Hi All, 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)? 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). The bottom line is I'm trying to measure phase and amplitude imbalance with just a spectrum analyzer, without having to buy an expensive VSA which can display the constellation, and give me EVM, etc. Thanks for any REAL help (which is Info from someone who isn't pretending to know more than they actually do!). Slick

# Fourier Transform (spectrum) of QPSK and BPSK?

Started by ●November 12, 2006

Posted by ●November 12, 2006

radio913@aol.com wrote:> Hi All, > > 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)? > > 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). > > The bottom line is I'm trying to measure phase > and amplitude imbalance with just a spectrum analyzer, > without having to buy an expensive VSA which can display > the constellation, and give me EVM, etc. > > Thanks for any REAL help (which is Info from someone > who isn't pretending to know more than they actually do!). > > > SlickIt will be harder to discern the imbalances from a QPSK SA display than with an ordinary tone. Try changing the modulation to a complex exponential: sin(wt) + j*cos(wt). You get a three-tone display on the SA. From left to right, the components are: opposite sideband (undesired), carrier (undesired), and desired signal. You can directly measure the levels in dBc of the undesired components. I'm not sure if you can relate them back mathematically to imbalances in the modulator, but what you can do is tweak the modulator until they are minimized. John

Posted by ●November 12, 2006

"radio913@aol.com" <radio913@aol.com> writes:> Hi All, > > 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)?From [proakiscomm], the spectrum expression for any linear modulation (such as QPSK) is Phi(f) = (1/T) * | G(f) |^2 * Phi_ii(f), where G(f) is the Fourier transform of the transmit pulse shape g(t) and Phi_ii(f) is the power spectral density of the information sequence I(n). T is the symbol period. So without knowing your pulse shape, we can't really give you a specific spectrum. --Randy @BOOK{proakiscomm, title = "{Digital Communications}", author = "John~G.~Proakis", publisher = "McGraw-Hill", edition = "fourth", year = "2001"} -- % Randy Yates % "Though you ride on the wheels of tomorrow, %% Fuquay-Varina, NC % you still wander the fields of your %%% 919-577-9882 % sorrow." %%%% <yates@ieee.org> % '21st Century Man', *Time*, ELO http://home.earthlink.net/~yatescr

Posted by ●November 12, 2006

radio913@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.> > 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). > > The bottom line is I'm trying to measure phase > and amplitude imbalance with just a spectrum analyzer, > without having to buy an expensive VSA which can display > the constellation, and give me EVM, etc.1. Run 01 10 01 10 01 10 ... and 11 00 11 00 ... patterns to measure the carrier suppression in I and Q channels. There should be no peak at the carrier frequency. 2. Run 00 01 11 10 00 01 11 10 .... pattern to measure the amplitude/phase balance between I and Q. There should be a symmetrical spectrum shifted up from the center frequency.> > 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. Vladimir Vassilevsky DSP and Mixed Signal Design Consultant http://www.abvolt.com

Posted by ●November 12, 2006

On Nov 12, 6:36=A0am, Randy Yates <y...@ieee.org> wrote:> "radio...@aol.com" <radio...@aol.com> writes: > > Hi All, > > > =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)?From [proakisco=mm], the spectrum expression for any linear modulation> (such as QPSK) is > > =A0 Phi(f) =3D (1/T) * | G(f) |^2 * Phi_ii(f), > > where G(f) is the Fourier transform of the transmit pulse shape g(t) > and Phi_ii(f) is the power spectral density of the information > sequence I(n). T is the symbol period. > > So without knowing your pulse shape, we can't really give you a > specific spectrum. >The data on the I and Q channels is just a square wave from a LVTTL source. The symbol rate is 200kHz, which is 400kBits/sec. The above equation doesn't look correct, because if you have an infinitely long symbol period, which is the same as staying in one quadrant forever, then we would still have amplitude at the carrier frequency. The above equation goes to zero for an infinitely long T. Slick

