Two-tone AM: spectrum, power, net index (50, 15, 20)

Rewrite the signal in standard AM form ($f_c = 1$ MHz, $f_1 = 1$ kHz, $f_2 = 200$ Hz, since $4\pi\times10^2 = 2\pi\times200$):

$$x(t) = 50\left[1 + 0.3\cos(2\pi\,10^3t) + 0.4\cos(2\pi\,200t)\right]\cos(2\pi\,10^6t)$$

so $A_c = 50$ V, $\mu_1 = 15/50 = 0.3$, $\mu_2 = 20/50 = 0.4$.

a) Spectrum

Using $\cos A\cos B = \frac{1}{2}[\cos(A+B)+\cos(A-B)]$:

$$x(t) = 50\cos\omega_ct + 7.5\cos(\omega_c \pm \omega_1)t + 10\cos(\omega_c \pm \omega_2)t$$

 Amp (V)
  50 |               |
     |               |
  10 |          |    |    |
 7.5 |     |    |    |    |    |
     +-----+----+----+----+----+--> f (kHz)
         999 999.8 1000 1000.2 1001

b) Total modulated power

Taking a 1 Ω load:

$$\begin{aligned} P_c &= \frac{50^2}{2} = 1250\ \text{W} \\ P_{1\,kHz\ pair} &= 2\times\frac{7.5^2}{2} = 56.25\ \text{W} \\ P_{200\,Hz\ pair} &= 2\times\frac{10^2}{2} = 100\ \text{W} \\ P_t &= 1250 + 56.25 + 100 = 1406.25\ \text{W} \end{aligned}$$

Check: $P_t = P_c\left(1 + \frac{\mu_1^2+\mu_2^2}{2}\right) = 1250(1.125) = 1406.25$ W.

c) Net modulation index

$$\mu = \sqrt{\mu_1^2 + \mu_2^2} = \sqrt{0.3^2 + 0.4^2} = \sqrt{0.25} = 0.5$$

Answer: spectrum as tabulated; $P_t = 1406.25$ W (1 Ω); net $\mu = 0.5$.