Signals and Systems

9.7 Analysis and Characterization of LTI Systems Using the Laplace Transform

Imron Rosyadi

Analysis and Characterization of LTI Systems Using the Laplace Transform

Welcome to this session on a crucial topic in Signals and Systems: using the Laplace Transform to analyze Linear Time-Invariant, or LTI, systems. We’ll move beyond just calculating transforms to understanding how they help us characterize fundamental system properties like causality and stability, and how they apply to real-world engineering problems.

The System Function: \(H(s)\)

The Laplace Transform is a powerful tool for analyzing LTI systems because it converts convolution in the time domain into multiplication in the s-domain.

\[ \begin{equation*} Y(s)=H(s) X(s) \tag{9.112} \end{equation*} \]

• \(X(s)\): Laplace Transform of the input \(x(t)\)
• \(Y(s)\): Laplace Transform of the output \(y(t)\)
• \(H(s)\): Laplace Transform of the impulse response \(h(t)\)

Note

\(H(s)\) is commonly referred to as the system function or transfer function. If the ROC of \(H(s)\) includes the imaginary axis, then \(H(j\omega)\) is the frequency response.

The equation \(Y(s) = H(s)X(s)\) is fundamental. It’s the Laplace transform counterpart to the convolution integral \(y(t) = h(t) * x(t)\). This transformation from convolution to multiplication dramatically simplifies system analysis. \(H(s)\) essentially describes how the system transforms an input. When we evaluate \(H(s)\) along the imaginary axis, i.e., \(s=j\omega\), we get the system’s frequency response, telling us how different frequency components are affected.

Causality of LTI Systems

A system is causal if its output at any time depends only on present and past inputs. For an LTI system, this means the impulse response \(h(t)=0\) for \(t<0\).

• General Rule: The ROC associated with the system function \(H(s)\) for a causal system is a right-half plane.

Important

For systems with a rational system function \(H(s)\): Causality \(\iff\) the ROC is the right-half plane to the right of the rightmost pole.

Causality is a critical property for real-time systems; you can’t predict the future. For LTI systems, this translates directly to the impulse response being zero before \(t=0\). The first rule is always true: a causal system’s ROC is a right-half plane. However, the second rule, which establishes a strong “if and only if” condition, is specifically for systems where \(H(s)\) is a rational function, meaning it can be expressed as a ratio of polynomials. This distinction is important, as we’ll see in an example.

Example: Causal System

Consider a system with impulse response:

\(h(t) = e^{-t}u(t)\)

The system is causal since \(h(t)=0\) for \(t<0\).

Its system function is:

\[ \begin{equation*} H(s)=\frac{1}{s+1}, \quad \operatorname{Re}\{s\}>-1 \tag{9.114} \end{equation*} \]

• Pole: at \(s=-1\)
• ROC: \(\operatorname{Re}\{s\}>-1\)

This ROC is to the right of the rightmost (and only) pole, confirming causality for this rational \(H(s)\).

Here we have a simple first-order system. The impulse response \(e^{-t}u(t)\) is zero for \(t<0\), so it’s causal. Its Laplace transform \(H(s)\) has a single pole at \(s=-1\). The Region of Convergence is \(\operatorname{Re}\{s\} > -1\), which is indeed a right-half plane extending to the right of the pole. This perfectly aligns with our rule for rational causal systems. The diagram visually represents the s-plane, showing the pole and the shaded ROC.

Example: Non-Causal System

Consider a system with impulse response:

\(h(t) = e^{-|t|}\)

This system is not causal since \(h(t) \neq 0\) for \(t<0\).

Its system function is:

\[ H(s)=\frac{-2}{s^{2}-1}, \quad-1<\operatorname{Re}\{s\}<+1 \]

• Poles: at \(s=-1\) and \(s=1\)
• ROC: \(-1 < \operatorname{Re}\{s\} < 1\)

The ROC is a strip, not a right-half plane to the right of the rightmost pole (\(s=1\)), which is consistent with the system being non-causal.

