When m=1 ,power carried by side bands is:

Options

A. 11.1% of the total power of AM wave
B. 22.2% of the total power of AM wave
C. 33.3% of the total power of AM wave
D. 44.4% of the total power of AM wave

Answer

To solve the question regarding the power carried by sidebands when \( m = 1 \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship**:
We need to use the relationship between the power carried by sidebands (\( P_s \)) and the total power of the AM wave (\( P_t \)). The formula is given by:
\[ \frac{P_s}{P_t} = \frac{m^2}{2 + m^2} \]

2. **Substitute the Given Value**:
Since we are given \( m = 1 \), we can substitute this value into the formula:
\[ \frac{P_s}{P_t} = \frac{1^2}{2 + 1^2} = \frac{1}{2 + 1} = \frac{1}{3} \]

3. **Calculate the Power of Sidebands**:
From the above relationship, we can express the power carried by sidebands in terms of total power:
\[ P_s = \frac{1}{3} P_t \]

4. **Convert to Percentage**:
To express the power carried by sidebands as a percentage of the total power:
\[ P_s = 0.33 P_t \quad \text{or} \quad P_s = 33.3\% \, \text{of} \, P_t \]

5. **Final Answer**:
Thus, the power carried by the sidebands when \( m = 1 \) is \( 33.3\% \) of the total power of the AM wave.