I have a question relating to demodulating QPSK.
QPSK can not be demodulated unless the exact freq and phase of the
unmodulated carrier is known.
Therefore, the classic way of demodulating QPSK is to Recover the
Carrier. This can be done with a Costas Loop or a Squaring
circuit---both of which require the presence of the "carrier" in each of
the bits being received. (Yes--I'm aware that QPSK is a suppressed
carrier system---BUT
what I am calling the "Carrier" is the RF signal present, in one fo the
the 4 phases, inside the actual bit).
OK---so to demodulated QPSK, into the I and Q mixers the LO is at 0
degrees
and 90 degrees phase, respectively. That is---not only is the LO
coherent with the incoming "carrier"---but it is In-phase with the I
"carrier" and in perfect quadrature with the Q "carrier"---an as such
there is zero crosstalk between the I and Q baseband outputs.
OK---but what happens if the LO just happens to be at exactly the
frequency of the incoming "carrier"---but NOT in-phase with the I
"carrier" and not in quadrature with the Q "Carrier". Then we will get
crosstalk---which is as situation where there is some I and some Q in
the designated "I" baseband ---and some I and some Q in the designated
"Q" baseband.
Therefore, there is really no "I" baseband--nor any "Q" baseband---for
they are drifting through each other. Note that this situation was
caused by a LO that was not in-phase and nor in-quadrature with the
incoming "carrier"s I and Q phases. What we have is an I channel that
has both I and Q basebands on top of each other---and a Q channel that
has both I and Q basebands on top of each other. The bit transitions
will be at the same time---so you will situations where the I and Q bits
either add together or cancel each other out. You will get only 2
states ---a "1" or a "0"---and you will not be able to separate the I
baseband from the Q baseband. That is how I see it.
Someone told me that the above situation is OK---that putting the above
described "I" and "Q" channels into a digital demod will result in a
separate I datastream and aseparate Q data stream. I don't see how that
is possible.
Would someone please either affirm what I say---or clue me in.
Thank you.
QPSK question
Started by ●February 23, 2009
Reply by ●February 23, 20092009-02-23
There are some very nice pictures on Wikipedia:
http://en.wikipedia.org/wiki/QPSK
that kind of show why it doesn't matter.
If you think about the Fourier Transform of those signals,
you'll see there are lot of higher harmonics. The shock of
changing phase forces that. So even if the LO drifts, those
higher harmonics are still present (you get sums and differences
of all frequencies), so those shock points still come through.
That is your base clock, which is different from the carrier.
You may get the exact opposite of what you want - so 00 is really
11 and 10 is really 01 (and vice versa) which can happen when your
clock is not locked correctly. But the I and Q streams are still
there, they start orthogonal and they stay orthogonal. It's not
like Schrodenger's cat!
Patience, persistence, truth,
Dr. mike
On Mon, 23 Feb 2009, P...@L-3Com.com wrote:
> I have a question relating to demodulating QPSK.
> QPSK can not be demodulated unless the exact freq and phase of the
> unmodulated carrier is known.
> Therefore, the classic way of demodulating QPSK is to Recover the
> Carrier. This can be done with a Costas Loop or a Squaring
> circuit---both of which require the presence of the "carrier" in each of
> the bits being received. (Yes--I'm aware that QPSK is a suppressed
> carrier system---BUT
> what I am calling the "Carrier" is the RF signal present, in one fo the
> the 4 phases, inside the actual bit).
>
> OK---so to demodulated QPSK, into the I and Q mixers the LO is at 0
> degrees
> and 90 degrees phase, respectively. That is---not only is the LO
> coherent with the incoming "carrier"---but it is In-phase with the I
> "carrier" and in perfect quadrature with the Q "carrier"---an as such
> there is zero crosstalk between the I and Q baseband outputs.
>
> OK---but what happens if the LO just happens to be at exactly the
> frequency of the incoming "carrier"---but NOT in-phase with the I
> "carrier" and not in quadrature with the Q "Carrier". Then we will get
> crosstalk---which is as situation where there is some I and some Q in
> the designated "I" baseband ---and some I and some Q in the designated
> "Q" baseband.