In this example, the impulse response \(e^{-|t|}\) is non-zero for \(t<0\), making the system non-causal. When we look at its system function, \(H(s)\), we find two poles, at \(s=-1\) and \(s=1\). The ROC is a vertical strip between these two poles. According to our rule, for a causal system with a rational \(H(s)\), the ROC must be to the right of the rightmost pole (which is \(s=1\) here). Since our ROC is not \(\operatorname{Re}\{s\}>1\), it correctly indicates that this system is not causal.

Causality Caveat: Non-Rational \(H(s)\)

The “rational \(H(s)\)” condition for causality is crucial.

Consider a system function:

\[ \begin{equation*} H(s)=\frac{e^{s}}{s+1}, \quad \operatorname{Re}\{s\}>-1 \tag{9.115} \end{equation*} \]

• Pole: at \(s=-1\)
• ROC: \(\operatorname{Re}\{s\}>-1\)

Here, the ROC is to the right of the pole.

However, \(H(s)\) is not rational due to the \(e^s\) term.

The impulse response is \(h(t) = e^{-(t+1)}u(t+1)\).

Since \(u(t+1)\) is non-zero for \(t < 0\) (specifically for \(-1 < t < 0\)), the system is not causal.

Caution

Causality implies the ROC is a right-half plane. The converse is not generally true unless the system function \(H(s)\) is rational.

This example highlights a common misconception. While a right-half plane ROC is a necessary condition for causality, it’s not always sufficient if \(H(s)\) is not a rational function. The \(e^s\) term in \(H(s)\) corresponds to a time shift in the time domain, specifically \(h(t) = e^{-(t+1)}u(t+1)\). Because this impulse response starts before \(t=0\), the system is non-causal, even though its ROC is a right-half plane. This emphasizes the importance of the “rational system function” condition in our second rule for causality.

Stability of LTI Systems

An LTI system is stable if every bounded input produces a bounded output (BIBO stability). This is equivalent to its impulse response being absolutely integrable: \(\int_{-\infty}^{\infty} |h(t)| dt < \infty\).

• General Rule: An LTI system is stable \(\iff\) the ROC of its system function \(H(s)\) includes the entire \(j\omega\)-axis (\(\operatorname{Re}\{s\}=0\)).

Important

For a causal system with a rational system function \(H(s)\): Stability \(\iff\) all poles of \(H(s)\) lie in the left-half of the s-plane (i.e., all poles have negative real parts).

Stability is another fundamental system property. A stable system won’t “blow up” or produce an unbounded output from a bounded input. The first rule for stability is very general: the ROC must encompass the entire imaginary axis. This directly links to the convergence of the Fourier Transform of \(h(t)\). The second rule is a powerful shortcut for a very common class of systems: causal systems with rational transfer functions. In this case, simply checking the pole locations in the s-plane tells you everything you need to know about stability.

Example: Stability & Multiple ROCs

Consider a system with \(H(s)=\frac{s-1}{(s+1)(s-2)}\). Poles at \(s=-1, s=2\). Zero at \(s=1\).

Different ROCs lead to different system properties:

Causal System

• Output depends only on present and past inputs.
• ROC: Right of rightmost pole.

(a) Causal, Unstable

ROC: \(\operatorname{Re}\{s\} > 2\) - This system does NOT include \(j\omega\)-axis.

Non-Causal System

• Output also depends on future inputs. Non-realizable in real-time.
• ROC: A strip between poles.

(b) Non-Causal, Stable

ROC: \(-1 < \operatorname{Re}\{s\} < 2\)

• Includes \(j\omega\)-axis.
• Not to the right of rightmost pole.

Non-Causal System

• Output depends only on future inputs, not present or past. Non-realizable in real-time.
• ROC: Left of leftmost pole.

(c) Anticausal, Unstable

ROC: \(\operatorname{Re}\{s\} < -1\)

• Left of leftmost pole.
• Does NOT include \(j\omega\)-axis.

