AM numerical: S(t) = 20(1 + 0.8cos2π1000t)cos(9424777.96t)

Compare the given wave with the standard single-tone AM equation:

$$s_{AM}(t) = A_c\,[1 + \mu \cos(2\pi f_m t)]\cos(2\pi f_c t)$$

So $A_c = 20$ V, $\mu = 0.8$, $f_m = 1000$ Hz and $\omega_c = 9424777.96$ rad/s, which gives

$$f_c = \frac{\omega_c}{2\pi} = \frac{9424777.96}{2\pi} = 1.5\times10^{6}\ \text{Hz} = 1.5\ \text{MHz}$$

Expanding the product:

$$s_{AM}(t) = 20\cos 2\pi f_c t + \frac{\mu A_c}{2}\cos 2\pi(f_c+f_m)t + \frac{\mu A_c}{2}\cos 2\pi(f_c-f_m)t$$

i) Amplitude of all frequency components

$$\frac{\mu A_c}{2} = \frac{0.8 \times 20}{2} = 8\ \text{V}$$

ii) Modulation index

The coefficient of the cosine term inside the bracket is the modulation index: $\mu = 0.8$ (80% modulation).

iii) Maximum and minimum amplitude

$$\begin{aligned} A_{max} &= A_c(1+\mu) = 20(1+0.8) = 36\ \text{V} \\ A_{min} &= A_c(1-\mu) = 20(1-0.8) = 4\ \text{V} \end{aligned}$$

Check: $\mu = \dfrac{A_{max}-A_{min}}{A_{max}+A_{min}} = \dfrac{36-4}{36+4} = 0.8$.

iv) Frequency of USB and LSB

$$\begin{aligned} f_{USB} &= f_c + f_m = 1500 + 1 = 1501\ \text{kHz} \\ f_{LSB} &= f_c - f_m = 1500 - 1 = 1499\ \text{kHz} \end{aligned}$$

Answer: carrier 20 V at 1.5 MHz, sidebands 8 V each; $\mu = 0.8$; $A_{max} = 36$ V, $A_{min} = 4$ V; USB = 1.501 MHz, LSB = 1.499 MHz.