Correct answer is (b) 1 + 0.9 \(e^{-j \frac{2π}{8} k6}\)
The explanation: Given N = 8.
Now, x[k] = \(∑_0^{N-1} x[n] e^{-j \frac{2π}{N} kn}\)
= \(∑_0^7 x[n] e^{-j \frac{2π}{8} kn}\)
= \(∑_{N=0}^7 (δ[n] + 0.9 δ [n – 6])e^{-j \frac{2π}{8} kn}\)
= 1 + 0.9 \(e^{-j \frac{2π}{8} kn}\)
Here, n = 6, from given question.
Hence, x[k] = 1 + 0.9 \(e^{-j \frac{2π}{8} k6}\).