This example beautifully illustrates how the same algebraic expression for \(H(s)\) can represent different systems depending on its ROC. We have poles at -1 and 2. For case (a), if the system is causal, its ROC must be \(\operatorname{Re}\{s\}>2\). This doesn’t include the \(j\omega\)-axis, so it’s unstable. For case (b), if the system is stable, its ROC must include the \(j\omega\)-axis, meaning it’s the strip \(-1 < \operatorname{Re}\{s\} < 2\). This system is stable, but because the ROC is not to the right of the rightmost pole, it’s non-causal. Finally, for case (c), if it’s anticausal, its ROC is \(\operatorname{Re}\{s\}<-1\). This is unstable as it doesn’t contain the \(j\omega\)-axis. The key takeaway: ROC is crucial for defining system properties.

Example: Causal System Stability

Consider causal LTI systems with rational system functions:

Stable System

• \(h(t) = e^{-t}u(t)\)
• \(H(s) = 1/(s+1)\)
• Pole: at \(s=-1\) (in the left-half s-plane)

This system is stable because its pole has a negative real part.

Unstable System

• \(h(t) = e^{2t}u(t)\)
• \(H(s) = 1/(s-2)\)
• Pole: at \(s=2\) (in the right-half s-plane)

This system is unstable because its pole has a positive real part.

Tip

For causal systems with rational \(H(s)\), simply check if all poles are in the LHP for stability.

These are straightforward examples applying the second rule for stability. For causal systems with rational transfer functions, stability is directly determined by the location of the poles. A pole in the left-half plane, like \(s=-1\), leads to a decaying exponential in the impulse response, which is absolutely integrable and thus stable. A pole in the right-half plane, like \(s=2\), leads to a growing exponential, which is not absolutely integrable, making the system unstable. This visual check in the s-plane is incredibly useful for engineers.

LTI Systems Characterized by Linear Constant-Coefficient Differential Equations

The Laplace Transform simplifies the analysis of LTI systems described by LCCDEs.

A general LCCDE:

\[ \begin{equation*} \sum_{k=0}^{N} a_{k} \frac{d^{k} y(t)}{d t^{k}}=\sum_{k=0}^{M} b_{k} \frac{d^{k} x(t)}{d t^{k}} \tag{9.131} \end{equation*} \]

Applying the Laplace Transform (using linearity and differentiation properties):

\[ \begin{equation*} \left(\sum_{k=0}^{N} a_{k} s^{k}\right) Y(s)=\left(\sum_{k=0}^{M} b_{k} s^{k}\right) X(s) \tag{9.132} \end{equation*} \]

The system function \(H(s) = Y(s)/X(s)\) is always rational:

\[ \begin{equation*} H(s)=\frac{\left\{\sum_{k=0}^{M} b_{k} s^{k}\right\}}{\left\{\sum_{k=0}^{N} a_{k} s^{k}\right\}} \tag{9.133} \end{equation*} \]

• Zeros: solutions to \(\sum_{k=0}^{M} b_{k} s^{k}=0\)
• Poles: solutions to \(\sum_{k=0}^{N} a_{k} s^{k}=0\)

Differential equations are a natural way to model many physical systems. The beauty of the Laplace transform is how it transforms these complex differential equations into simple algebraic equations in the s-domain. This allows us to easily find the system function \(H(s)\) as a ratio of polynomials, which makes it a rational function. The roots of the numerator polynomial are the zeros of the system, and the roots of the denominator polynomial are the poles. Remember, the differential equation itself doesn’t specify the ROC; additional information like causality or stability is needed.

Example: First-Order LCCDE

Consider an LTI system where input \(x(t)\) and output \(y(t)\) satisfy:

\[ \begin{equation*} \frac{d y(t)}{d t}+3 y(t)=x(t) \tag{9.126} \end{equation*} \]

1. Apply Laplace Transform: \(sY(s) + 3Y(s) = X(s)\) \((s+3)Y(s) = X(s)\)

2. Determine System Function: \[ \begin{equation*} H(s)=\frac{Y(s)}{X(s)}=\frac{1}{s+3} \tag{9.128} \end{equation*} \]

3. Infer ROC (with additional info):

   • If causal: ROC is \(\operatorname{Re}\{s\}>-3\). (System is stable).
   • If anticausal: ROC is \(\operatorname{Re}\{s\}<-3\). (System is unstable).

