AM numerical: u(t) = [20 + 2cos3000πt + 10cos6000πt]cos2πfct

Expand the given wave ($f_c = 10^5$ Hz = 100 kHz; message tones $f_1 = 1500$ Hz and $f_2 = 3000$ Hz):

$$\begin{aligned} u(t) &= 20\cos 2\pi f_c t + 2\cos(2\pi\,1500t)\cos 2\pi f_ct + 10\cos(2\pi\,3000t)\cos 2\pi f_ct \\ &= 20\cos 2\pi f_ct + \cos 2\pi(f_c \pm 1500)t + 5\cos 2\pi(f_c \pm 3000)t \end{aligned}$$

(each $\pm$ means two terms, one with + and one with −).

i) Spectrum (one-sided amplitude spectrum)

 Amp (V)
  20 |              |
     |              |
     |              |
   5 |    |         |         |
   1 |    |    |    |    |    |
     +----+----+----+----+----+----> f (kHz)
         97  98.5  100 101.5 103

ii) Power in each component

Taking a 1 Ω load, power of a cosine of amplitude $A$ is $A^2/2$.

$$\begin{aligned} P_c &= \frac{20^2}{2} = 200\ \text{W} \\ P_{98.5} = P_{101.5} &= \frac{1^2}{2} = 0.5\ \text{W each} \\ P_{97} = P_{103} &= \frac{5^2}{2} = 12.5\ \text{W each} \\ P_t &= 200 + 2(0.5) + 2(12.5) = 226\ \text{W} \end{aligned}$$

iii) Efficiency

$$\eta = \frac{P_{sidebands}}{P_t} = \frac{26}{226} = 0.115 = 11.5\%$$

(Check with $\mu_1 = 2/20 = 0.1$, $\mu_2 = 10/20 = 0.5$: $\mu^2 = 0.26$, $\eta = \frac{0.26}{2+0.26} = 11.5\%$.)

iv) Transmission bandwidth

The highest message frequency is 3 kHz, so

$$B_T = 2f_{max} = 2 \times 3 = 6\ \text{kHz} \quad (97\ \text{to}\ 103\ \text{kHz})$$

Answer: carrier 200 W, 1.5 kHz sidebands 0.5 W each, 3 kHz sidebands 12.5 W each (total 226 W); efficiency 11.5%; bandwidth 6 kHz.