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As shown in fig., In this scheme input signal is square and passed through a narrow(high Q) bandpass filter tuned to \$ w_c\$.Finally the ouput comes to be a sinusoid of frequency \$ w_c\$.

But to implement this scheme you need prior information of \$ w_c\$ because you need to tuned your bandpass filter at \$ w_c\$.

My question is that if you know \$ w_c\$ then why not just use a crystal oscillator to generate carrier of same. why use all PLL, filters etc. ?

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You have to use a coherent detector and not any old arbitrary oscillator module or crystal that is stamped with the correct frequency. It has to be phase coherent i.e. exactly the same frequency and a stable (relatively) phase. This is why squaring or Costas loops are used. Costas loop: -

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Without the continuous presence of a carrier (i.e. the carrier is suppressed) you have to use these methods.

The band-pass filter used in a squaring coherent carrier detection system does need to be broadly centred at the carrier frequency but it must have a pass-band that is wide enough so that errors are not significant.

A filter does not make a local oscillator signal phase coherent - the act of squaring the signal and then removing DC and double frequency components gives you phase coherence. The PLL is superfluous to a squaring coherent detector BTW.

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  • \$\begingroup\$ But in costa's receiver quiescent frequency of the oscillator is same as carrier in modulated signal y(t). Let's say y(t) can have any frequency between ( 90 -107 MHz ). Is it required to change quiescent frequency of the oscillator whenever \$ w_c\$ changes. \$\endgroup\$
    – Virange
    Nov 17, 2015 at 11:10
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    \$\begingroup\$ A costas loop works after the IF stage when the input frequency covers a significant range. See this: books.google.co.uk/… this indeed means that the the VCO has to shift in order to detect a different carrier range. Ditto the low pass and high pass filters. \$\endgroup\$
    – Andy aka
    Nov 17, 2015 at 11:18
  • \$\begingroup\$ The same is true of the BPF in a squaring loop. \$\endgroup\$
    – Andy aka
    Nov 17, 2015 at 11:20

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