A carrier wave of peak voltage 12 V is used to transmit a message signal. What should be the peak voltage of the modulating signal in order to have a modulation index of 75%?

Options

A. 5V
B. 9V
C. 12 V
D. 15 V

Answer

To find the peak voltage of the modulating signal (Em) required to achieve a modulation index (m) of 75% with a carrier wave peak voltage (Ec) of 12 V, we can use the formula for the modulation index in amplitude modulation:

\[ m = \frac{E_{max} - E_{min}}{E_{max} + E_{min}} \]

Where:
- \( E_{max} \) is the peak voltage of the carrier wave plus the peak voltage of the modulating signal.
- \( E_{min} \) is the peak voltage of the carrier wave minus the peak voltage of the modulating signal.

Given that:
- \( E_c = 12 \, V \) (the peak voltage of the carrier wave)
- \( m = 0.75 \) (75% modulation index)

### Step-by-step Solution:

1. **Understand the modulation index formula**:
\[ m = \frac{E_{max} - E_{min}}{E_{max} + E_{min}} \]
Here, we can express \( E_{max} \) and \( E_{min} \) in terms of \( E_c \) and \( E_m \):
\[ E_{max} = E_c + E_m \]
\[ E_{min} = E_c - E_m \]

2. **Substituting into the modulation index formula**:
Substitute \( E_{max} \) and \( E_{min} \) into the modulation index formula:
\[ m = \frac{(E_c + E_m) - (E_c - E_m)}{(E_c + E_m) + (E_c - E_m)} \]
Simplifying this gives:
\[ m = \frac{2E_m}{2E_c} \]

3. **Simplifying the equation**:
The equation simplifies to:
\[ m = \frac{E_m}{E_c} \]

4. **Solving for \( E_m \)**:
Rearranging the equation to solve for \( E_m \):
\[ E_m = m \cdot E_c \]

5. **Substituting the known values**:
Substitute \( m = 0.75 \) and \( E_c = 12 \, V \):
\[ E_m = 0.75 \cdot 12 \]

6. **Calculating \( E_m \)**:
\[ E_m = 9 \, V \]

### Final Answer:
The peak voltage of the modulating signal should be **9 V**.