Signals and Systems

10.1 The Z-Transform

Imron Rosyadi

10.1 What is the Z-Transform?

Definition

The z-transform of a discrete-time signal \(x[n]\) is defined as:

\[ X(z) \triangleq \sum_{n=-\infty}^{+\infty} x[n] z^{-n} \]

Where \(z\) is a complex variable.

We denote this relationship as: \[ x[n] \stackrel{\mathcal{Z}}{\longleftrightarrow} X(z) \]

• It’s a generalization of the Discrete-Time Fourier Transform (DTFT).
• It converts a discrete-time sequence \(x[n]\) into a function \(X(z)\) in the complex \(z\)-plane.

10.1 What is the Z-Transform?

Connection to DTFT

Express \(z\) in polar form: \(z = r e^{j \omega}\)

Substituting into the definition: \[ X\left(r e^{j \omega}\right)=\sum_{n=-\infty}^{+\infty}\left\{x[n] r^{-n}\right\} e^{-j \omega n} \]

This implies: \[ X\left(r e^{j \omega}\right)=\mathfrak{F}\left\{x[n] r^{-n}\right\} \]

Crucially, when \(r=1\) (i.e., \(|z|=1\)): \[ \left.X(z)\right|_{z=e^{j \omega}}=X\left(e^{j \omega}\right)=\mathfrak{F}\{x[n]\} \]

The Z-transform is a cornerstone for analyzing discrete-time systems. It transforms a sequence from the time domain to the complex frequency domain. Notice the similarity to the Laplace transform for continuous-time signals, but with \(z^{-n}\) instead of \(e^{-st}\).

The key takeaway here is its direct relationship with the DTFT. By setting the magnitude of \(z\) to 1, i.e., \(z = e^{j\omega}\), the Z-transform directly gives us the DTFT. This means the DTFT is a special case of the Z-transform, evaluated on the unit circle in the complex z-plane.

The Z-Plane and Unit Circle

Complex \(z\)-plane

• The variable \(z\) is complex, \(z = r e^{j \omega}\).
• The real part of \(z\) is \(r \cos(\omega)\).
• The imaginary part of \(z\) is \(r \sin(\omega)\).

The \(z\)-plane has a real axis and an imaginary axis.

Important

Unit Circle:

The Z-transform reduces to the Fourier transform when the magnitude of \(z\) is unity (\(|z|=1\)). This corresponds to a circle with radius 1 in the \(z\)-plane, known as the unit circle.

Note

Analogy to Laplace Transform:

For continuous-time signals, the Laplace transform reduces to the Fourier transform on the imaginary axis (where \(\text{Re}(s)=0\)). For discrete-time, the Z-transform reduces to the Fourier transform on the unit circle (where \(|z|=1\)).

Figure 10.1 Complex z-plane.

Visualizing the Z-plane is critical. Imagine a 2D plane where the horizontal axis is the real part of \(z\) and the vertical axis is the imaginary part. The unit circle is a very important boundary on this plane.

When we talk about the Fourier transform of a discrete-time signal, we are essentially looking at the Z-transform’s behavior only on this unit circle. This is a key difference from the Laplace transform, where the Fourier transform is found on the imaginary axis.

Region of Convergence (ROC)

Convergence Condition

For the Z-transform \(X(z) = \sum_{n=-\infty}^{+\infty} x[n] z^{-n}\) to converge, the sum must be finite.

This means that the Fourier transform of the sequence \(x[n] r^{-n}\) must converge.

\[ \sum_{n=-\infty}^{+\infty}\left|x[n] r^{-n}\right| < \infty \]

• The set of all \(z\) values for which \(X(z)\) converges is called the Region of Convergence (ROC).

Importance of ROC

• Uniqueness: The Z-transform and its ROC uniquely define a signal \(x[n]\). Different signals can have the same algebraic expression for \(X(z)\) but different ROCs.
• Fourier Transform Existence: If the ROC includes the unit circle, then the Discrete-Time Fourier Transform (DTFT) of \(x[n]\) converges.
• System Properties: The ROC provides insights into system causality and stability.

Warning

Always specify the ROC when defining a Z-transform!

Just like with the Laplace transform, the ROC is not just an arbitrary mathematical detail; it carries crucial physical information about the signal or system. Without the ROC, the Z-transform is ambiguous.

For example, a right-sided signal will have an ROC that extends outwards from the outermost pole, while a left-sided signal will have an ROC that extends inwards from the innermost pole. This distinction is vital for inverse Z-transform and understanding system behavior.

Example 10.1: Right-Sided Exponential

Signal: \(x[n] = a^n u[n]\)

Where \(u[n]\) is the unit step function.

