Question

2) find the 8-point DFT of the sequence
x(n) = cos(2πkn/N)​

Answer 1

Step-by-step explanation:

X1(k) and X2(k) are the N-point DFTs of X1(n) and x2(n) respectively, then what is the N-point DFT of x(n)=ax1(n)+bx2(n)?

a) X1(ak)+X2(bk)

b) aX1(k)+bX2(k)

c) eakX1(k)+ebkX2(k)

d) None of the mentioned

View Answer

Answer: b

Explanation: We know that, the DFT of a signal x(n) is given by the expression

X(k)=∑N−1n=0x(n)e−j2πkn/N

Given x(n)=ax1(n)+bx2(n)

=>X(k)= ∑N−1n=0(ax1(n)+bx2(n))e−j2πkn/N

=a∑N−1n=0x1(n)e−j2πkn/N+b∑N−1n=0x2(n)e−j2πkn/N

=>X(k)=aX1(k)+bX2(k).