A carrier wave with amplitude of 250, text{V} | vedclass

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(A) The maximum amplitude of the amplitude modulated wave is given by $A_{\max} = A_c + A_m$,where $A_c$ is the carrier amplitude and $A_m$ is the sig

Vedclass · Accepted Answer

$200$

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(A) The maximum amplitude of the amplitude modulated wave is given by $A_{\max} = A_c + A_m$,where $A_c$ is the carrier amplitude and $A_m$ is the signal amplitude.$A_{\max} = 250 + 150 = 400\, 	ext{V}$.The minimum amplitude of the amplitude modulated wave is given by $A_{\min} = A_c - A_m$.$A_{\min} = 250 - 150 = 100\, 	ext{V}$.The ratio of minimum amplitude to maximum amplitude is $\frac{A_{\min}}{A_{\max}} = \frac{100}{400} = \frac{1}{4}$.Given that the ratio is $50:x$,we have $\frac{50}{x} = \frac{1}{4}$.Solving for $x$,we get $x = 50 	imes 4 = 200$.

$A$ carrier wave with amplitude of $250\, \text{V}$ is amplitude modulated by a sinusoidal baseband signal of amplitude $150\, \text{V}$. The ratio of minimum amplitude to maximum amplitude for the amplitude modulated wave is $50:x$,then the value of $x$ is ....... .

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