Signals and Systems

9.1 The Laplace Transform

Imron Rosyadi

The Laplace Transform

Unlocking System Dynamics in ECE

Welcome to this session on the Laplace Transform. This powerful mathematical tool is fundamental in Electrical and Computer Engineering, especially for analyzing linear time-invariant systems. Today, we’ll explore its definition, its relationship to the Fourier Transform, and how it helps us understand system behavior through poles, zeros, and the region of convergence.

What is the Laplace Transform?

The Laplace Transform extends the Fourier Transform to analyze a broader class of signals and systems, particularly those that are unstable or non-causal.

It transforms a time-domain signal \(x(t)\) into a complex frequency-domain function \(X(s)\), where \(s\) is a complex variable.

LTI System Response to Complex Exponentials

For an LTI system with impulse response \(h(t)\), the response to \(e^{st}\) is:

\[ y(t) = H(s) e^{st} \quad \text{(9.1)} \]

where \(H(s)\) is the Laplace Transform of \(h(t)\):

\[ H(s) = \int_{-\infty}^{\infty} h(t) e^{-st} dt \quad \text{(9.2)} \]

What is the Laplace Transform? (cont.)

General Definition

The Laplace Transform of a general signal \(x(t)\) is defined as:

\[ X(s) \triangleq \int_{-\infty}^{+\infty} x(t) e^{-st} dt \quad \text{(9.3)} \]

We denote this relationship as \(x(t) \stackrel{\mathfrak{L}}{\longleftrightarrow} X(s)\).

The Laplace Transform is a cornerstone of signals and systems analysis. Unlike the Fourier Transform, which is limited to absolutely integrable signals, the Laplace Transform uses a complex exponential \(e^{-st}\) as its basis function, where \(s\) is a complex variable \(\sigma + j\omega\). This generalization allows us to analyze a much wider range of signals and systems, including unstable ones or those that grow exponentially. The transform converts differential equations into algebraic equations, simplifying analysis significantly.

Laplace Transform and the Fourier Transform

The complex variable \(s\) can be written as \(s = \sigma + j\omega\), where \(\sigma\) is the real part and \(\omega\) is the imaginary part.

Substituting \(s = \sigma + j\omega\) into the Laplace Transform definition:

\[ X(\sigma+j\omega) = \int_{-\infty}^{+\infty} x(t) e^{-(\sigma+j\omega)t} dt \quad \text{(9.7)} \]

This can be rewritten as:

\[ X(\sigma+j\omega) = \int_{-\infty}^{+\infty} [x(t) e^{-\sigma t}] e^{-j\omega t} dt \quad \text{(9.8)} \]

Note

The Laplace Transform \(X(s)\) is essentially the Fourier Transform of \(x(t)e^{-\sigma t}\)! When \(s=j\omega\) (i.e., \(\sigma=0\)), the Laplace Transform becomes the Fourier Transform: \(X(j\omega) = \int_{-\infty}^{+\infty} x(t) e^{-j\omega t} dt = \mathfrak{F}\{x(t)\}\)

Region of Convergence (ROC)

The Laplace Transform integral (Eq 9.3) does not always converge for all values of \(s\).

The Region of Convergence (ROC) is the set of all \(s\) values for which the integral converges.

Why is ROC Important?

• The algebraic expression for \(X(s)\) alone is not sufficient to uniquely define the time-domain signal \(x(t)\).
• Different time-domain signals can have the same algebraic \(X(s)\) but vastly different ROCs.
• The ROC determines properties of the signal, such as causality and stability.

The s-Plane

The complex variable \(s = \sigma + j\omega\) is visualized on the s-plane, with \(\operatorname{Re}\{s\}\) (the \(\sigma\)-axis) as the horizontal axis and \(\operatorname{Im}\{s\}\) (the \(j\omega\)-axis) as the vertical axis.

Important

The ROC is always a strip or a half-plane in the s-plane.

The Region of Convergence is perhaps the most critical concept when dealing with the Laplace Transform. Without specifying the ROC, the algebraic expression of \(X(s)\) is ambiguous. I’ll emphasize that the ROC is not just a mathematical detail; it provides vital information about the nature of the time-domain signal. We’ll see in upcoming examples how two very different signals can share the same algebraic Laplace Transform expression, but their ROCs will be distinct, allowing us to differentiate them. The s-plane provides a visual map for understanding where the transform converges.

