Which statement best describes peak envelope power (PEP) and average power in AM transmissions?

• A. PEP is the maximum instantaneous RF envelope power; average power is the mean over time and depends on modulation index and carrier amplitude.
• B. PEP is the average power in all cases.
• C. PEP is the minimum instantaneous power during a cycle.
• D. PEP equals the carrier power for all modulation levels.

Answer: (A)

Peak envelope power is the highest instantaneous RF power in an AM signal, reached when the envelope of the carrier is at its maximum. For an AM wave s(t) = A_c [1 + m cos(ω_m t)] cos(ω_c t), the envelope is A_c [1 + m cos(ω_m t)]. The maximum envelope occurs when cos(ω_m t) = 1, giving an envelope of A_c(1 + m). Since power is proportional to the square of the envelope, the peak envelope power is PEP = (A_c^2 / 2R) (1 + m)^2, where R is the load resistance. The average power, found by averaging s^2(t) over time, is P_avg = (A_c^2 / 2R) (1 + m^2/2). This is generally less than PEP and shows how average power depends on the modulation index and carrier amplitude. For no modulation, both PEP and P_avg reduce to the carrier power P_carrier = A_c^2 / (2R). At full modulation (m = 1), PEP is four times the carrier power. So the statement that PEP is the maximum instantaneous RF envelope power and that average power is the mean over time depending on modulation index and carrier amplitude is correct. The other notions—PEP being the average power, or the minimum instantaneous power, or equal to carrier power for all modulation levels—do not fit AM behavior.