If X(k) is the N-point DFT of a sequence x(n), then what is the DFT of x*(n)?Select one:1. X(N-k)2. None of the mentioned3. X*(k)4. X*(N-k)

To determine the DFT of $x^*(n)$ where $x^*(n)$ is the conjugate of the sequence $x(n)$, we can use the properties of the Discrete Fourier Transform (DFT).

Step 1: Understanding DFT Properties

The DFT of a signal $x(n)$ is given by the equation: $X(k) = \sum_{n=0}^{N-1} x(n) e^{-j \frac{2\pi kn}{N}}$ where $k = 0, 1, 2, \ldots, N-1$.

Step 2: Conjugate of the Sequence

If we take the conjugate of the time-domain sequence $x(n)$, we have: $x^*(n)$

Step 3: DFT of the Conjugated Sequence

The DFT of $x^*(n)$ is computed as follows: $X^*(k) = \sum_{n=0}^{N-1} x^*(n) e^{-j \frac{2\pi kn}{N}}$ Using the property of conjugates in the exponentials: $X^*(k) = \sum_{n=0}^{N-1} x^*(n) e^{-j \frac{2\pi kn}{N}} = \sum_{n=0}^{N-1} x(n) e^{j \frac{2\pi kn}{N}} = X(N-k)$

Step 4: Verification and Conclusion

Thus, we have established that: $\text{DFT of } x^*(n) = X(N-k)$

Final Answer

The answer to the question is:

1. X(N-k)