On Nov 12, 3:36=A0pm, Randy Yates <y...@ieee.org> wrote:> "radio...@aol.com" <radio...@aol.com> writes: > > On Nov 12, 11:18=A0am, Vladimir Vassilevsky <antispam_bo...@hotmail.com> > > wrote: > >> radio...@aol.com wrote: > >> > =A0 =A0 =A0 =A0 Does anyone have a mathematical derivation > >> > =A0of the expected spectrum of a QPSK signal, given a > >> > known symbol rate (which is 1/2 the data rate for QPSK)?If the data =is random, then the spectum is the Fourier transform of a> >> single pulse. A textbook like Proakis or Sklar should have that. > > > =A0 =A0 =A0 =A0A single pulse would have a continuous spectrum, > > which is not what i'm measuring at all.Hi, > > Vladimir said, "*IF* the data is random, then the spectrum is the > Fourier transform of a single pulse." Your data isn't random - it's > highly correlated. His statement agrees 100 percent with the equation > I posted from Proakis.Then why does your equation go to zero if the period T is infinitely long? Surely the carrier will still be there if we stay in one quadrant!> > > =A0 =A0 =A0 =A0And although i agree the data stream will affect > > the spectrum (i.e., a bunch of "00"s or "11"s will be just > > the carrier freq.), making the data even psuedo- > > random will not give you a continuus spectrum.I'm curious why you think=so. It is pretty basic knowledge that the> power spectrum of a random signal is the transform of its > autocorrelation function (the Wiener-Khinchine theorem), and it's > pretty easy to see that a *continuous* flat spectrum is produced by a > sequence with a Kronecker delta, and that it requires an uncorrelated > sequence to produce a Kronecker delta autocorrelation. >The Kronecker delta is related to the Delta-dirac impulse function, which indeed has a "white noise" continuous spectrum. But that is not what i'm modulating my carrier with. Even if the data is random, the period of one symbol is not infinitely short. And if we make the period fairly large (staying in one quadrant for a very long time), then we will definitely not have a continuous spectrum.> So Proakis's expression makes a lot of sense: the total spectrum is > the cascade of the information sequence spectrum and the transmit > pulse spectrum. > > In any case, you must argue the point with John Proakis, who has > written a textbook on the subject, because that is what he claims. >Perhaps he would argue that you aren't applying his formula correctly.> >> > =A0 =A0 =A0 =A0 =A0 Thanks for any REAL help (which is Info from som=eone> >> > who isn't pretending to know more than they actually do!).Your majes=ty's "Thank you" means soo much. You don't have to thank me,> >> $100 will be just all right. > > > =A0 =A0 =A0 You didn't earn it! > > > =A0 =A0 =A0 My remark refers to people just like you, > > but I don't expect a C++ programmer to know > > the Fourier of QPSK, PhD or not.Vladimir is no mere C++ programmer. Fro=m what I've seen of him though> his posts over the past months/years, he is brilliant in many topics > on DSP and digital communications.I don't doubt that he may be very knowledgable, but he's not answering my question here. Slick
Fourier Transform (spectrum) of QPSK and BPSK?
