AM numerical: S(t) = 10(1 + 0.8cos25132.74t)cos(9424777.96t)

Compare with $s_{AM}(t) = A_c[1 + \mu\cos\omega_mt]\cos\omega_ct$:

• $A_c = 10$ V, $\mu = 0.8$
• $f_m = \dfrac{25132.74}{2\pi} = 4000$ Hz = 4 kHz
• $f_c = \dfrac{9424777.96}{2\pi} = 1.5\times10^6$ Hz = 1.5 MHz

Expanding:

$$s_{AM}(t) = 10\cos\omega_ct + 4\cos(\omega_c+\omega_m)t + 4\cos(\omega_c-\omega_m)t$$

i) Amplitude of all frequency components

Sideband amplitude $= \mu A_c/2 = 0.8\times10/2 = 4$ V.

ii) Modulation index

$\mu = 0.8$ (80%).

iii) Maximum and minimum amplitude

$$\begin{aligned} A_{max} &= A_c(1+\mu) = 10(1.8) = 18\ \text{V} \\ A_{min} &= A_c(1-\mu) = 10(0.2) = 2\ \text{V} \end{aligned}$$

iv) Bandwidth

$$B = 2f_m = 2\times4 = 8\ \text{kHz}$$

v) Power spectrum

Normalized power (1 Ω load), $P = A^2/2$ for each cosine:

$$\begin{aligned} P_c &= \frac{10^2}{2} = 50\ \text{W} \\ P_{USB} = P_{LSB} &= \frac{4^2}{2} = 8\ \text{W} \\ P_t &= 50 + 8 + 8 = 66\ \text{W} \end{aligned}$$

Check: $P_t = P_c(1+\mu^2/2) = 50(1.32) = 66$ W.

In two-sided form, each power is split equally between $+f$ and $-f$: impulses of weight 25 W at $\pm1500$ kHz and 4 W at $\pm1496$ and $\pm1504$ kHz.

 P (W)
  50 |          |
     |          |
   8 |     |    |    |
     +-----+----+----+------> f (kHz)
         1496 1500 1504
 (one-sided power spectrum)

Answer: carrier 10 V, sidebands 4 V; $\mu = 0.8$; $A_{max} = 18$ V, $A_{min} = 2$ V; $B = 8$ kHz; powers 50 W, 8 W, 8 W (total 66 W, 1 Ω).