Correct Option: B

Given, x(n) = 6e	j2πn
	4

or x(n) = 6	lim s → 0		cos	2πn	+ j sin	2πn
				4		4

The signal is periodic with period

N =	2π	m = 4
	2π
	4

with m = 1

x(0) = 6		cos	2π	· 0 + j sin	2π	· 0		= 6[1 + 0] = 6
			4		4

x(1) = 6		cos	2π	· 1 + j sin	2π	· 1
			4		4

= 6		cos	π	+ j sin	π			= j6
			2		2

x(2) = 6		cos	2π	· 2 + j sin	2π	· 1
			4		4

= 6 [cos π + j sin π] = 6 [– 1 + j·0] = – 6

x(3) = 6		cos	2π	· 3 + j sin	2π	· 3
			4		4

= 6		cos	3π	+ j sin	3π			= 6[– j] = – j6.
			2		2

Now, the power of the signal x(n)

	N - 1
P = 1/N	∑	| x(n)|2
	n = 0

	4 - 1
= 1/4	∑	|x(n)|2
	n = 0

=	1	[|x(0)|2 + |x(1)|2 + |x(2)|2 + |x(3)|2]
	4

=	1	[|6|2 + |j6|2 + |6|2+ |– 6|2]
	4

=	1	[36 + 36 + 36 + 36]
	4

= 36 watt
Hence, alternative (B) is the correct choice.