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Continuous phase modulation has been described to me very similar to wikipedia's definition, which basically indicated that the CPM is different from QPSK, for example, because QPSK abruptly shifts phase on symbol transitions. But this isn't true. Every real system is bandlimited, and as such there is no such thing as an abrupt or instantaneous change. In QPSK systems, the bandwidth is limited by a nyquist filter such as a root raised cosine, to be 2*(1+ the excess bandwidth)* the symbol rate.

Per wikipedia; in CPM, "Each symbol is modulated by gradually changing the phase of the carrier from the starting value to the final value, over the symbol duration." This is exactly how PSK works.

My question is, every real system is bandlimited, which means that symbol transitions occur over a finite time period (1/2T ?), which means the instantanous phase of the waveform is continous. That said, what does Continous Phase Modulation mean to imply is done differently than in any other phase modulated waveform

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My understanding of the difference is as follows:

With CPM the change in phase happens over the entire time of the symbol. With QPSK it happens at the beginning. Agreed, it's not an abrupt change from one phase to another, but it still takes considerably less time than the time for the symbol.

So in QPSK there is a change phase, followed by a hold phase. In CPM there is just the change phase, and no hold phase.

Say a symbol lasts for 4 wavelengths, and you want to change from 0° to 90°. In QPSK the first wave would be the transition from 0° to 90° (it would typically take less than 1 wave to change phase), and the next 3 would be holding at 90°. With CPM each wave would be a different phase from 0° to 90° - say 22.5°, 45°, 67.5° and 90°.

Consequently, CPM encodes extra information that's not present in QPSK. If you go from 0° to 180°, which way did it go? Clockwise, or counter-clockwise? QPSK has no way of knowing that. With CPM that direction of change is also encoded, along with the amount of change.

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  • \$\begingroup\$ So, I have a digital QPSK modem that is operating on complex baseband I/Q data at 2 samples per symbol. Assuming my all my loops are working, 1 sample is the data sample which gets fed to the slicer, and 1 sample is the transition or timing sample which can be used for timing recovery. In this case, the timing sample is the phase halfway between the previous and next symbol (or through zero). \$\endgroup\$ – Jotorious Oct 21 '14 at 13:58
  • \$\begingroup\$ perhaps, when running at 2 samples/symbol, CPM and QPSK converge. \$\endgroup\$ – Jotorious Oct 21 '14 at 14:20
  • \$\begingroup\$ Possibly, but only if it takes an entire wavelength to change phase. \$\endgroup\$ – Majenko Oct 21 '14 at 14:27
  • \$\begingroup\$ I don't understand how the number of waveperiods affects this. To demodulate, you bring the signal to complex baseband, and there is no wavelength to be concerned with. The signal in baseband QAM-Space transitions between the symbol points as fast as the RX pulse shaping filter lets it. I'm still not getting the difference, other than the fact that it is being said that QPSK "can" be transmitted with excess bandwidth. If I'm running at nyquist or close to nyquist sampling rates, the transitions between symbols are as bandwidth efficient as is physically possible, with the transitions as slow \$\endgroup\$ – Jotorious Oct 21 '14 at 19:55
  • \$\begingroup\$ as are physically possible for that symbol rate. \$\endgroup\$ – Jotorious Oct 21 '14 at 19:56

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