Fourier Transform (spectrum) of QPSK and BPSK?

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

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

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

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