Two-tone AM: index, power, spectrum (5cos, 10cos, 15cos)

Given $m(t) = 5\cos(2\pi\,1000\,t) + 10\cos(2\pi\,500\,t)$ and $c(t) = 15\cos(2\pi\times10^6t)$.

So $A_{m1} = 5$ V at 1 kHz, $A_{m2} = 10$ V at 500 Hz, $A_c = 15$ V, $f_c = 1$ MHz. Assume standard AM $s(t) = [A_c + m(t)]\cos\omega_ct$ and $R = 1\ \Omega$.

$$s(t) = 15\left[1 + \tfrac13\cos(2\pi10^3t) + \tfrac23\cos(\pi10^3t)\right]\cos(2\pi10^6t)$$

i) Modulation index

$$\mu_1 = \frac{5}{15} = 0.333, \quad \mu_2 = \frac{10}{15} = 0.667$$ $$\mu_t = \sqrt{\mu_1^2 + \mu_2^2} = \sqrt{\frac19 + \frac49} = \sqrt{0.5556} = 0.7454$$

ii) Total power

$$\begin{aligned} P_c &= \frac{A_c^2}{2R} = \frac{15^2}{2} = 112.5\ \text{W} \\ P_t &= P_c\left(1 + \frac{\mu_t^2}{2}\right) = 112.5\left(1 + \frac{0.5556}{2}\right) = 143.75\ \text{W} \end{aligned}$$

iii) Spectrum

Side amplitudes $A_m/2$: 2.5 V (1 kHz tone), 5 V (500 Hz tone).

  A(V)
  15 |                  |
     |                  |
   5 |         |        |        |
 2.5 |    |    |        |        |    |
     +----+----+--------+--------+----+---> f (kHz)
        999  999.5    1000   1000.5 1001

Answer: $\mu_t = 0.745$; $P_t = 143.75$ W; bandwidth 2 kHz.