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Q: Phase-coherence vs phase-continuous

M

miso

Jan 1, 1970
0
Well, the article does say that phase coherence is preferred for some
services.

The theory is that if you have two independent generators (their A and B
signals) that you switch between, then it is easier to do coherent
demodulation, which gains you a few dB of performance in bit error rate
vs. SNR.

I'm not sure whether they're glossing over or I didn't read it, but for
certain kinds of FSK (MSK comes to mind) you can do coherent demodulation
from a phase-continuous transmitted signal, thereby getting your more
favorable performance vs. SNR while still minimizing bandwidth.
If you are not continuous phase, you need more bandwidth. Noise is
proportional to the square root of the bandwidth, so I can't see phase
coherence being beneficial in performance. I suppose if you went GMSK
and reduced the bandwidth, then your point is valid.
 
M

miso

Jan 1, 1970
0
Could be, or alternatively it could be reduced inter-symbol interference
due to not having a bunch of random-height step transients at every
transition.

Cheers

Phil Hobbs

CPFSK eliminates the jumps, but the author is espousing the virtues of
PCFSK, i.e. just the opposite.
 
J

John S

Jan 1, 1970
0
My guess is that some FSK detectors may operate by locking a pair of
PLLs to the two tones in the signal, with the PLLs having a narrow
capture range and a low rolloff in their loop filter. If the
transmission is phase-coherent, then each PLL would be "right on
target" after a state change in the input signal and would not have to
drift around by a variable amount and re-lock to a different signal
phase.

I'm not sure whether this would be an issue for demodulators using
quadrature detection.

Excellent point! I had not considered the use to two PLLs. Maybe at high
data rates that would be important.

Thanks.
 
J

John S

Jan 1, 1970
0
My guess is that some FSK detectors may operate by locking a pair of
PLLs to the two tones in the signal, with the PLLs having a narrow
capture range and a low rolloff in their loop filter. If the
transmission is phase-coherent, then each PLL would be "right on
target" after a state change in the input signal and would not have to
drift around by a variable amount and re-lock to a different signal
phase.

Yes, but how do you distinguish a frequency shift using two slow PLLs?
 
S

SoothSayer

Jan 1, 1970
0
Yes, but how do you distinguish a frequency shift using two slow PLLs?

Very slowly, methodically, and carefully.

Sub hertz sample rates. ;-) Hehehehehe!
 
L

Les Cargill

Jan 1, 1970
0
Dave said:
My guess is that some FSK detectors may operate by locking a pair of
PLLs to the two tones in the signal, with the PLLs having a narrow
capture range and a low rolloff in their loop filter. If the
transmission is phase-coherent, then each PLL would be "right on
target" after a state change in the input signal and would not have to
drift around by a variable amount and re-lock to a different signal
phase.

I'm not sure whether this would be an issue for demodulators using
quadrature detection.

Using a PLL to make a matched filter seems iffy. Couple
of resonant filters and a comparator sounds better to my ear. 'Course,
there's a lot of dance-management too...
 
J

josephkk

Jan 1, 1970
0
CPFSK eliminates the jumps, but the author is espousing the virtues of
PCFSK, i.e. just the opposite.

I am fairly sure you disunderstand. Can you (or anybody here) explain the
difference CPFSK and PCFSK clearly(; and the S/N effects)?

?-)
 
J

josephkk

Jan 1, 1970
0
Coherent demodulation.

If you've got two PLLs, each one of which has locked to one of the two
tones in both frequency and phase (e.g. during a training sequence),
then you can simply multiply the incoming signal by each of the two
PLL outputs, and low-pass-filter the results.

When the input is receiving the "low" tone, the input and low-tone PLL
will be in-phase sinusoids, the product of the multiplication will
always be non-negative, and the low-pass-filtered version will be
positive. The input and high-tone PLL will be sinusoids of different
frequencies, of opposite polarity half of the time, the product of the
two will average out to zero and the filtered product will be close to
zero most of the time. Run the two filtered products into a
comparator, and the output will be the original data signal (prior to
the FSK modulation).

You have to set the low-pass filter appropriately, of course, based on
the baud rate of the data signal.

Just for grins can you set up a simulation to play with? I don't quite
follow yet.

?-)
 
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