Discussion
At an index of unity, the fraction of total AM power in the sidebands is
Answer: A
With \( \mu = 1 \) the total is \( 1.5P_c \) and the sidebands carry \( 0.5P_c \), so their share is
\[ \frac{0.5}{1.5} = \frac{1}{3} \]
which is a third, the best full-carrier AM can achieve. Two thirds of the transmitter's output goes into a carrier that carries no information at all, which is the central criticism of the scheme and the motive for suppressed-carrier systems. A half or two thirds would require an index above unity, and a quarter corresponds to an index of about 0.82.
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