>
> Therefore, there is really no "I" baseband--nor any "Q" baseband---for
> they are drifting through each other. Note that this situation was
> caused by a LO that was not in-phase and nor in-quadrature with the
> incoming "carrier"s I and Q phases. What we have is an I channel that
> has both I and Q basebands on top of each other---and a Q channel that
> has both I and Q basebands on top of each other. The bit transitions
> will be at the same time---so you will situations where the I and Q bits
> either add together or cancel each other out. You will get only 2
> states ---a "1" or a "0"---and you will not be able to separate the I
> baseband from the Q baseband. That is how I see it.
>
> Someone told me that the above situation is OK---that putting the above
> described "I" and "Q" channels into a digital demod will result in a
> separate I datastream and aseparate Q data stream. I don't see how that
> is possible.
> Would someone please either affirm what I say---or clue me in.
>
> Thank you.
>
http://en.wikipedia.org/wiki/QPSK
that kind of show why it doesn't matter.
If you think about the Fourier Transform of those signals,
you'll see there are lot of higher harmonics. The shock of
changing phase forces that. So even if the LO drifts, those
higher harmonics are still present (you get sums and differences
of all frequencies), so those shock points still come through.
That is your base clock, which is different from the carrier.
You may get the exact opposite of what you want - so 00 is really
11 and 10 is really 01 (and vice versa) which can happen when your
clock is not locked correctly. But the I and Q streams are still
there, they start orthogonal and they stay orthogonal. It's not
like Schrodenger's cat!
Patience, persistence, truth,
Dr. mike
On Mon, 23 Feb 2009, P...@L-3Com.com wrote:
> I have a question relating to demodulating QPSK.
> QPSK can not be demodulated unless the exact freq and phase of the
> unmodulated carrier is known.
> Therefore, the classic way of demodulating QPSK is to Recover the
> Carrier. This can be done with a Costas Loop or a Squaring
> circuit---both of which require the presence of the "carrier" in each of
> the bits being received. (Yes--I'm aware that QPSK is a suppressed
> carrier system---BUT
> what I am calling the "Carrier" is the RF signal present, in one fo the
> the 4 phases, inside the actual bit).
>
> OK---so to demodulated QPSK, into the I and Q mixers the LO is at 0
> degrees
> and 90 degrees phase, respectively. That is---not only is the LO
> coherent with the incoming "carrier"---but it is In-phase with the I
> "carrier" and in perfect quadrature with the Q "carrier"---an as such
> there is zero crosstalk between the I and Q baseband outputs.
>
> OK---but what happens if the LO just happens to be at exactly the
> frequency of the incoming "carrier"---but NOT in-phase with the I
> "carrier" and not in quadrature with the Q "Carrier". Then we will get
> crosstalk---which is as situation where there is some I and some Q in
> the designated "I" baseband ---and some I and some Q in the designated
> "Q" baseband.
>
> Therefore, there is really no "I" baseband--nor any "Q" baseband---for
> they are drifting through each other. Note that this situation was
> caused by a LO that was not in-phase and nor in-quadrature with the
> incoming "carrier"s I and Q phases. What we have is an I channel that
> has both I and Q basebands on top of each other---and a Q channel that
> has both I and Q basebands on top of each other. The bit transitions
> will be at the same time---so you will situations where the I and Q bits
> either add together or cancel each other out. You will get only 2
> states ---a "1" or a "0"---and you will not be able to separate the I
> baseband from the Q baseband. That is how I see it.
>
> Someone told me that the above situation is OK---that putting the above
> described "I" and "Q" channels into a digital demod will result in a
> separate I datastream and aseparate Q data stream. I don't see how that
> is possible.
> Would someone please either affirm what I say---or clue me in.
>
> Thank you.
>
Reply by ●February 25, 20092009-02-25
Dr. Mike,
Thanks for your reply--but I'm not sure whether you are agreeing with my
position or not.
I won't repeat my original statement ad nauseum---but what I am saying
is that it is only possible to separate I data from Q data
(ie, no crosstalk) with the assistance of the carrier phase.
If we convert to "baseband" in one fell swoop, we make the LO not only
coherent with the incoming "carrier"---but also in-phase with the I
"carrier" and in quadrature with the Q "carrier:
If we do that---then we have gone directly down to I and Q baseband
(with no crosstalk--ie no i in Q,---and no Q in I). Thus we have
perfectly demodulated the I and Q data streams on the QPSK signal.
