Since the FS coefficient repeat every N. Thus
X[1] = X[15], X[2] = X[16], X[3] = X[17]
The signal real and odd, the FS coefficient X[k] will be purely imaginary and odd. Therefore X[0] = 0 X[-1] = -X[1], X[-2] = -X[2], X[-3] = -X[-3]
Therefore (D) is correct option.
This discussion on A real and odd periodic signal x[n] has fundamental period N = 7 and FS coefficients X [k]. Given that X [15] = j, X [16] =2j, X [17] = 3j. The values of X [0],X [-1], X [-2], and X [-3] will bea)0, j,2j, 3jb)1, -1, -2, -3c)1,1,2,3d)0, -j , -2 j, -3 jCorrect answer is option 'D'. Can you explain this answer? is done on EduRev Study Group by Electronics and Communication Engineering (ECE) Students. The Questions and
Answers of A real and odd periodic signal x[n] has fundamental period N = 7 and FS coefficients X [k]. Given that X [15] = j, X [16] =2j, X [17] = 3j. The values of X [0],X [-1], X [-2], and X [-3] will bea)0, j,2j, 3jb)1, -1, -2, -3c)1,1,2,3d)0, -j , -2 j, -3 jCorrect answer is option 'D'. Can you explain this answer? are solved by group of students and teacher of Electronics and Communication Engineering (ECE), which is also the largest student
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A real and odd periodic signal x[n] has fundamental period N = 7 and FS coefficients X [k]. Given that X [15] = j, X [16] =2j, X [17] = 3j. The values of X [0],X [-1], X [-2], and X [-3] will bea)0, j,2j, 3jb)1, -1, -2, -3c)1,1,2,3d)0, -j , -2 j, -3 jCorrect answer is option 'D'. Can you explain this answer? over here on EduRev! Apart from being the largest Electronics and Communication Engineering (ECE) community, EduRev has the largest solved
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