0
$\begingroup$

What does the phase discriminator portion of the Costas Receiver do mathematically?

enter image description here

  1. The output of the $I$-channel is $ \frac{1}{2}A_C \cos \phi \, m(t) $. Which means for small deviation of phase $ \phi $ , $ \frac{1}{2}A_C \cos \phi \, m(t) \approx \frac{1}{2}A_C \, m(t) $. Or it can be said that $ \frac{1}{2}A_C \, m(t) $ would be attenuated by a small amount. However, we have to keep in mind as the text says (as it not might be as innocuous as it seems):

    enter image description here

  2. The output of the $Q$-channel is $ \frac{1}{2}A_C \sin \phi \, m(t) $. Which means for small deviation of phase $ \phi $ , $ \frac{1}{2}A_C \sin \phi \, m(t) \approx \frac{1}{2}A_C \phi \, m(t) $.

  3. As per the text, the phase discriminator consists of a multiplier followed by a low pass filter.

  4. Which means multiplication of $ \frac{1}{2}A_C \cos \phi \, m(t) $ and $ \frac{1}{2}A_C \phi \, m(t) $ would yield:

    $ g(t)=\frac{1}{2}A_C \cos \phi \, m(t) * \frac{1}{2}A_C \phi \, m(t) $

    $ g(t)= \frac{1}{4}A^2_C \phi \cos \phi \, m^2(t)$

    Now, let's say $ m(t) $ is band limited to $ M $. then the term $ m^2(t) $ in the frequency domain would spread across $ -2M $ to $ +2M $ centered around $ f=0 $ and the term $ \frac{1}{4}A^2_C \phi \cos \phi $ is a constant.

  5. Now, if $ g(t) $ is subjected to a LPF, then the term $ \frac{1}{4}A^2_C \phi \cos \phi \, m^2(t) $ would be retained as it is(if the cutoff frequency of the LPF is slightly > $ 2M $).

  6. So, what does phase discriminator in Costas receiver do mathematically?

    Also, if the phase error is significant (that is we have no idea at all about the probable phase of the carrier), then how does the analysis change? Because this is a possibility that we might have no idea about the phase of the carrier.

Text Used: Communication Systems By Simon Haykin

$\endgroup$
0
$\begingroup$

As the name "phase discriminator" suggests its purpose is to provide a VCO input that is proportional to the phase error between the upper and lower legs.

The product of the signals on the upper and lower legs, i.e., the inputs of the discriminator, is proportional to $ 2\textrm{sin}(\phi) \textrm{cos}(\phi) m^2(t) \approx \textrm{sin}(2\phi)$ because $m(t)$ is a binary modulation and then $m^2(t)\approx 1$.

When the phase error is small, that is the upper and lower legs are in phase quadrature, the VCO input is nearly zero and the carrier loop is locked. If the error is large then it is possible to show that the feedback is negative and the loop is always driven to $\phi \approx k\pi$, for some $k=0, \pm 1, \pm 2, ...$. This also shows that the stable phases and thus the demodulated symbols will have $0, \pi$ ambiguity as a consequence of the restoring feedback being proportional to $\textrm{sin}(2\phi)$. This ambiguity will have to be resolved by some other means in the modem.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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