Curioustab Logo Curioustab
Home » Electronics and Communication Engineering » Signals and Systems

For a continuous-time signal with highest frequency component f, what should the sampling time T be to satisfy the Nyquist sampling criterion (i.e., avoid aliasing)?

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:

  • Signal has a maximum (one-sided) frequency component at f hertz.
  • Sampling is uniform with sampling period T seconds and sampling frequency fs = 1 / T.
  • No antialias filter imperfections are considered (idealized case).


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)
Next Question→

More Questions from Signals and Systems

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion