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I read something about phase memory but didn't get proper information. My specific doubt is, let's say phase memory is L, then after L '1s', will we have a phase discontinuity and will the frequency restart from the beginning?

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  • \$\begingroup\$ What makes you think that the frequency keeps increasing? It is the phase which is increasing. But since \$cos(x) = cos(x+2n\pi)\$ with \$n = 1,2,3...\$ the increasing phase has no effect. \$\endgroup\$ – Bimpelrekkie Jun 27 '18 at 12:17
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You seem to misunderstand the mathematical formula given here.

In a system with two symbols, there will only be two frequencies, one indicating a '1' has been sent, one indicating that a '0' has been sent. In that system, the integral increases the argument of the cosine by $$D_{f} \cdot t$$ when sending a '1', which gives a cosine with frequency $$f_c + D_f $$ compared to a cosine with frequency $$f_c$$ when sending a '0'.

In a more general system with N symbols, there will be N possible frequencies.

There will never be a phase discontinuity, which is why it's called "Continuous Phase FSK".

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Continuous phase frequency shift keying is frequency shift keying without discontinuities in the final waveform: -

enter image description here

Picture source.

So, with FSK you have two frequencies; one for logic 0 and one for logic 1 and frequency remains at one of these two values whilever the logic level is constant.

Discontinuous FSK has a poor spectral shape due to the sudden changes in phase: -

enter image description here

Picture source.

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  • \$\begingroup\$ In discontinuous phase FSK, what if we choose f1 and f2 (frequencies) to be integral multiple of symbol time? Is the problem with ensuring that both the oscillators maintain the required initial phase synchronization? \$\endgroup\$ – Aniruddha Paturkar Jun 30 '18 at 16:33
  • \$\begingroup\$ Andy aka Sir, am I thinking in the right direction now? \$\endgroup\$ – Aniruddha Paturkar Jul 1 '18 at 14:37
  • \$\begingroup\$ @AniruddhaPaturkar yes, that is definitely a way of synchronizing the phase shift to avoid discontinuity. \$\endgroup\$ – Andy aka Jul 1 '18 at 18:21

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