Question

A carrier wave of peak voltage 15 V is used to transmit a message signal. If the modulation index is 60%, then the peak voltage of the modulating signal is

• A. 3 V
• B. 6 V
• C. 9 V
• D. 12 V

Hint:

The modulation index relates the peak voltage of the modulating signal to the carrier signal. Multiply the modulation index with the peak carrier voltage to determine the peak modulating signal voltage.

Answer 1

The Correct Option is C

The modulation index (\( m \)) is defined as the ratio of the peak voltage of the modulating signal (\( V_m \)) to the peak voltage of the carrier signal (\( V_c \)): \[ m = \frac{V_m}{V_c} \] Given: \[ V_c = 15 \, \text{V}, \quad m = 60% = 0.6 \] To find \( V_m \), use the formula: \[ V_m = m \times V_c \] Substitute the values: \[ V_m = 0.6 \times 15 = 9 \, \text{V} \] Thus, the peak voltage of the modulating signal is \( \boxed{9 \, \text{V}} \).

Answer 2

Step 1: Understand amplitude modulation (AM).
In amplitude modulation, the modulation index (μ) is defined as the ratio of the peak voltage of the modulating signal (\(V_m\)) to the peak voltage of the carrier wave (\(V_c\)):
\[ \mu = \frac{V_m}{V_c} \]

Step 2: Given data.
- Peak voltage of the carrier wave, \(V_c = 15 \, \text{V}\)
- Modulation index, \(\mu = 60\% = 0.6\)

Step 3: Use the formula to find \(V_m\).
\[ V_m = \mu \times V_c = 0.6 \times 15 = 9 \, \text{V} \]

Step 4: Conclusion.
The peak voltage of the modulating signal is 9 V.