Question

In Amplitude Modulation, the message signal $V_m(t) = 10 \sin(2\pi \times 10^5 t)$ volts and Carrier signal $V_c(t) = 20 \sin(2\pi \times 10^7 t)$ volts. The modulated signal now contains the message signal with lower side band and upper side band frequency, therefore the bandwidth of modulated signal is $\alpha \text{ kHz}$. The value of $\alpha$ is :

• A. 200 kHz
• B. 100 kHz
• C. 50 kHz
• D. 0

Hint:

Bandwidth in AM is always twice the highest frequency component of the modulating (message) signal.
The carrier frequency does not affect the bandwidth magnitude itself, only its location in the spectrum.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
In Amplitude Modulation (AM), the bandwidth is the difference between the maximum and minimum frequency components of the modulated signal.
The frequency components are the carrier frequency (\(f_c\)), the upper sideband (\(f_c + f_m\)), and the lower sideband (\(f_c - f_m\)).
Step 2: Key Formula or Approach:
Bandwidth (\(BW\)) of an AM signal = \( (f_c + f_m) - (f_c - f_m) = 2 f_m \).
Where \(f_m\) is the frequency of the message signal.
Step 3: Detailed Explanation:
The given message signal is \(V_m(t) = 10 \sin(2\pi \times 10^5 t)\).
The standard form is \(V_m(t) = A_m \sin(2\pi f_m t)\).
Comparing the two, we get message frequency \(f_m = 10^5 \text{ Hz}\).
Convert Hz to kHz:
\[ f_m = \frac{10^5}{10^3} \text{ kHz} = 100 \text{ kHz} \]
The bandwidth of the modulated signal is:
\[ BW = 2 f_m = 2 \times 100 \text{ kHz} = 200 \text{ kHz} \]
Given bandwidth is \(\alpha \text{ kHz}\), so \(\alpha = 200\).
Step 4: Final Answer:
The value of \(\alpha\) is 200.