Posted by ●November 12, 2006

radio913@aol.com skrev:> Hi All,> Thanks for any REAL help (which is Info from someone > who isn't pretending to know more than they actually do!).Do you have any particular person in mind? If not, how do you tell the difference between pretenders and others? Rune

Posted by ●November 12, 2006

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.A single pulse would have a continuous spectrum, which is not what i'm measuring at all. 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.> > > > =A0 =A0 =A0 =A0 The signal i am measuring doesn't have any > > =A0sideband carrier suppression at all, =A0which 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). > > > =A0 =A0 =A0 =A0 =A0The bottom line is I'm trying to measure phase > > and amplitude imbalance with just a spectrum analyzer, > > without having to buy an expensive VSA which can display > > the constellation, and give me EVM, etc.1. Run 01 10 01 10 01 10 ... an=d 11 00 11 00 ... patterns to measure the> carrier suppression in I and Q channels. There should be no peak at the > carrier frequency.That would be BPSK in the cases you mention.> > 2. Run 00 01 11 10 00 01 11 10 .... pattern to measure the > amplitude/phase balance between I and Q. There should be a symmetrical > spectrum shifted up from the center frequency.That's the pattern i'm using. The center carrier is suppressed, but the sidebands are very close in amplitude, so unless my orthogonality is way off, I don't see any indication of amplitude/phase imbalance.> > > > > =A0 =A0 =A0 =A0 =A0 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. Slick

Posted by ●November 12, 2006

On Nov 12, 12:21=A0pm, "Rune Allnor" <all...@tele.ntnu.no> wrote:> radio...@aol.com skrev: > > > Hi All, > > =A0 =A0 =A0 =A0 =A0 Thanks for any REAL help (which is Info from someone > > who isn't pretending to know more than they actually do!).Do you have a=ny particular person in mind? If not, how do you> tell the difference between pretenders and others? >=20I had YOU in mind, judging from your other posts!

Posted by ●November 12, 2006

radio913@aol.com skrev:> On Nov 12, 12:21?pm, "Rune Allnor" <all...@tele.ntnu.no> wrote: > > radio...@aol.com skrev: > > > > > Hi All, > > > ? ? ? ? ? Thanks for any REAL help (which is Info from someone > > > who isn't pretending to know more than they actually do!).Do you have anyparticular person in mind? If not, how do you> > tell the difference between pretenders and others? > > > > I had YOU in mind, judging from your other > posts!Well, that's usenet for you; you get exactly what you pay for. As others already hinted at, you might be better off contacting people whose knowledge you acknowledge, and pay them. Rune

Posted by ●November 12, 2006

"radio913@aol.com" <radio913@aol.com> writes:> On Nov 12, 11:18�am, Vladimir Vassilevsky<antispam_bo...@hotmail.com>> wrote: >> radio...@aol.com wrote: >> > � � � � Does anyone have amathematical 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.> 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. 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.>> > � � � � � Thanksfor any REAL help (which is Info from someone>> > who isn't pretending to know more than they actually do!).Your majesty's "Thankyou" 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. -- % Randy Yates % "I met someone who looks alot like you, %% Fuquay-Varina, NC % she does the things you do, %%% 919-577-9882 % but she is an IBM." %%%% <yates@ieee.org> % 'Yours Truly, 2095', *Time*, ELO http://home.earthlink.net/~yatescr

