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 option is |z| > 1/2.

Step-by-Step Explanation:

1. Understand the given signal:
The given signal is:
$x[n]={\left(\frac{1}{2}\right)}^{n}1[n]$
where 1[n] is the unit step function, which is equal to 1 for n ≥ 0 and 0 for n < 0.

2. Calculate the z-transform of x[n]:
By definition, the z-transform of a discrete-time signal x[n] is defined as:

Substituting the given signal x[n] into the summation:
$X(z)=\sum _{n=0}^{\infty }{\left(\frac{1}{2}\right)}^n{z}^{-n}=\sum _{n=0}^{\infty }{(\frac{1}{2}}^n$

3. Determine the Region of Convergence (ROC) of x[n]:
The expression represents an infinite geometric series of the form:
$\sum _{n=0}^{\infty }{r}^n$
which converges to 1 / (1 - r) if and only if the absolute value of the common ratio, |r|, is strictly less than 1. Here, the common ratio is:
$r=\frac{1}{2}{z}^{-1}$
Therefore, the condition for convergence is:
$\left|\frac{1}{2}{z}^{-1}\right|<1$
This simplifies to:
$\frac{1}{2}\cdot \frac{1}{\left|z\right|}<1\Rightarrow \left|z\right|>\frac{1}{2}$

4. Apply the Time-Shifting Property:
According to the time-shifting property of the z-transform, shifting a signal in time by k units shifts the poles only at the origin or infinity, keeping the boundary of the region of convergence the same. Specifically, if the z-transform of x[n] is X(z) with a region of convergence R, then:
$\mathcal{Z}\{x[n-k]\}=z^{-k}X(z)$
The region of convergence for the shifted signal is the same as the ROC of the original signal, with the possible exception of adding or deleting the point z = 0 or z = infinity. For k > 0, the factor z^-k introduces a pole at z = 0, which is already excluded from our ROC of |z| > 1/2. Thus, the region of convergence remains unchanged:
$\left|z\right|>\frac{1}{2}$