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(A) For a detector circuit (envelope detector),the condition for proper demodulation is $\frac{1}{f_c} < RC$,where $f_c$ is the carrier frequency.

Vedclass · Accepted Answer

$<< 10^9\, Hz$

Vedclass · Answer

(A) For a detector circuit (envelope detector),the condition for proper demodulation is $\frac{1}{f_c} Given $R = 1000\, \Omega$ and $C = 1000\, pF = 1000 	imes 10^{-12}\, F = 10^{-9}\, F$.The time constant $RC = 1000 	imes 10^{-9} = 10^{-6}\, s$.Therefore,$\frac{1}{f_c} > 10^6\, Hz$.However,in standard detector design,the carrier frequency must be much greater than the signal frequency but limited by the $RC$ time constant to follow the envelope. Based on the provided options and standard textbook problems,the condition is $f_c Calculating $\frac{1}{RC} = \frac{1}{10^{-9}} = 10^9\, Hz$.Thus,the carrier frequency should be $<< 10^9\, Hz$.

In the given detector circuit,the suitable value of carrier frequency is

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