Question

The z-transform of the discrete time signal $x(n) = u(n) * u(n)$ with '$*$' being convolution is

• A. $\frac{1}{[1-z^{-1}]^2}$, ROC: $|z| < 1$
• B. $\frac{1}{[1-z^{-1}]^2}$, ROC: $|z| > 1$
• C. $\frac{1}{[1-z]^2}$, ROC: $|z| > 1$
• D. $\frac{1}{[1-z]^2}$, ROC: $|z| < 1$

Hint:

Convolution in the time domain is equivalent to multiplication in the z-domain.
Since the unit step \( u(n) \) is a causal, right-sided sequence, its ROC must be exterior to a circle, i.e., \( |z| > 1 \).
The convolution of two causal sequences must also be causal, hence the ROC remains \( |z| > 1 \).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks to find the z-transform and the ROC of the convolution of a unit step sequence \( u(n) \) with itself.

Step 2: Key Formula or Approach:
The convolution property of the z-transform states that:
\[ Z\{x_1(n) * x_2(n)\} = X_1(z) \cdot X_2(z) \] The resulting ROC is the intersection of the ROCs of \( X_1(z) \) and \( X_2(z) \):
\[ \text{ROC} = \text{ROC}_1 \cap \text{ROC}_2 \]

Step 3: Detailed Explanation:

• Find the z-transform of the unit step signal \( u(n) \):
\[ U(z) = Z\{u(n)\} = \sum_{n=0}^{\infty} z^{-n} = \frac{1}{1-z^{-1}} \] The Region of Convergence for this right-sided sequence is \( |z| > 1 \).

• Apply the convolution property to \( x(n) = u(n) * u(n) \):
\[ X(z) = U(z) \cdot U(z) \] \[ X(z) = \left( \frac{1}{1-z^{-1}} \right) \cdot \left( \frac{1}{1-z^{-1}} \right) = \frac{1}{[1-z^{-1}]^2} \]
• Determine the common region of convergence:
The ROC of the first term is \( |z| > 1 \).
The ROC of the second term is \( |z| > 1 \).
The intersection of these two regions is \( |z| > 1 \).

• Thus, the final z-transform is:
\[ X(z) = \frac{1}{[1-z^{-1}]^2}, \quad \text{ROC: } |z| > 1 \]