Poles, ROC, Stability, and Causality

Hello! Welcome to your third lesson in the "System Analysis in the s-Domain" module.

In our last lesson, we formally derived the bilateral Laplace transform from the Fourier transform, establishing that it is defined by both an algebraic expression and a Region of Convergence (ROC). We saw that the Fourier transform exists only if the ROC includes the $j ω$ -axis.

Today, we get to the practical payoff of all that theory. We will explore how the geometry of the s-plane—specifically the locations of a system's poles relative to its ROC—provides a powerful visual method for determining two of the most fundamental properties of any LTI system: causality and stability. By the end of this lesson, you will be able to look at a system's pole-zero plot and ROC and immediately deduce its core behaviors.

Poles, Zeros, and the Structure of the s-Plane

As we saw, many signals of interest give rise to Laplace transforms that are rational functions, meaning they are a ratio of two polynomials in $s$ :

H (s) = \frac{N ( s )}{D ( s )}

The roots of the numerator polynomial $N (s)$ are called the zeros of the system. At these values of $s$ , $H (s) = 0$ .
The roots of the denominator polynomial $D (s)$ are called the poles of the system. At these values of $s$ , $H (s) \to \infty$ .

Poles and zeros are the critical landmarks of the s-plane. In the last lesson's readings, we encountered several properties of the ROC. These properties form the rules of our analysis, so let's consolidate the most important ones.

THE LAPLACE TRANSFORM

To begin, please review the key properties of the Region of Convergence. This section from the text 'THE LAPLACE TRANSFORM' provides a formal and complete summary of these rules, which are the foundation for everything we will do today.

Please read Section 9.2, 'THE REGION OF CONVERGENCE FOR LAPLACE TRANSFORMS' (pages 662-670). Focus on Properties 1, 2, 4, 5, 6, and 8. These rules explain how the ROC is constrained by the signal's type (right-sided, left-sided, etc.) and by the locations of the poles.

To summarize the most critical rules for our purposes:

The ROC consists of vertical strips in the s-plane and cannot contain any poles.
If a signal $h (t)$ is right-sided (zero before some time $T_{1}$ ), its ROC is a right-half plane. If $H (s)$ is rational, the ROC is to the right of the rightmost pole.
If a signal $h (t)$ is left-sided (zero after some time $T_{2}$ ), its ROC is a left-half plane. If $H (s)$ is rational, the ROC is to the left of the leftmost pole.
If a signal $h (t)$ is two-sided, its ROC is a vertical strip between two poles (if it exists at all).

With these rules, we are ready to analyze system properties.

Causality and the ROC

A system is causal if its output depends only on past and present inputs. For an LTI system, this is equivalent to its impulse response $h (t)$ being zero for all negative time:

h (t) = 0 for t < 0

By definition, a causal impulse response is a right-sided signal. Applying the rules we just reviewed, we arrive at our first major conclusion:

For a causal LTI system, the ROC of its system function $H (s)$ must be a right-half plane.

Furthermore, if $H (s)$ is rational, we know the ROC is bounded by poles. This leads to a more specific rule:

A rational LTI system is causal if and only if its ROC is the half-plane to the right of the rightmost pole.

The textbook section below formalizes this exact point.

THE LAPLACE TRANSFORM

This section of the textbook, 'Causality,' directly connects the time-domain definition of causality with the s-plane properties we've established. Pay attention to Example 9.19, which is a great case study showing why the 'rational' condition is important.

Please read Section 9.7.1, 'Causality' (pages 693-695). The core argument is presented in the first few paragraphs. The examples illustrate how to apply the rule.

Test your understanding!

An LTI system has a rational transfer function $H (s)$ with poles at $s = - 3$ and $s = 1$ . The system is known to be causal. What is the ROC of $H (s)$ ?

Show answer

Since the system is causal, its ROC must be a right-half plane to the right of the rightmost pole. The poles are at $σ = - 3$ and $σ = 1$ . The rightmost pole is at $s = 1$ . Therefore, the ROC must be $Re {s} > 1$ .

Stability and the ROC

The second key property is Bounded-Input, Bounded-Output (BIBO) stability. A system is BIBO stable if every bounded input produces a bounded output. For LTI systems, this is equivalent to the impulse response being absolutely integrable:

\int_{- \infty}^{\infty} ∣ h (t) ∣ d t < \infty

As we established in the last lesson, this is precisely the condition for the Fourier transform $H (j ω)$ to exist. And the condition for the Fourier transform to exist is that the ROC of the Laplace transform must include the $j ω$ -axis ( $σ = 0$ ). This gives us our second major conclusion:

An LTI system is stable if and only if the ROC of its system function $H (s)$ includes the imaginary axis ( $j ω$ -axis).

The textbook section below confirms this and provides an excellent example of how knowing about stability allows you to identify the correct ROC from multiple possibilities.

THE LAPLACE TRANSFORM

The section 'Stability' formally states this crucial condition and then uses an example with three possible ROCs to demonstrate how the stability requirement uniquely determines the system's impulse response.

Please read Section 9.7.2, 'Stability' (pages 695-697). Focus on the main statement and Example 9.20 with Figure 9.25, which clearly shows the three possible ROCs for a given set of poles.

The Causal and Stable System: The Left-Half Plane

Most real-world systems we design are both causal and stable. Let's combine our two rules for a rational system function $H (s)$ :

Causality implies: The ROC is the right-half plane to the right of the rightmost pole.
Stability implies: The ROC must include the $j ω$ -axis.

