Laplace Transforms

Laplace Transforms

Learning Objectives

Define the Laplace Transform and identify its region of convergence.
Apply standard transforms to common mathematical functions.
Use derivative properties to transform and solve initial value problems.
Perform Inverse Laplace Transforms using Partial Fraction Decomposition.
Utilize step functions and the Dirac delta function to model discontinuous inputs.
Apply the Convolution Theorem to evaluate inverse transforms.

The Laplace Transform is a powerful integral transform used to switch a function from the time domain ( $t$ ) to the frequency domain ( $s$ ). It effectively converts linear differential equations involving derivatives into simple algebraic equations involving polynomials, which are easier to solve. Once solved algebraically, the inverse transform returns the solution back to the time domain.

Laplace Transform Definition

An integral transform that converts a function of a real variable $t$ (often time) to a function of a complex variable $s$ (complex frequency).

Laplace Transform

The formal definition of the one-sided Laplace transform for a function defined for t \ge 0.

\mathcal{L}\{f(t)\} = F(s) = \int_0^\infty e^{-st} f(t) \, dt

Variables

Symbol	Description	Unit
$\mathcal{L}\{f(t)\}$	Laplace transform of the function	-
$F(s)$	Transformed function in the s-domain	-
$f(t)$	Original function in the time domain	-
$s$	Complex frequency variable	-
$t$	Time variable	s

Common Transforms Table

Standard Transforms

$f(t)$	$F(s) = \mathcal{L}\{f(t)\}$	Region of Convergence
$1$	$\frac{1}{s}$	$s > 0$
$t^n$ ( $n \in \mathbb{Z}^+$ )	$\frac{n!}{s^{n+1}}$	$s > 0$
$e^{at}$	$\frac{1}{s-a}$	$s > a$
$\sin(kt)$	$\frac{k}{s^2 + k^2}$	$s > 0$
$\cos(kt)$	$\frac{s}{s^2 + k^2}$	$s > 0$
$\sinh(kt)$	$\frac{k}{s^2 - k^2}$	$s > \|k\|$
$\cosh(kt)$	$\frac{s}{s^2 - k^2}$	$s > \|k\|$

Properties for Solving IVPs

The main application is solving linear DEs with constant coefficients subject to initial conditions.

Derivative Properties

The Laplace Transform converts differentiation in the time domain into multiplication by $s$ in the frequency domain, directly incorporating initial conditions:

First Derivative: $\mathcal{L}\{f'(t)\} = sF(s) - f(0)$
Second Derivative: $\mathcal{L}\{f''(t)\} = s^2F(s) - sf(0) - f'(0)$
Nth Derivative: $\mathcal{L}\{f^{(n)}(t)\} = s^nF(s) - s^{n-1}f(0) - \dots - f^{(n-1)}(0)$

Inverse Laplace Transforms & Partial Fractions

After solving algebraically for $F(s)$ , it usually appears as a complex rational function $P(s)/Q(s)$ . Finding the Inverse Laplace Transform, $f(t) = \mathcal{L}^{-1}\{F(s)\}$ , almost always requires decomposing this fraction into simpler parts that match the standard table using Partial Fraction Decomposition.

Partial Fraction Decomposition Strategies

Given a rational expression $\frac{P(s)}{Q(s)}$ where the degree of polynomial $P$ is less than $Q$ :

Factor the Denominator: Factor $Q(s)$ completely into linear factors $(s-a)$ and irreducible quadratic factors $(s^2 + bs + c)$ .
Set up the Decomposition Form:
- For each distinct linear factor $(s-a)$ , add $\frac{A}{s-a}$ .
- For repeated linear factors $(s-a)^n$ , add $\frac{A_1}{(s-a)} + \frac{A_2}{(s-a)^2} + \dots + \frac{A_n}{(s-a)^n}$ .
- For each irreducible quadratic factor $(s^2 + bs + c)$ , add $\frac{As + B}{s^2 + bs + c}$ . Completing the square is usually required after this step to match shifted sine/cosine transforms.
- For repeated irreducible quadratic factors $(s^2 + bs + c)^2$ , add $\frac{A_1 s + B_1}{s^2 + bs + c} + \frac{A_2 s + B_2}{(s^2 + bs + c)^2}$ .
Solve for the Unknown Constants ( $A, B, C...$ ): Multiply both sides by the original denominator $Q(s)$ to clear fractions. Then, either pick strategic values for $s$ to eliminate terms, or expand both sides and equate the coefficients of matching powers of $s$ .
Apply the Inverse Transform: Use the linearity property $\mathcal{L}^{-1}\{aF(s) + bG(s)\} = a\mathcal{L}^{-1}\{F(s)\} + b\mathcal{L}^{-1}\{G(s)\}$ and the standard table to invert term-by-term.

