Fraction of total power carried by the side bands `(P_(S)//P_(T))` is given by

Options

A. `m^(2)`
B. `(2 + m^(2))/(m^(2))`
C. `(2 +m)/(m)`
D. `(m^(2))/(2 + m^(2))`

Answer

To solve the problem of finding the fraction of total power carried by the sidebands \( \frac{P_s}{P_t} \), we can follow these steps:

### Step 1: Understand the Definitions
- **Total Power (\( P_t \))**: In amplitude modulation (AM), the total power is given by the formula:
\[ P_t = P_c \left(1 + \frac{\mu^2}{2}\right) \]
where \( P_c \) is the carrier power and \( \mu \) is the modulation index.

### Step 2: Identify Sideband Power (\( P_s \))
- The sideband power (\( P_s \)) can be expressed as:
\[ P_s = P_c \frac{\mu^2}{2} \]

### Step 3: Set Up the Ratio
- We need to find the ratio of sideband power to total power:
\[ \frac{P_s}{P_t} = \frac{P_c \frac{\mu^2}{2}}{P_c \left(1 + \frac{\mu^2}{2}\right)} \]

### Step 4: Simplify the Ratio
- Cancel \( P_c \) from the numerator and denominator:
\[ \frac{P_s}{P_t} = \frac{\frac{\mu^2}{2}}{1 + \frac{\mu^2}{2}} \]

### Step 5: Further Simplification
- To simplify further, multiply both the numerator and denominator by 2:
\[ \frac{P_s}{P_t} = \frac{\mu^2}{2 + \mu^2} \]

### Final Result
- Thus, the fraction of total power carried by the sidebands is:
\[ \frac{P_s}{P_t} = \frac{\mu^2}{2 + \mu^2} \]