELO
Reply by Les Cargill August 15, 20092009-08-15
Randy Yates wrote:
> karl bezzoto <karl.bezzoto@googlemail.com> writes: > >> Hello, >> Per definition a white noise signal has a zero mean value. > > Really? By whos definition? Neither of two definitions I have > found require the mean to be zero. I examine this question in > some detail here: > > http://www.digitalsignallabs.com/white.pdf >
Can't you say : " If white noise has a nonzero mean, then white noise is a signal" ( has nonzero information-content ) , leading to a contradiction? -- Les Cargill
Reply by Randy Yates August 15, 20092009-08-15
karl bezzoto <karl.bezzoto@googlemail.com> writes:

> Hello, > Per definition a white noise signal has a zero mean value.
Really? By whos definition? Neither of two definitions I have found require the mean to be zero. I examine this question in some detail here: http://www.digitalsignallabs.com/white.pdf -- Randy Yates % "Remember the good old 1980's, when Digital Signal Labs % things were so uncomplicated?" mailto://yates@ieee.org % 'Ticket To The Moon' http://www.digitalsignallabs.com % *Time*, Electric Light Orchestra
Reply by zzbu...@netscape.net August 15, 20092009-08-15
On Aug 12, 7:55&#4294967295;am, karl bezzoto <karl.bezz...@googlemail.com> wrote:
> Hello, > Per definition a white noise signal has a zero mean value. yet by > definition also , it has a flat power density, i,e. every frequency > including zero has the same power which means it has a DC value, hence > its mean is not equal to zero. > I am confused. any help please?
Well, you should be, since they defined white noise with the mathematical magic of delta functions and generalized functions, not anything to do with reality. So, the people who actually understand engineering mostly these days work on Electronic Books, Miniature External Hard Disks, Laser Disk Libraries, Flat Screen Software Debuggers, Multiplexed Fiber Optics Signal and Conrol Systems, C++, USB, XML, All-In-One Printers, HDTV, Home Broadband, Cyber Batteries, Pv Cell Energy, Distributed Processing Software Holograms, Cell Phones, Microcomputers, mp3, mpeg, PGP, Desktop Publishing, On-Line Banking, On-Line Shopping, Blue Ray, HDTV, On-Line Publishing, Atomic Clock Wristwatches, Light Sticks, Self-Assembling Robots, and Self-Replicating Machines and let just the Quantum Mechanics, GM, and GE cranks worry about what power means.
Reply by Piergiorgio Sartor August 14, 20092009-08-14
On 08/12/2009 01:55 PM, karl bezzoto wrote:
> Hello, > Per definition a white noise signal has a zero mean value. yet by
Is "per definition" or is it a consequence of the definition of white noise? bye, -- piergiorgio
Reply by Jerry Avins August 14, 20092009-08-14
dvsarwate@yahoo.com wrote:
> On Aug 14, 9:38 am, Jerry Avins <j...@ieee.org> responded: > > >>> So it is similar to statistics with a continuous variable given pdf >>> (x) the probability at any individual point is zero? >>> Don't you just love mathematics :) > >> With an infinite number of probability points and a finite sum, what >> would you expect? > > > A (countably) infinite number of points *can* have probabilities > that sum to 1, but these points *cannot* all have *equal* probability. > The sum is, of course, defined in the sense of a limit (this is the > topic that causes most people to fall asleep in their calculus > classes :-) ) and a sum of the form c + c + c + ... + c does not > converge to a finite value as the number of terms in the sum > increases. (Sums of the form c + c(1-c) + c(1-c)^2 + ... + c(1-c)^ > {n-1} > = 1-(1-c)^n do converge to 1 provided that c is in (0, 1] but the > terms > are not equal ....)
For a given distribution D, the probability of an event e occurring between p1 and p2 is p2 ( ) Integral(p(D)dp) e=p1 ( ) The probability of an event occurring at exactly p3 is p3 ( ) Integral(p(D)dp) e=p3 ( ) Surely, nobody here needs help evaluating that!
> For a continuous random variable, the number of possible value is > *uncountably* infinite, and the notion of a *sum* of all such values > is not defined in the above sense; the corresponding notion is that > of an integral or area under the pdf, which we should remember > stands for probability *density* function: it is measured in units > of probability mass per unit length, and we don't get a probability > from the pdf unless we "multiply" by a length or integrate the pdf > over an interval. The "reason" that the probability that a continuous > variable X equals c is 0 is that the "point" c has zero length (or > width if you prefer) and so multiplying the pdf value by the length > (or doing an integral if you prefer) gives 0; there is no *area* under > the curve above the point of zero width. In short, a good reason for > loving mathematics is its insistence that c times 0 is 0 for any > real number c (and no, "infinity" is not a real number.....).
I don't think we're disagreeing. You put it more rigorously. Jerry -- Engineering is the art of making what you want from things you can get. &#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;