Inverse Laplace Transform Of 1 S 1 2

Inverse Laplace Transform of 1/(s(s+1)^2)

The inverse Laplace transform is a crucial operation in engineering and applied mathematics, allowing us to translate functions in the Laplace domain back to the time domain. This article delves into the detailed calculation and understanding of the inverse Laplace transform of the function 1/(s(s+1)²). We will explore multiple approaches to solving this problem, highlighting the strengths and weaknesses of each method, and providing a comprehensive explanation for a thorough understanding.

Understanding the Problem: 1/(s(s+1)²)

Our objective is to find the inverse Laplace transform of the function F(s) = 1/(s(s+1)²). This function represents a transfer function in the s-domain, common in control systems and signal processing. Finding its inverse Laplace transform, f(t), will give us the corresponding time-domain representation. The presence of a repeated root (s = -1) in the denominator necessitates the use of partial fraction decomposition, a pivotal technique for solving this type of problem.

Method 1: Partial Fraction Decomposition

This is the most common and generally preferred method for solving this type of problem. The process involves decomposing the rational function into simpler fractions whose inverse Laplace transforms are known.

Step 1: Partial Fraction Expansion

We assume the partial fraction decomposition has the form:

1/(s(s+1)²) = A/s + B/(s+1) + C/(s+1)²

where A, B, and C are constants to be determined.

Step 2: Determining the Constants

To find A, B, and C, we can employ several methods. A common approach is to multiply both sides of the equation by the denominator (s(s+1)²):

1 = A(s+1)² + Bs(s+1) + Cs

Now we can solve for A, B, and C using several techniques:

• Method of Convenient Values: Let's choose values of 's' that simplify the equation:

  • If s = 0, then 1 = A(1)² => A = 1
  • If s = -1, then 1 = C(-1) => C = -1

  To find B, we can choose any other value of 's' (say s = 1):

  1 = A(2)² + B(1)(2) + C(1) = 4 + 2B -1 => 2B = -2 => B = -1

• Equating Coefficients: We can expand the equation and equate the coefficients of the powers of 's':

  1 = A(s² + 2s + 1) + B(s² + s) + Cs

  1 = (A + B)s² + (2A + B + C)s + A

  Equating coefficients:

  • s²: A + B = 0
  • s: 2A + B + C = 0
  • Constant: A = 1

  Solving this system of equations yields A = 1, B = -1, and C = -1.

Step 3: Inverse Laplace Transform

Now that we have the partial fraction decomposition:

1/(s(s+1)²) = 1/s - 1/(s+1) - 1/(s+1)²

We can apply the inverse Laplace transform to each term individually:

• L⁻¹{1/s} = 1
• L⁻¹{1/(s+1)} = e⁻ᵗ
• L⁻¹{1/(s+1)²} = te⁻ᵗ

Therefore, the inverse Laplace transform is:

f(t) = 1 - e⁻ᵗ - te⁻ᵗ

Method 2: Convolution Theorem

The convolution theorem provides an alternative approach. It states that the inverse Laplace transform of the product of two functions in the s-domain is the convolution of their individual inverse Laplace transforms in the time domain.

We can rewrite F(s) as a product:

F(s) = 1/s * 1/(s+1)²

Let's define:

• G(s) = 1/s => g(t) = L⁻¹{G(s)} = 1
• H(s) = 1/(s+1)² => h(t) = L⁻¹{H(s)} = te⁻ᵗ

According to the convolution theorem:

f(t) = (g * h)(t) = ∫₀ᵗ g(τ)h(t-τ) dτ = ∫₀ᵗ 1 * (t-τ)e⁻⁽ᵗ⁻τ⁾ dτ

Solving this integral (using integration by parts) yields:

f(t) = 1 - e⁻ᵗ - te⁻ᵗ

This confirms the result obtained using the partial fraction method.

Comparison of Methods

Both the partial fraction decomposition and the convolution theorem lead to the same result. However, the partial fraction decomposition is generally easier and more straightforward for this specific problem. The convolution theorem can be more complex, especially for more intricate functions. The choice of method depends on the complexity of the function and the solver's familiarity with each technique.

Applications and Significance

The inverse Laplace transform of 1/(s(s+1)²) has numerous applications in various fields:

• Control Systems: This function represents the response of a second-order system with a pole at the origin and a repeated pole at s = -1. Analyzing this response helps in designing and tuning control systems for desired performance.

• Signal Processing: It can represent the impulse response of a linear time-invariant system. Understanding the impulse response allows for the prediction of the system's output to any input signal.

• Mechanical Systems: The function could model the displacement or velocity of a damped mechanical system subjected to an impulsive force.

• Electrical Circuits: It can represent the current or voltage response of an RC circuit subjected to a step input.

Conclusion

The inverse Laplace transform of 1/(s(s+1)²) is elegantly solved using partial fraction decomposition, yielding the time-domain function f(t) = 1 - e⁻ᵗ - te⁻ᵗ. The convolution theorem offers an alternative approach, although it might be more computationally intensive for this specific problem. Understanding this transform is crucial in many engineering and scientific disciplines, providing a powerful tool for analyzing and understanding the behavior of dynamic systems. The methods discussed here provide a solid foundation for tackling more complex inverse Laplace transforms encountered in practical applications. Mastering these techniques is essential for anyone working with linear systems and their analysis. Further exploration into advanced techniques, such as the use of tables of Laplace transforms and complex variable theory, can expand the range of problems solvable using these powerful mathematical tools.