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A carrier wave with amplitude of 250 V is amplitude modulated by a sinusoidal base band signal of amplitude 150 V. The ratio of minimum amplitude to maximum amplitude for the amplitude modulated wave is $50 : x$ , then value of x is ___________.

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answer is 200.

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Detailed Solution

Given, Amplitude of carrier wave, $A_{c} = 250 V$
Amplitude of signal, $A_{m} = 150 V$
Maximum amplitude $A_{max} = A_{c} + A_{m} = 250 + 150 = 400$
Minimum amplitude,

$A_{\min} = A_{c} - A_{m} = 250 - 150 = 100 ∴ \frac{A_{min}}{A_{max}} = \frac{100}{400} = \frac{1}{4} = \frac{50}{200} = \frac{50}{x}$
$∴ x = 200$

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