In AM wave total power of side bands is given by :

Options

A. `P_(S ) =(E_C ^(2 ))/(4R )`
B. `P_(s )=(4E _(C )^(2 ))/(R )`
C. `P_(s ) =(m^2 E _(c )^(2))/(4R )`
D. `P_(s )=(E _(c )^(2) )/(4m ^(2 R)`

Answer

To find the total power of the sidebands in an amplitude modulation (AM) wave, we can follow these steps:

### Step 1: Understanding the AM Wave
In amplitude modulation, we have a carrier wave that is modulated by a message signal. The modulated wave consists of the carrier frequency and two sidebands: the upper sideband (USB) and the lower sideband (LSB).

### Step 2: Define the Peak Voltage
Let the peak voltage of the carrier wave be denoted as \( E_C \). The root mean square (RMS) voltage of the carrier wave can be expressed as:
\[ E_{RMS} = \frac{E_C}{\sqrt{2}} \]

### Step 3: Determine the Sideband Voltages
In AM, the sidebands are created by the modulation process. The voltages for the lower sideband (LSB) and upper sideband (USB) can be expressed as:
\[ E_{LSB} = \frac{M E_C}{2\sqrt{2}} \quad \text{and} \quad E_{USB} = \frac{M E_C}{2\sqrt{2}} \]
where \( M \) is the modulation index.

### Step 4: Calculate Power of Each Sideband
The power of an electrical signal can be calculated using the formula:
\[ P = \frac{V^2}{R} \]
where \( V \) is the voltage and \( R \) is the resistance.

For the lower sideband:
\[ P_{LSB} = \frac{E_{LSB}^2}{R} = \frac{\left(\frac{M E_C}{2\sqrt{2}}\right)^2}{R} = \frac{M^2 E_C^2}{8R} \]

For the upper sideband:
\[ P_{USB} = \frac{E_{USB}^2}{R} = \frac{\left(\frac{M E_C}{2\sqrt{2}}\right)^2}{R} = \frac{M^2 E_C^2}{8R} \]

### Step 5: Total Power of Sidebands
To find the total power of the sidebands, we add the power of the lower sideband and the upper sideband:
\[ P_{total} = P_{LSB} + P_{USB} = \frac{M^2 E_C^2}{8R} + \frac{M^2 E_C^2}{8R} = \frac{M^2 E_C^2}{4R} \]

### Conclusion
Thus, the total power of the sidebands in an amplitude modulation wave is given by:
\[ P_{total} = \frac{M^2 E_C^2}{4R} \]