ODE: Inverse Laplace Transform

The Laplace transform converts differential equations into algebraic equations, making them easier to solve. However, the solution is only useful if you can interpret it back in the original time domain. The inverse Laplace transform is the essential operation that recovers the time-domain function $f(t)$ from its Laplace-domain representation $F(s)$. Mastering this technique is fundamental for engineers analyzing systems, from circuit responses to mechanical vibrations, as it bridges the gap between a convenient mathematical model and a tangible, time-based result.

Defining the Operation and Its Core Property

Formally, if the Laplace transform of a function $f(t)$ is $F(s)$, denoted by $\mathcal{L}\{f(t)\}=F(s)$, then the inverse Laplace transform is denoted by ${\mathcal{L}}^{-1}\{F(s)\}=f(t)$. The operator ${\mathcal{L}}^{-1}$ is the inverse of $\mathcal{L}$.

The most powerful and frequently used property is the linearity of the inverse transform. This means that if you have a sum of Laplace-domain functions multiplied by constants, you can take the inverse transform of each piece separately. Mathematically, for constants $\alpha$ and $\beta$, $${\mathcal{L}}^{-1}\{\alpha F(s)+\beta G(s)\}=\alpha {\mathcal{L}}^{-1}\{F(s)\}+\beta {\mathcal{L}}^{-1}\{G(s)\}=\alpha f(t)+\beta g(t).$$ This linearity allows you to break down complex expressions $F(s)$ into simpler, recognizable components whose inverses you can find individually. The entire systematic procedure for finding inverses rests on this foundational property.

Uniqueness: Lerch's Theorem

A critical question arises: could two different time functions have the same Laplace transform? If so, the inverse operation would be ambiguous. Lerch's theorem (also called the uniqueness theorem) provides the answer. It states that if two functions $f(t)$ and $g(t)$ have the same Laplace transform $F(s)$, and if both functions are continuous for $t\ge 0$, then $f(t)=g(t)$ for all $t\ge 0$.

In practical terms, this means that for the continuous functions you typically encounter in engineering contexts, the inverse Laplace transform is unique. You do not need to worry about multiple possible answers. This theorem justifies all the table lookup and decomposition methods you will use, as they will lead you to the one correct time-domain function. It's important to note that the theorem holds for continuous functions; functions with discontinuities (modeled with step functions) are handled piecewise but are still unique in the sense of generalized functions.

The Foundation: Using Transform Tables

Your primary tool is a table of common Laplace transform pairs. A typical entry shows a time function $f(t)$ and its corresponding transform $F(s)$. To use it for the inverse, you read the table from right to left: find an $F(s)$ in your expression that matches a table entry, and the corresponding $f(t)$ is its inverse.

For example, from a standard table you know $\mathcal{L}\{e^{at}\}=\frac{1}{s-a}$. Therefore, the inverse is ${\mathcal{L}}^{-1}\{\frac{1}{s-a}\}=e^{at}$. Similarly, ${\mathcal{L}}^{-1}\{\frac{s}{s^2+{\omega }^2}\}=\cos (\omega t)$.

The challenge in practice is that $F(s)$ is often a complex rational function that does not match a table entry directly. This is where algebraic manipulation becomes crucial. The goal of all systematic procedures is to manipulate $F(s)$ into a sum of simpler terms, each of which appears in your transform table.

Systematic Procedure: Partial Fraction Decomposition

When $F(s)$ is a proper rational function (the degree of the numerator is less than the degree of the denominator), partial fraction decomposition is the standard method. This technique expresses $F(s)$ as a sum of simpler fractions whose inverses are easily found.

The procedure depends on the factors in the denominator:

1. Distinct Linear Factors: For a term like $\frac{A}{s-p}$, the inverse is $A{e}^{pt}$.
2. Repeated Linear Factors: For a term like $\frac{A}{(s-p{)}^n}$, you use the transform pair ${\mathcal{L}}^{-1}\{\frac{1}{(s-p{)}^n}\}=\frac{t^{n-1}}{(n-1)!}{e}^{pt}$.
3. Irreducible Quadratic Factors: For a term like $\frac{As+B}{s^2+bs+c}$, you often need the completing the square technique to find its inverse.

