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

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

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

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

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

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

"radio913@aol.com" <radio913@aol.com> writes:

> On Nov 13, 9:41&#4294967295;pm, Randy Yates <y...@ieee.org> wrote:
>> "radio...@aol.com" <radio...@aol.com> writes:
>> > [...]
>> > &#4294967295; &#4294967295; &#4294967295; 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

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

"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

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. =A0

      Especially 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