Discrete Fourier Transform Properties MCQ Quiz – Objective Question with Answer for Discrete Fourier Transform Properties

We know that the expression for an DFT is given as

X(k)=\(\sum_{n=0}^{N-1} x(n)e^{-j2πkn/N}\)

x(n)=x(n+N)=>X1(k)=\(\sum_{n=0}^{N-1} x(n+N)e^{-j2πkn/N}\)

Let n+N=l=>X1(k)=\(\sum_{l=N}^0 x(l)e^{-j2πkl/N}\)=X(k)

Therefore, we got x(n)=x(n+N)

We know that

x(n)=\(\frac{1}{N}\sum_{k=0}^{N-1} x(k)e^{j2πkn/N}\)

Let X(k)=X(k+N)

=>x1(n)=\(\frac{1}{N} \sum_{k=0}^{N-1}X(k+N)e^{j2πkn/N}\)=x(n)

Therefore, we have X(k)=X(k+N)

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

X(k)=\(\sum_{n=0}^{N-1}x(n)e^{-j2πkn/N}\)

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

=>X(k)= \(\sum_{n=0}^{N-1}(ax_1(n)+bx_2(n))e^{-j2πkn/N}\)

=\(a\sum_{n=0}^{N-1} x_1(n)e^{-j2πkn/N}+b\sum_{n=0}^{N-1}x_2(n)e^{-j2πkn/N}\)

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

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,

XR(k)=\(\sum_{n=0}^{N-1} x_R (n) cos⁡\frac{2πkn}{N}+x_I (n) sin⁡\frac{2πkn}{N}\)

We know that

X(k)=\(\sum_{n=0}^{N-1} x(n)e^{-j2πkn/N}\)

Now X(N-k)=\(\sum_{n=0}^{N-1} x(n)e^{-j2π(N-k)n/N}\)=X*(k)=X(-k)

Therefore,

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

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

X(k)=\(\sum_{n=0}^{N-1} x(n) cos⁡\frac{2πkn}{N}\) ;0 ≤ k ≤ N-1

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

\(x(n)=j \frac{1}{N} \sum_{k=0}^{N-1} x(k) sin⁡\frac{2πkn}{N}\)

If X1(n), x2(n) and x3(m) are three sequences each of length N whose DFTs are given as X1(k), x2(k) and X3(k) respectively and X3(k)=X1(k).X2(k), then according to the multiplication property of DFT we have x3(m) is the circular convolution of X1(n) and x2(n).

That is x3(m) = \(\sum_{n=0}^{N-1}x_1 (n) x_2 (m-n)_N \).

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

x(m)= \(\sum_{n=0}^{N-1}x_1 (n) x_2 (m-n)_N\)

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}.

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}.