Worked Example: Find ${\mathcal{L}}^{-1}\{\frac{s+5}{s^2+3s+2}\}$. First, factor the denominator: $s^2+3s+2=(s+1)(s+2)$. Perform partial fraction decomposition: $$\frac{s+5}{(s+1)(s+2)}=\frac{A}{s+1}+\frac{B}{s+2}.$$ Solving for $A$ and $B$: $s+5=A(s+2)+B(s+1)$. Let $s=-1$: $4=A(1)$ $\Rightarrow$ $A=4$. Let $s=-2$: $3=B(-1)$ $\Rightarrow$ $B=-3$. Thus, $F(s)=\frac{4}{s+1}-\frac{3}{s+2}$. Using linearity and the table: ${\mathcal{L}}^{-1}\{F(s)\}=4e^{-t}-3e^{-2t}$ for $t\ge 0$.

Systematic Procedure: Completing the Square

This technique is vital for handling quadratic factors in the denominator that have complex roots, which correspond to oscillatory responses like sines and cosines (or damped sinusoids) in the time domain. You complete the square to rewrite the quadratic in the form $(s+\alpha {)}^2+{\omega }^2$, which matches the denominator of sine and cosine transforms.

Worked Example: Find ${\mathcal{L}}^{-1}\{\frac{s+2}{s^2+4s+13}\}$. The denominator does not factor with real numbers. Complete the square: $s^2+4s+13=(s^2+4s+4)+9=(s+2{)}^2+3^2$. Rewrite the numerator to match the shift: $s+2=(s+2)$. Now the expression becomes: $$F(s)=\frac{s+2}{(s+2{)}^2+3^2}.$$ This now matches the standard form $\frac{s-a}{(s-a{)}^2+{\omega }^2}$, which has an inverse of $e^{at}\cos (\omega t)$. Here, $a=-2$ and $\omega =3$. Therefore, ${\mathcal{L}}^{-1}\{F(s)\}=e^{-2t}\cos (3t)$.

If the numerator were a constant, you would manipulate it to create the form for a sine function. For example, $\frac{5}{s^2+4s+13}=\frac{5}{(s+2{)}^2+3^2}=\frac{5}{3}\cdot \frac{3}{(s+2{)}^2+3^2}$, whose inverse is $\frac{5}{3}{e}^{-2t}\sin (3t)$.

Common Pitfalls

1. Misapplying Linearity Before Full Decomposition: A common error is to try to take the inverse of a product by taking the inverse of each factor. The linearity property only applies to sums, not products. That is, ${\mathcal{L}}^{-1}\{F(s)G(s)\}\ne {\mathcal{L}}^{-1}\{F(s)\}\cdot {\mathcal{L}}^{-1}\{G(s)\}$. You must first use algebra (like partial fractions) to convert any product-of-transforms form into a sum-of-transforms form.

1. Ignoring the Region of Convergence and Uniqueness Conditions: While Lerch's theorem guarantees uniqueness for continuous functions, it's easy to forget the implicit assumption that $t\ge 0$. The inverse Laplace transform is always defined for $t\ge 0$, and we typically take $f(t)=0$ for $t<0$. Also, when decomposing, ensure your final answer is only stated for $t\ge 0$.

1. Errors in Partial Fraction Decomposition or Completing the Square: A simple algebraic mistake in solving for coefficients or in the completing-the-square step will lead to an entirely incorrect time function. Always double-check your algebra. For completing the square, verify your work by expanding your result: $(s+\alpha {)}^2+{\omega }^2$ should equal the original quadratic $s^2+bs+c$.

1. Overlooking the Need to Make the Expression Proper: The partial fraction method requires $F(s)$ to be a proper rational function. If the numerator degree is equal to or greater than the denominator degree (an improper fraction), you must first perform polynomial long division to get a polynomial plus a proper fraction. The polynomial term corresponds to derivatives of the delta function $\delta (t)$ in the time domain, which are significant in system analysis.

Summary

• The inverse Laplace transform ${\mathcal{L}}^{-1}$ recovers the time-domain function $f(t)$ from its transform $F(s)$. Its linearity is the property that makes decomposition methods possible.
• Lerch's theorem assures the uniqueness of the inverse for continuous functions, meaning your solution is the only correct one.
• The primary strategy is to manipulate $F(s)$ into a sum of terms that appear in a Laplace transform table, using partial fraction decomposition for rational functions and completing the square for quadratics with complex roots.
• Partial fraction decomposition systematically breaks down proper rational functions based on the nature of the denominator's factors (linear, repeated linear, or quadratic).
• The completing the square technique is essential for inverting terms that correspond to damped sinusoidal responses, transforming them into standard forms involving $e^{at}\sin (\omega t)$ and $e^{at}\cos (\omega t)$.
• Avoid critical mistakes by remembering you can only invert sums linearly (not products), always stating your answer for $t\ge 0$, and meticulously checking your algebraic manipulations in decomposition and square completion.