A carrier wave v_{c}(t) = 160 sin(2 pi times | vedclass

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(C) The maximum voltage of an amplitude-modulated wave is given by $V_{\max} = A_{c} + A_{m}$,where $A_{c}$ is the amplitude of the carrier wave and $

Vedclass · Accepted Answer

$40$

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(C) The maximum voltage of an amplitude-modulated wave is given by $V_{\max} = A_{c} + A_{m}$,where $A_{c}$ is the amplitude of the carrier wave and $A_{m}$ is the peak voltage of the modulating signal.Given,$A_{c} = 160 	ext{ V}$ and $V_{\max} = 200 	ext{ V}$.Substituting these values into the equation:$200 = 160 + A_{m}$$A_{m} = 200 - 160$$A_{m} = 40 	ext{ V}$.Alternatively,using the minimum voltage formula: $V_{\min} = A_{c} - A_{m}$.$120 = 160 - A_{m}$$A_{m} = 160 - 120 = 40 	ext{ V}$.Thus,the peak voltage of the modulating signal is $40 	ext{ V}$.

$A$ carrier wave $v_{c}(t) = 160 \sin(2 \pi \times 10^{6} t) \text{ V}$ is made to vary between $V_{\max} = 200 \text{ V}$ and $V_{\min} = 120 \text{ V}$ by a message signal $v_{m}(t) = A_{m} \sin(2 \pi \times 10^{3} t) \text{ V}$. The peak voltage $A_{m}$ of the modulating signal is ..... .

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