Skip to main content

Sampling; Nyquist of 1 + cos200πt + sin4000πt

Short answer

Sampling converts a continuous-time signal to a discrete sequence x[n]=x(nT) at intervals T. For x(t)=1+cos(200πt)+sin(4000πt), the frequencies are 0 Hz (DC), 100 Hz, and 2000 Hz. The highest frequency is 2000 Hz, so Nyquist rate = 2×2000 = 4000 Hz.

What is sampling? Determine the Nyquist rate for the following signal: x(t) = 1 + cos(200πt) + sin(4000πt)

Sampling

Sampling is the conversion of a continuous-time signal x(t)x(t) into a discrete-time sequence x[n]=x(nT)x[n] = x(nT) by taking its values at regular intervals TT (sampling period). The sampling frequency is fs=1/Tf_s = 1/T. By the sampling theorem, a band-limited signal can be recovered from its samples if fs>2fMf_s > 2f_M, where fMf_M is its highest frequency. Sampling is the first step of analog-to-digital conversion.

Nyquist rate of x(t)=1+cos⁡(200πt)+sin⁡(4000πt)x(t) = 1 + \cos(200\pi t) + \sin(4000\pi t)

Termω\omega (rad/s)ff (Hz)
110000 (DC)
cos⁡(200πt)\cos(200\pi t)200π200\pi100100
sin⁡(4000πt)\sin(4000\pi t)4000π4000\pi20002000

fM=2000f_M = 2000 Hz.

fN=2fM=4000 Hzf_N = 2 f_M = 4000\ \text{Hz}

Answer: Nyquist rate = 4000 Hz (4000 samples/s).

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Signals and Systems chapter 4 questions →