But if we just use a non-coherent LO to produce an IF--we then have a
rotating I and Q constellation ---and if we digitize that signal and
feed
itinto a DSP---then the DSP can perform the carrier recovery
and in-phase and quadrature phasing. The DSP can do that ONLY because
it is an IF that we are feeding to it. I'm saying that there HAS to be
a "carrier" accompanying the I and Q.
Au contraire, if we were to employ direct conversion (zero-IF) using
"0 degrees" and "90 degrees" mixers -- then I claim that we are fooling
ourselves by claiming that the mixer outputs represent "I" and "Q" data.
No--rather---what we have is a rolling baseband---with I and Q rolled
together in the same baseband channels. I don't see how a DSP can do
ANYTHING regarding demodulating this---because there is no longer a
"carrier"---and no longer a possibility of recovering a "carrier".
Am i correct in this?
Regards,
Phil
-----Original Message-----
From: Mike Rosing [mailto:e...@eskimo.com]
Sent: Monday, February 23, 2009 10:09 PM
To: Dragonetti, Philip G. @ CSW-NOVA
Cc: A...
Subject: Re: [adsp] QPSK question
There are some very nice pictures on Wikipedia:
http://en.wikipedia.org/wiki/QPSK
that kind of show why it doesn't matter.
If you think about the Fourier Transform of those signals,
you'll see there are lot of higher harmonics. The shock of
changing phase forces that. So even if the LO drifts, those
higher harmonics are still present (you get sums and differences
of all frequencies), so those shock points still come through.
That is your base clock, which is different from the carrier.
You may get the exact opposite of what you want - so 00 is really
11 and 10 is really 01 (and vice versa) which can happen when your
clock is not locked correctly. But the I and Q streams are still
there, they start orthogonal and they stay orthogonal. It's not
like Schrodenger's cat!
Patience, persistence, truth,
Dr. mike
On Mon, 23 Feb 2009, P...@L-3Com.com wrote:
> I have a question relating to demodulating QPSK.
> QPSK can not be demodulated unless the exact freq and phase of the
> unmodulated carrier is known.
> Therefore, the classic way of demodulating QPSK is to Recover the
> Carrier. This can be done with a Costas Loop or a Squaring
> circuit---both of which require the presence of the "carrier" in each
of
> the bits being received. (Yes--I'm aware that QPSK is a suppressed
> carrier system---BUT
> what I am calling the "Carrier" is the RF signal present, in one fo
the
> the 4 phases, inside the actual bit).
>
> OK---so to demodulated QPSK, into the I and Q mixers the LO is at 0
> degrees
> and 90 degrees phase, respectively. That is---not only is the LO
> coherent with the incoming "carrier"---but it is In-phase with the I
> "carrier" and in perfect quadrature with the Q "carrier"---an as such
> there is zero crosstalk between the I and Q baseband outputs.
>
> OK---but what happens if the LO just happens to be at exactly the
> frequency of the incoming "carrier"---but NOT in-phase with the I
> "carrier" and not in quadrature with the Q "Carrier". Then we will
get
> crosstalk---which is as situation where there is some I and some Q in
> the designated "I" baseband ---and some I and some Q in the designated
> "Q" baseband.
>
> Therefore, there is really no "I" baseband--nor any "Q" baseband---for
> they are drifting through each other. Note that this situation was
> caused by a LO that was not in-phase and nor in-quadrature with the
> incoming "carrier"s I and Q phases. What we have is an I channel that
> has both I and Q basebands on top of each other---and a Q channel that
> has both I and Q basebands on top of each other. The bit transitions
> will be at the same time---so you will situations where the I and Q
bits
> either add together or cancel each other out. You will get only 2
> states ---a "1" or a "0"---and you will not be able to separate the I
> baseband from the Q baseband. That is how I see it.
>
> Someone told me that the above situation is OK---that putting the
above
> described "I" and "Q" channels into a digital demod will result in a
> separate I datastream and aseparate Q data stream. I don't see how
that
> is possible.
> Would someone please either affirm what I say---or clue me in.
>
> Thank you.
>
Thanks for your reply--but I'm not sure whether you are agreeing with my
position or not.
I won't repeat my original statement ad nauseum---but what I am saying
is that it is only possible to separate I data from Q data
(ie, no crosstalk) with the assistance of the carrier phase.
