Inverse Laplace Transform

Precise definition

The inverse Laplace transform $\mathcal L^{-1}\{F(s)\}=f(t)$ recovers a time-domain function whose Laplace transform is $F(s)$. It is unique up to equality almost everywhere within standard classes such as piecewise-continuous functions of exponential order. Inversion usually uses known pairs and transform properties rather than evaluating the Bromwich integral.

Notation and mathematical language

Core pairs include $1/s\leftrightarrow1$, $1/(s-a)\leftrightarrow e^{at}$, $s/(s^2+\omega^2)\leftrightarrow\cos\omega t$, and $\omega/(s^2+\omega^2)\leftrightarrow\sin\omega t$. The shift $F(s-a)$ corresponds to $e^{at}f(t)$; a factor $e^{-as}$ corresponds to $u(t-a)f(t-a)$.

Conceptual picture

Partial fractions separate a rational transform into recognizable dynamical modes. Completing a square reveals damped sine and cosine terms. A delay factor does not merely shift the graph horizontally; the unit step also keeps the response zero before the delay.

Conditions and key results

Rational decomposition requires a proper fraction or polynomial division first. Repeated and irreducible quadratic factors need their correct numerator forms. Transform pairs assume a causal time domain $t\ge0$ in the usual ODE convention.

A reliable strategy

1. Simplify $F(s)$, divide if improper, and factor the denominator over the reals.
2. Use partial fractions or complete the square, retaining shift and delay factors.
3. Match each term to a transform pair and apply linearity, frequency scaling, and shifts exactly.
4. Take the forward transform of the result to verify every coefficient and delay.

Fully worked example

Interpretation and application

Inverse transforms solve linear initial value problems with impulses, switching, and delays. The recovered function exactly corresponds to the algebraic transform under its assumptions; an engineering input represented by an ideal step or impulse is itself an approximation to a physical signal.

Common mistakes

Verification and reasonableness

• Apply the forward Laplace transform term-by-term and recover $F(s)$.
• Check large-$s$ behaviour against expected initial values when the initial-value theorem applies.
• For delayed functions, verify the time-domain expression is zero before the delay.

Practice

1. Find $\mathcal L^{-1}\{1/(s-4)\}$.
2. Find $\mathcal L^{-1}\{2/(s^2+4)\}$.
3. What does $e^{-3s}F(s)$ produce?

Further deduction

Convolution handles products that do not simplify well: if $F=\mathcal L\{f\}$ and $G=\mathcal L\{g\}$, then $\mathcal L^{-1}\{FG\}=(f*g)(t)=\int_0^t f(\tau)g(t-\tau)d\tau$. For example, $1/[s(s^2+1)]$ is the transform of $\int_0^t\sin(t-\tau)d\tau=1-\cos t$. Partial fractions give the same answer, providing an independent check.

For repeated linear factors, $1/(s-a)^n$ corresponds to $t^{n-1}e^{at}/(n-1)!$. Thus a double pole creates a factor $t$ and often signals resonance or repeated modes in an ODE. Pole locations determine exponential growth or decay, while multiplicity contributes polynomial factors; partial fractions expose both features.

Related topics

• Laplace Transform
• Step Functions
• Second-Order Linear Differential Equations