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Aliasing; Nyquist rate of 1 + cos + sin

Short answer

Aliasing is spectral overlap when fs < 2fm, making high frequencies appear as false low frequencies. Component frequencies are DC, 1000 Hz, and 2000 Hz, so fm = 2000 Hz. Nyquist rate = 4000 samples/s (4 kHz), Nyquist interval = 0.25 ms.

What is aliasing in sampling? Determine the Nyquist rate for the continuous time signal x(t) = 1 + cos(2,000πt) + sin(4,000πt)

Aliasing

Aliasing is the overlap of the shifted spectra of a sampled signal when the sampling frequency is less than twice the highest frequency (fs<2fmf_s < 2f_m). High-frequency components then appear as false low frequencies ("aliases"), and the original signal cannot be recovered. For example, a 7 kHz tone sampled at 10 kHz looks like a 3 kHz tone.

Nyquist rate of x(t)=1+cos⁡(2000πt)+sin⁡(4000πt)x(t) = 1 + \cos(2000\pi t) + \sin(4000\pi t)

Find the frequency of each term using ω=2πf\omega = 2\pi f:

Termω\omega (rad/s)ff (Hz)
1100 (DC)
cos⁡(2000πt)\cos(2000\pi t)2000π2000\pi1000
sin⁡(4000πt)\sin(4000\pi t)4000π4000\pi2000

Highest frequency: fm=2000f_m = 2000 Hz.

Nyquist rate fN=2fm=2×2000=4000 HzNyquist interval TN=1fN=14000=0.25 ms\begin{aligned} \text{Nyquist rate } f_N &= 2f_m = 2 \times 2000 = 4000\ \text{Hz} \\ \text{Nyquist interval } T_N &= \frac{1}{f_N} = \frac{1}{4000} = 0.25\ \text{ms} \end{aligned}

Answer: Nyquist rate =4000= 4000 samples/s (4 kHz), i.e. ωs=8000π\omega_s = 8000\pi rad/s; Nyquist interval =0.25= 0.25 ms.

✓ Answer checked by Claude Opus 5.5 (high effort)

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