If we convert to "baseband" in one fell swoop, we make the LO not only
coherent with the incoming "carrier"---but also in-phase with the I
"carrier" and in quadrature with the Q "carrier:
If we do that---then we have gone directly down to I and Q baseband
(with no crosstalk--ie no i in Q,---and no Q in I). Thus we have
perfectly demodulated the I and Q data streams on the QPSK signal.
But if we just use a non-coherent LO to produce an IF--we then have a
rotating I and Q constellation ---and if we digitize that signal and
feed
itinto a DSP---then the DSP can perform the carrier recovery
and in-phase and quadrature phasing. The DSP can do that ONLY because
it is an IF that we are feeding to it. I'm saying that there HAS to be
a "carrier" accompanying the I and Q.
Au contraire, if we were to employ direct conversion (zero-IF) using
"0 degrees" and "90 degrees" mixers -- then I claim that we are fooling
ourselves by claiming that the mixer outputs represent "I" and "Q" data.
No--rather---what we have is a rolling baseband---with I and Q rolled
together in the same baseband channels. I don't see how a DSP can do
ANYTHING regarding demodulating this---because there is no longer a
"carrier"---and no longer a possibility of recovering a "carrier".
Am i correct in this?
Regards,
Phil
-----Original Message-----
From: Mike Rosing [mailto:e...@eskimo.com]
Sent: Monday, February 23, 2009 10:09 PM
To: Dragonetti, Philip G. @ CSW-NOVA
Cc: A...
Subject: Re: [adsp] QPSK question
There are some very nice pictures on Wikipedia:
http://en.wikipedia.org/wiki/QPSK
that kind of show why it doesn't matter.
If you think about the Fourier Transform of those signals,
you'll see there are lot of higher harmonics. The shock of
changing phase forces that. So even if the LO drifts, those
higher harmonics are still present (you get sums and differences
of all frequencies), so those shock points still come through.
That is your base clock, which is different from the carrier.
You may get the exact opposite of what you want - so 00 is really
11 and 10 is really 01 (and vice versa) which can happen when your
clock is not locked correctly. But the I and Q streams are still
there, they start orthogonal and they stay orthogonal. It's not
like Schrodenger's cat!
Patience, persistence, truth,
Dr. mike
On Mon, 23 Feb 2009, P...@L-3Com.com wrote:
> I have a question relating to demodulating QPSK.
> QPSK can not be demodulated unless the exact freq and phase of the
> unmodulated carrier is known.
> Therefore, the classic way of demodulating QPSK is to Recover the
> Carrier. This can be done with a Costas Loop or a Squaring
> circuit---both of which require the presence of the "carrier" in each
of
> the bits being received. (Yes--I'm aware that QPSK is a suppressed
> carrier system---BUT
> what I am calling the "Carrier" is the RF signal present, in one fo
the
> the 4 phases, inside the actual bit).
>
> OK---so to demodulated QPSK, into the I and Q mixers the LO is at 0
> degrees
> and 90 degrees phase, respectively. That is---not only is the LO
> coherent with the incoming "carrier"---but it is In-phase with the I
> "carrier" and in perfect quadrature with the Q "carrier"---an as such
> there is zero crosstalk between the I and Q baseband outputs.
>
> OK---but what happens if the LO just happens to be at exactly the
> frequency of the incoming "carrier"---but NOT in-phase with the I
> "carrier" and not in quadrature with the Q "Carrier". Then we will
get
> crosstalk---which is as situation where there is some I and some Q in
> the designated "I" baseband ---and some I and some Q in the designated
> "Q" baseband.
>
> Therefore, there is really no "I" baseband--nor any "Q" baseband---for
> they are drifting through each other. Note that this situation was
> caused by a LO that was not in-phase and nor in-quadrature with the
> incoming "carrier"s I and Q phases. What we have is an I channel that
> has both I and Q basebands on top of each other---and a Q channel that
> has both I and Q basebands on top of each other. The bit transitions
> will be at the same time---so you will situations where the I and Q
bits
> either add together or cancel each other out. You will get only 2
> states ---a "1" or a "0"---and you will not be able to separate the I
> baseband from the Q baseband. That is how I see it.
>
> Someone told me that the above situation is OK---that putting the
above
> described "I" and "Q" channels into a digital demod will result in a
> separate I datastream and aseparate Q data stream. I don't see how
that
> is possible.
> Would someone please either affirm what I say---or clue me in.
>
> Thank you.