Derivation: \[ X(z)=\sum_{n=-\infty}^{+\infty} a^{n} u[n] z^{-n} \] Since \(u[n]\) is 0 for \(n<0\): \[ X(z)=\sum_{n=0}^{\infty} (a z^{-1})^{n} \] This is a geometric series. For convergence, we require \(|a z^{-1}| < 1\), which means \(|z| > |a|\).

Thus, the sum converges to: \[ X(z)=\frac{1}{1-a z^{-1}} = \frac{z}{z-a}, \quad |z|>|a| \]

Example 10.1: Right-Sided Exponential

Pole-Zero Plot and ROC

• Zero: \(z=0\) (from numerator \(z\))
• Pole: \(z=a\) (from denominator \(z-a\))

The ROC is the region outside the circle of radius \(|a|\).

Figure 10.2 Pole-zero plot and region of convergence for Example 10.1 for \(0<a<1\).

Here we have the most common Z-transform pair. The signal \(a^n u[n]\) is a right-sided exponential, meaning it starts at \(n=0\) and extends to positive infinity.

The derivation relies on the geometric series formula. The condition for this series to converge directly gives us the ROC: \(|z| > |a|\). This means the ROC is an exterior region, extending outwards from the pole at \(z=a\). If \(|a|<1\), the unit circle is included in the ROC, implying the DTFT exists. If \(|a| \ge 1\), the DTFT does not converge.

Interactive: Visualizing ROC for \(a^n u[n]\)

Let’s explore how the pole location affects the ROC. Adjust the value of ‘a’ below.

viewof a_val = Inputs.range([0.1, 1.5], {value: 0.5, step: 0.1, label: "Magnitude of 'a'"});

This interactive plot helps visualize the ROC. As you change ‘a’, observe how the pole moves and the ROC boundary changes. The green circle represents the boundary of the ROC. For a right-sided signal like \(a^n u[n]\), the ROC is always the region outside this boundary.

Notice the condition for the DTFT to exist: the unit circle (dashed gray) must be within the green ROC region. This happens when the magnitude of ‘a’ is less than 1.

Example 10.2: Left-Sided Exponential

Signal: \(x[n] = -a^n u[-n-1]\)

This is a left-sided exponential, extending to negative infinity.

Derivation: \[ X(z)=-\sum_{n=-\infty}^{+\infty} a^{n} u[-n-1] z^{-n} \] Since \(u[-n-1]\) is 1 for \(n \le -1\): \[ X(z)=-\sum_{n=-\infty}^{-1} a^{n} z^{-n} \] Let \(m = -n\). As \(n\) goes from \(-\infty\) to \(-1\), \(m\) goes from \(\infty\) to \(1\). \[ X(z)=-\sum_{m=1}^{\infty} a^{-m} z^{m} = -\sum_{m=1}^{\infty} (a^{-1} z)^{m} \] This is a geometric series. For convergence, we require \(|a^{-1} z| < 1\), which means \(|z| < |a|\).

Example 10.2: Left-Sided Exponential

Thus, the sum converges to: \[ X(z)=-( \frac{a^{-1}z}{1-a^{-1}z} ) = \frac{-z}{a-z} = \frac{z}{z-a}, \quad |z|<|a| \]

Pole-Zero Plot and ROC

• Zero: \(z=0\)
• Pole: \(z=a\)

The ROC is the region inside the circle of radius \(|a|\).

Figure 10.3 Pole-zero plot and region of convergence for Example 10.2 for \(0<a<1\).

Now consider a left-sided exponential. The algebraic form of the Z-transform is identical to the right-sided case! This highlights the critical role of the ROC.

The key difference is in the convergence condition for the geometric series, which now leads to \(|z| < |a|\). This means the ROC for a left-sided signal is an interior region, inside the circle defined by the pole at \(z=a\).

ROC Differences: Right-Sided vs. Left-Sided

Right-Sided Signal: \(x[n] = a^n u[n]\)

• \(X(z) = \frac{z}{z-a}\)
• ROC: \(|z| > |a|\) (exterior region)
• Pole at \(z=a\).
• ROC is outside the outermost pole.

Left-Sided Signal: \(x[n] = -a^n u[-n-1]\)

• \(X(z) = \frac{z}{z-a}\)
• ROC: \(|z| < |a|\) (interior region)
• Pole at \(z=a\).
• ROC is inside the innermost pole.

Caution

The algebraic expression for \(X(z)\) alone is insufficient to uniquely determine the discrete-time signal \(x[n]\). The Region of Convergence (ROC) is essential!

This slide summarizes the crucial distinction. Even though \(a^n u[n]\) and \(-a^n u[-n-1]\) have the same algebraic Z-transform, their ROCs are entirely different. This is a common point of confusion for students, so emphasize it.

The ROC directly tells us about the causality of the signal (right-sided implies ROC extending outwards, left-sided implies ROC extending inwards). This also relates to system stability and causality for LTI systems.

