Total transmitted power of an AM signal at 50 percent modulation
BlinkNBuild practice problem · GATE standard, authored and verified in-house
A carrier of power $100\ \text{W}$ is amplitude modulated by a single tone to a modulation index $\mu = 0.5$.
The total transmitted power, in watts, is ________.
Show the step-by-step derivation
Answer
112.5 W W
Step-by-step derivation
- A tone-modulated AM signal is $$s(t) = A_c\big[1 + \mu\cos(2\pi f_m t)\big]\cos(2\pi f_c t),$$ which expands into a carrier plus two sidebands at $f_c \pm f_m$, each of amplitude $\mu A_c/2$.
- Power is proportional to the square of amplitude. Taking the carrier power as $P_c = A_c^2/2$, each sideband carries $$P_{SB} = \frac{(\mu A_c/2)^2}{2} = \frac{\mu^2}{4}\cdot\frac{A_c^2}{2} = \frac{\mu^2}{4}P_c.$$
- There are two sidebands, so the total is $$P_t = P_c + 2 \cdot \frac{\mu^2}{4}P_c = P_c\left(1 + \frac{\mu^2}{2}\right).$$
- Substitute $P_c = 100$ W and $\mu = 0.5$: $$P_t = 100\left(1 + \frac{0.25}{2}\right) = 100\,(1 + 0.125) = \mathbf{112.5\ W}.$$
- The efficiency point. Only the sidebands carry information: $$\eta = \frac{P_t - P_c}{P_t} = \frac{12.5}{112.5} = 11.1\%.$$ Even at full modulation ($\mu = 1$) efficiency only reaches $33.3\%$ - two thirds of a broadcast AM transmitter's power is spent on a carrier that conveys nothing. That is the entire motivation for DSB-SC and SSB.
The trap this question is built around
Writing $P_t = P_c(1+\mu^2)$ and answering $125$ W. The factor of two in the denominator comes from each sideband holding $\mu^2/4$ of the carrier power, not $\mu^2/2$.