A modulating signal 2 sin (6.28 times 10^{6} | vedclass

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(C) Given modulating signal frequency $\omega_{m} = 6.28 	imes 10^{6} 	ext{ rad/s}$.Frequency $f_{m} = \frac{\omega_{m}}{2\pi} = \frac{6.28 	imes 1

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(C) Given modulating signal frequency $\omega_{m} = 6.28 	imes 10^{6} 	ext{ rad/s}$.Frequency $f_{m} = \frac{\omega_{m}}{2\pi} = \frac{6.28 	imes 10^{6}}{2 	imes 3.14} = 1 	imes 10^{6} 	ext{ Hz} = 1 	ext{ MHz}$.Carrier signal frequency $\omega_{c} = 12.56 	imes 10^{9} 	ext{ rad/s}$.Frequency $f_{c} = \frac{\omega_{c}}{2\pi} = \frac{12.56 	imes 10^{9}}{2 	imes 3.14} = 2 	imes 10^{9} 	ext{ Hz} = 2000 	ext{ MHz}$.The square law device produces frequencies: $2f_{c}, f_{c}+f_{m}, f_{c}, f_{c}-f_{m}, 2f_{m}, f_{m}$.The band pass filter centered at $f_{c}$ allows the frequencies $f_{c}-f_{m}, f_{c}, f_{c}+f_{m}$ to pass.The bandwidth of the output signal is $(f_{c}+f_{m}) - (f_{c}-f_{m}) = 2f_{m}$.Bandwidth $= 2 	imes 1 	ext{ MHz} = 2 	ext{ MHz}$.

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