In amplitude modulation carrier wave is `C=4 sin 2000t` and message wave `M=2 sin 200t`, Find out modulation index.

Options

A. 9
B. 3
C. `(1)/(9)`
D. `(1)/(3)`

Answer

To find the modulation index in amplitude modulation, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Carrier and Message Waves:**
- The carrier wave \( C \) is given as:
\[ C = 4 \sin(2000t) \]
- The message wave \( M \) is given as:
\[ M = 2 \sin(200t) \]

2. **Determine the Peak Amplitudes:**
- The peak amplitude of the carrier wave \( V_c \) is:
\[ V_c = 4 \]
- The peak amplitude of the message wave \( V_m \) is:
\[ V_m = 2 \]

3. **Calculate the Modulation Index:**
- The modulation index \( m \) is defined as the ratio of the peak amplitude of the message wave to the peak amplitude of the carrier wave:
\[ m = \frac{V_m}{V_c} \]
- Substituting the values we have:
\[ m = \frac{2}{4} = \frac{1}{2} \]

4. **Alternative Calculation Using Maximum and Minimum Values:**
- The maximum value of the modulated wave \( V_{max} \) is given by \( V_c + V_m = 4 + 2 = 6 \).
- The minimum value of the modulated wave \( V_{min} \) is given by \( V_c - V_m = 4 - 2 = 2 \).
- The modulation index can also be calculated using:
\[ m = \frac{V_{max} - V_{min}}{V_{max} + V_{min}} \]
- Substituting the values:
\[ m = \frac{6 - 2}{6 + 2} = \frac{4}{8} = \frac{1}{2} \]

5. **Final Result:**
- Therefore, the modulation index \( m \) is:
\[ m = \frac{1}{2} \]