Energy or power of DT cos(3πk/16)

x[k] = cos(3πk/16) for -10 ≤ k ≤ 0, and 0 otherwise.

The signal is non-zero for only 11 samples (finite duration), so its energy is finite and its average power is zero. It is an energy signal.

Energy

E = Σ |x[k]|² = Σ_{k=-10}^{0} cos²(3πk/16)

Using cos²θ = (1 + cos 2θ)/2:

E = ½ Σ 1 + ½ Σ cos(3πk/8) = 11/2 + ½ Σ_{k=-10}^{0} cos(3πk/8)

Evaluating the 11 terms numerically, Σ cos(3πk/8) = 0.3244, so

E = 5.5 + 0.1622 = 5.662

Power

P = lim (N→∞) 1/(2N+1) Σ_{k=-N}^{N} |x[k]|² = 5.662 / ∞ = 0

Answer: x[k] is an energy signal with E ≈ 5.66 and P = 0.