Started by ●November 12, 2006
Reply by ●November 12, 20062006-11-12
Reply by ●November 12, 20062006-11-12
"radio913@aol.com" <radio913@aol.com> writes:> [...] > I don't doubt that he may be very knowledgable, but he's not > answering my question here.Oh but he did. You're not hearing him. Now perhaps there are a few things you don't understand, but until you can omit the impertinence and ask your questions in a more respectful and constructive tone, our conversation is over. -- % Randy Yates % "She's sweet on Wagner-I think she'd die for Beethoven. %% Fuquay-Varina, NC % She love the way Puccini lays down a tune, and %%% 919-577-9882 % Verdi's always creepin' from her room." %%%% <yates@ieee.org> % "Rockaria", *A New World Record*, ELO http://home.earthlink.net/~yatescr
Reply by ●November 12, 20062006-11-12
radio913@aol.com wrote:> > On Nov 12, 3:36?pm, Randy Yates <y...@ieee.org> wrote: > >>"radio...@aol.com" <radio...@aol.com> writes: >> >>>On Nov 12, 11:18?am, Vladimir Vassilevsky <antispam_bo...@hotmail.com> >>>wrote: >>> >>>>radio...@aol.com wrote: >>>> >>>>>? ? ? ? Does anyone have a mathematical derivation >>>>>?of the expected spectrum of a QPSK signal, given a >>>>>known symbol rate (which is 1/2 the data rate for QPSK)?If the data is random, then the spectum is the Fourier transform of a >>>> >>>>single pulse. A textbook like Proakis or Sklar should have that. >> >>>? ? ? ?A single pulse would have a continuous spectrum, >>>which is not what i'm measuring at all.Hi, >> >>Vladimir said, "*IF* the data is random, then the spectrum is the >>Fourier transform of a single pulse." Your data isn't random - it's >>highly correlated. His statement agrees 100 percent with the equation >>I posted from Proakis. > > > Then why does your equation go to zero if > the period T is infinitely long? > > Surely the carrier will still be there if we > stay in one quadrant! > > > > >>>? ? ? ?And although i agree the data stream will affect >>>the spectrum (i.e., a bunch of "00"s or "11"s will be just >>>the carrier freq.), making the data even psuedo- >>>random will not give you a continuus spectrum.I'm curious why you think so. It is pretty basic knowledge that the >> >>power spectrum of a random signal is the transform of its >>autocorrelation function (the Wiener-Khinchine theorem), and it's >>pretty easy to see that a *continuous* flat spectrum is produced by a >>sequence with a Kronecker delta, and that it requires an uncorrelated >>sequence to produce a Kronecker delta autocorrelation. >> > > > The Kronecker delta is related to the Delta-dirac > impulse function, which indeed has a "white noise" continuous spectrum. > But that is not what i'm modulating > my carrier with. Even if the data is random, the period > of one symbol is not infinitely short. And if we make > the period fairly large (staying in one quadrant for a very long time), > then we will definitely not have a continuous spectrum. > > > > >>So Proakis's expression makes a lot of sense: the total spectrum is >>the cascade of the information sequence spectrum and the transmit >>pulse spectrum. >> >>In any case, you must argue the point with John Proakis, who has >>written a textbook on the subject, because that is what he claims. >> > > > Perhaps he would argue that you aren't applying > his formula correctly. > > > >>>>>? ? ? ? ? Thanks for any REAL help (which is Info from someone >>>>>who isn't pretending to know more than they actually do!).Your majesty's "Thank you" means soo much. You don't have to thank me, >>>> >>>>$100 will be just all right. >> >>>? ? ? You didn't earn it! >> >>>? ? ? My remark refers to people just like you, >>>but I don't expect a C++ programmer to know >>>the Fourier of QPSK, PhD or not.Vladimir is no mere C++ programmer. From what I've seen of him though >> >>his posts over the past months/years, he is brilliant in many topics >>on DSP and digital communications. > > > I don't doubt that he may be very knowledgable, > but he's not answering my question here. > > Slick >"which is Info from someone who isn't pretending to know more than they actually do!" What a marvelous excuse for not bothering to work at understanding someone's answers! Vladimir is exactly on target here. I couldn't have answered the question better myself. Some subjects cannot be answered easily. Were you to ask "How do I do brain surgery", for instance, the question would take nearly a decade to answer well enough so that you can actually go do it, and you'd have to work your butt off. Perhaps you should stop pretending to be dumber than you really are, and actually work through the reason that the answer is correct. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Posting from Google? See http://cfaj.freeshell.org/google/ "Applied Control Theory for Embedded Systems" came out in April. See details at http://www.wescottdesign.com/actfes/actfes.html
