An audio signal `V_(m)=5sin 6 pixx10^(3)t` is to be modulated on a carrier wave given by `V_(c)=15sin 2pi xx 10^(5)t` The frequencies of side bands and band width

Options

A. 103 KHz , 97 KHz , 6 KHz
B. 130 KHz , 70 KHz , 6 KHz
C. 130 KHz , 97 KHz , 3 KHz
D. 103 KHz , 97 KHz , 3 KHz

Answer

To solve the problem, we need to determine the frequencies of the sidebands and the bandwidth of the modulated signal. Let's break it down step by step.

### Step 1: Identify the frequencies of the audio signal and the carrier wave

The audio signal is given by:
\[ V_m = 5 \sin(6 \pi \times 10^3 t) \]

From this, we can identify the frequency \( f_m \) of the audio signal:
\[ f_m = \frac{6 \pi \times 10^3}{2 \pi} = 3 \times 10^3 \text{ Hz} = 3 \text{ kHz} \]

The carrier wave is given by:
\[ V_c = 15 \sin(2 \pi \times 10^5 t) \]

From this, we can identify the frequency \( f_c \) of the carrier wave:
\[ f_c = 10^5 \text{ Hz} = 100 \text{ kHz} \]

### Step 2: Calculate the frequencies of the sidebands

The frequencies of the sidebands are given by:
- Lower Sideband (LSB):
\[ f_{LSB} = f_c - f_m \]
- Upper Sideband (USB):
\[ f_{USB} = f_c + f_m \]

Now, substituting the values:
1. For the Lower Sideband:
\[ f_{LSB} = 100 \text{ kHz} - 3 \text{ kHz} = 97 \text{ kHz} \]

2. For the Upper Sideband:
\[ f_{USB} = 100 \text{ kHz} + 3 \text{ kHz} = 103 \text{ kHz} \]

### Step 3: Calculate the bandwidth

The bandwidth \( BW \) of the modulated signal is the difference between the frequencies of the upper and lower sidebands:
\[ BW = f_{USB} - f_{LSB} \]

Substituting the values we found:
\[ BW = 103 \text{ kHz} - 97 \text{ kHz} = 6 \text{ kHz} \]

### Final Results
- Frequencies of the sidebands:
- Lower Sideband: \( 97 \text{ kHz} \)
- Upper Sideband: \( 103 \text{ kHz} \)
- Bandwidth: \( 6 \text{ kHz} \)