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 find the fraction of total power carried by the sidebands \((P_S/P_T)\) in amplitude modulation, we can follow these steps:

### Step 1: Understand the Total Power in Amplitude Modulation
In amplitude modulation, the total power \(P_T\) can be expressed as:
\[ P_T = P_C + \frac{P_C \mu^2}{2} \]
where \(P_C\) is the carrier power and \(\mu\) is the modulation index.

### Step 2: Identify the Sideband Power
The sideband power \(P_S\) is given by the term:
\[ P_S = \frac{P_C \mu^2}{2} \]

### Step 3: Substitute the Values into the Fraction
Now, we can substitute \(P_S\) and \(P_T\) into the fraction \(\frac{P_S}{P_T}\):
\[ \frac{P_S}{P_T} = \frac{\frac{P_C \mu^2}{2}}{P_C + \frac{P_C \mu^2}{2}} \]

### Step 4: Simplify the Fraction
We can factor out \(P_C\) from the denominator:
\[ \frac{P_S}{P_T} = \frac{\frac{P_C \mu^2}{2}}{P_C \left(1 + \frac{\mu^2}{2}\right)} \]
This simplifies to:
\[ \frac{P_S}{P_T} = \frac{\mu^2/2}{1 + \frac{\mu^2}{2}} \]

### Step 5: Further Simplification
Now, we can multiply the numerator and denominator by 2 to eliminate the fraction in the denominator:
\[ \frac{P_S}{P_T} = \frac{\mu^2}{2 + \mu^2} \]

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