This example demonstrates the process of finding the system function from a differential equation. We take the Laplace transform of both sides, use the differentiation property, and then simply rearrange the terms to solve for \(Y(s)/X(s)\). The result, \(H(s) = 1/(s+3)\), has a pole at \(s=-3\). As discussed, the differential equation alone doesn’t give the ROC. If we add the condition of causality, we know the ROC must be \(\operatorname{Re}\{s\}>-3\), which also makes the system stable since the pole is in the LHP. If it were anticausal, the ROC would be \(\operatorname{Re}\{s\}<-3\), making it unstable.

Application: RLC Circuit Analysis

An RLC circuit is a classic example of an LTI system described by an LCCDE.

Consider the series RLC circuit where \(x(t)\) is the source voltage and \(y(t)\) is the capacitor voltage.

The differential equation is: \[ \begin{equation*} L C \frac{d^{2} y(t)}{d t^{2}}+R C \frac{d y(t)}{d t}+y(t)=x(t) \tag{9.136} \end{equation*} \]

Applying Laplace Transform: \[ \begin{equation*} H(s)=\frac{1 / L C}{s^{2}+(R / L) s+(1 / L C)} \tag{9.137} \end{equation*} \]

Note

For positive values of \(R, L, C\), the poles of this system function will always have negative real parts, ensuring the system is stable.

Here’s a practical application. A series RLC circuit is a fundamental circuit in ECE. If we take the input as the source voltage and the output as the voltage across the capacitor, we can derive a second-order differential equation describing its behavior. Using the Laplace transform, we can easily find its system function \(H(s)\). For physically realizable components (positive R, L, C), the poles of this system will always reside in the left-half of the s-plane, meaning the RLC circuit is inherently stable. This is a powerful conclusion derived quickly using Laplace transforms.

Example: Deduce System from I/O Pair

Suppose we know that for an LTI system:

• Input: \(x(t)=e^{-3 t} u(t)\)
• Output: \(y(t)=\left[e^{-t}-e^{-2 t}\right] u(t)\)

Let’s find \(H(s)\) and system properties:

1. Laplace Transforms of I/O: \(X(s)=\frac{1}{s+3}, \quad \operatorname{Re}\{s\}>-3\) \(Y(s)=\frac{1}{(s+1)(s+2)}, \quad \operatorname{Re}\{s\}>-1\)

2. System Function: \(H(s)=\frac{Y(s)}{X(s)}=\frac{s+3}{(s+1)(s+2)}\)

3. ROC of \(H(s)\): The ROC of \(Y(s)\) must include the intersection of ROCs of \(X(s)\) and \(H(s)\). Given \(X(s)\) has \(\operatorname{Re}\{s\}>-3\) and \(Y(s)\) has \(\operatorname{Re}\{s\}>-1\), the only consistent ROC for \(H(s)\) is \(\operatorname{Re}\{s\}>-1\).

4. Deduced System Properties:

   • Causal: Yes (ROC is RHP to the right of rightmost pole at \(s=-1\)).
   • Stable: Yes (All poles at \(s=-1, -2\) are in LHP, ROC includes \(j\omega\)-axis).
   • Differential Equation: \(\frac{d^{2} y(t)}{d t^{2}}+3 \frac{d y(t)}{d t}+2 y(t)=\frac{d x(t)}{d t}+3 x(t)\)

This example showcases the power of the Laplace transform to reverse-engineer system properties from a single input-output pair. By transforming \(x(t)\) and \(y(t)\) to the s-domain, we can easily calculate \(H(s)\). Crucially, the ROC of \(H(s)\) must be compatible with both \(X(s)\) and \(Y(s)\). From this, we can deduce causality, stability, and even the underlying differential equation. This is a common problem-solving technique in system identification.

Application: Butterworth Filters

Butterworth filters are a widely used class of LTI systems known for their maximally flat passband frequency response.