Reply by ●November 13, 20062006-11-13
On Nov 12, 7:00=A0pm, Randy Yates <y...@ieee.org> wrote:> "radio...@aol.com" <radio...@aol.com> writes: > > [...] > > I don't doubt that he may be very knowledgable, but he's not > > answering my question here.Oh but he did. You're not hearing him. >How so? He never gave me the Fourier transform of a QPSK signal with a known symbol rate, and how the side-band suppression could indicate amplitude and phase imbalance: http://www.rfcafe.com/references/electrical/quad_mod.htm The sidebands were level on the mixer i tested, but staying statically within each quadrant, the amplitudes were within 1dB, and the phase imbalance was within 6 degrees. So where is the sideband suppression?> Now perhaps there are a few things you don't understand, but until you > can omit the impertinence and ask your questions in a more respectful > and constructive tone, our conversation is over.We all have things we don't understand, but don't give me a B.S. answer if you don't even understand what i'm asking! And you never answered me about that equation going to zero when the period T goes to infinity. Slick
Reply by ●November 13, 20062006-11-13
On Nov 12, 7:44=A0pm, Tim Wescott <t...@seemywebsite.com> wrote:> radio...@aol.com wrote: > > > On Nov 12, 3:36?pm, Randy Yates <y...@ieee.org> wrote: > > >>"radio...@aol.com" <radio...@aol.com> writes: > > >>>On Nov 12, 11:18?am, Vladimir Vassilevsky <antispam_bo...@hotmail.com> > >>>wrote: > > >>>>radio...@aol.com wrote: > > >>>>>? ? ? ? Does anyone have a mathematical derivation > >>>>>?of the expected spectrum of a QPSK signal, given a > >>>>>known symbol rate (which is 1/2 the data rate for QPSK)?If the data =is random, then the spectum is the Fourier transform of a> > >>>>single pulse. A textbook like Proakis or Sklar should have that. > > >>>? ? ? ?A single pulse would have a continuous spectrum, > >>>which is not what i'm measuring at all.Hi, > > >>Vladimir said, "*IF* the data is random, then the spectrum is the > >>Fourier transform of a single pulse." Your data isn't random - it's > >>highly correlated. His statement agrees 100 percent with the equation > >>I posted from Proakis. > > > =A0 =A0 =A0Then why does your equation go to zero if > > the period T is infinitely long? > > > =A0 =A0 =A0Surely the carrier will still be there if we > > stay in one quadrant! > > >>>? ? ? ?And although i agree the data stream will affect > >>>the spectrum (i.e., a bunch of "00"s or "11"s will be just > >>>the carrier freq.), making the data even psuedo- > >>>random will not give you a continuus spectrum.I'm curious why you thin=k so. It is pretty basic knowledge that the> > >>power spectrum of a random signal is the transform of its > >>autocorrelation function (the Wiener-Khinchine theorem), and it's > >>pretty easy to see that a *continuous* flat spectrum is produced by a > >>sequence with a Kronecker delta, and that it requires an uncorrelated > >>sequence to produce a Kronecker delta autocorrelation. > > > =A0 =A0 =A0 =A0The Kronecker delta is related to the Delta-dirac > > impulse function, which indeed has a "white noise" continuous spectrum. > > =A0But that is not what i'm modulating > > my carrier with. =A0Even if the data is random, the period > > of one symbol is not infinitely short. =A0And if we make > > the period fairly large (staying in one quadrant for a very long time), > > then we will definitely not have a continuous spectrum. > > >>So Proakis's expression makes a lot of sense: the total spectrum is > >>the cascade of the information sequence spectrum and the transmit > >>pulse spectrum. > > >>In any case, you must argue the point with John Proakis, who has > >>written a textbook on the subject, because that is what he claims. > > > =A0 =A0 =A0 Perhaps he would argue that you aren't applying > > his formula correctly. > > >>>>>? ? ? ? ? Thanks for any REAL help (which is Info from someone > >>>>>who isn't pretending to know more than they actually do!).Your majes=ty's "Thank you" means soo much. You don't have to thank me,> > >>>>$100 will be just all right. > > >>>? ? ? You didn't earn it! > > >>>? ? ? My remark refers to people just like you, > >>>but I don't expect a C++ programmer to know > >>>the Fourier of QPSK, PhD or not.Vladimir is no mere C++ programmer. Fr=om what I've seen of him though> > >>his posts over the past months/years, he is brilliant in many topics > >>on DSP and digital communications. > > > =A0 =A0 =A0 I don't doubt that he may be very knowledgable, > > but he's not answering my question here. > > > Slick"which is Info from someone who isn't pretending to know more than=they> actually do!" > > What a marvelous excuse for not bothering to work at understanding > someone's answers! > > Vladimir is exactly on target here. =A0I couldn't have answered the > question better myself.I'd agree with that, as your specialty is in control systems.>=A0Some subjects cannot be answered easily. =A0Especially when the person answering doesn't know either!> Perhaps you should stop pretending to be dumber than you really are, and > actually work through the reason that the answer is correct. >Perhaps you should stop pretending to be smarter than you really are, and stop answering questions you clearly don't understand. Stick with control systems. Slick
