AM numerical: 750 kHz carrier, 110 V and 90 V amplitudes

Given: $f_c = 750$ kHz, $f_m = 325$ Hz, $A_{max} = 110$ V, $A_{min} = 90$ V. Assume load $R = 1\ \Omega$ (normalised).

$$\begin{aligned} A_c &= \frac{A_{max} + A_{min}}{2} = \frac{110 + 90}{2} = 100\ \text{V} \\ \mu &= \frac{A_{max} - A_{min}}{A_{max} + A_{min}} = \frac{20}{200} = 0.1 \end{aligned}$$

a) Waveform

 v(t)
 110 -  _   _   _   _   _   _   _   _      envelope
  90 - / \_/ \_/ \_/ \_/ \_/ \_/ \_/ \_    varies 90..110 V
       |||||||||||||||||||||||||||||||     carrier 750 kHz
   0 --+--------------------------------> t
       |||||||||||||||||||||||||||||||
 -90 - \_/ \_/ \_/ \_/ \_/ \_/ \_/ \_/
-110 -
       |<------ 1/325 s = 3.08 ms ---->|

The carrier swings between ±110 V at envelope peaks and ±90 V at envelope troughs; the envelope period is $1/325 = 3.08$ ms. Only 10% modulation, so the envelope ripple is shallow.

b) Expression

$$s(t) = 100\,[1 + 0.1\cos(2\pi\,325\,t)]\cos(2\pi\times750\times10^3\,t)\ \text{V}$$

c) Total power and efficiency

$$\begin{aligned} P_c &= \frac{A_c^2}{2R} = \frac{100^2}{2} = 5000\ \text{W} \\ P_t &= P_c\left(1 + \frac{\mu^2}{2}\right) = 5000(1 + 0.005) = 5025\ \text{W} \\ \eta &= \frac{\mu^2}{2 + \mu^2} = \frac{0.01}{2.01} = 0.4975\% \end{aligned}$$

Answer: $s(t) = 100[1 + 0.1\cos(2\pi325t)]\cos(2\pi750\times10^3t)$ V; $P_t = 5025$ W (for 1 Ω); $\eta \approx 0.50\%$.