Example 9.1: Causal Exponential

Consider the signal \(x(t)=e^{-at}u(t)\).

The Laplace Transform is:

\[ X(s) = \int_{0}^{\infty} e^{-at} e^{-st} dt = \int_{0}^{\infty} e^{-(s+a)t} dt \quad \text{(9.10)} \]

For convergence, we require \(\operatorname{Re}\{s+a\} > 0\), or \(\operatorname{Re}\{s\} > -a\).

Thus,

\[ X(s) = \frac{1}{s+a}, \quad \operatorname{Re}\{s\} > -a \quad \text{(9.13)} \]

Tip

For a causal signal \(x(t)u(t)\), the ROC is always a right-half plane.

Example 9.1: Causal Exponential (cont.)

Interactive Demonstration: Convergence of \(x(t)e^{-\sigma t}\)

Adjust ‘a’ and ‘sigma’ (\(\operatorname{Re}\{s\}\)) to observe how \(x(t)e^{-\sigma t}\) changes and whether it converges.

viewof a_val = Inputs.range([-2, 2], {value: 1, step: 0.1, label: "Parameter 'a'"});
viewof sigma_val = Inputs.range([-3, 3], {value: 0.5, step: 0.1, label: "sigma (Re{s})"});

Here, we take a common signal, the causal exponential \(e^{-at}u(t)\). We derive its Laplace Transform and identify its ROC. The interactive plot demonstrates the term \(x(t)e^{-\sigma t}\). Notice that for the integral to converge, this modified signal must decay to zero as \(t\) approaches infinity. By adjusting ‘a’ and ‘sigma’, you can visually observe when this decay condition is met, directly correlating to the \(\operatorname{Re}\{s\} > -a\) condition for convergence. If \(a+\sigma\) is positive, the exponential decays, and the integral converges. If \(a+\sigma\) is negative, it grows, and the integral diverges.

Example 9.2: Anti-Causal Exponential

Consider the signal \(x(t)=-e^{-at}u(-t)\).

The Laplace Transform is:

\[ X(s) = -\int_{-\infty}^{0} e^{-at} e^{-st} dt = -\int_{-\infty}^{0} e^{-(s+a)t} dt \quad \text{(9.17)} \]

For convergence, we require \(\operatorname{Re}\{s+a\} < 0\), or \(\operatorname{Re}\{s\} < -a\).

Thus,

\[ X(s) = \frac{1}{s+a}, \quad \operatorname{Re}\{s\} < -a \quad \text{(9.19)} \]

Tip

For an anti-causal signal \(x(t)u(-t)\), the ROC is always a left-half plane.

Example 9.2: Anti-Causal Exponential (cont.)

Comparing ROCs

Example 9.1: \(e^{-at}u(t)\) \(X(s) = \frac{1}{s+a}\), \(\operatorname{Re}\{s\} > -a\)

The ROC is to the right of the pole.

Example 9.2: \(-e^{-at}u(-t)\) \(X(s) = \frac{1}{s+a}\), \(\operatorname{Re}\{s\} < -a\)

The ROC is to the left of the pole.

This example is crucial for understanding the importance of the ROC. Here, we encounter an anti-causal exponential, \(-e^{-at}u(-t)\). Notice that its algebraic Laplace Transform form, \(1/(s+a)\), is identical to that of the causal exponential from Example 9.1. However, the condition for convergence is entirely different: \(\operatorname{Re}\{s\} < -a\).

The side-by-side comparison clearly shows that the same algebraic expression can correspond to two distinct time-domain signals, differentiated solely by their ROCs. This emphasizes that to fully specify a signal’s Laplace Transform, you must provide both the algebraic form and its corresponding ROC.

Poles, Zeros, and the s-Plane

For many practical signals, especially those from LTI systems described by differential equations, the Laplace Transform \(X(s)\) is a rational function:

\[ X(s) = \frac{N(s)}{D(s)} \quad \text{(9.31)} \]

where \(N(s)\) and \(D(s)\) are polynomials in \(s\).