Linearity Property

Statement:

If \(x_1[n] \stackrel{\mathcal{Z}}{\longleftrightarrow} X_1(z)\) with ROC \(R_1\), and \(x_2[n] \stackrel{\mathcal{Z}}{\longleftrightarrow} X_2(z)\) with ROC \(R_2\),

Then for constants \(A\) and \(B\): \[ A x_1[n] + B x_2[n] \stackrel{\mathcal{Z}}{\longleftrightarrow} A X_1(z) + B X_2(z) \]

ROC of the combined signal: The ROC for \(A X_1(z) + B X_2(z)\) will be at least the intersection of \(R_1\) and \(R_2\).

\[ R_{new} \supseteq R_1 \cap R_2 \]

Why “at least”?

Sometimes, pole-zero cancellations can occur when summing transforms, potentially expanding the ROC.

For example, if a pole from \(X_1(z)\) is cancelled by a zero from \(X_2(z)\), the overall system might converge in a larger region than the intersection.

Tip

Linearity is extremely useful for finding the Z-transform of complex signals by breaking them down into simpler components.

Linearity is one of the most powerful properties of transforms. It allows us to decompose complex problems into simpler, manageable parts. We can find the Z-transform of individual components and then sum them up.

The “at least” part of the ROC intersection is important. While the common region of convergence is guaranteed, sometimes the combined transform might have a larger ROC if poles and zeros cancel out. This is analogous to the behavior in Laplace transforms.

Example 10.3: Sum of Exponentials

Signal: \[ x[n]=7\left(\frac{1}{3}\right)^{n} u[n]-6\left(\frac{1}{2}\right)^{n} u[n] \]

Using linearity and results from Example 10.1:

1. For \(7\left(\frac{1}{3}\right)^{n} u[n]\): \(X_1(z) = \frac{7}{1-\frac{1}{3} z^{-1}}\), with ROC \(R_1: |z| > \frac{1}{3}\).

2. For \(-6\left(\frac{1}{2}\right)^{n} u[n]\): \(X_2(z) = \frac{-6}{1-\frac{1}{2} z^{-1}}\), with ROC \(R_2: |z| > \frac{1}{2}\).

Example 10.3: Sum of Exponentials

Combined Z-Transform: \[ X(z) = X_1(z) + X_2(z) = \frac{7}{1-\frac{1}{3} z^{-1}}-\frac{6}{1-\frac{1}{2} z^{-1}} \] \[ X(z) = \frac{7(1-\frac{1}{2} z^{-1}) - 6(1-\frac{1}{3} z^{-1})}{(1-\frac{1}{3} z^{-1})(1-\frac{1}{2} z^{-1})} \] \[ X(z) = \frac{7 - \frac{7}{2} z^{-1} - 6 + \frac{6}{3} z^{-1}}{(1-\frac{1}{3} z^{-1})(1-\frac{1}{2} z^{-1})} \] \[ X(z) = \frac{1 - (\frac{7}{2} - 2) z^{-1}}{(1-\frac{1}{3} z^{-1})(1-\frac{1}{2} z^{-1})} = \frac{1 - \frac{3}{2} z^{-1}}{(1-\frac{1}{3} z^{-1})(1-\frac{1}{2} z^{-1})} \] Expressing in powers of \(z\): \[ X(z) = \frac{z(z-\frac{3}{2})}{(z-\frac{1}{3})(z-\frac{1}{2})} \]

Example 10.3: Sum of Exponentials

ROC and Pole-Zero Plot

• Poles: \(z = \frac{1}{3}\) and \(z = \frac{1}{2}\).
• Zeros: \(z = 0\) and \(z = \frac{3}{2}\).

The ROC for \(X(z)\) is the intersection of \(R_1\) and \(R_2\).

\(R_1: |z| > \frac{1}{3}\)

\(R_2: |z| > \frac{1}{2}\)

Therefore, the combined ROC is \(R: |z| > \frac{1}{2}\).

(a) \(1 /\left(1-\frac{1}{3} z^{-1}\right),|z|>\frac{1}{3}\);

(b) \(1 /\left(1-\frac{1}{2} z^{-1}\right),|z|>\frac{1}{2}\);

(c) \(7 /\left(1-\frac{1}{3} z^{-1}\right)-6 /\left(1-\frac{1}{2} z^{-1}\right),|z|>\frac{1}{2}\).

Figure 10.4 Pole-zero plot and region of convergence for the individual terms and the sum in Example 10.3

This example demonstrates the power of linearity. We can combine two individual Z-transforms to find the transform of their sum.

Since both original signals are right-sided, their ROCs are exterior regions. The intersection of these two exterior regions is simply the larger of the two. Hence, the combined ROC is \(|z| > 1/2\). The pole-zero plot shows the locations of all poles and zeros, and the ROC is the region outside the outermost pole.

