In an amplitude modulator circuit,the carrier | vedclass

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Question

(A) The modulation index $\mu$ is defined as the ratio of the amplitude of the modulating signal $(A_m)$ to the amplitude of the carrier wave $(A_c)$:

Vedclass · Accepted Answer

$0.5$ and $9 	ext{ kHz}$

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(A) The modulation index $\mu$ is defined as the ratio of the amplitude of the modulating signal $(A_m)$ to the amplitude of the carrier wave $(A_c)$: $\mu = \frac{A_m}{A_c} = \frac{2}{4} = 0.5$. The carrier frequency $f_c$ is obtained from $2 \pi f_c = 20000 \pi$,which gives $f_c = 10000 	ext{ Hz} = 10 	ext{ kHz}$. The modulating frequency $f_m$ is obtained from $2 \pi f_m = 2000 \pi$,which gives $f_m = 1000 	ext{ Hz} = 1 	ext{ kHz}$. The lower sideband frequency $(LSB)$ is given by $f_c - f_m = 10 	ext{ kHz} - 1 	ext{ kHz} = 9 	ext{ kHz}$. Therefore,the modulation index is $0.5$ and the lower sideband frequency is $9 	ext{ kHz}$.

In an amplitude modulator circuit,the carrier wave is given by $C(t) = 4 \sin(20000 \pi t)$ and the modulating signal is given by $m(t) = 2 \sin(2000 \pi t)$. The values of the modulation index and the lower sideband frequency are:

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