Zeros

The roots of the numerator polynomial \(N(s)\) are called the zeros of \(X(s)\). At these values of \(s\), \(X(s)=0\).

Poles

The roots of the denominator polynomial \(D(s)\) are called the poles of \(X(s)\). At these values of \(s\), \(X(s)\) becomes infinite.

Poles, Zeros, and the s-Plane (cont.)

Pole-Zero Plot

• A pole-zero plot graphically represents the locations of poles and zeros in the s-plane.
• Poles are typically marked with an ‘x’.
• Zeros are typically marked with an ‘o’.
• The pole-zero plot, combined with the ROC, fully characterizes a rational Laplace Transform.

Important

The ROC is always bounded by poles and never contains any poles.

Rational Laplace Transforms are very common in ECE, particularly when dealing with systems described by linear constant-coefficient differential equations. The concepts of poles and zeros provide a powerful way to characterize these transforms. Poles are the values of ‘s’ where the transform “blows up” or goes to infinity, while zeros are where it becomes zero. Plotting these critical points on the s-plane, along with the ROC, gives us a complete visual signature of the system or signal in the Laplace domain. This pole-zero representation is fundamental for understanding system stability, frequency response, and transient behavior.

Example 9.3: Sum of Two Causal Exponentials

Consider \(x(t)=3e^{-2t}u(t) - 2e^{-t}u(t)\).

Using linearity and results from Example 9.1:

\[ \mathcal{L}\{3e^{-2t}u(t)\} = \frac{3}{s+2}, \quad \operatorname{Re}\{s\} > -2 \]

\[ \mathcal{L}\{-2e^{-t}u(t)\} = \frac{-2}{s+1}, \quad \operatorname{Re}\{s\} > -1 \]

For \(X(s)\) to converge, both individual transforms must converge. The intersection of their ROCs is \(\operatorname{Re}\{s\} > -1\).

Combining terms:

\[ X(s) = \frac{3}{s+2} - \frac{2}{s+1} = \frac{3(s+1) - 2(s+2)}{(s+2)(s+1)} = \frac{s-1}{s^2+3s+2} \]

Example 9.3: Sum of Two Causal Exponentials (cont.)

Thus,

\[ X(s) = \frac{s-1}{(s+2)(s+1)}, \quad \operatorname{Re}\{s\} > -1 \quad \text{(9.23)} \]

Pole-Zero Plot

• Poles: \(s = -1, s = -2\) (roots of \(D(s)\))
• Zeros: \(s = 1\) (root of \(N(s)\))

graph TD
    S_Plane_Ex3("s-Plane for Example 9.3")
    subgraph Locations
        Pole_1("X: s = -1")
        Pole_2("X: s = -2")
        Zero_1("O: s = 1")
    end
    S_Plane_Ex3 --> Pole_1
    S_Plane_Ex3 --> Pole_2
    S_Plane_Ex3 --> Zero_1
    ROC_Ex3("ROC: Re{s} > -1 (Right of the rightmost pole)")
    S_Plane_Ex3 -- is bounded by --> ROC_Ex3

In this example, we combine two causal exponentials. Due to linearity, the Laplace Transform of the sum is the sum of the individual Laplace Transforms. The crucial point here is how the ROC is determined: it must be the intersection of the ROCs of all individual terms. Since both terms are causal exponentials, their ROCs are right-half planes. The overall ROC is the right-half plane to the right of the rightmost pole, which is at \(s=-1\). We identify the poles at \(s=-1\) and \(s=-2\), and a zero at \(s=1\).

Example 9.4: Complex Exponentials

Consider \(x(t)=e^{-2t}u(t) + e^{-t}(\cos 3t)u(t)\).

Using Euler’s relation, \(\cos 3t = \frac{1}{2}(e^{j3t} + e^{-j3t})\), so:

\[ x(t) = \left[e^{-2t} + \frac{1}{2}e^{-(1-j3)t} + \frac{1}{2}e^{-(1+j3)t}\right]u(t) \]

The individual Laplace Transforms are:

1. \(\mathcal{L}\{e^{-2t}u(t)\} = \frac{1}{s+2}, \quad \operatorname{Re}\{s\} > -2\)
2. \(\mathcal{L}\{\frac{1}{2}e^{-(1-j3)t}u(t)\} = \frac{1/2}{s+(1-j3)}, \quad \operatorname{Re}\{s\} > -1\)
3. \(\mathcal{L}\{\frac{1}{2}e^{-(1+j3)t}u(t)\} = \frac{1/2}{s+(1+j3)}, \quad \operatorname{Re}\{s\} > -1\)

The common ROC is \(\operatorname{Re}\{s\} > -1\).