Step Functions and Dirac Delta

These functions allow us to model discontinuous inputs (like turning a switch on or off) and impulses (like a hammer blow to a mass-spring system).

Unit Step Function u(t-a)

Also known as the Heaviside step function, it models a switch turning on at a specific time.

Unit Step Function Definition

The piecewise definition of the Heaviside step function.

\begin{aligned} \mathcal{U}(t-a) = \begin{cases} 0 & 0 \le t < a \\ 1 & t \ge a \end{cases} \end{aligned}

Variables

Symbol	Description	Unit
$\mathcal{U}$	Unit step function	-
$t$	Time	s
$a$	Time at which the step occurs	s

Transform and Shift Property

The Laplace transform of the unit step function and the related shift property.

\mathcal{L}\{\mathcal{U}(t-a)\} = \frac{e^{-as}}{s}

\mathcal{L}\{f(t-a)\mathcal{U}(t-a)\} = e^{-as}F(s)

Variables

Symbol	Description	Unit
$\mathcal{L}$	Laplace Transform operator	-
$\mathcal{U}(t-a)$	Unit step function shifted by a	-
$s$	Complex frequency variable	-
$a$	Time delay	s

Dirac Delta Function

Represents an instantaneous, infinitely strong impulse at $t=a$ , such that its integral over all time is 1.

Dirac Delta Transform

The Laplace transform of the Dirac delta function.

\mathcal{L}\{\delta(t-a)\} = e^{-as}

Variables

Symbol	Description	Unit
$\delta(t-a)$	Dirac delta function centered at a	-
$a$	Time of the impulse	s
$s$	Complex frequency variable	-

Convolution Theorem

The Convolution Theorem allows us to find the inverse Laplace Transform of a product of two functions $F(s)G(s)$ . The convolution of two functions $f(t)$ and $g(t)$ is denoted by $f * g$ and defined as an integral. The Convolution Theorem states that multiplication in the frequency domain is equivalent to convolution in the time domain.

Convolution Integral and Theorem

The definition of convolution and the transform theorem.

(f * g)(t) = \int_0^t f(\tau)g(t-\tau) \, d\tau

\mathcal{L}\{f * g\} = F(s)G(s)

\mathcal{L}^{-1}\{F(s)G(s)\} = (f * g)(t)

Variables

Symbol	Description	Unit
$f * g$	Convolution of f and g	-
$t$	Time	s
$\tau$	Integration variable (dummy variable for time)	s
$F(s), G(s)$	Laplace transforms of f(t) and g(t)	-

Visualizing Convolution

Interactive Simulation

Interact with the simulation below to visualize functions in both the time and complex frequency domains.

Laplace Transform Pairs

Select a function to see its representation in the time domain $t$ and the complex frequency domain $s$ .

Transform Pair

\mathcal{L}\{e^{-at} \quad (a=1)\} = \frac{1}{s+1}

Time Domain $f(t)$

t = [0, 10]

s-Domain $F(s)$

s = [0, 5]

Key Takeaways

Laplace Transforms convert integral-differential equations (calculus) into algebraic equations. They are exceptionally useful for Initial Value Problems because the initial conditions $f(0), f'(0)$ are baked directly into the derivative properties.
Linearity: $\mathcal{L}\{af(t) + bg(t)\} = aF(s) + bG(s)$ .
Inverse Transform: Typically relies heavily on Partial Fraction Decomposition and completing the square to manipulate algebraic expressions into recognizable table forms.
Shift Theorems: Essential for handling exponentials ( $e^{at}f(t) \leftrightarrow F(s-a)$ ) and step functions ( $f(t-a)\mathcal{U}(t-a) \leftrightarrow e^{-as}F(s)$ ).
Convolution: Evaluates $\mathcal{L}^{-1}\{F(s)G(s)\}$ analytically without needing partial fractions, by computing the integral $\int_0^t f(\tau)g(t-\tau) \, d\tau$ .