If xin(t) = sin(2$$\pi $$ × 4000t) + 0.75sin(2$$\pi $$ × 5000t + $$\frac{\pi }{4}$$) is sampled with Fs = 16000 Hz. Calculate X(0) if $$X\left( m \right) = \sum\nolimits_{n = 0}^{N - 1} {x\left( n \right){e^{ - \frac{{j2\pi nm}}{N}}}} .$$ When N = 8, where x(n) = xin(nts)
A. 0.0 - j4.0
B. 0.0 - j0.0
C. 1.414 + j1.414
D. 0.0 + j4.0
Answer: Option B

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