Question

For a vector $\bar{x} = [x[0], x[1], ..., x[7]]$, the 8–point DFT is $\bar{X} = DFT(\bar{x}) = [X[0], X[1], ..., X[7]]$ where \[ X[k] = \sum_{n=0}^{7} x[n]\exp\left(-j\,\frac{2\pi}{8}nk\right). \] If $\bar{x} = [1, 0, 0, 0, 2, 0, 0, 0]$ and $\bar{y} = DFT(DFT(\bar{x}))$, the value of $y[0]$ is __________ (rounded off to one decimal place).

Hint:

The DC term of a DFT equals the sum of sequence elements. Applying DFT twice produces a scaled reversal.

Answer 1

Correct Answer: 7.9

Given \[ x = [1, 0, 0, 0, 2, 0, 0, 0]. \] First compute the DFT. For any 8–point sequence, the DFT at index $k$ is: \[ X[k] = 1 + 2\,e^{-j\pi k}. \] Since \[ e^{-j\pi k} = (-1)^k, \] we have: \[ X[k] = 1 + 2(-1)^k. \] Thus, \[ X[k] = \begin{cases} 3, & k\ \text{even} \\ -1, & k\ \text{odd} \end{cases} \] Now compute $y = DFT(X)$. The DC term $y[0]$ is: \[ y[0] = \sum_{k=0}^{7} X[k] = (3 + 3 + 3 + 3) + (-1 - 1 - 1 - 1) \] \[ y[0] = 12 - 4 = 8. \] Thus, \[ \boxed{8.0} \quad (\text{acceptable range: } 7.9\text{–}8.1) \]