Discussion
A signal \(x_{1}(t)\) is bandlimited to 2 kHz while \(x_{2}(t)\) is bandlimited to 3 kHz. The Nyquist sampling rate for signal \(x_{1}(t)\, x_{2}(t)\) shall be :
Answer: A
The Nyquist rate is 10 kHz.
Multiplying two signals in time convolves their spectra in frequency, and the bandwidth of the product is the sum of the two bandwidths:
\[ \begin{aligned} B &= 2 + 3 = 5\ \text{kHz} \\ f_{s} &= 2B = 10\ \text{kHz} \end{aligned} \]
Using only the larger bandwidth, 3 kHz, gives 6 kHz, and only the smaller gives 4 kHz; both ignore the widening caused by multiplication. The value 12 kHz is the product 2 times 3 doubled, which has no basis. For a sum of the two signals, the Nyquist rate would indeed be 6 kHz.
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