91Ó°ÊÓ

Partial fraction expansion is a method that helps break down complex rational expressions into simpler terms. This technique is especially useful in mathematical fields like calculus and signal processing, where it allows us to work with simpler expressions instead of dealing with complex equations all at once. In the given problem, we start with a rational function:

• \(X(z) = \frac{1-\frac{1}{3} z^{-1}}{(1-z^{-1})(1+2z^{-1})}\)

The aim is to express this function as a sum of simpler fractions. We rewrite the rational function:

• \(X(z) = \frac{A}{1-z^{-1}} + \frac{B}{1+2z^{-1}}\)

Here, \(A\) and \(B\) are constants determined by equating coefficients from both sides of the equation. This step simplifies the function and makes it easier to then apply inverse transformations.

Finding the inverse z-transform means converting a function from the frequency domain back to the time domain. The frequency domain analysis provides insight into how different frequency components make up the signal, while the time domain tells us what the signal looks like as a function of time.To solve for the inverse z-transform, we exploit known pairs and theorems from the transform tables. Once we have our partial fraction expansion:

• \(X(z) = \frac{\frac{2}{9}}{1-z^{-1}} + \frac{\frac{7}{9}}{1+2z^{-1}}\)

We use known inverse transform pairs:

• \(\frac{1}{1-az^{-1}} \leftrightarrow a^n u[n]\)

By applying this, we find:

• \(\frac{2}{9} \cdot \frac{1}{1-z^{-1}} \leftrightarrow \frac{2}{9}u[n]\)
• \(\frac{7}{9} \cdot \frac{1}{1-(-2)z^{-1}} \leftrightarrow \frac{7}{9}(-2)^n u[n]\)

Bringing it together, the resulting time-domain expression is:

• \(x[n] = \frac{2}{9}u[n] + \frac{7}{9}(-2)^n u[n]\)

The unit step function, often denoted as \(u[n]\), plays a key role in signal processing by acting as a function that is zero for negative inputs and one for non-negative inputs. It essentially 'turns on' a signal at a specify time, making it very useful in defining signals that begin at a specific point in time.In time-domain analysis, you will often come across functions like \(a^n u[n]\), where the unit step function determines that the signal only exists for \(n \geq 0\). In our inverse z-transform problem:

• \(\frac{2}{9}u[n]\) represents a constant signal starting from \(n=0\).
• \(\frac{7}{9}(-2)^n u[n]\) represents an alternating signal starting from \(n=0\), amplified or attenuated by negative and positive powers of \(-2\).

Frequency domain analysis provides a powerful lens through which signals can be examined based on their frequency content. The z-transform is a crucial tool that allows us to represent discrete-time signals in the frequency domain, giving us insights into their behavior that are not easily visible in the time domain.In this exercise, we start with \(X(z)\) in the frequency domain. Through partial fraction expansion and inverse transform techniques, we analyze and eventually convert it back to the time domain while preserving the characteristics defined in the frequency domain. By seeing how our signal components map to time signals:

• The structure of \(\frac{A}{1-z^{-1}}\) and \(\frac{B}{1+2z^{-1}}\) show the individual frequency components that make up \(X(z)\).

This dual-domain analysis complements our understanding, allowing complex manipulations and insights that can be very powerful, especially in fields like control systems and communications.