Re: Demodulating QPSK


"Vladimir Vassilevsky" <antispam_bogus@xxxxxxxxxxx> wrote in message news:Dvy0k.2088$ZE5.1498@xxxxxxxxxxxxxxxxxxxxxxx


John E. Hadstate wrote:


After extensive tests, I've hit upon one robust, sure-fire system for demodulating long-duration QPSK. (This system is not especially useful for signals that occur in short bursts.)

Why reinventing a wheel?


Your wheel doesn't work very well.



Demodulation of the long duration QPSK without noise, fading and frequency wander is a piece of cake task, and there is 10000 solutions for that.


Yes, but 9,998 of them don't work at all and the other two didn't work well enough to use in my application. In fact, mine doesn't work well enough to use in my application.



As always with this type of system, you will probably need to re-map your recovered dibits based on some knowledge of the signal's contents and internal structure.

In my implementation, all filters were two-pole IIR filters. The low-pass filters were designed by applying the BLT to Butterworth pole placements from the analog domain.

Minor technical detail. Why does it matter?


It matters a great deal to anyone who is trying to design and/or implement a highly responsive feedback loop, but then you're not competent in that area, are you?




The band-pass filters were designed with complex-conjugate poles just inside the unit circle and on the radial lines corresponding to the desired center frequency. All of the critical frequencies were either a multiple or a fraction of the baud rate of the dibits.

Too many of the clever scientific words. Understood nothing.


That says it all, and explains a lot of the rest of the bumbling incompetence that pervades your posts. If you'd had any brains, you'd have stopped right there, but of course, you were on a roll. I'm sorry if this is all too technical for you, you know, "Too many of the clever scientific words" and all that, but I have a strong hunch that there are people out there who can use the information.





Here's the block diagram in text form:

1. What is new about it?


See, what I mean? You can't complain that I'm creating a new solution to a solved problem and then simultaneously complain that there's nothing new about it. But then, logic has never been your strong suit, has it?



2. Why this is better then a classic approach?



It works, and if you had ever read any of the literature on the subject, you'd have already grasped that it is one of the classic approaches.


1. Re-tune t QPSK signal so that its frequency is an integral multiple of the baud rate.

Having all frequencies coupled is inflexible and inconvenient.


How? The inconvenient truth is that it is necessary. Why should I care if it inconveniences you?



To do this, you need to have some estimate of frequency of the incoming signal. You don't necessarily know the exact frequency of the original signal, but you must estimate "close enough". Initialize a frequency accumulator to the re-tuned frequency.

2. Square the signal from (1). Bandpass filter around 2x the nominal (re-tuned) frequency.

3. Square the signal from (2). Bandpass filter around 4x the nominal (re-tuned) frequency.

4-th order loop classics.


Yep. That's how we make it work, but I thought that one of your complaints was that it's not a "classical" approach.



4. Multiply the signal from (3) by a sin and cos from a quadrature Numerically Controlled Oscillator (NCO) running at 4x the frequency indicated by the frequency accumulator.

5. Low-pass filter the I and Q arms to remove the 8x component.

Trivial implementation of a bandpass in complex domain.


Yep. That's one way to make it work. Thanks for your "help".



Note: At this point you are dealing with frequencies of at least 8x the re-tuned signal (which might be higher than the frequency of your original signal). Make sure your sample rate is adequate to handle these frequencies.

Another stupid limitation.


Blame Nyquist. I didn't make the rules, but I do have brains enough to know which rules I have to live by, and point them out as pitfalls for the unwary.



6. Use the first difference of atan2(Q,I) to estimate the rate-of-change of phase angle between I and Q. Unwrap this result so it stays between -Pi and +Pi. Using this technique makes your PLL's loop gain less vulnerable to changes in signal level.

And less noise resistant, too.


Oh, really? Obviously you're not familiar with the literature on the subject.



7. Filter the phase rate-of-change, multiply by a gain constant and subtract it from the frequency accumulator. (This is the same frequency accumulator that is multiplied by 4 and used as the control frequency of the NCO in step (4).

This could be done much simpler.


Yep, probably so. But my method works. (For another method that would probably work, see Rick Lyons book. I won't tell you which one, or where, because you could obviously stand to read the whole thing.)



At this point, if you've got all the filters right and the gain right, you will have constructed a PLL that will lock-up fairly quickly at a frequency of 4x the re-tuned frequency and it will follow changes in the carrier frequency of the re-tuned signal.

Not any different from the 4-th order Costas Loop.


A lot different, especially because mine works. This remark suggests that you don't even know what a Costas loop is.



The double squaring will have wiped out most of the effect of modulation on the frequency estimation.

Classics. So what is new?


I'm beginning to see a pattern in your posts: there is a strong demonstration of the fact that those who can, do, and those who can't, crap on those who do. In case you've any doubts, you're one of those who can't, and you've got crap all over yourself.



8. Apply the frequency from the frequency accumulator in (7) directly to another quadrature NCO. It is this NCO that produces the reconstructed carrier.

So?


So if you try demodding the signal with any other carrier frequency, the constellation will spin like a top.


9. Multiply the re-tuned signal from (1) by the sin and cos from the NCO in (8).

So?


So it's called a product mixer, a very useful tool for getting at phase.



10. Low-pass filter the I and Q arms from (9) to eliminate the 2x frequency.

So?

11. Look at the signs of the signal on the I and Q arms in (10) to decide which quadrant the signal is in. You have to invent the necessary logic to synchronize, decimate, and recover the demodulated dibits. Hint: it helps if you sampled the incoming signal at a rate that was an integral multiple of the baud rate (in addition to being at least 16x the re-tuned frequency).

What is the point of this very trivial exersize?


Vladimir Vassilevsky
DSP and Mixed Signal Design Consultant
http://www.abvolt.com


At this point, you've made it very clear that you haven't the background or education to benefit from any information I can offer, and I don't feel the need to waste any more time with you.

I do hope that before contracting for your "Consulting Services", your potential clients stumble upon some of your posts here. It will save them a lot of time and money.


.

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