Two point DFT of a sequence x[n] is X[k] = [6, 2], compute its inverse.

Solution

The DFT of a signal x[n] is given by

X[k] = \( \sum_{n=0}^{N-1} x[n] e^{\frac{-j2\pi f kn}{N}} \)

The IDFT to get signal x[n] is given by

x[n] = \(\frac{1}{N} \sum_{n=0}^{N-1} X[k] e^{\frac{+j2\pi f kn}{N}} \)

Here , N is the length of the sequence

Calculation

Given X[k] = [6, 2] ,

X[0] = 6, X[1] = 2

So N = 2

Now x[n] = \(\frac{1}{N} \sum_{n=0}^{N-1} X[k] e^{\frac{+j2\pi f kn}{N}} \)

x[0] = \(\frac{1}{2} \sum_{n=0}^{1} X[k] \) = \(\frac{X[0] + X[1]}{2}\)

x[0] = 4

Also , x[1] = \(\frac{1}{2} \sum_{n=0}^{1} X[k] e^{\frac{+j2\pi f k}{2}} \)

x[1] = \(\frac {1}{2} [X[0] + X[1] e^{{+j\pi f }{}}] \)

x[1] = \(\frac {1}{2} [X[0] - X[1] ]\)

x[1] = 2

Therefore x[n] = [ 4, 2 ] 

The correct answer is option 4