Posted by ●November 15, 2006

julius wrote:> radio913@aol.com wrote: > > > > An ideal IQ mixer has channels that are perfectly > > orthogonal to each other, or offset at a perfect 90 degrees. > > Also, their amplitudes are the same, so that for each of the > > 4 symbol states of QPSK, all amplitudes are the same, and > > all phase deltas are 90 degrees (or multiples of 90). > > > > Usually, people will measure this imbalance with a > > VSA, which has the cabability of displaying the constellation, > > along with error vector magnitude, rms phase error and amplitude > > imbalance, as well as DC leakage, etc. > > > > This link supposedly describes a way to get a figure of merit of > > these > > imbalances with a spectrum analyzer alone: > > > > http://www.rfcafe.com/references/electrical/quad_mod.htm > > > > Thanks for the explanation. Unfortunately, that article is unreadable > to me. Are you saying that due to IQ imbalance, your QPSK > constellation will not look square anymore, but rather like a > parallelogram? > > In that case, in complex baseband you can still write: > > x(t) = \sum_n I_n p(t-nT), .... (1) > > but in this new case your symbols I_n take values on the edges of a > parallelogram. > > Now, for any modulation scheme of form (1), the spectrum is given by > what Randy posted previously. All that is left is for you to find the > spectrum > of the "information sequence", which in this case should be called the > symbol sequence. > > Further, care should be given when comparing something that is measured > to something that is derived analytically. The power spectrum is the > *average* power at different frequencies, it is not exact. The less > amount > of randomness you have within your observation window, the less likely > it > is to look like its average realization. Well, under some assumption > anyway. Now, that is maybe what you saw when you make your symbol > rate really large, because now you will have to have a proportionally > larger > observation window to see the same thing. > > What do you think? >I think i'm asking the wrong group. S

Posted by ●November 15, 2006

radio913@aol.com wrote:> > An ideal IQ mixer has channels that are perfectly > orthogonal to each other, or offset at a perfect 90 degrees. > Also, their amplitudes are the same, so that for each of the > 4 symbol states of QPSK, all amplitudes are the same, and > all phase deltas are 90 degrees (or multiples of 90). > > Usually, people will measure this imbalance with a > VSA, which has the cabability of displaying the constellation, > along with error vector magnitude, rms phase error and amplitude > imbalance, as well as DC leakage, etc. > > This link supposedly describes a way to get a figure of merit of > these > imbalances with a spectrum analyzer alone: > > http://www.rfcafe.com/references/electrical/quad_mod.htm >Thanks for the explanation. Unfortunately, that article is unreadable to me. Are you saying that due to IQ imbalance, your QPSK constellation will not look square anymore, but rather like a parallelogram? In that case, in complex baseband you can still write: x(t) = \sum_n I_n p(t-nT), .... (1) but in this new case your symbols I_n take values on the edges of a parallelogram. Now, for any modulation scheme of form (1), the spectrum is given by what Randy posted previously. All that is left is for you to find the spectrum of the "information sequence", which in this case should be called the symbol sequence. Further, care should be given when comparing something that is measured to something that is derived analytically. The power spectrum is the *average* power at different frequencies, it is not exact. The less amount of randomness you have within your observation window, the less likely it is to look like its average realization. Well, under some assumption anyway. Now, that is maybe what you saw when you make your symbol rate really large, because now you will have to have a proportionally larger observation window to see the same thing. What do you think? Julius

Posted by ●November 14, 2006

juliusk@gmail.com wrote:> 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"? >An ideal IQ mixer has channels that are perfectly orthogonal to each other, or offset at a perfect 90 degrees. Also, their amplitudes are the same, so that for each of the 4 symbol states of QPSK, all amplitudes are the same, and all phase deltas are 90 degrees (or multiples of 90). Usually, people will measure this imbalance with a VSA, which has the cabability of displaying the constellation, along with error vector magnitude, rms phase error and amplitude imbalance, as well as DC leakage, etc. This link supposedly describes a way to get a figure of merit of these imbalances with a spectrum analyzer alone: http://www.rfcafe.com/references/electrical/quad_mod.htm S This article

Posted by ●November 14, 2006

Randy Yates wrote:> "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? > --More than you! S

Posted by ●November 14, 2006

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

Posted by ●November 14, 2006

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

Posted by ●November 14, 2006

"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 thatequation>> > 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

Posted by ●November 14, 2006

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

Posted by ●November 14, 2006

"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

Posted by ●November 13, 2006

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