Z-Transform Calculation for a Left-Sided Signal

The z-transform is a mathematical tool used in digital signal processing to convert discrete-time signals from the time domain to the complex frequency domain (z-domain). It is particularly useful for analyzing and designing discrete-time systems.

For a discrete-time signal \(x(n)\), the unilateral z-transform is defined as:

And the bilateral z-transform is defined as:

In this problem, we are given the signal \(x(n)=\left(\frac{1}{2}\right)^n\ u(-n)\). Since the signal involves \(u(-n)\), which is non-zero for negative values of \(n\), we must use the bilateral z-transform definition.

Signal Analysis: Understanding \(x(n) = \left(\frac{1}{2}\right)^n u(-n)\)

Let's analyze the given discrete-time signal \(x(n)=\left(\frac{1}{2}\right)^n\ u(-n)\):

• The term \(\left(\frac{1}{2}\right)^n\) is an exponential sequence.
• The term \(u(-n)\) is a unit step function that is defined as: \[ u(-n) = \begin{cases} 1 & \text{for } n \le 0 \\ 0 & \text{for } n > 0 \end{cases} \]
• Because of \(u(-n)\), the signal \(x(n)\) is only non-zero for \(n \le 0\). This means \(x(n)\) is a left-sided signal.

Applying the Z-Transform Definition for \(x(n)\)

Now, we substitute \(x(n)\) into the bilateral z-transform formula:

Since \(x(n) = \left(\frac{1}{2}\right)^n u(-n)\) is non-zero only for \(n \le 0\), the summation limits change from \(-\infty\) to \(0\):

We can combine the terms inside the summation:

Transform Substitution and Simplification

To convert this summation into a standard geometric series form (which typically sums from \(0\) to \(\infty\)), we can perform a change of variable. Let \(m = -n\). When \(n=-\infty\), \(m=\infty\). When \(n=0\), \(m=0\). So the summation becomes:

Using the property \((a^b)^{-c} = a^{-bc}\) and \( (A^{-1})^{-1} = A \), we can simplify the term inside the parenthesis:

So the z-transform expression becomes:

Geometric Series Application for Z-Transform

This is now in the form of a standard infinite geometric series \(\sum_{k=0}^{\infty} r^k\), which converges to \(\frac{1}{1-r}\) provided that \(|r| < 1\).

In our case, \(r = 2Z\). Therefore, applying the geometric series formula:

ROC Determination for Z-Transform

For the geometric series to converge, we must have \(|r| < 1\).

So, \(|2Z| < 1\).

This implies \(|Z| < \frac{1}{2}\).

This Region of Convergence (ROC) is the interior of a circle centered at the origin with radius \(\frac{1}{2}\). This is consistent with a left-sided signal, whose ROC always extends inward from the innermost pole to the origin, or includes the origin if the signal is stable and finite in duration.

Final Z-Transform Result

The z-transform of \(x(n)=\left(\frac{1}{2}\right)^n\ u(-n)\) is \(X(Z) = \frac{1}{1 - 2Z}\) with a Region of Convergence (ROC) of \(|Z| < \frac{1}{2}\).