An \(N\)-th order lowpass Butterworth filter has a frequency response magnitude squared given by:

\[ \begin{equation*} |B(j \omega)|^{2}=\frac{1}{1+\left(j \omega / j \omega_{c}\right)^{2 N}} \tag{9.140} \end{equation*} \]

To find the system function \(B(s)\):

1. Use the property \(|B(j \omega)|^{2}=B(j \omega) B^{*}(j \omega)\).

2. For real impulse response, \(B^{*}(j \omega)=B(-j \omega)\). So, \(|B(j \omega)|^{2}=B(j \omega) B(-j \omega)\).

3. Substitute \(s=j\omega\):

   \[ \begin{equation*} B(s) B(-s)=\frac{1}{1+\left(s / j \omega_{c}\right)^{2 N}} \tag{9.144} \end{equation*} \]

Tip

From \(B(s)B(-s)\), we find the poles of the filter. To ensure a causal and stable filter, we select the poles that lie in the left-half of the s-plane.

Butterworth filters are a cornerstone of analog and digital filter design due to their flat passband and monotonic rolloff. We begin with the definition of its squared magnitude response. The key step here is to relate \(|B(j\omega)|^2\) to \(B(s)B(-s)\) by replacing \(j\omega\) with \(s\). This gives us an expression for \(B(s)B(-s)\), from which we can find all \(2N\) poles. To construct the stable and causal filter \(B(s)\), we must select only those poles that lie in the left-half of the s-plane. This process directly links the frequency domain specification to the s-domain system function.

Butterworth Filter Pole Locations (Interactive)

The poles of \(B(s)B(-s)\) are located on a circle of radius \(\omega_c\) in the s-plane.

Specifically, \(s_p = \omega_c \exp\left(j\left[\frac{\pi(2k+1)}{2N}+\frac{\pi}{2}\right]\right)\) for \(k=0, \dots, 2N-1\).

Butterworth Filter Pole Locations (Interactive)

Use the sliders to explore how the pole locations change with filter order \(N\) and cutoff frequency \(\omega_c\).

viewof N_butter = Inputs.range([1, 6], {value: 2, step: 1, label: "Filter Order (N)"});
viewof wc_butter = Inputs.range([0.5, 2.0], {value: 1.0, step: 0.1, label: "Cutoff Frequency (ωc)"});

This interactive plot allows you to visualize the pole locations for Butterworth filters. The red ‘x’ marks show all \(2N\) poles of \(B(s)B(-s)\), which are symmetrically distributed on a circle of radius \(\omega_c\). The blue circles represent the \(N\) poles selected for \(B(s)\) to ensure causality and stability – these are the poles exclusively in the left-half of the s-plane. Notice how increasing \(N\) adds more poles, making the filter sharper, and changing \(\omega_c\) scales the entire pole constellation.

Butterworth Transfer Functions & Differential Equations

Once the poles for \(B(s)\) are identified, the transfer function can be constructed.

• N=1: \(B(s)=\frac{\omega_{c}}{s+\omega_{c}}\) Differential Equation: \(\frac{d y(t)}{d t}+\omega_{c} y(t)=\omega_{c} x(t)\)

• N=2: \(B(s)=\frac{\omega_{c}^{2}}{s^{2}+\sqrt{2} \omega_{c} s+\omega_{c}^{2}}\) Differential Equation: \(\frac{d^{2} y(t)}{d t^{2}}+\sqrt{2} \omega_{c} \frac{d y(t)}{d t}+\omega_{c}^{2} y(t)=\omega_{c}^{2} x(t)\)

• N=3: \(B(s)=\frac{\omega_{c}^{3}}{s^{3}+2 \omega_{c} s^{2}+2 \omega_{c}^{2} s+\omega_{c}^{3}}\) Differential Equation: \(\frac{d^{3} y(t)}{d t^{3}}+2 \omega_{c} \frac{d^{2} y(t)}{d t^{2}}+2 \omega_{c}^{2} \frac{d y(t)}{d t}+\omega_{c}^{3} y(t)=\omega_{c}^{3} x(t)\)

Note

Higher-order filters (larger \(N\)) result in more complex differential equations but offer sharper cutoff characteristics in the frequency domain.

Here we see the actual transfer functions and their corresponding differential equations for the first few orders of Butterworth filters. As the order \(N\) increases, the denominator polynomial gets more complex, leading to higher-order differential equations. This directly translates to more components in an analog circuit implementation or more complex algorithms in a digital filter. The trade-off is improved filter performance, such as a steeper roll-off in the stopband, at the cost of increased system complexity.