Difficulty: Easy
Correct Answer: T ≤ 1 / (2f)
Explanation:
Introduction / Context:Sampling theory is foundational in signals and systems, digital signal processing, and communications. The Nyquist–Shannon sampling theorem tells us how fast to sample a continuous-time signal so that it can be reconstructed without aliasing. This question asks for the correct relationship between sampling time T and the highest frequency component f in the analog signal.
Given Data / Assumptions:
Concept / Approach:For perfect reconstruction, the sampling frequency must satisfy fs ≥ 2f. This is commonly called the Nyquist rate. Since fs = 1 / T, the condition translates to 1 / T ≥ 2f, or equivalently T ≤ 1 / (2f). Choosing T strictly smaller than 1 / (2f) gives extra margin against nonidealities.
Step-by-Step Solution:
Let fs = 1 / T.Nyquist condition: fs ≥ 2f.Substitute fs: 1 / T ≥ 2f.Rearrange: T ≤ 1 / (2f).Verification / Alternative check:
If T = 1 / (2f), then fs = 2f (Nyquist rate). Any T larger than 1 / (2f) makes fs < 2f and causes aliasing. Any T smaller than 1 / (2f) increases safety margin.Why Other Options Are Wrong:
T ≥ 1 / (2f): makes fs ≤ 2f and risks aliasing.T = 1 / f or T = 2 / f: these give fs = f or fs = f/2, both below Nyquist.T = 1 / (4f): acceptable (stricter) but not the general inequality; the correct condition is T ≤ 1 / (2f).Common Pitfalls:
Confusing rad/s with Hz; Nyquist uses Hz (cycles/s). Also mixing up T and fs.Final Answer:
T ≤ 1 / (2f)
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