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(B) For an $AM$ wave,the maximum voltage is given by $V_{max} = V_c + V_m$,where $V_c$ is the carrier amplitude and $V_m$ is the message signal amplit

Vedclass · Accepted Answer

$\frac{V_{max} - V_{min}}{V_{max} + V_{min}}$

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(B) For an $AM$ wave,the maximum voltage is given by $V_{max} = V_c + V_m$,where $V_c$ is the carrier amplitude and $V_m$ is the message signal amplitude.The minimum voltage is given by $V_{min} = V_c - V_m$.Adding these two equations: $V_{max} + V_{min} = 2V_c \implies V_c = \frac{V_{max} + V_{min}}{2}$.Subtracting these two equations: $V_{max} - V_{min} = 2V_m \implies V_m = \frac{V_{max} - V_{min}}{2}$.The modulation index $\mu$ is defined as $\mu = \frac{V_m}{V_c}$.Substituting the values: $\mu = \frac{(V_{max} - V_{min})/2}{(V_{max} + V_{min})/2} = \frac{V_{max} - V_{min}}{V_{max} + V_{min}}$.

If the maximum and minimum voltage of an $AM$ wave are $V_{max}$ and $V_{min}$,then the modulation index is given by:

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