Q: Phase-coherence vs phase-continuous

I've looked for info on this and found:

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The article states:

Phase-coherence is preferred because the phase is remembered. Phase-continuous is lower bandwidth.

They do not say why phase-coherence is preferred.

If one is doing FSK signaling, why is phase-coherence preferred?

Thanks.

Reply to
John S
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Oops! This was my conclusion ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

Reply to
John S

phase_coherent.html

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.

--
Tim Wescott 
Control system and signal processing consulting 
www.wescottdesign.com
Reply to
Tim Wescott

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.

--
Dave Platt                                    AE6EO 
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Reply to
Dave Platt

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.

Reply to
miso

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

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
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hobbs at electrooptical dot net 
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Reply to
Phil Hobbs

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

Reply to
miso

MSK demodulates like PSK. Reportedly, you can demodulate continuous- phase Bell 202 (I think it's Bell 202: it's the one with a 2:3 ratio between frequency separation and baud rate) like a mutant PSK, too.

MSK certainly gives you SNR advantages, and with some of the bandwidth advantages of GMSK (which is, after all, a modified form of MSK).

--
Tim Wescott 
Control system and signal processing consulting 
www.wescottdesign.com
Reply to
Tim Wescott

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

Thanks.

Reply to
John S

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

Reply to
John S

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Very slowly, methodically, and carefully.

Sub hertz sample rates. ;-) Hehehehehe!

Reply to
SoothSayer

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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...

--
Les Cargill
Reply to
Les Cargill

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.

--
Dave Platt                                    AE6EO 
Friends of Jade Warrior home page:  http://www.radagast.org/jade-warrior 
  I do _not_ wish to receive unsolicited commercial email, and I will 
     boycott any company which has the gall to send me such ads!
Reply to
Dave Platt

nt.html

Formal consistency check failed. Please resubmit.

?-)

Reply to
josephkk

interference

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

?-)

Reply to
josephkk

to

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

?-)

Reply to
josephkk

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