A message single of frequency 100 MHz and peak voltage 100 V is used to execute amplitude modulation on a carrier wave of frequency 300 GHz and peak voltage 400 V. The modulation index and difference between the two side band frequencies are:

Options

A. `4, 1 xx 10^(8)` Hz
B. `0.25, 2 xx 10^(8)` Hz
C. `4.2 xx 10^(8)` Hz
D. `0.25, 1 xx 10^(8)` Hz

Explanation

To solve the problem, we need to find two things: the modulation index and the difference between the two sideband frequencies in an amplitude modulation (AM) system.

### Step 1: Identify the given values
- Frequency of the message signal (fm) = 100 MHz = \(100 \times 10^6\) Hz
- Peak voltage of the message signal (Vm) = 100 V
- Frequency of the carrier wave (fc) = 300 GHz = \(300 \times 10^9\) Hz
- Peak voltage of the carrier wave (Vc) = 400 V

### Step 2: Calculate the modulation index (μ)
The modulation index (μ) in amplitude modulation is defined as the ratio of the peak voltage of the message signal to the peak voltage of the carrier wave. The formula is given by:

\[ \mu = \frac{V_m}{V_c} \]

Substituting the given values:

\[ \mu = \frac{100 \, \text{V}}{400 \, \text{V}} = 0.25 \]

### Step 3: Calculate the difference between the two sideband frequencies
The difference between the two sideband frequencies in amplitude modulation is given by:

\[ \Delta f = f_{max} - f_{min} = 2f_m \]

Where:
- \(f_{max}\) is the maximum frequency component (carrier frequency + message frequency)
- \(f_{min}\) is the minimum frequency component (carrier frequency - message frequency)

Substituting the value of \(f_m\):

\[ \Delta f = 2 \times f_m = 2 \times 100 \times 10^6 \, \text{Hz} = 200 \times 10^6 \, \text{Hz} = 200 \, \text{MHz} \]

### Final Results
- Modulation index (μ) = 0.25
- Difference between the two sideband frequencies (Δf) = 200 MHz