Example 9.4: Complex Exponentials (cont.)

Combining these over a common denominator yields:

\[ X(s)=\frac{2s^2+5s+12}{(s^2+2s+10)(s+2)}, \quad \operatorname{Re}\{s\}>-1 \quad \text{(9.30)} \]

Pole-Zero Plot

• Poles: \(s = -2\), and \(s = -1 \pm j3\) (roots of \(s^2+2s+10=0\))
• Zeros: Roots of \(2s^2+5s+12=0\) (complex conjugate pair)

graph TD
    S_Plane_Ex4("s-Plane for Example 9.4")
    subgraph Locations
        Pole_1("X: s = -2")
        Pole_2("X: s = -1 + j3")
        Pole_3("X: s = -1 - j3")
        Zero_1("O: s = complex (conjugate pair)")
    end
    S_Plane_Ex4 --> Pole_1
    S_Plane_Ex4 --> Pole_2
    S_Plane_Ex4 --> Pole_3
    S_Plane_Ex4 --> Zero_1
    ROC_Ex4("ROC: Re{s} > -1 (Right of the rightmost poles)")
    S_Plane_Ex4 -- is bounded by --> ROC_Ex4

This example introduces complex exponentials, which are crucial for representing sinusoidal signals. By using Euler’s formula, we decompose the cosine into two complex exponentials. Each exponential has its own Laplace Transform and ROC. Again, the overall ROC is the intersection of these individual ROCs, leading to \(\operatorname{Re}\{s\} > -1\). This results in complex conjugate poles and zeros on the s-plane, which always appear in pairs for real-valued signals. The location of these complex poles and zeros provides insight into the oscillatory behavior of the signal.

Example 9.5: Unit Impulse and Exponentials

Consider \(x(t)=\delta(t) - \frac{4}{3}e^{-t}u(t) + \frac{1}{3}e^{2t}u(t)\).

1. \(\mathcal{L}\{\delta(t)\} = 1\). The ROC is the entire s-plane.
2. \(\mathcal{L}\{-\frac{4}{3}e^{-t}u(t)\} = \frac{-4/3}{s+1}, \quad \operatorname{Re}\{s\} > -1\).
3. \(\mathcal{L}\{\frac{1}{3}e^{2t}u(t)\} = \frac{1/3}{s-2}, \quad \operatorname{Re}\{s\} > 2\).

The overall ROC is the intersection of all three, which is \(\operatorname{Re}\{s\} > 2\).

Combining terms:

\[ X(s) = 1 - \frac{4/3}{s+1} + \frac{1/3}{s-2} = \frac{(s+1)(s-2) - \frac{4}{3}(s-2) + \frac{1}{3}(s+1)}{(s+1)(s-2)} \]

\[ X(s) = \frac{s^2-s-2 - \frac{4}{3}s + \frac{8}{3} + \frac{1}{3}s + \frac{1}{3}}{(s+1)(s-2)} = \frac{s^2 - \frac{6}{3}s + \frac{3}{3}}{(s+1)(s-2)} = \frac{s^2-2s+1}{(s+1)(s-2)} \]

\[ X(s) = \frac{(s-1)^2}{(s+1)(s-2)}, \quad \operatorname{Re}\{s\} > 2 \quad \text{(9.35)} \]

Example 9.5: Unit Impulse and Exponentials (cont.)

Pole-Zero Plot and Fourier Transform

• Poles: \(s = -1, s = 2\)
• Zeros: \(s = 1\) (a second-order zero, as it’s repeated)

graph TD
    S_Plane_Ex5("s-Plane for Example 9.5")
    subgraph Locations
        Pole_1("X: s = -1")
        Pole_2("X: s = 2")
        Zero_1("O: s = 1 (2nd order)")
    end
    S_Plane_Ex5 --> Pole_1
    S_Plane_Ex5 --> Pole_2
    S_Plane_Ex5 --> Zero_1
    ROC_Ex5("ROC: Re{s} > 2 (Right of the rightmost pole)")
    S_Plane_Ex5 -- is bounded by --> ROC_Ex5

Warning

Since the ROC does not include the \(j\omega\)-axis (\(\operatorname{Re}\{s\}=0\)), the Fourier Transform for this signal does not converge. This is due to the growing exponential term \(e^{2t}u(t)\).

