The mathematical form of an AM wave is given | vedclass

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(C) The standard equation for an $AM$ wave is $e = E_c(1 + m_a \sin(\omega_m t)) \sin(\omega_c t)$.Comparing the given equation $e = 50(1 + 0.5 \sin(2

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$4.995 	ext{ MHz}, 5.005 	ext{ MHz}$

Vedclass · Answer

(C) The standard equation for an $AM$ wave is $e = E_c(1 + m_a \sin(\omega_m t)) \sin(\omega_c t)$.Comparing the given equation $e = 50(1 + 0.5 \sin(2\pi 	imes 5 	imes 10^3 t)) \sin(31.4 	imes 10^6 t)$ with the standard form:$1$. The modulating frequency $f_m$ is obtained from $\omega_m = 2\pi f_m = 2\pi 	imes 5 	imes 10^3 	ext{ rad/s}$,so $f_m = 5 	imes 10^3 	ext{ Hz} = 5 	ext{ kHz} = 0.005 	ext{ MHz}$.$2$. The carrier frequency $f_c$ is obtained from $\omega_c = 2\pi f_c = 31.4 	imes 10^6 	ext{ rad/s}$. Since $31.4 \approx 10\pi$,we have $2\pi f_c = 10\pi 	imes 10^6$,so $f_c = 5 	imes 10^6 	ext{ Hz} = 5 	ext{ MHz}$.$3$. The Lower Sideband $(LSB)$ frequency is $f_c - f_m = 5 	ext{ MHz} - 0.005 	ext{ MHz} = 4.995 	ext{ MHz}$.$4$. The Upper Sideband $(USB)$ frequency is $f_c + f_m = 5 	ext{ MHz} + 0.005 	ext{ MHz} = 5.005 	ext{ MHz}$.Therefore,the $LSB$ is $4.995 	ext{ MHz}$ and the $USB$ is $5.005 	ext{ MHz}$.

The mathematical form of an $AM$ wave is given by $e = 50 (1 + 0.5 \sin(2\pi \times 5 \times 10^3 t)) \sin(31.4 \times 10^6 t) \text{ V}$. The frequencies of the $LSB$ and $USB$ in the given $AM$ wave are,respectively,....... and ....... .

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