Nyquist sampling for a multitone analog signal: For m(t) = 4 cos(100π t) + 8 sin(200π t) + cos(300π t), what is the Nyquist sampling interval (s)?

Introduction / Context:Nyquist sampling theory states that to avoid aliasing, a continuous-time signal with maximum frequency f_max must be sampled at a rate f_s ≥ 2 f_max. Some exams present choices as sampling intervals in seconds (T_s = 1/f_s) rather than sampling rates in hertz; you must read carefully to select the correct unit.

Given Data / Assumptions:

• m(t) = 4 cos(100π t) + 8 sin(200π t) + cos(300π t).
• Frequencies present: 100π, 200π, and 300π radians/second.
• Nyquist criterion: f_s ≥ 2 f_max; sampling interval T_s = 1/f_s.

Concept / Approach:

Convert angular frequencies ω to hertz using f = ω/(2π). Then identify the highest component frequency and compute the Nyquist sampling rate f_N = 2 f_max. Finally, invert to obtain the sampling interval T_s = 1/f_N.

Step-by-Step Solution:

Verification / Alternative check:

Sampling at 300 Hz ensures at least two samples per period for the 150 Hz component. Any slower rate risks aliasing of the highest frequency term into a lower frequency band.

Why Other Options Are Wrong:

• 1/100 or 1/200: correspond to 100 Hz or 200 Hz sampling, below Nyquist for 150 Hz maximum frequency.
• 1/600: corresponds to 600 Hz sampling interval? No—1/600 s is faster (600 Hz), which is acceptable but the question asks the Nyquist sampling interval (minimum that satisfies criterion); 1/300 is the tightest bound.
• 1/150: corresponds to sampling at 150 Hz, clearly below Nyquist.

Common Pitfalls:

• Confusing angular frequency ω with hertz f.
• Choosing the sampling rate instead of sampling interval; here options are in seconds.

Final Answer:

1/300