Z-Transform

Z-Transform#

The Z-transform is an essential tool for analyzing discrete-time signals and systems. Similar to the Laplace transform for continuous systems, the Z-transform converts a time-domain sequence into a complex frequency-domain representation. This transformation simplifies the analysis and design of digital filters, control systems, and other discrete-time processes.

Definition 14

Z-Transform: The Z-transform of a discrete-time sequence \(x[n]\) is defined as:

\[ X(z) = \mathcal{Z}\{x[n]\} = \sum_{n=-\infty}^{\infty} x[n] z^{-n}, \quad z \in \mathbb{C} \]

where:

  • \(x[n]\) is a discrete-time sequence,

  • \(n\) is an integer representing discrete time,

  • \(z\) is a complex variable, often expressed in polar form as \(z = re^{j\omega}\),

  • \(\mathcal{Z}\{\cdot\}\) denotes the Z-transform operator.

The Z-transform is widely used to solve linear difference equations, analyze stability, and perform frequency analysis in digital signal processing.

Properties of the Z-Transform#

Similar to the Laplace transform, the Z-transform has several properties that are valuable when analyzing discrete-time systems. These properties provide insights into the time and frequency behavior of the sequences.

Theorem 7

Linearity of the Z-Transform: If \(a\) and \(b\) are constants, and \(x[n]\) and \(y[n]\) are discrete-time sequences, then the Z-transform of a linear combination is given by:

\[ \mathcal{Z}\{a x[n] + b y[n]\} = a X(z) + b Y(z) \]

This property shows that the Z-transform of a sum of scaled sequences is equal to the sum of their individual transforms, each scaled accordingly.

→ The Z-transform converts linear difference equations into algebraic equations, simplifying the process of solving them in the frequency domain.

Theorem 8

Time Shifting Property: If \(\mathcal{Z}\{x[n]\} = X(z)\), then the Z-transform of a shifted sequence \(x[n-k]\) is given by:

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

This property is useful for analyzing delayed sequences and determining how a time shift in the sequence affects its frequency-domain representation.

→ Time shifting is critical in applications like digital filters, where delays are a fundamental operation.

Theorem 9

Convolution Property: If \(x[n]\) and \(h[n]\) are two discrete-time sequences with Z-transforms \(X(z)\) and \(H(z)\), then the Z-transform of their convolution \((x * h)[n]\) is given by:

\[ \mathcal{Z}\{(x * h)[n]\} = X(z) H(z) \]

This property states that the convolution of two sequences in the time domain corresponds to multiplication in the frequency domain.

This property is particularly useful in digital signal processing, as it simplifies filtering operations by converting them into simple multiplications.

→ Convolution in the time domain is equivalent to multiplication in the frequency domain, making the Z-transform a powerful tool for analyzing linear systems.

Region of Convergence (ROC)#

One of the unique features of the Z-transform is the Region of Convergence (ROC). This region determines the range of \(z\) values for which the Z-transform converges. The ROC is crucial in determining the stability and causality of the system.

Theorem 10

Region of Convergence: Given a discrete-time sequence \(x[n]\) and its Z-transform \(X(z)\), the Region of Convergence (ROC) is the set of values of \(z\) for which \(X(z)\) converges:

\[ \text{ROC} = \{ z \in \mathbb{C} \mid \sum_{n=-\infty}^{\infty} x[n] z^{-n} < \infty \} \]

The ROC provides information on the stability and behavior of the sequence.

→ A sequence is stable if the ROC includes the unit circle \(|z| = 1\), and causal if the ROC extends outwards from the outermost pole.

Understanding the ROC is fundamental when using the Z-transform to analyze system stability and causality.

Example: Inverse Z-Transform Using Partial Fractions#

Let’s consider an example where we find the inverse Z-transform using partial fraction decomposition. Suppose we have:

\[ H(z) = \frac{3z^2 + 2z + 1}{(z-1)(z-2)} \]

Step-by-Step Solution:#

  1. Decompose into Partial Fractions:

We express \(H(z)\) as a sum of simpler fractions:

\[ H(z) = \frac{A}{z-1} + \frac{B}{z-2} \]

  1. Solve for Constants:

Multiply through by the common denominator \((z-1)(z-2)\) and equate coefficients:

\[ 3z^2 + 2z + 1 = A(z-2) + B(z-1) \]

By expanding and matching coefficients, we find:

  • \(A = 5\)

  • \(B = -2\)

  1. Rewrite the Fraction:

Now we can write:

\[ H(z) = \frac{5}{z-1} - \frac{2}{z-2} \]

  1. Inverse Z-Transform:

Using standard Z-transform pairs:

  • \(\mathcal{Z}^{-1}\left\{\frac{5}{z-1}\right\} = 5 (1)^n u[n]\)

  • \(\mathcal{Z}^{-1}\left\{\frac{-2}{z-2}\right\} = -2 (2)^n u[n]\)

  1. Final Solution:

The time-domain sequence \(h[n]\) is:

\[ h[n] = 5 (1)^n u[n] - 2 (2)^n u[n] \]

This example demonstrates how partial fraction decomposition can simplify the inverse Z-transform process, allowing us to retrieve the original discrete-time sequence.

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