An amplitude modulated wave is modulated to 50%. What is the saving in power if carrier as well as one of the side bands are suppressed.

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answer is 94.

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Detailed Solution

$P_{s b} = P_{c} {(\frac{m_{a}}{2})}^{2} = P_{c} \frac{{(0.5)}^{2}}{4} = 0.0625 P_{c} A l s o P = P_{c} (1 + \frac{m_{a}^{2}}{2}) = P_{c} (1 + \frac{{(0.5)}^{2}}{2}) = 1.125 P_{c}$

$∴ % s a v i n g = \frac{(1.125 p_{c} - 0.0625 p_{c})}{1.125 p_{c}} X 100 = 94.4 % .$

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