An analog signal m(t) = 3 sin(500π t) + 2 sin(700π t) is sampled at 600 samples per second. What is the folding (Nyquist) frequency of this sampled system?

Difficulty: Easy

Correct Answer: 300 Hz

Explanation:


Introduction / Context:
In sampling theory for signals and systems, the folding frequency (also called the Nyquist frequency) is a crucial limit that separates positive and negative frequency images after sampling. It determines where aliasing reflections occur in the discrete-time spectrum. This question evaluates recognition of the Nyquist frequency when the sampling frequency is specified explicitly.


Given Data / Assumptions:

  • m(t) = 3 sin(500π t) + 2 sin(700π t).
  • Sampling frequency fs = 600 samples per second (600 Hz).
  • Standard radian–frequency to hertz conversion: ω = 2π f.


Concept / Approach:
The Nyquist frequency fN is one-half of the sampling frequency: fN = fs / 2. Spectral components above fN will fold (alias) back into the first Nyquist zone upon sampling. The individual analog components have ω1 = 500π rad/s and ω2 = 700π rad/s, corresponding to f1 = ω1 / (2π) = 250 Hz and f2 = ω2 / (2π) = 350 Hz; however, the folding frequency is determined solely by the sampling frequency, not by the signal content.


Step-by-Step Solution:

Compute Nyquist frequency: fN = fs / 2.Substitute fs = 600 Hz → fN = 600 / 2 = 300 Hz.Therefore, the folding frequency is 300 Hz.


Verification / Alternative check:

Component at 250 Hz lies below fN, so it will not alias. Component at 350 Hz lies above fN, so it will fold around 300 Hz to 300 − (350 − 300) = 250 Hz in the sampled spectrum, consistent with folding about 300 Hz.


Why Other Options Are Wrong:

1400 Hz, 700 Hz, 500 Hz, 200 Hz: These do not equal fs / 2 for fs = 600 Hz. The Nyquist frequency only depends on fs, not on individual sinusoid frequencies.


Common Pitfalls:

Confusing a component frequency with the Nyquist frequency; the folding frequency is always fs / 2.


Final Answer:

300 Hz

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