For both conditions to hold simultaneously, the right-half plane ROC must contain the $j ω$ -axis. This can only happen if the boundary of that ROC—the rightmost pole—is to the left of the $j ω$ -axis. If the rightmost pole is in the left-half plane, then all poles must be in the left-half plane. This leads to the single most important criterion in analog system analysis:

A causal LTI system with a rational system function $H (s)$ is stable if and only if all of its poles lie in the left-half of the s-plane.

The following video provides an excellent, concise summary of this argument.

Why are Poles Negative in Laplace Transform of Real Stable LTI Systems?

This video from Iain Explains brings together the concepts of causality and stability to derive the 'left-half plane' rule. It's a perfect synthesis of what we've just discussed.

Please watch the video from 00:15 to 04:50. The presenter clearly walks through the three properties (causal -> right-handed ROC, stable -> ROC includes jω-axis, ROC bounded by poles) and combines them to show why poles must be in the left-half plane for a system to be both causal and stable.

A visual summary of the conditions for a causal and stable system. The ROC is a right-half plane that includes the jω-axis, which forces all poles (marked by 'x') to lie in the left-half plane.

Analyzing System Scenarios

Let's now put this into practice. Given a system function's algebraic expression, we can determine all possible impulse responses and their properties by considering all valid ROCs.

The following video from Neso Academy walks through this analytical process. It also introduces the concepts of anti-causal and non-causal systems and their corresponding stability criteria, which completes the picture.

ROC and Stability of Causal LTI Systems

This video, 'ROC and Stability of Causal LTI Systems,' provides clear definitions and graphical examples for analyzing different system types. It reinforces the rules for causal systems and extends them to anti-causal and non-causal scenarios.

Please watch from the beginning to 05:35. Pay attention to how the type of system (causal, anti-causal) dictates the ROC, and how stability is always determined by whether that ROC includes the jω-axis.

This analysis gives us a powerful framework. Let's consider a system with the transfer function:

H (s) = \frac{s - 1}{( s + 1 ) ( s - 2 )}

This system has poles at $s = - 1$ and $s = 2$ , and a zero at $s = 1$ . Based on the rule that the ROC cannot contain poles, there are three possible ROCs for this system:

ROC 1: $Re {s} > 2$
- This is a right-half plane to the right of the rightmost pole. The system is causal.
- This ROC does not include the $j ω$ -axis. The system is unstable.
ROC 2: $- 1 < Re {s} < 2$
- This is a vertical strip. The corresponding impulse response $h (t)$ would be two-sided (non-zero for $t > 0$ and $t < 0$ ). The system is non-causal.
- This ROC does include the $j ω$ -axis. The system is stable.
ROC 3: $Re {s} < - 1$
- This is a left-half plane to the left of the leftmost pole. The system is anti-causal ( $h (t) = 0$ for $t > 0$ ).
- This ROC does not include the $j ω$ -axis. The system is unstable.

This example, which is the same as Example 9.20 from your reading, perfectly illustrates how the ROC is the key to interpreting a system's properties.

Test your understanding!

An LTI system has a transfer function $H (s) = \frac{1}{s ^{2} + 3 s + 2}$ .

Find the poles of the system.
If the system is known to be causal, is it stable? Justify your answer.
If the system is known to be stable, what is its ROC, and is it causal?

Show answer

The denominator is $s^{2} + 3 s + 2 = (s + 1) (s + 2)$ . The poles are at $s = - 1$ and $s = - 2$ .
Yes, the system is stable. Since the system is causal, its ROC must be to the right of the rightmost pole, which is $Re {s} > - 1$ . This region includes the $j ω$ -axis, so the system is stable. Alternatively, since the system is causal and both its poles ( $- 1, - 2$ ) are in the left-half plane, it must be stable.
To be stable, the ROC must contain the $j ω$ -axis. The only valid ROC for poles at -1 and -2 that contains the $j ω$ -axis is the vertical strip $- 2 < Re {s} < - 1$ . Since this is not a right-half plane to the right of the rightmost pole, the system corresponding to this ROC is not causal.

Conclusion

In this lesson, we established the profound connection between the s-plane representation of a system and its fundamental time-domain properties. By understanding the rules governing the Region of Convergence, we can now determine a system's causality and stability simply by inspecting the location of its poles.

Key Takeaways:

Causality Rule: An LTI system is causal if and only if the ROC of its rational system function $H (s)$ is the right-half plane to the right of the rightmost pole.
Stability Rule: An LTI system is stable if and only if the ROC of $H (s)$ includes the imaginary ( $j ω$ ) axis.
Causal & Stable Rule: A causal LTI system with a rational system function is stable if and only if all of its poles are in the left-half of the s-plane.
For any given rational $H (s)$ , different ROCs correspond to different impulse responses with distinct causality and stability properties.

Preview of the Next Lesson:
So far, we have used the bilateral Laplace transform, which is ideal for system analysis. However, when it comes to solving practical problems like finding the response of a circuit to an input when there are initial conditions (e.g., a charged capacitor), a slightly different tool is more convenient. In our next lesson, we will introduce the unilateral Laplace transform and see how it is perfectly suited for solving linear time-invariant (LTI) differential equations with initial conditions.

Recommended resources

Poles, Zeros, and the Structure of the s-Plane

Causality and the ROC

Stability and the ROC

The Causal and Stable System: The Left-Half Plane

Analyzing System Scenarios

Conclusion