Test: DFT Properties - Electronics and Communication Engineering (ECE) MCQ

 According to the complex conjugate property of DFT, we have if X(k) is the N-point DFT of a sequence x(n), then what is the DFT of x*(n) is X*(N-k).

Explanation: According to the circular time shift property of a sequence, If X(k) is the N-point DFT of a sequence x(n), then the N-pint DFT of x((n-l))_N is X(k)e^-j2πkl/N.

Explanation: Given x1(n)={2,1,2,1}=>X1(k)=[6,0,2,0] Given x2(n)={1,2,3,4}=>X2(k)=[10,-2+j2,-2,-2-j2] when we multiply both DFTs we obtain the product
X(k)=X1(k).X2(k)=[60,0,-4,0] By applying the IDFT to the above sequence, we get
x(n)={14,16,14,16}.

Explanation: We know that the circular convolution of two sequences is given by the expression

For m=0,x2((-n))4={1,4,3,2}
For m=1,x2((1-n))4={2,1,4,3}
For m=2,x2((2-n))4={3,2,1,4}
For m=3,x2((3-n))4={4,3,2,1}
Now we get x(m)={14,16,14,16}.

Explanation: If x(n) is real and odd, that is x(n)=-x(N-n), then XR(k)=0. Hence X(k) is purely imaginary and odd. Since XR(k) reduces to zero, the IDFT reduces to

Explanation: Given x(n) is real and even, that is x(n)=x(N-n)
We know that XI(k)=0. Hence the DFT reduces to

Explanation: We know that

Therefore,
X(N-k)=X*(k)=X(-k)

Explanation: Given x(n)=xR(n)+jxI(n)=> xR(n)=1/2(x(n)+x*(n))
Substitute the above equation in the DFT expression
Thus we get,

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

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