Question

The maximum permissible distance between two samples of a 2 kHz signal is

• A. 250 \(\mu\) sec
• B. 500 \(\mu\) sec
• C. 250 m sec
• D. 500 m sec

Hint:

Nyquist Sampling Theorem: \(f_s \ge 2f_{max}\).
Nyquist Rate = \(2f_{max}\).
Maximum sampling interval (Nyquist Interval) \(T_s = 1/(2f_{max})\).
\(1 \text{ kHz} = 10^3 \text{ Hz}\). \(1 \text{ msec} = 10^{-3} \text{ sec}\). \(1 \mu\text{sec} = 10^{-6} \text{ sec}\).

Answer 1

The Correct Option is A

This question refers to the Nyquist sampling theorem. To perfectly reconstruct a bandlimited signal with maximum frequency \(f_{max}\), the sampling frequency \(f_s\) must be at least twice the maximum frequency: \[ f_s \ge 2f_{max} \] This minimum sampling frequency, \(2f_{max}\), is called the Nyquist rate. The maximum permissible time interval between samples (sampling period), \(T_s\), is the reciprocal of the Nyquist rate: \[ T_s \le \frac{1}{2f_{max}} \] Given the signal has a maximum frequency \(f_{max} = 2 \text{ kHz} = 2 \times 10^3 \text{ Hz}\). The Nyquist rate is \(2f_{max} = 2 \times (2 \times 10^3 \text{ Hz}) = 4 \times 10^3 \text{ Hz} = 4 \text{ kHz}\). The maximum permissible distance (time interval) between two samples is: \[ T_{s,max} = \frac{1}{2f_{max}} = \frac{1}{4 \times 10^3 \text{ Hz}} = \frac{1}{4000} \text{ sec} \] \[ T_{s,max} = \frac{1}{4} \times 10^{-3} \text{ sec} = 0.25 \times 10^{-3} \text{ sec} \] To express this in microseconds (\(\mu\)sec): \(10^{-3} \text{ sec} = 1000 \times 10^{-6} \text{ sec} = 1000 \mu\text{sec}\). So, \(T_{s,max} = 0.25 \times 1000 \mu\text{sec} = 250 \mu\text{sec}\). This matches option (a). \[ \boxed{250 \text{ } \mu \text{sec}} \]