Question:

A carrier wave of peak voltage \(60\,\text{V}\) is used to transmit a message signal. Then the peak voltage of the modulating signal in order to have a modulation index of \(90\%\) is

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In amplitude modulation, \[ m=\frac{V_m}{V_c}. \] For proper transmission without distortion, \[ 0\le m\le 1. \] If \(m>1\), over-modulation occurs and the received signal becomes distorted.
Updated On: Jun 18, 2026
  • \(30\,\text{V}\)
  • \(54\,\text{V}\)
  • \(45\,\text{V}\)
  • \(60\,\text{V}\)
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The Correct Option is B

Solution and Explanation

Step 1: Write the expression for modulation index.
For amplitude modulation (AM), the modulation index is \[ m=\frac{V_m}{V_c} \] where \[ V_m=\text{peak voltage of modulating signal} \] and \[ V_c=\text{peak voltage of carrier signal}. \]

Step 2: Substitute the given values.

Given, \[ m=90\%=0.9 \] and \[ V_c=60\,\text{V}. \] Therefore, \[ 0.9=\frac{V_m}{60} \] \[ V_m=0.9\times 60 \] \[ V_m=54\,\text{V}. \]

Step 3: Verify the result.

Substituting \[ V_m=54\,\text{V} \] and \[ V_c=60\,\text{V}, \] \[ m=\frac{54}{60} \] \[ m=0.9=90\%. \] Hence, the value satisfies the required modulation index.

Step 4: Final conclusion.

Therefore, the peak voltage of the modulating signal is \[ \boxed{54\,\text{V}} \]
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