Nyquist rate for a product of two cosines
BlinkNBuild practice problem · GATE standard, authored and verified in-house
Consider the signal $$x(t) = 10\cos(2\pi \cdot 2000\,t)\,\cos(2\pi \cdot 500\,t).$$ The Nyquist rate for $x(t)$, in kHz, is ________.
Show the step-by-step derivation
Answer
5 kHz kHz
Step-by-step derivation
- A product of sinusoids is not band-limited to the higher of the two frequencies. Expand it first using the product-to-sum identity $$\cos A \cos B = \tfrac{1}{2}\left[\cos(A-B) + \cos(A+B)\right].$$
- With $A = 2\pi(2000)t$ and $B = 2\pi(500)t$: $$x(t) = \frac{10}{2}\Big[\cos\big(2\pi(2000-500)t\big) + \cos\big(2\pi(2000+500)t\big)\Big].$$
- So $$x(t) = 5\cos(2\pi \cdot 1500\,t) + 5\cos(2\pi \cdot 2500\,t),$$ a sum of two sinusoids at $1500$ Hz and $2500$ Hz.
- The highest frequency present is $$f_{max} = 2500\ \text{Hz} = 2.5\ \text{kHz}.$$
- The Nyquist rate is twice the maximum frequency: $$f_{Nyq} = 2 f_{max} = 2 \times 2500 = 5000\ \text{Hz} = \mathbf{5\ kHz}.$$
The trap this question is built around
Reading $2000$ Hz off the first factor and answering $4$ kHz. Multiplication in time is convolution in frequency, so the spectrum spreads - the product contains energy at $2500$ Hz, above either original component. Always expand a product before looking for $f_{max}$.