To solve the problem, we need to find the modulation index and the difference between the two sideband frequencies in an amplitude modulation (AM) scenario.
### Given Data:
- Frequency of the message signal, \( f_m = 100 \, \text{MHz} = 100 \times 10^6 \, \text{Hz} \)
- Peak voltage of the message signal, \( A_m = 100 \, \text{V} \)
- Frequency of the carrier wave, \( f_c = 300 \, \text{GHz} = 300 \times 10^9 \, \text{Hz} \)
- Peak voltage of the carrier wave, \( A_c = 400 \, \text{V} \)
### Step 1: Calculate the Modulation Index
The modulation index \( \mu \) for amplitude modulation is given by the formula:
\[
\mu = \frac{A_m}{A_c}
\]
Substituting the values:
\[
\mu = \frac{100 \, \text{V}}{400 \, \text{V}} = \frac{1}{4} = 0.25
\]
### Step 2: Calculate the Difference Between the Two Sideband Frequencies
In amplitude modulation, the frequencies of the upper and lower sidebands are given by:
- Upper Sideband Frequency (USB): \( f_c + f_m \)
- Lower Sideband Frequency (LSB): \( f_c - f_m \)
The difference between the two sideband frequencies is:
\[
\Delta f = (f_c + f_m) - (f_c - f_m) = f_m + f_m = 2f_m
\]
Substituting the value of \( f_m \):
\[
\Delta f = 2 \times 100 \times 10^6 \, \text{Hz} = 200 \times 10^6 \, \text{Hz} = 200 \, \text{MHz}
\]
### Final Results:
- Modulation Index \( \mu = 0.25 \)
- Difference between the two sideband frequencies \( \Delta f = 200 \, \text{MHz} \)
### Summary:
The modulation index is \( 0.25 \) and the difference between the two sideband frequencies is \( 200 \, \text{MHz} \).
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