
For an AM-system the total power of modulated signal is 600W and that of carrier is 400W, the modulation index is
(A) 0.25
(B) 0.36
(C) 0.54
(D) 1
Answer
151.5k+ views
Hint The modulation index for the total power of the modulated signal and carrier signal is related to total power and carrier power is given by ${P_T} = {P_C}\left( {1 + \dfrac{{{m_a}^2}}{2}} \right)$ . Substitute the power values and simplify to get the modulation index.
Complete step-by-step answer
The process of changing the amplitude of a carrier wave in accordance with the amplitude of the audio frequency signal is known as amplitude modulation (AM).

Power modulation index is given by,
${P_T} = {P_C}\left( {1 + \dfrac{{{m_a}^2}}{2}} \right)$
Where, ${P_T}$ is the total power modulated, ${P_C}$ is the carrier power, and ${m_a}$ is the modulation index.
It is given in the question that,
Total power of modulated signal$ = 600W$
Carrier power$ = 400W$
Substitute the given data in the above expression.
$600 = 400\left( {1 + \dfrac{{m_a^2}}{2}} \right)$
$\dfrac{3}{2} = 1 + \dfrac{{m_a^2}}{2}$
$\dfrac{{m_a^2}}{2} = \dfrac{1}{2}$
${m_a} = 1$
Hence, the modulation index is 1 and the correct option is D.
Note Maximum power in the AM without distortion will occur when ${m_a} = 1$
If ${I_C}$is unmodulated current and${I_T}$ is total modulated current then,
$\dfrac{{{P_T}}}{{{P_C}}} = {\left( {\dfrac{{{I_T}}}{{{I_C}}}} \right)^2}$
The ratio of change of amplitude of carrier wave to the amplitude of original carrier wave is called modulation factor or degree of modulation to modulation index. It is also called modulation depth.
Complete step-by-step answer
The process of changing the amplitude of a carrier wave in accordance with the amplitude of the audio frequency signal is known as amplitude modulation (AM).

Power modulation index is given by,
${P_T} = {P_C}\left( {1 + \dfrac{{{m_a}^2}}{2}} \right)$
Where, ${P_T}$ is the total power modulated, ${P_C}$ is the carrier power, and ${m_a}$ is the modulation index.
It is given in the question that,
Total power of modulated signal$ = 600W$
Carrier power$ = 400W$
Substitute the given data in the above expression.
$600 = 400\left( {1 + \dfrac{{m_a^2}}{2}} \right)$
$\dfrac{3}{2} = 1 + \dfrac{{m_a^2}}{2}$
$\dfrac{{m_a^2}}{2} = \dfrac{1}{2}$
${m_a} = 1$
Hence, the modulation index is 1 and the correct option is D.
Note Maximum power in the AM without distortion will occur when ${m_a} = 1$
If ${I_C}$is unmodulated current and${I_T}$ is total modulated current then,
$\dfrac{{{P_T}}}{{{P_C}}} = {\left( {\dfrac{{{I_T}}}{{{I_C}}}} \right)^2}$
The ratio of change of amplitude of carrier wave to the amplitude of original carrier wave is called modulation factor or degree of modulation to modulation index. It is also called modulation depth.
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