**AM Modulation Problem**A modulating signal \( m(t) = 10 \cos(2\pi \times 10^3 t) \) is amplitude modulated with a carrier signal \( c(t) = 50 \cos(2\pi \times 10^5 t) \). Find the modulation index, the carrier power, and the power required for transmitting the AM wave.**Solution Explanation**1. **Modulation Index (\( \mu \))**: The modulation index is calculated using the formula: \[ \mu = \frac{A_m}{A_c} \] where \( A_m = 10 \) is the amplitude of the modulating signal and \( A_c = 50 \) is the amplitude of the carrier signal. \[ \mu = \frac{10}{50} = 0.2 \]2. **Carrier Power (\( P_c \))**: The carrier power is given by: \[ P_c = \frac{A_c^2}{2R} \] Assuming the resistance \( R = 1 \, \Omega \), the carrier power is: \[ P_c = \frac{50^2}{2 \times 1} = 1250 \, \text{watts} \]3. **Total Power in AM Wave (\( P_t \))**: The total power transmitted in an AM wave is: \[ P_t = P_c \left(1 + \frac{\mu^2}{2}\right) \] Substituting the values: \[ P_t = 1250 \left(1 + \frac{0.2^2}{2}\right) = 1250 \left(1 + \frac{0.04}{2}\right) = 1250 \times 1.02 = 1275 \, \text{watts} \]Therefore, the modulation index is 0.2, the carrier power is 1250 watts, and the total power for transmitting the AM wave is 1275 watts.

The modulation index for amplitude modulation is defined as the ratio of the amplitude of the…

A modulating signal m (t) = 10 cos (27 x 10°t) is amplitude modulated with a carrier signal c(t) = 50 cos (27 x 10t) . Find the modulation index, the carrier power, and the power required for transmitting AM wave.

A modulating signal m (t) = 10 cos (27 x 10°t) is amplitude modulated with a carrier signal c(t) = 50 cos (27 x 10t) . Find the modulation index, the carrier power, and the power required for transmitting AM wave.

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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