Reply by ●November 14, 20062006-11-14
"radio913@aol.com" <radio913@aol.com> writes:> [...] > And you never answered me about that equation > going to zero when the period T goes to infinity.What about this bothers you? When T goes to infinity, you have a single pulse g(t) of energy, i.e., a finite-energy signal. Thus you have a zero-power signal. Any power signal necessarily has infinite energy. -- % Randy Yates % "She tells me that she likes me very much, %% Fuquay-Varina, NC % but when I try to touch, she makes it %%% 919-577-9882 % all too clear." %%%% <yates@ieee.org> % 'Yours Truly, 2095', *Time*, ELO http://home.earthlink.net/~yatescr
Reply by ●November 14, 20062006-11-14
On Nov 13, 9:41=A0pm, Randy Yates <y...@ieee.org> wrote:> "radio...@aol.com" <radio...@aol.com> writes: > > [...] > > =A0 =A0 =A0 And you never answered me about that equation > > going to zero when the period T goes to infinity.What about this bother=s you? When T goes to infinity, you have a> single pulse g(t) of energy, i.e., a finite-energy signal. Thus you > have a zero-power signal.T is the symbol period, so if it's infinity, then you stay in only one of the quadrants of QPSK, so you are essentially a CW carrier.>Any power signal necessarily has infinite > energy.=20 Where your paper on this one?=20 Slick
Reply by ●November 14, 20062006-11-14
"radio913@aol.com" <radio913@aol.com> writes:> On Nov 13, 9:41�pm, Randy Yates <y...@ieee.org> wrote: >> "radio...@aol.com" <radio...@aol.com> writes: >> > [...] >> > � � � And you never answered me about that equation >> > going to zero when the period T goes to infinity.What about this bothers you? When T goes to infinity, you have a >> single pulse g(t) of energy, i.e., a finite-energy signal. Thus you >> have a zero-power signal. > > T is the symbol period, so if it's infinity, then > you stay in only one of the quadrants of QPSK, so > you are essentially a CW carrier. > > > >>Any power signal necessarily has infinite >> energy. > > > Where your paper on this one?And people are actually paying you to work in this field? -- % Randy Yates % "She has an IQ of 1001, she has a jumpsuit %% Fuquay-Varina, NC % on, and she's also a telephone." %%% 919-577-9882 % %%%% <yates@ieee.org> % 'Yours Truly, 2095', *Time*, ELO http://home.earthlink.net/~yatescr
Reply by ●November 14, 20062006-11-14
radio913@aol.com wrote:> The signal i am measuring doesn't have any > sideband carrier suppression at all, which is supposed > to be a figure of merit for phase and amplitude imbalance > of your IQ mixer (how orthogonal or in-quadrature the I and Q > channels are). >What is "phase and amplitude imbalance of your IQ mixer"? Thanks, Julius
Reply by ●November 14, 20062006-11-14
Randy Yates <yates@ieee.org> writes:> "radio913@aol.com" <radio913@aol.com> writes: >> [...] >> And you never answered me about that equation >> going to zero when the period T goes to infinity. > > What about this bothers you? When T goes to infinity, you have a > single pulse g(t) of energy, i.e., a finite-energy signal.For the sake of others reading this thread, this statement is not necessarily true. The energy in g(t) may scale with increasing period T, so as T approaches infinity you may not have have a finite-energy signal. A perfect example is the raised cosine pulse. It does scale with the period. In fact, if you use root-raised-cosine, then there is a factor of sqrt(T) in the expression for the spectrum, so it works out that the 1/T in the total signal spectrum cancels the (sqrt(T))^2 in |G(f)|^2. The other theoretical situation is when you have a rectangular pulse shape. In that case, the total signal spectrum actually blows up (at least at f=0) since |G(f)|^2 has a factor T^2. Again, to re-orient, we were "discussing" the 1/T factor in the following expression for the total signal spectrum of a linear modulation: Phi(f) = (1/T) * | G(f) |^2 * Phi_ii(f), -- % Randy Yates % "The dreamer, the unwoken fool - %% Fuquay-Varina, NC % in dreams, no pain will kiss the brow..." %%% 919-577-9882 % %%%% <yates@ieee.org> % 'Eldorado Overture', *Eldorado*, ELO http://home.earthlink.net/~yatescr
