Worked Problems: Inverse Laplace Transform with Repeated and Irreducible Factors

📚 Understanding Inverse Laplace Transforms

The inverse Laplace transform is a powerful tool for solving linear differential equations. It allows us to convert a function in the s-domain (the result of a Laplace transform) back to a function in the time-domain. When dealing with rational functions containing repeated and irreducible factors, the process involves partial fraction decomposition and recognizing standard inverse Laplace transforms.

🕰️ History and Background

The Laplace transform was developed by Pierre-Simon Laplace in the late 18th century. It became a fundamental technique in engineering and physics for analyzing dynamic systems. The inverse Laplace transform is crucial for obtaining time-domain solutions from frequency-domain representations.

🗝️ Key Principles

• 🔍 Partial Fraction Decomposition: The first step is to decompose the rational function into simpler fractions. The form of the decomposition depends on the nature of the factors in the denominator.
• 🧱 Repeated Factors: For a repeated factor $(s-a)^n$, include terms of the form $\frac{A_1}{s-a} + \frac{A_2}{(s-a)^2} + ... + \frac{A_n}{(s-a)^n}$.
• 📈 Irreducible Quadratic Factors: For an irreducible quadratic factor $as^2 + bs + c$, include a term of the form $\frac{As + B}{as^2 + bs + c}$. Complete the square in the denominator to facilitate finding the inverse Laplace transform.
• 🧮 Standard Inverse Transforms: Recognize common Laplace transform pairs. For example, $\mathcal{L}^{-1}\left{\frac{1}{s-a}\right} = e^{at}$ and $\mathcal{L}^{-1}\left{\frac{n!}{(s-a)^{n+1}}\right} = t^n e^{at}$.

📝 Worked Example 1: Repeated Factor

Find the inverse Laplace transform of $F(s) = \frac{2s + 1}{(s-1)^3}$.

1. 🧩 Decomposition: $\frac{2s + 1}{(s-1)^3} = \frac{A}{s-1} + \frac{B}{(s-1)^2} + \frac{C}{(s-1)^3}$.
2. ➕ Solve for Coefficients: Multiplying through by $(s-1)^3$ gives $2s + 1 = A(s-1)^2 + B(s-1) + C$. Setting $s = 1$ yields $C = 3$. Differentiating once gives $2 = 2A(s-1) + B$, so $B = 2$. Differentiating again gives $0 = 2A$, so $A = 0$.
3. ✅ Inverse Transform: Thus, $F(s) = \frac{2}{(s-1)^2} + \frac{3}{(s-1)^3}$. Therefore, $f(t) = \mathcal{L}^{-1}{F(s)} = 2te^{t} + \frac{3}{2}t^2e^{t}$.

🧪 Worked Example 2: Irreducible Quadratic Factor

Find the inverse Laplace transform of $F(s) = \frac{3s + 5}{s^2 + 2s + 5}$.

1. 🏗️ Complete the Square: $s^2 + 2s + 5 = (s + 1)^2 + 4 = (s + 1)^2 + 2^2$.
2. ✍️ Rewrite the Numerator: $3s + 5 = 3(s + 1) + 2$. Thus, $F(s) = \frac{3(s + 1) + 2}{(s + 1)^2 + 2^2} = 3\frac{s + 1}{(s + 1)^2 + 2^2} + \frac{2}{(s + 1)^2 + 2^2}$.
3. ✅ Inverse Transform: $f(t) = \mathcal{L}^{-1}{F(s)} = 3e^{-t}\cos(2t) + e^{-t}\sin(2t)$.

💡 Tips and Tricks

• 🔢 Practice: Work through numerous examples to build familiarity.
• 📚 Tables: Keep a table of common Laplace transform pairs handy.
• ✔️ Verification: Check your answer by taking the Laplace transform of your result to see if you get back the original function in the s-domain.

🌍 Real-World Applications

Inverse Laplace transforms are used extensively in electrical engineering (circuit analysis), mechanical engineering (vibration analysis), and control systems. For instance, determining the time-domain response of a circuit to a step input requires the inverse Laplace transform.

⭐ Conclusion

Mastering inverse Laplace transforms with repeated and irreducible factors requires a solid understanding of partial fraction decomposition and familiarity with common Laplace transform pairs. By practicing various examples, you'll develop the skills needed to solve complex problems in engineering and physics.