A standard AM transmission, sinusoidally modulated to depth of 40%, produces sideband frequencies of 6.824 and 6.854 Mhz. The amplitude of each sideband frequency is 50V. Determine the amplitude and frequency of the carrier.
Hello Aspirant,
This question can be solved in the following way.
Given-
The amplitude of each sideband = 50 V
The upper sideband frequency is 6.854 MHz.
The lower sideband frequency is 6.825 MHz.
To find- Frequency and amplitude of carrier wave I.e., fc, and Ac.
Solution- It is given that AM wave was modulated to a depth of 40%.
Hence, the modulation index, m = 40/100 = 0.4
We know that amplitude of sideband =(m×Ac)/2
=> 50 V = (0.4×Ac)/2 => (50×20)/4 = Ac
=> Ac= 250 V
We have the upper and lowerband frequency as 6.854 MHz and 6.825 MHz.
We know that fs is the frequency of signal and fc is the frequency of the carrier wave.
fs + fc = 6.854 MHz ---------------(1)
fs - fc = 6.825 MHz ---------------(2)
Adding equation (1) and (2), we get
=> 2fc = 13.679 => fc = 6.839 MHz
Hence the amplitude and frequency of carrier wave is 62.5 V and 6.839 MHz respectively.
i hope it helps.
Thank you