Example 10.4: Sinusoidal Signal

Signal: \[ x[n] = \left(\frac{1}{3}\right)^{n} \sin \left(\frac{\pi}{4} n\right) u[n] \]

Recall Euler’s formula: \(\sin(\theta) = \frac{e^{j\theta} - e^{-j\theta}}{2j}\). \[ x[n] = \frac{1}{2j}\left(\frac{1}{3}\right)^{n} (e^{j \frac{\pi}{4} n} - e^{-j \frac{\pi}{4} n}) u[n] \] \[ x[n] = \frac{1}{2j}\left(\frac{1}{3} e^{j \frac{\pi}{4}}\right)^{n} u[n] - \frac{1}{2j}\left(\frac{1}{3} e^{-j \frac{\pi}{4}}\right)^{n} u[n] \]

Using linearity and Example 10.1: \(X(z) = \frac{1}{2j} \frac{1}{1-\frac{1}{3} e^{j \pi / 4} z^{-1}} - \frac{1}{2j} \frac{1}{1-\frac{1}{3} e^{-j \pi / 4} z^{-1}}\)

Example 10.4: Sinusoidal Signal

After algebraic manipulation, this simplifies to: \[ X(z)=\frac{\frac{1}{3 \sqrt{2}} z}{\left(z-\frac{1}{3} e^{j \pi / 4}\right)\left(z-\frac{1}{3} e^{-j \pi / 4}\right)} \]

ROC and Pole-Zero Plot

• Poles: At \(z = \frac{1}{3} e^{j \pi / 4}\) and \(z = \frac{1}{3} e^{-j \pi / 4}\). These are complex conjugate poles.
  • \(\frac{1}{3} (\cos(\pi/4) + j \sin(\pi/4)) = \frac{1}{3} (\frac{\sqrt{2}}{2} + j \frac{\sqrt{2}}{2})\)
  • \(\frac{1}{3} (\cos(-\pi/4) + j \sin(-\pi/4)) = \frac{1}{3} (\frac{\sqrt{2}}{2} - j \frac{\sqrt{2}}{2})\)
• Zeros: \(z=0\) (from numerator \(z\)).

The ROC for each exponential term is \(|z| > |\frac{1}{3} e^{j \pi / 4}| = \frac{1}{3}\). Thus, the combined ROC is \(|z| > \frac{1}{3}\).

Figure 10.5 Pole-zero plot and ROC for the \(z\)-transform in Example 10.4.

This example shows how real-valued sinusoidal signals lead to complex conjugate poles in the Z-plane. This is a common pattern: for any real-valued signal, its complex poles and zeros must appear in conjugate pairs.

The ROC is determined similarly to the previous examples. Since it’s a right-sided signal (\(u[n]\)), the ROC is an exterior region, outside the circle defined by the magnitude of the poles. In this case, both poles have a magnitude of \(1/3\), so the ROC is \(|z| > 1/3\).

Key Takeaways and Applications

Key Takeaways

• Definition: The Z-transform is \(\sum x[n] z^{-n}\).
• DTFT Connection: The DTFT is the Z-transform evaluated on the unit circle (\(|z|=1\)).
• ROC is Crucial: The Region of Convergence (ROC) is essential for uniquely defining the signal and understanding its properties.
  • Right-sided signals: ROC is an exterior region (\(|z| > R\)).
  • Left-sided signals: ROC is an interior region (\(|z| < R\)).
  • Two-sided signals: ROC is an annular region (\(R_1 < |z| < R_2\)).
• Pole-Zero Plots: Visual representation of \(X(z)\)’s singularities. Poles and zeros provide insights into system behavior.
• Linearity: Simplifies analysis of combined signals.

Key Takeaways and Applications

Engineering Applications

• Digital Filter Design:
  • Analyzing frequency response.
  • Designing IIR (Infinite Impulse Response) and FIR (Finite Impulse Response) filters.
• Discrete-Time System Analysis:
  • Determining system stability (poles within unit circle).
  • Assessing causality.
  • Finding the transfer function \(H(z)\) of LTI systems.
• Digital Control Systems:
  • Designing controllers for discrete-time systems.
  • Analyzing system response in the discrete-time domain.
• Signal Processing:
  • Spectrum analysis.
  • Deconvolution.

Important

The Z-transform is fundamental for understanding and designing virtually all discrete-time systems in ECE!

To wrap up, the Z-transform is a powerful mathematical tool that bridges the time domain and the complex frequency domain for discrete-time signals and systems. It’s not just a theoretical concept; it has direct and profound implications in many areas of ECE.

From designing the digital filters in your phone to analyzing the stability of a control system, the Z-transform provides the mathematical framework. Mastering this concept is crucial for your journey in ECE.