Consider a signal , $x\left[n\right]={\left(\frac{1}{2}\right)}^{n}1\left[n\right]$where 1[n] = 0 if n < 0, and 1[n] = 1 if n ≥ 0. The z-transform of x[n - k], k > 0 is $\frac{z^{-k}}{1-\frac{1}{2}{z}^{-1}}$with region of convergence being

The correct answer is |z| > 1/2.

Step-by-Step Explanation:

1. Understanding the Base Signal:
We are given a discrete-time signal defined as:

$x[n]={\left(\frac{1}{2}\right)}^{n}u[n]$

where $1[n]$ (or $u[n]$) is the unit step signal, which equals 1 for $n\ge 0$ and 0 for $n<0$.

2. z-Transform of the Base Signal:
The standard z-transform of a causal exponential sequence $a^{n}u[n]$ is given by:

$\mathcal{Z}\{a^{n}u[n]\}=\frac{1}{1-a{z}^{-1}}$

For this infinite geometric series to converge, the common ratio magnitude must be less than 1, which yields the Region of Convergence (ROC):

$\left|a{z}^{-1}\right|<1\Rightarrow \left|z\right|>\left|a\right|$

Substituting $a=\frac{1}{2}$, the z-transform of $x[n]$ is:

$X(z)=\frac{1}{1-\frac{1}{2}{z}^{-1}}$

with the Region of Convergence:

$\left|z\right|>\frac{1}{2}$

3. Time-Shifting Property of z-Transform:
According to the time-shifting property of the bilateral z-transform, shifting a signal in time by $k$ samples yields:

$\mathcal{Z}\{x[n-k]\}=z^{-\mathrm{k}}X(z)$

Multiplying by $z^{-k}$ for finite $k>0$ only adds or removes poles/zeros at $z=0$. It does not change the radial bound of convergence established by the pole at $z=\frac{1}{2}$.

Therefore, the region of convergence remains unchanged:

$\left|z\right|>\frac{1}{2}$