Question

The Z-transform of a discrete signal \(x[n]\) is \[ X(z) = \frac{4z}{\left(z-\tfrac{2}{3}\right)(z-3)}, \text{with ROC} = R. \] Which one of the following statements is true?

• A. Discrete-time Fourier transform (DTFT) of \(x[n]\) converges if ROC is \(|z| > 3\).
• B. Discrete-time Fourier transform of \(x[n]\) converges if ROC is \(\tfrac{2}{3} < |z| < 3\).
• C. Discrete-time Fourier transform of \(x[n]\) converges if ROC is such that \(x[n]\) is a left-sided sequence.
• D. Discrete-time Fourier transform of \(x[n]\) converges if ROC is such that \(x[n]\) is a right-sided sequence.

Hint:

For the DTFT to converge, the unit circle \(|z|=1\) must lie within the ROC of the Z-transform.

Answer 1

The Correct Option is B

Step 1: Identify poles.
Denominator is \((z-\tfrac{2}{3})(z-3)\). Thus poles are: \[ z = \tfrac{2}{3}, z = 3. \]

Step 2: Possible ROCs.
- If \(x[n]\) is right-sided, ROC is outside outermost pole: \(|z| > 3\). - If \(x[n]\) is left-sided, ROC is inside innermost pole: \(|z| < \tfrac{2}{3}\). - If \(x[n]\) is two-sided, ROC is between the poles: \(\tfrac{2}{3} < |z| < 3\).

Step 3: DTFT condition.
For DTFT to exist, the unit circle \(|z|=1\) must lie within the ROC. This happens only if: \[ \tfrac{2}{3} < 1 < 3. \] So the ROC must be \(\tfrac{2}{3} < |z| < 3\).

Step 4: Eliminate wrong options.
- (A) \(|z|>3\): unit circle \(|z|=1\) not included. Wrong. - (C) Left-sided ROC: \(|z|<\tfrac{2}{3}\). Unit circle not included. Wrong. - (D) Right-sided ROC: \(|z|>3\). Again unit circle not included. Wrong. Only option (B) satisfies DTFT existence.

Final Answer: \[ \boxed{\tfrac{2}{3} < |z| < 3} \]