Fraction of total power carried by side bands is given by:

Options

A. `(P_(S))/P_(T)=m^(2)`
B. `(P_(S))/P_(T)=(1)/(m^(2))`
C. `(P_(S))/P_(T)=(2+m^(2))/(m^(2)`
D. `(P_(S))/P_(T)=(m^(2))/(2+m^(2))`

Answer

To find the fraction of total power carried by side bands in an Amplitude Modulated (AM) wave, we can follow these steps:

### Step 1: Understand the Total Power of AM Wave
The total power \( P_T \) of an AM wave is given by the formula:
\[ P_T = \frac{m^2 E_c^2}{4R} + \frac{E_c^2}{2R} \]
where:
- \( m \) is the modulation index,
- \( E_c \) is the peak value of the carrier wave,
- \( R \) is the resistance.

### Step 2: Calculate the Power of Side Bands
The power \( P_S \) carried by the side bands is given by:
\[ P_S = \frac{m^2 E_c^2}{4R} \]

### Step 3: Find the Fraction of Power in Side Bands
To find the fraction of total power carried by the side bands, we use the formula:
\[ \text{Fraction} = \frac{P_S}{P_T} \]
Substituting the expressions for \( P_S \) and \( P_T \):
\[ \text{Fraction} = \frac{\frac{m^2 E_c^2}{4R}}{\frac{m^2 E_c^2}{4R} + \frac{E_c^2}{2R}} \]

### Step 4: Simplify the Expression
Now, we simplify the expression:
1. The denominator can be rewritten by finding a common denominator:
\[ P_T = \frac{m^2 E_c^2}{4R} + \frac{2E_c^2}{4R} = \frac{m^2 E_c^2 + 2E_c^2}{4R} = \frac{(m^2 + 2)E_c^2}{4R} \]
2. Now substituting this back into the fraction:
\[ \text{Fraction} = \frac{\frac{m^2 E_c^2}{4R}}{\frac{(m^2 + 2)E_c^2}{4R}} \]
3. The \( \frac{1}{4R} \) cancels out:
\[ \text{Fraction} = \frac{m^2 E_c^2}{(m^2 + 2)E_c^2} \]
4. The \( E_c^2 \) also cancels out:
\[ \text{Fraction} = \frac{m^2}{m^2 + 2} \]

### Conclusion
Thus, the fraction of total power carried by the side bands is:
\[ \frac{m^2}{m^2 + 2} \]