Could someone explain to me BFSK Modulation on the Simulink model below?

Binary frequency shift keying (BFSK)

The simulation above simulates a binary FSK system with a correlation (or matched filter) receiver.

I have to understand the general idea about it.

  • Could someone explain this?
  • Moreover I cannot understand the meaning of correlator and tone frequency fn the above simulator?
  • $\begingroup$ Is the signal to the multiplier complex or real? I have seen FSK demodulators with phase rotators and can explain that case if that is indeed what is going on (that would not be clear from the block diagram as shown). If the signal is real and the two references are indeed just combined can you then comment on what frequency each reference is and what frequency deviation of the BFSK signal is for the two FSK symbols? $\endgroup$ Mar 28, 2017 at 20:55
  • $\begingroup$ I wish people would at least explain why they down-vote a question when it is down-voted. Not sure what was wrong with this question?? $\endgroup$ Mar 28, 2017 at 23:32

1 Answer 1


So, ignoring noise, the input into the receiver multiplier coming from the channel is either $\cos(2\pi f_0t)$ or $\cos(2\pi f_1t)$. The other multiplier input is $\cos(2\pi f_0t)-\cos(2\pi f_1t)$. All signals are defined from 0 to $T$. Let's assume the transmitted signal has frequency $f_0$. Then you have an integrator:

\begin{align} &\int_0^T \cos(2\pi f_0t) \left[\cos(2\pi f_0t)-\cos(2\pi f_1t)\right]dt \\ &=\int_0^T \cos(2\pi f_0t)\cos(2\pi f_0t) dt - \int_0^T \cos(2\pi f_0t)\cos(2\pi f_1t)dt. \end{align}

The frequencies $f_0$ and $f_1$ are chosen so that the cosines are orthogonal for $0<t<T$; that means that $$\int_0^T \cos(2\pi f_0t)\cos(2\pi f_1t)dt =0.$$ The integrate and dump block outputs the result of the definite integral at $t=T$. You can see that the result is positive if frequency $f_0$ was transmitted and negative if $f_1$ was transmitted.

The sign block is a quantizer, and the lookup table converts the numbers back to bits 0 and 1.

  • $\begingroup$ Very good @MBaz, so this implies a coherent demod is required, correct? Carrier recovery could be extracted from the multiplier output so I see how that could work well. $\endgroup$ Mar 28, 2017 at 22:55
  • $\begingroup$ @DanBoschen Yep, this would be a coherent receiver. It's also assumed that the I&D block is synchronized to the symbol time. $\endgroup$
    – MBaz
    Mar 28, 2017 at 23:03
  • $\begingroup$ very nice- just wanted to be sure I understood. Thanks for the explanation! $\endgroup$ Mar 28, 2017 at 23:23
  • $\begingroup$ @DanBoschen You're very welcome! $\endgroup$
    – MBaz
    Mar 28, 2017 at 23:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.