Question

Ina detector, output circuit consists of `R=10kOmega` and `C=100 pF`, the frequency (in MHz) of a carrier signal it can detect must be greater than:

Answer 1

Here, `R=10Omega=10^(4)Omega,C=100pF=10^(-10)F`
Time constant of the cirucit.
`tau=RC=10^(4)xx10^(-10)=10^(-6)s`
For demodulation, `(1)/(v_(C))ltltRC` or `v_(c)gtgt1//RC`
or `v_(C)gtgt(1)/(10^(-6))=10^(6)Hz=1MHz`
`therefore` Frequency of carrier signal must be greater than 1 MHz.