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My knowledge of signals and signal processing is limited, so I apologize in advance for my naïve and possibly confusing question.

In the case of BFSK, the frequencies are altered to represent the data (1's and 0's). The frequency transitions are abrupt. You can use Continuous Phase BFSK to "line up" the transitions in frequency.

Now, for my question: lets say I am using BPSK which uses changes in phase to represent the data. If the phase changes from 0 to 180 (or vice versa), wouldn't that have a similar abruptness as BFSK? Is there something I can do, similar to CP BFSK to smooth the transition out?

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You're right that the transitions can be abrupt, but you can always choose to use pulse shaping, i.e., instead of using a rectangular pulse, use some smoother function $p(t)$. The corresponding baseband signal is then

$$s(t)=\sum_kA_kp(t-kT)\tag{1}$$

where $A_k$ are the symbols, and $T$ is the symbol length. A smooth function $p(t)$ will avoid abrupt transitions, and will therefore reduce spectral sidelobe levels. I think this paper uses that idea.

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  • $\begingroup$ Reducing the side lobes (mentioned in the linked paper), is definitely what we want to achieve. So that I am clear, what is k? $\endgroup$ – Ironchef Dec 20 '18 at 16:28
  • $\begingroup$ @Ironchef: That's just the symbol index. $\endgroup$ – Matt L. Dec 20 '18 at 17:03
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If we use this definition of instantaneous frequency, you can construct a BFSK signal that is continuous in phase but is discontinuous in frequency.

If we denote the two phase signals as $\phi_1(t)$ and $\phi_2(t)$, corresponding to frequencies $f_1,f_2$, and define them such that: $$ \phi_1(t) = 2\pi f_1t + k_1 \\ \phi_2(t) = 2\pi f_2t + k_2 $$ and denote the symbol transition as occurring at $t = T_s$, then we just need to ensure that $$ \phi_1(T_s) = \phi_2(T_s) $$ for the phase to be continuous, which can be done by setting $k_1,k_2$ appropriately. (These constants will change each symbol time.)

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  • $\begingroup$ To my understanding the question was about BP(!)SK ... $\endgroup$ – Matt L. Dec 20 '18 at 15:48
  • $\begingroup$ I like your clear explanation, but I want to do something similar with BPSK, if that is possible. But maybe it is the same? My phase transitions are 0 and 180. Does the frequency get adjusted (e.g. by k1 in your formula) such that the phase transition is 0 (or 180)? $\endgroup$ – Ironchef Dec 20 '18 at 16:07

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