Question 6A377

For 100% amplitude modulation (AM), the total average power of the transmitted signal increases by 50% compared to the unmodulated carrier power. This additional power is contained within the sidebands and is supplied by the modulator. 1. **Modulator AC Power:** At 100% modulation, the AC power supplied by the modulator must be equal to half of the unmodulated carrier power ($P_c$). So, $P_{AC\_mod} = 0.5 \cdot P_c$. 2. **RF Amplifier DC Input Power:** The DC power input to the modulated RF amplifier is what establishes the unmodulated carrier power ($P_c$), assuming a high-efficiency Class C amplifier. So, $P_{DC\_RF} \approx P_c$. Combining these, we get $P_{AC\_mod} = 0.5 \cdot P_{DC\_RF}$. This means the AC power from the modulation is half the DC power input to the RF amplifier. The question asks for the relationship of the **DC power input to the RF amplifier** relative to the **AC power from the modulation**. Based on the derivation above, $P_{DC\_RF} = 2 \cdot P_{AC\_mod}$. However, the provided correct answer is B) "Be 1/2 the AC power from the modulation". This option suggests that $P_{DC\_RF} = 0.5 \cdot P_{AC\_mod}$. This would imply $P_c = 0.5 \cdot (0.5 \cdot P_c)$, which simplifies to $P_c = 0.25 \cdot P_c$, a mathematical contradiction. Given the standard theory, there appears to be an inversion in the phrasing of option B relative to the question. A common form of this question correctly states that the *AC power from the modulation* should be 1/2 the DC power input (carrier power). If the intent was to highlight the "1/2" relationship commonly associated with 100% modulation, it is likely that option B is presenting an inverted or misphrased version of this fundamental principle.