Given an Amplitude Modulation (AM) signal $s(t) = 100[1 + m(t)] \cos(2\pi f_c t)$, where $f_c = 800 \text{ MHz}$ and $m(t) = \sin(3000\pi t) + 0.5 \cos(5000\pi t)$, we will solve the following:

(i) Average power in the carrier and in the sidebands

1. Carrier power:

The AM signal can be written as: $s(t) = A_c[1 + m(t)]\cos(2\pi f_c t)$ where $A_c = 100$ is the carrier amplitude. The carrier frequency $f_c = 800 \text{ MHz}$.

The average power in the carrier $P_c$ is given by: $P_c = \frac{A_c^2}{2R}$ where $R = 50 \, \Omega$ is the load impedance. Substituting the values: $P_c = \frac{100^2}{2 \times 50} = \frac{10000}{100} = 100 \text{ W}$

2. Sideband power:

To find the power in the sidebands, we need to calculate the total modulation index $\mu$. The modulation index is determined by the peak value of $m(t)$.

Given $m(t) = \sin(3000\pi t) + 0.5 \cos(5000\pi t)$, the peak value of $m(t)$ is: $\text{Peak value of } m(t) = \sqrt{1^2 + 0.5^2} = \sqrt{1 + 0.25} = \sqrt{1.25} \approx 1.118$ Thus, the modulation index is $\mu = 1.118$.

The total power in the sidebands is given by: $P_{SB} = P_c \times \frac{\mu^2}{2}$ Substituting the values: $P_{SB} = 100 \times \frac{1.118^2}{2} = 100 \times \frac{1.25}{2} = 100 \times 0.625 = 62.5 \text{ W}$

Thus, the average power in the sidebands is $P_{SB} = 62.5 \text{ W}$.

(ii) Modulation index and Peak Power Delivered to the Load

1. Modulation index:

From the previous step, we already determined that the modulation index $\mu \approx 1.118$.

2. Peak power:

The peak power is the maximum power delivered to the load. Since the total signal is $100[1 + m(t)] \cos(2\pi f_c t)$, the peak amplitude of $s(t)$ is: $A_{\text{peak}} = A_c (1 + \mu) = 100 \times (1 + 1.118) = 100 \times 2.118 = 211.8$ The peak power is given by: $P_{\text{peak}} = \frac{A_{\text{peak}}^2}{2R}$ Substituting the values: $P_{\text{peak}} = \frac{211.8^2}{2 \times 50} = \frac{44860.84}{100} = 448.6 \text{ W}$

Final Answers:

• Carrier Power: $100 \text{ W}$
• Sideband Power: $62.5 \text{ W}$
• Modulation Index: $1.118$
• Peak Power Delivered to the Load: $448.6 \text{ W}$

Would you like any specific details or further clarifications?

Here are 5 questions related to the topic:

1. How is the modulation index calculated in AM signals?
2. Why is the carrier power independent of the modulation signal?
3. What happens if the modulation index exceeds 1 in an AM signal?
4. How do sidebands in AM carry information from the baseband signal?
5. What are the practical limitations of AM in terms of power efficiency?

Tip: Always ensure the modulation index remains below 1 to avoid distortion, ensuring clear transmission in AM systems.