>
Reply by ●February 25, 20092009-02-25
On Tue, 24 Feb 2009, P...@L-3com.com wrote:
> Dr. Mike,
> Thanks for your reply--but I'm not sure whether you are agreeing with my
> position or not.
I guess I disagree, but you are not wrong :-)
> But if we just use a non-coherent LO to produce an IF--we then have a
> rotating I and Q constellation ---and if we digitize that signal and
> feed
> itinto a DSP---then the DSP can perform the carrier recovery
> and in-phase and quadrature phasing. The DSP can do that ONLY because
> it is an IF that we are feeding to it. I'm saying that there HAS to be
> a "carrier" accompanying the I and Q.
But we can ignore the carrier. The mixture gives us a sum and difference,
the difference is the IF. The DSP does the same thing, the multiplication
by a cosine at the fundamental gives a sum and difference - and we just
keep the IF. If the DSP is fast enough that is.
> Au contraire, if we were to employ direct conversion (zero-IF) using
> "0 degrees" and "90 degrees" mixers -- then I claim that we are fooling
> ourselves by claiming that the mixer outputs represent "I" and "Q" data.
> No--rather---what we have is a rolling baseband---with I and Q rolled
> together in the same baseband channels. I don't see how a DSP can do
> ANYTHING regarding demodulating this---because there is no longer a
> "carrier"---and no longer a possibility of recovering a "carrier".
>
> Am i correct in this?
The mixers create an IF, that's the point. If you just look at the input
wave and try to decode it directly then I think you are right - it would
be very difficult to tell which phase you are detecting. But from a DQPSK
perspective, you could always tell when you change the phase, so you could
at least get some data out.
I think the key here is that in the math we drop the fundamental frequency
- the carrier - because it divides out from all the equations. You can't
just have a "0 degree" mixer - you have to have a carrier there! But when
you write down the math the only thing left after you divide out
exp(i*w*t) is the fixed phase. In a phasor diagram, things are fixed.
But in the real world, it's cos(w*t) always.
That's why I mean you are not wrong, but you aren't quite looking at all
the details either. I and Q are definitly there, so is the base carrier,
and so is the lack of which phase has what meaning. And if your DSP is
fast enough, you don't need an IF, but you do have to work harder!
Patience, persistence, truth,
Dr. mike
> Dr. Mike,
> Thanks for your reply--but I'm not sure whether you are agreeing with my
> position or not.
I guess I disagree, but you are not wrong :-)
> But if we just use a non-coherent LO to produce an IF--we then have a
> rotating I and Q constellation ---and if we digitize that signal and
> feed
> itinto a DSP---then the DSP can perform the carrier recovery
> and in-phase and quadrature phasing. The DSP can do that ONLY because
> it is an IF that we are feeding to it. I'm saying that there HAS to be
> a "carrier" accompanying the I and Q.
But we can ignore the carrier. The mixture gives us a sum and difference,
the difference is the IF. The DSP does the same thing, the multiplication
by a cosine at the fundamental gives a sum and difference - and we just
keep the IF. If the DSP is fast enough that is.
> Au contraire, if we were to employ direct conversion (zero-IF) using
> "0 degrees" and "90 degrees" mixers -- then I claim that we are fooling
> ourselves by claiming that the mixer outputs represent "I" and "Q" data.
> No--rather---what we have is a rolling baseband---with I and Q rolled
> together in the same baseband channels. I don't see how a DSP can do
> ANYTHING regarding demodulating this---because there is no longer a
> "carrier"---and no longer a possibility of recovering a "carrier".
>
> Am i correct in this?
The mixers create an IF, that's the point. If you just look at the input
wave and try to decode it directly then I think you are right - it would
be very difficult to tell which phase you are detecting. But from a DQPSK
perspective, you could always tell when you change the phase, so you could
at least get some data out.
I think the key here is that in the math we drop the fundamental frequency
- the carrier - because it divides out from all the equations. You can't
just have a "0 degree" mixer - you have to have a carrier there! But when
you write down the math the only thing left after you divide out
exp(i*w*t) is the fixed phase. In a phasor diagram, things are fixed.
But in the real world, it's cos(w*t) always.
That's why I mean you are not wrong, but you aren't quite looking at all
the details either. I and Q are definitly there, so is the base carrier,
and so is the lack of which phase has what meaning. And if your DSP is
fast enough, you don't need an IF, but you do have to work harder!
Patience, persistence, truth,
Dr. mike
