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Nyquist rate of 5cos3000πt·sin1000πt

Short answer

x(t) = 5cos(3000πt)sin(1000πt) = 2.5 sin(4000πt) − 2.5 sin(2000πt), giving frequencies 2000 Hz and 1000 Hz. Maximum frequency is 2000 Hz, so Nyquist rate = 4000 Hz.

Determine the Nyquist rate for the signal x(t) = 5cos(3,000πt)sin(1,000πt).

The product must be written as a sum of sinusoids. Use sin⁡Acos⁡B=12[sin⁡(A+B)+sin⁡(A−B)]\sin A\cos B = \tfrac12[\sin(A+B) + \sin(A-B)] with A=1000πtA = 1000\pi t, B=3000πtB = 3000\pi t:

x(t)=5cos⁡(3000πt)sin⁡(1000πt)=52[sin⁡(4000πt)+sin⁡(−2000πt)]=2.5sin⁡(4000πt)−2.5sin⁡(2000πt)\begin{aligned} x(t) &= 5\cos(3000\pi t)\sin(1000\pi t) \\ &= \frac52\big[\sin(4000\pi t) + \sin(-2000\pi t)\big] \\ &= 2.5\sin(4000\pi t) - 2.5\sin(2000\pi t) \end{aligned}
Termω\omega (rad/s)ff (Hz)
2.5sin⁡(4000πt)2.5\sin(4000\pi t)4000π4000\pi20002000
−2.5sin⁡(2000πt)-2.5\sin(2000\pi t)2000π2000\pi10001000

fM=2000f_M = 2000 Hz, so

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

Answer: Nyquist rate = 4000 Hz (4000 samples/s, ωN=8000π\omega_N = 8000\pi rad/s). Note it is not twice 1500 Hz: multiplication creates sum and difference frequencies.

✓ Answer checked by Claude Opus 5.5 (high effort)

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