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

Here, `v_c = 20 MHz = 20 xx 10^6 Hz, 1/v_c = 1/(20 xx 10^6) = 0.5 xx 10^(-7) s`
Band width `= 2v_m = 3kHz = 3 xx 10^3 Hz , v_m = 1.5 xx 10^3 Hz :. 1/ v_m = 1/(1.5 xx 10^(3)) = 0.7 xx 10^(-3) s`
Demodulation by a diode detector is possible if the condition `1/v_c lt lt RC lt 1/v_m` is satisfied.
`(i) RC = 1 k Omega xx 0.01 mu F = 10^3 Omega xx (0.01 xx 10^(-6) F) = 10^(-5) s`
Here condition `1/v_c lt lt RC lt 1/v_m ` is satisfied. Hence it can be demodulated.
(ii) `RC = 10kOmega xx 0.01 muF = 10^4 Omega xx 10^(-8) F = 10^(-4) s `
Here condition `1/v_c lt lt RC lt 1/v_m` is satisfied. Hence it can be demodulated.
(iii) `RC = 10 k Omega xx 1mu muF = 10^4 Omega xx 10^(-12) F = 10^(-8) s`
Here, condition `1/v_c gt RC`, so this cannot be demodulated.