Question 1. Assume x(t) has a bandwidth of 40 kHz. 1. What is the minimum sampling rate for this signal? 2. What is the minimum sampling rate if a guard band of 10 kHz is required? 3. What is the maximum sampling interval for the signal x1 (t)=x (t) cos(80,000nt)?

Question

Answer

1. Minimum Sampling Rate for x(t)
To determine the minimum sampling rate for a signal with a bandwidth of 40 kHz, we apply the Nyquist-Shannon sampling theorem. The theorem states that to avoid aliasing, the sampling rate must be at least twice the maximum frequency present in the signal.
- Bandwidth of x(t): 40 kHz
- Minimum Sampling Rate: $2 \times 40 \text{ kHz} = 80 \text{ kHz}$
2. Minimum Sampling Rate with Guard Band
When a guard band is required, the effective bandwidth increases. In this case, we add the guard band of 10 kHz to the original bandwidth.
- Effective Bandwidth: $40 \text{ kHz} + 10 \text{ kHz} = 50 \text{ kHz}$
- Minimum Sampling Rate: $2 \times 50 \text{ kHz} = 100 \text{ kHz}$
3. Maximum Sampling Interval for x1(t)
The signal $x_1(t) = x(t) \cos(80,000 \pi t)$ represents amplitude modulation where $x(t)$ is modulated by a cosine wave. The frequency of the cosine wave is critical for determining the sampling interval.
- Carrier Frequency of Cosine: $f_c = 80,000 \text{ Hz} = 80 \text{ kHz}$
- Bandwidth of x(t): 40 kHz implies frequencies up to 40 kHz.
- Total Bandwidth Consideration: The modulated signal has frequency components from $0$ to $80 \text{ kHz} + 40 \text{ kHz} = 120 \text{ kHz}$.
Using the Nyquist criterion again, the maximum sampling rate for $x_1(t)$ must be:
- Sampling Rate: $2 \times 120 \text{ kHz} = 240 \text{ kHz}$
Thus, the maximum sampling interval $T$ is:
- Maximum Sampling Interval: $T = \frac{1}{240,000} \approx 4.17 \text{ μs}$

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