Question

For the sequence x[n] = {1, -1, 1, -1}, with n = 0, 1, 2, 3, the DFT is computed as \(X\left( k \right) = \mathop \sum \limits_{n = 0}^3 x\left[ n \right]{e^{ - j\frac{{2\pi }}{4}nk}}\), for k = 0, 1, 2, 3. The value of k for which X(k) is not zero is

Answer 1

Correct Answer - Option 3 : 2

Analysis:

Given,

x(n) = {1, -1, 1, -1}

\(X\left( k \right) = \mathop \sum \limits_{h = 0}^3 x\left( n \right){e^{ - j}}\frac{{2\pi }}{4}nk\)

We can now write:

\(X\left( 0 \right) = \mathop \sum \limits_{n = 0}^3 x\left( n \right) \cdot 1\)

\( = x\left( 0 \right) + x\left( 1 \right) + x\left( 2 \right) + x\left( 3 \right) \)

X(0) = 0

\(X\left( 1 \right) = \mathop \sum \limits_{n = 0}^3 x\left( n \right){\left( { - j} \right)^n} \)

\(= 1 + j - 1 - j \)

X(1) = 0

\(X\left( 2 \right) = \mathop \sum \limits_{n = 0}^3 x\left( n \right){\left( { - 1} \right)^n} \)

\(= 1 + 1 + 1 + 1\)

X(2) = 4

\(X\left( 3 \right) = \mathop \sum \limits_{n = 0}^3 x\left( n \right){\left( j \right)^n} \)

\(= 1 - j - 1 + j\)

X(3) = 0

Hence,

X(k) = [0, 0, 4, 0]

We observe that the term for k = 2 is not zero.

Short Trick:

X(k) = X* (N – K) where N = 4

x(0) = x(0) + x(1) + x(2) + x(3) = 0

x(N/2) = x(2) = x(0) – x(1) + x(2) – x(3) = 4

Hence for k = 2, X(k) ≠ 0