Question

The z-transform of the discrete time signal $x(n) = u(-n)$ is

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

Hint:

Left-sided sequences in the time domain always result in an ROC of the form \( |z| < r \).
Right-sided sequences result in an ROC of the form \( |z| > r \).
Since \( u(-n) \) is left-sided, we can immediately narrow down the options to those with \( |z| < 1 \).

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question asks for the z-transform and the associated Region of Convergence (ROC) of the discrete-time sequence \[ x(n)=u(-n). \] 

Step 2: Key Formula or Approach:
The bilateral z-transform of a discrete-time signal \(x(n)\) is defined as \[ X(z)=\sum_{n=-\infty}^{\infty}x(n)z^{-n}. \] 
Step 3: Detailed Explanation: 

• The given signal is \[ x(n)=u(-n), \] which is a left-sided unit step sequence defined by \[ u(-n)= \begin{cases} 1, & n\le 0,\\ 0, & n>0. \end{cases} \] 
• Substitute \(x(n)\) into the z-transform definition: \[ X(z) = \sum_{n=-\infty}^{0}z^{-n}. \] 
• Let \[ m=-n. \] Then, - when \(n=-\infty\), \(m=\infty\), - when \(n=0\), \(m=0\). Therefore, \[ X(z) = \sum_{m=0}^{\infty}z^{m}. \] 
• This is an infinite geometric series with common ratio \[ r=z. \] An infinite geometric series converges only if \[ |z|<1. \] Hence, the Region of Convergence (ROC) is \[ |z|<1. \] 
• Using the sum of an infinite geometric series, \[ \sum_{m=0}^{\infty}r^m=\frac{1}{1-r}, \qquad |r|<1, \] we obtain \[ X(z) = \frac{1}{1-z}. \] 
• Therefore, the z-transform and its Region of Convergence are \[ \boxed{ X(z)=\frac{1}{1-z}, \qquad \text{ROC: }|z|<1. } \]