This final example demonstrates a signal that includes a unit impulse and a growing exponential. The unit impulse has an ROC that covers the entire s-plane, meaning its Laplace Transform always converges. However, the growing exponential \(e^{2t}u(t)\) requires \(\operatorname{Re}\{s\} > 2\) for its transform to converge. The overall ROC is thus dictated by the most restrictive condition, \(\operatorname{Re}\{s\} > 2\).

Notice the second-order zero at \(s=1\), meaning \((s-1)\) appears twice in the numerator. Critically, because the ROC does not include the \(j\omega\)-axis (the imaginary axis where \(\operatorname{Re}\{s\}=0\)), this signal does not possess a Fourier Transform. This highlights a key advantage of the Laplace Transform: its ability to analyze signals for which the Fourier Transform does not exist, thanks to its broader convergence properties.

Application: System Analysis in ECE

The Laplace Transform is a cornerstone in ECE for analyzing Linear Time-Invariant (LTI) systems.

Why use Laplace Transform?

1. Simplifies Differential Equations: Transforms complex differential equations into algebraic equations, which are much easier to solve.
2. Frequency Domain Analysis: Provides insights into system behavior across a range of complex frequencies, not just real frequencies.
3. Stability Analysis: Pole locations directly indicate system stability.
4. Circuit Analysis: Ideal for analyzing RLC circuits, filters, and control systems.

Application: System Analysis in ECE

LTI System Analysis Workflow

flowchart LR
    A["Input Signal x(t)"] --> B{"LTI System <br> h(t)"}
    B --> C["Output Signal y(t)"]

    A_LT["X(s)"] --> B_LT{"System Function <br> H(s)"}
    B_LT --> C_LT["Y(s)"]

    subgraph "Time Domain"
        A
        B
        C
    end

    subgraph "Laplace Domain"
        A_LT
        B_LT
        C_LT
    end

    A -- "Laplace Transform" --> A_LT
    C_LT -- "Inverse Laplace Transform" --> C
    B_LT -- "$$H(s)=Y(s)/X(s)$$" --> B

This slide summarizes the practical utility of the Laplace Transform in ECE. Its ability to convert differential equations into algebraic ones is a game-changer for analyzing complex systems like filters, control systems, and communication channels. The system function \(H(s)\), which is the Laplace Transform of the impulse response \(h(t)\), becomes the algebraic representation of the system. By solving for \(Y(s) = H(s)X(s)\) in the Laplace domain and then performing an inverse Laplace Transform, we can find the time-domain output \(y(t)\). This systematic approach makes analyzing LTI systems much more tractable and intuitive than working directly with convolution in the time domain or differential equations. Poles and zeros of \(H(s)\) directly tell us about the system’s stability and frequency response characteristics.

Key Takeaways

1. Generalization: The Laplace Transform is a generalization of the Fourier Transform, allowing analysis of a wider class of signals.
2. Complex Frequency \(s\): Uses a complex variable \(s = \sigma + j\omega\), where \(\sigma\) influences convergence.
3. Region of Convergence (ROC): Crucial for uniquely identifying a signal and understanding its properties (causality, stability). The ROC is always bounded by poles.
4. Poles & Zeros: These are the roots of the denominator and numerator polynomials of \(X(s)\), respectively, and define its algebraic form.

Note

Practical Impact: The Laplace Transform simplifies the analysis of LTI systems by converting differential equations into algebraic ones, making it indispensable for circuit analysis, control systems, and signal processing.

To wrap up, remember these core ideas: the Laplace Transform is a powerful extension, the complex ‘s’ variable and the ROC are fundamental to its meaning, and poles and zeros are its fingerprint. This tool is not just theoretical; it’s a practical workhorse in ECE. Continue exploring its properties and applications to truly master it.