A carrier signal of frequency v_1 and peak vo | vedclass

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(B) The modulation index $m$ is defined as the ratio of the peak voltage of the message signal $(V_2)$ to the peak voltage of the carrier signal $(V_1

Vedclass · Accepted Answer

$v_1 = \frac{v_{+} + v_{-}}{2}$

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(B) The modulation index $m$ is defined as the ratio of the peak voltage of the message signal $(V_2)$ to the peak voltage of the carrier signal $(V_1)$, so $m = \frac{V_2}{V_1}$.In amplitude modulation, the sideband frequencies are given by $v_{+} = v_1 + v_2$ and $v_{-} = v_1 - v_2$.Adding these two equations: $v_{+} + v_{-} = (v_1 + v_2) + (v_1 - v_2) = 2v_1$.Therefore, the carrier frequency is $v_1 = \frac{v_{+} + v_{-}}{2}$.Thus, both option $B$ and the definition of $m$ are relevant, but based on standard physics problems of this type, $v_1 = \frac{v_{+} + v_{-}}{2}$ is a standard identity.

$A$ carrier signal of frequency $v_1$ and peak voltage of $V_1$ is modulated by a message signal of frequency $v_2$ and peak voltage of $V_2$. Let $m$ be the modulation index and $v_{+}, v_{-}$ be side bands produced. The correct statement is

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