Two-tone AM: net modulation index, power, spectrum

Rewrite the signal in standard two-tone AM form:

$$s(t) = 50\left[1 + 0.4\cos(2\pi10^3t) + 0.24\cos(2\pi\,200\,t)\right]\cos(2\pi10^6t)$$

since $\dfrac{20}{50} = 0.4$, $\dfrac{12}{50} = 0.24$ and $4\pi10^2t = 2\pi(200)t$.

So $A_c = 50$ V, $f_c = 1$ MHz, $\mu_1 = 0.4$ at $f_{m1} = 1$ kHz, $\mu_2 = 0.24$ at $f_{m2} = 200$ Hz.

a) Net modulation index

$$\mu_t = \sqrt{\mu_1^2 + \mu_2^2} = \sqrt{0.16 + 0.0576} = \sqrt{0.2176} = 0.4665$$

Answer: $\mu_t \approx 0.466$ (46.6%).

b) Total modulated power

Assume $R = 1\ \Omega$ (normalised).

$$\begin{aligned} P_c &= \frac{A_c^2}{2R} = \frac{50^2}{2} = 1250\ \text{W} \\ P_t &= P_c\left(1 + \frac{\mu_t^2}{2}\right) = 1250\left(1 + \frac{0.2176}{2}\right) \\ &= 1250 \times 1.1088 = 1386\ \text{W} \end{aligned}$$

Answer: $P_t = 1386$ W (sideband power 136 W; efficiency $= 136/1386 = 9.81\%$).

c) Spectrum

Side-frequency amplitudes $= \mu A_c/2$: for 1 kHz, $20/2 = 10$ V; for 200 Hz, $12/2 = 6$ V.

 A (V)
 50 |                 |
    |                 |
 10 |  |              |              |
  6 |  |     |        |        |     |
    +--+-----+--------+--------+-----+---> f (kHz)
     999  999.8     1000   1000.2  1001

Bandwidth $= 2 \times 1\text{ kHz} = 2$ kHz.