GATE 2017 EC – Question 60
In binary frequency shift keying (FSK), the given signal waveforms are
$u_0(t) = 5\cos(20000\pi t)$; $0 \leq t \leq T$, and
$u_1(t) = 5\cos(22000\pi t)$; $0 \leq t \leq T$,
where $T$ is the bit-duration interval and $t$ is in seconds. Both $u_0(t)$ and $u_1(t)$ are zero outside the interval $0 \leq t \leq T$. With a matched filter (correlator) based receiver, the smallest positive value of $T$ (in ms) required to have $u_0(t)$ and $u_1(t)$ uncorrelated is
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Correct answer: (B) 0.5 ms
Explanation
The two frequencies are $f_0 = 10$ kHz and $f_1 = 11$ kHz, so they differ by 1 kHz. The signals are uncorrelated over $[0, T]$ when the difference frequency completes a whole number of half cycles, $(f_1 - f_0)T = \frac{k}{2}$. The smallest positive value is $T = \frac{1}{2 \times 1000} = 0.5$ ms. At 0.5 ms the sum-frequency term also vanishes, since $21\ \text{kHz} \times 0.5\ \text{ms}$ is 10.5 cycles.