
A transmitter radiates $10kW$ of power with the carrier unmodulated and $11.8kW$ with the carrier sinusoidally modulated. The modulation factor is:
A) $56\% $.
B) $60\% $.
C) $72\% $.
D) $84\% $.
Answer
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Hint:This problem includes the terms modulated and unmodulated and power and therefore the formula for power AM signal modulation will be required to solve this problem. Modulation index in AM is the ratio of maximum change in the carrier amplitude to the original carrier amplitude.
Formula used: The formula of the power of AM signal modulation is given by,
${P_t} = {P_c}\left( {1 + \dfrac{{{m^2}}}{2}} \right)$Where${P_t}$ is the power of modulated carrier signal and ${P_c}$ is the power of unmodulated carrier signal and $m$ is the modulation index.
Complete step by step answer:
As it is given that a transmitter radiates a carrier unmodulated signal of power of $10kW$and the transmitter radiates a modulated carrier signal of$11.8kW$.
As the unmodulated and modulated signal of the carrier have a relation given by,
${P_t} = {P_c}\left( {1 + \dfrac{{{m^2}}}{2}} \right)$………eq. (1)
Where ${P_t}$is the power of a modulated carrier signal and ${P_c}$ is the power of an unmodulated carrier signal and $m$ is the modulation index.
Put the value of ${P_t}$ and ${P_c}$in the equation (1).
${P_t} = {P_c}\left( {1 + \dfrac{{{m^2}}}{2}} \right)$
Replace ${P_c} = 11.8$ and ${P_t} = 10$ in this relation.
$
{P_t} = {P_c}\left( {1 + \dfrac{{{m^2}}}{2}} \right) \\
\Rightarrow 11.8 = 10\left( {1 + \dfrac{{{m^2}}}{2}} \right) \\
\Rightarrow \left( {1 + \dfrac{{{m^2}}}{2}} \right) = 1.18 \\
\Rightarrow \dfrac{{{m^2}}}{2} = 0.18 \\
\Rightarrow {m^2} = 0.36 \\
\Rightarrow m = \sqrt {0.36} \\
\Rightarrow m = 0.6 \\
$
Converting in percentage $m = 60\% $.
So the correct answer for this problem is option B.
Note: The students should remember this formula in which the modulation index, the power of the modulated carrier power and unmodulated carrier power are included. The carrier signal is modulated in order to send signals over a band-pass frequency range.
Formula used: The formula of the power of AM signal modulation is given by,
${P_t} = {P_c}\left( {1 + \dfrac{{{m^2}}}{2}} \right)$Where${P_t}$ is the power of modulated carrier signal and ${P_c}$ is the power of unmodulated carrier signal and $m$ is the modulation index.
Complete step by step answer:
As it is given that a transmitter radiates a carrier unmodulated signal of power of $10kW$and the transmitter radiates a modulated carrier signal of$11.8kW$.
As the unmodulated and modulated signal of the carrier have a relation given by,
${P_t} = {P_c}\left( {1 + \dfrac{{{m^2}}}{2}} \right)$………eq. (1)
Where ${P_t}$is the power of a modulated carrier signal and ${P_c}$ is the power of an unmodulated carrier signal and $m$ is the modulation index.
Put the value of ${P_t}$ and ${P_c}$in the equation (1).
${P_t} = {P_c}\left( {1 + \dfrac{{{m^2}}}{2}} \right)$
Replace ${P_c} = 11.8$ and ${P_t} = 10$ in this relation.
$
{P_t} = {P_c}\left( {1 + \dfrac{{{m^2}}}{2}} \right) \\
\Rightarrow 11.8 = 10\left( {1 + \dfrac{{{m^2}}}{2}} \right) \\
\Rightarrow \left( {1 + \dfrac{{{m^2}}}{2}} \right) = 1.18 \\
\Rightarrow \dfrac{{{m^2}}}{2} = 0.18 \\
\Rightarrow {m^2} = 0.36 \\
\Rightarrow m = \sqrt {0.36} \\
\Rightarrow m = 0.6 \\
$
Converting in percentage $m = 60\% $.
So the correct answer for this problem is option B.
Note: The students should remember this formula in which the modulation index, the power of the modulated carrier power and unmodulated carrier power are included. The carrier signal is modulated in order to send signals over a band-pass frequency range.
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