If carrier wave is given by `y_c = A_c sin(omega_c t )` and message signal is `y_s = A_s sin( omega_s t)` find bandwidth of AM wave (in hz)

Options

A. `omega_s/pi`
B. `omega_s /(2 pi)`
C. `(omega_c- omega_s)/pi`
D. `2(omega_c- omega_s)/pi`

Answer

To find the bandwidth of an Amplitude Modulated (AM) wave given the carrier wave and message signal, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information**:
- The carrier wave is given by:
\[ y_c = A_c \sin(\omega_c t) \]
- The message signal is given by:
\[ y_s = A_s \sin(\omega_s t) \]

2. **Understand the Concept of Bandwidth**:
- The bandwidth of an AM wave can be defined as the difference between the upper sideband (USB) and the lower sideband (LSB).
- The formula for bandwidth (BW) is:
\[ \text{BW} = \text{USB} - \text{LSB} \]

3. **Determine the Upper Sideband (USB) and Lower Sideband (LSB)**:
- The upper sideband frequency is given by:
\[ \text{USB} = \omega_c + \omega_s \]
- The lower sideband frequency is given by:
\[ \text{LSB} = \omega_c - \omega_s \]

4. **Calculate the Bandwidth**:
- Now, substituting the values of USB and LSB into the bandwidth formula:
\[ \text{BW} = (\omega_c + \omega_s) - (\omega_c - \omega_s) \]
- Simplifying this gives:
\[ \text{BW} = \omega_c + \omega_s - \omega_c + \omega_s = 2\omega_s \]

5. **Convert Angular Frequency to Frequency**:
- The relationship between angular frequency (\(\omega\)) and frequency (\(f\)) is given by:
\[ \omega = 2\pi f \]
- Therefore, the frequency of the signal wave (\(f_s\)) can be expressed as:
\[ f_s = \frac{\omega_s}{2\pi} \]

6. **Substitute to Find Bandwidth in Hz**:
- Now substituting \(f_s\) into the bandwidth equation:
\[ \text{BW} = 2\omega_s = 2 \times (2\pi f_s) = 4\pi f_s \]

7. **Final Expression for Bandwidth**:
- Thus, the bandwidth of the AM wave in terms of the frequency of the signal wave is:
\[ \text{BW} = 2f_s \]