An AM signal is generated by modulating the carrier fc=800MHz by the signal
m(t) = sin 3000πt+0.5cos5000πt. the AM signal s(t)=100[1+m(t)]cos2πfct is fed to a 50
ohm load. (i)Determine the average power in the carrier and in the sidebands. (ii) Find
the modulation index and peak power delivered to the load.

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.

An AM signal is generated by modulating the carrier fc=800MHz by the signal m(t) = sin 3000πt+0.5cos5000πt. The AM signal s(t)=100[1+m(t)]cos2πfct is fed to a 50 ohm load. (i)Determine the average power in the carrier and in the sidebands. (ii) Find the modulation index and peak power delivered to the load.

Explore the detailed solution for determining the average power in the carrier and sidebands of an AM signal modulating a 800MHz carrier frequency. Learn how to calculate the modulation index and peak power delivered to a 50-ohm load with clear steps. Ideal for undergraduate students in signal processing.

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