Laplace Transform
From time domain to s-domain — solving differential equations algebraically
1. Building Intuition
You already know that the Fourier Transform decomposes a signal into pure sine waves (eiωt, constant amplitude). But many real signals — the voltage as a capacitor charges, the oscillation of a shock absorber, the step response of a control system — either grow or decay over time. Pure sine waves can't represent them compactly.
The Laplace Transform fixes this by using damped oscillations e-(σ+jω)tas basis functions instead of eiωt. The extra real part σ lets the basis functions grow or decay, so any causal signal (one that starts at t = 0) can be represented — even exponentially growing ones.
The second, equally important superpower: it turns differential equations into algebra.When you apply the Laplace Transform to an ODE, every derivative d/dt becomes a multiplication by s. What was a differential equation in t becomes a polynomial equation in s — high-school algebra. Solve for Y(s), look up the inverse in a table, done. No guessing solution forms, no variation of parameters.
The s-plane maps system behavior. Poles (×) determine response character — their real part controls decay rate, imaginary part controls oscillation frequency. Zeros (○) shape the response. Poles in the left half-plane mean stability.
2. The Mathematics
The Transform: What It Does
A transform is a tool that converts a problem into a different (easier) form, solves it there, then converts back. The Laplace Transform converts a function of time f(t) into a function F(s) of the complex variable s = σ + jω:
The Region of Convergence (ROC) is the set of s values for which this integral converges. For most engineering signals, this is the right half-plane Re(s) > σ₀ for some real σ₀. The imaginary axis (σ = 0) corresponds to the Fourier Transform — Laplace is strictly more general.
The factor e-st = e-σt·e-jωt does two things: e-jωtoscillates (like Fourier), while e-σt is a decaying envelope that forces the integral to converge even for growing functions like et or t². This is what makes Laplace more powerful than Fourier for transient and unstable signals.
Key Properties — Why It Simplifies ODEs
The core reason Laplace is used for ODEs is the derivative property: every derivative in time becomes a multiplication by s. Initial conditions enter automatically as constants, so you never need to set up the homogeneous+particular solution separately.
Integration in the time domain becomes division by s. Convolution becomes multiplication (same as Fourier). These algebraic operations are why solving a 2nd-order ODE with Laplace requires no guessing of particular solutions.
Common Transform Pairs
After solving for Y(s) algebraically, you use a table (or partial fractions) to invert. Memorizing these handful of pairs covers 90% of engineering problems:
| f(t) (t ≥ 0) | F(s) |
|---|---|
| δ(t) (unit impulse) | 1 |
| u(t) (unit step) | 1/s |
| t | 1/s² |
| tⁿ | n! / s^{n+1} |
| e^{at} | 1/(s−a) |
| sin(ωt) | ω/(s²+ω²) |
| cos(ωt) | s/(s²+ω²) |
| e^{at} sin(ωt) | ω/((s−a)²+ω²) |
| e^{at} cos(ωt) | (s−a)/((s−a)²+ω²) |
| tⁿ e^{at} | n!/(s−a)^{n+1} |
Partial Fractions — How to Invert in Practice
After solving an ODE algebraically you get Y(s) as a rational function P(s)/Q(s). To invert, factor the denominator and decompose into simple fractions, each matching a table entry:
Multiply both sides by (s+1)(s+2) and equate coefficients, or use the cover-up method: set s = −1 to get A = 2; set s = −2 to get B = −1. Then invert term-by-term using the table.
Transfer Functions and the s-Domain
For a linear time-invariant (LTI) system with zero initial conditions, the transfer functionH(s) = Y(s)/X(s) completely characterizes the system — independent of the input. The poles of H(s) (roots of the denominator) determine the natural response: poles in the left half-plane → stable (decay), right half-plane → unstable (grow), imaginary axis → sustained oscillation.
3. Applications
Fourier analysis uses pure sinusoids (top). Laplace uses exponentially damped sinusoids (bottom), capturing both oscillation frequency ω and decay rate σ. The yellow envelope shows e-σt.
Control Systems
Transfer functions, Bode plots, stability analysis via pole placement. The foundation of PID controller design.
Circuit Analysis
Impedance of capacitors (1/sC) and inductors (sL) in s-domain. Solve circuits algebraically.
Mechanical Systems
Spring-mass-damper models become simple polynomials in s. Natural frequency and damping ratio from pole locations.
Signal Processing
Filter design (Butterworth, Chebyshev) is done in the s-domain, then discretized for digital implementation.
4. Worked Examples
Example 1: Solving y'' + 3y' + 2y = 0
Given initial conditions y(0) = 1, y'(0) = 0. Apply Laplace Transform:
Inverse transform using the table (L-1{1/(s+a)} = e-at):
Example 2: RC Circuit Step Response
An RC circuit with R = 1kΩ, C = 1μF has transfer function:
For a unit step input (Vin = 1/s), the output is:
Example 3: Spring-Mass-Damper System
A mass m on a spring (constant k) with damper (coefficient b): mx'' + bx' + kx = F(t)
The poles s = (-b ± √(b²-4mk)) / 2m determine the behavior: underdamped (b² < 4mk, oscillatory decay), critically damped (b² = 4mk, fastest non-oscillatory), overdamped (b² > 4mk, slow exponential decay).