In a simple AM transmitter, the tuned circuit uses a coil of `40muH` and a shunt capacitor of 1 n F. If the oscillator output is modulated by audio frequencies upto 5kHz, what is the frequency range occupied by side bands?

Answer

791 to 801 kHz

Explanation

Here, `L = 40muH = 40 xx 10^(-6)H`,
`C = 1 nF = 10^(-9) F, v_m = 5 kHz`.
`v_c = 1/(2pi sqrt(LC)) = 1/(2pisqrt(40 xx 10^(-6) xx 10^(-9))) = 10^7/(4pi)`
`= 796 kHz`.
`:.` Frequency range of side bands
`v_C +- v_m = (796 +-5) kHz = 791 kHz "to" 801kHz`.