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Nyquist rate of 20sin500πt + 10cos300πt − 50cos1000πt

Short answer

Sampling theorem: a band-limited signal with maximum frequency fm is exactly recoverable from samples if fs ≥ 2fm; the minimum rate is the Nyquist rate, and 1/(2fm) is the Nyquist interval. Component frequencies are 250, 150, and 500 Hz, so fm = 500 Hz. Nyquist rate = 1000 samples/s, interval = 1 ms.

Define sampling theorem and Nyquist criteria for sampling. Find the Nyquist rate for the signal: x(t) = 20 sin(500πt) + 10 cos(300πt) − 50 cos(1000πt)

Sampling theorem

A continuous-time signal that is band-limited to fmf_m Hz (no frequency components above fmf_m) is completely described by its samples taken at a uniform rate fs≥2fmf_s \ge 2f_m. The signal can then be exactly recovered from the samples by passing them through an ideal low-pass filter with cutoff between fmf_m and fs−fmf_s - f_m.

Nyquist criterion

  • The minimum sampling rate fN=2fmf_N = 2f_m is the Nyquist rate; its reciprocal TN=12fmT_N = \frac{1}{2f_m} is the Nyquist interval (maximum gap between samples).
  • If fs<2fmf_s < 2f_m, the spectral copies overlap and aliasing occurs. In practice fsf_s is kept above 2fm2f_m (e.g. audio: fm=20f_m = 20 kHz, fs=44.1f_s = 44.1 kHz).

Nyquist rate of x(t)=20sin⁡(500πt)+10cos⁡(300πt)−50cos⁡(1000πt)x(t) = 20\sin(500\pi t) + 10\cos(300\pi t) - 50\cos(1000\pi t)

Termω\omega (rad/s)f=ω/2πf = \omega/2\pi (Hz)
20sin⁡(500πt)20\sin(500\pi t)500π500\pi250
10cos⁡(300πt)10\cos(300\pi t)300π300\pi150
−50cos⁡(1000πt)-50\cos(1000\pi t)1000π1000\pi500

Highest frequency fm=500f_m = 500 Hz (the amplitudes do not matter).

fN=2fm=2×500=1000 HzTN=11000=1 ms\begin{aligned} f_N &= 2f_m = 2\times500 = 1000\ \text{Hz} \\ T_N &= \frac{1}{1000} = 1\ \text{ms} \end{aligned}

Answer: Nyquist rate =1000= 1000 samples/s (1 kHz, or 2000π2000\pi rad/s); Nyquist interval =1= 1 ms.

✓ Answer checked by Claude Opus 5.5 (high effort)

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