Question:

Let \( x[n] \) be a discrete-time signal whose \( z \)-transform is \( X(z) \). Which of the following statements is/are TRUE?

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The DTFT exists when the region of convergence of the \( z \)-transform contains the unit circle. For a signal to have a DTFT, it must be well-behaved at all frequencies.
Updated On: Apr 15, 2025
  • The discrete-time Fourier transform (DTFT) of \( x[n] \) always exists
  • The region of convergence (RoC) of \( X(z) \) contains neither poles nor zeros
  • The discrete-time Fourier transform (DTFT) exists if the region of convergence (RoC) contains the unit circle
  • If \( x[n] = \alpha \delta[n] \), where \( \delta[n] \) is the unit impulse and \( \alpha \) is a scalar, then the region of convergence (RoC) is the entire \( z \)-plane
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The Correct Option is C, D

Solution and Explanation

(A) The discrete-time Fourier transform (DTFT) of \( x[n] \) always exists:
This statement is false because the DTFT only exists when the region of convergence (RoC) includes the unit circle. For some signals, the RoC may not include the unit circle, so the DTFT does not always exist.

(B) The region of convergence (RoC) of \( X(z) \) contains neither poles nor zeros:
This statement is incorrect. The RoC of the \( z \)-transform can contain poles, but it cannot contain zeros.

(C) The discrete-time Fourier transform (DTFT) exists if the region of convergence (RoC) contains the unit circle:
This statement is true. The DTFT is the \( z \)-transform evaluated on the unit circle, and for the DTFT to exist, the RoC must include the unit circle.

(D) If \( x[n] = \alpha \delta[n] \), where \( \delta[n] \) is the unit impulse and \( \alpha \) is a scalar, then the region of convergence (RoC) is the entire \( z \)-plane:
This is true. The \( z \)-transform of \( \delta[n] \) is 1 for all \( z \), so the RoC is the entire \( z \)-plane for such a signal.

Thus, the correct answers are (C) and (D).
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