An audio signal is modulated by a carrier wave of 20 MHz such that the bandwidth required for modulation is 3 kHz. Could this wave be demodulated by a diode detector which has the values of R and C as (i) `R = 1 k Omega, C = 0.01 mu F (ii) R = 10 k Omega, C = 0.01 muF (iii) R = 10k Omega , C = 1 mu muF`.

Explanation

Given, carrier wave frequency `f_(c)=20MHz`
`=20 xx 10^(6)HZ`
Bandwidth required for modulation is
`2f_(m)=3kHz = 3 xx 10^(3)Hz`
`implies" "f_(m)=(3xx10^(3))/(2)=1.5 xx 10^(3)Hz`
Demodulation by a diode is possible if the condition `(1)/(f_(c))ltltRC lt (1)/(f_(m))` is satisfied
Thus, `(1)/(f_(c))=(1)/(20xx10^(6))=0.5 xx 10^(-7)" ...(i)"`
and `(1)/(f_(m))=(1)/(1.5 xx 10^(3))Hz = 0.7 xx 10^(-3)s" ...(ii)"`
Now, gain throgh all the options of R and C one by one, we get
(i) `RC = 1 kOmega xx0.01 muF = 10^(3) Omega xx(0.01 xx 10^(-6)F)=10^(-5)S`
Here, condition `(1)/(f_(c))ltlt RC lt(1)/(f_(m))` is satisfied.
Hence it can be demodulated.
(ii) `RC = 10 kOmega xx 0.01 mu F = 10^(4) Omega xx 10^(-8)F = 10^(-4) S `
Here condition `(1)/(f_(c))ltlt RC lt(1)/(f_(m))` is satisfied.
Hence, it can be demodulated.
(iii) `RC=10 k Omega xx 1mumuF = 10^(4) Omega xx 10^(-12) F = 10^(-8)S`
Here, condition `(1)/(f_(c)) gt RC`, so this cannot be demodulated.