Problem 25 - Differential Equations and Linear Algebra

Question

Determine the inverse Laplace transform of \(F.\) $$F(s)=\frac{2 e^{-2 s}}{(s-1)\left(s^{2}+1\right)}$$.

Step-by-Step Solution

Answer

The inverse Laplace transform of the given function is: \(f(t)=2\Big(u(t-2)\ast\frac{1}{2}e^{t}u(t)+u(t-2)\ast cos(t)u(t)-u(t-2)\ast sin(t)u(t)\Big)\)

Step 1: Rewrite the given fraction in the correct form for partial fraction decomposition

Write the given function as the sum of a shifted term and a non-shifted term: \(F(s)=2 e^{-2 s}\frac{1}{(s-1)(s^{2}+1)}\)

Step 2: Perform partial fraction decomposition

Decompose the non-shifted term into partial fractions: \(\frac{A}{s-1}+\frac{Bs + C}{s^2+1}\) First, find the constant A: \((s-1)\frac{1}{(s-1)(s^{2}+1)}=\frac{A}{s-1}\) At s = 1, we have: \(A=\frac{1}{1^2 + 1}= \frac{1}{2}\) To find B and C, rewrite the decomposed function and combine terms: \(\frac{1}{(s-1)(s^{2}+1)} = \frac{1/2}{s-1} + \frac{Bs + C}{s^2+1}\) \(\frac{1}{(s-1)(s^{2}+1)} = \frac{1}{2}\frac{2s^2 - 2s + 2 + 2Bs + 2C}{(s-1)(s^{2}+1)}\) Equating the coefficients, we have: - For s^2: 0 = 2 - 2B - For s: 0 = -2 - 2C - For constants: 1 = 2 + 2C Solving these equations, we find: B = 1, C = -1 Now, plug back the values, and rewrite the decomposed function: \(F(s)=2 e^{-2 s}\left(\frac{1/2}{s-1}+\frac{s -1}{s^2+1}\right)\)

Step 3: Perform the inverse Laplace transform

Now, apply the inverse Laplace transform to each term: \(f(t) =2[u(t-2) \ast \mathcal{L}^{-1}\left(\frac{1/2}{s-1}+\frac{s -1}{s^2+1}\right)\] \(f(t) = 2[u(t-2) \ast \left(\frac{1}{2}e^{t}\Theta(t)+ cos(t)\Theta(t)-sin(t)\Theta(t)\right)\]

Step 4: Calculate the inverse Laplace transform with the impulse function

Apply the impulse function using the convolution property of the Laplace transform: \(f(t)=2\Big([u(t-2)\ast\frac{1}{2}e^{t-u(t)}u(t)]+[u(t-2)\ast cos(t)u(t)]-[u(t-2)\ast sin(t)u(t)]\Big)\) Finally, our inverse Laplace Transform is: \(f(t)=2\Big(u(t-2)\ast\frac{1}{2}e^{t}u(t)+u(t-2)\ast cos(t)u(t)-u(t-2)\ast sin(t)u(t)\Big)\)

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a crucial technique for breaking down complex rational expressions into simpler terms. This method makes it easier to apply mathematical operations like the Laplace transform. In our exercise, we started with the function:\[ F(s)=\frac{2 e^{-2 s}}{(s-1)(s^2+1)} \]The goal was to express \( F(s) \) as a sum of simpler fractions. Here's how you can do it:

• Identify each factor of the denominator. In this case, we have \( (s-1) \) and \( (s^2+1) \).
• Write the expression as a sum of fractions where each denominator is one of these factors: \[ \frac{A}{s-1} + \frac{Bs + C}{s^2+1} \]
• Solve for constants \( A \), \( B \), and \( C \) by substituting convenient values of \( s \) and equating coefficients of corresponding powers of \( s \).

Using these steps, we found that \( A = \frac{1}{2} \), \( B = 1 \), and \( C = -1 \). Substituting back gives us the decomposed form. This decomposition simplifies the process of finding the inverse Laplace transform.

Convolution Theorem

The Convolution Theorem is a powerful tool in the context of Laplace transforms, allowing you to handle products of transforms more easily. The theorem states that the inverse Laplace transform of a product of two Laplace transforms \( F(s) \cdot G(s) \) is the convolution of their respective inverse transforms:\[ g(t) = f_1(t) \ast f_2(t) \]Convolution is essentially a process of integrating the product of two functions with a sliding parameter. In our problem, after simplifying using partial fraction decomposition, the function to transform was tackled through convolution. Here's how it helps:

• It allows for handling of the multiplication involved in the time-shifting theorem due to \( e^{-2s} \).
• Convolution separates the procedural difficulty of transforming complex signals, breaking it down into manageable pieces related to simpler functions.

The Convolution Theorem transforms what can be a difficult integration problem into a more straightforward product, simplifying complex calculations significantly.

Impulse Function

The impulse function, often denoted by \( \delta(t) \), is a significant concept in engineering and control theory, especially in the context of Laplace transforms. It represents a theoretical pulse of infinite height and infinitesimal duration, essentially acting like a 'kick' to a system at \( t=0 \). This function is useful in simplifying operations involving convolutions. In our exercise, the use of the unit step function \( u(t-2) \) rather than the impulse function was key:

• The unit step function shifts the resulting waveform, modeling a time delay (here, by 2 units).
• When convolved with other functions, it influences their start time, unaffected by other operations.

In practical terms, while the impulse function itself doesn't appear directly in the calculations here, understanding its role and relation to the unit step function helps in manipulating signals and simplifying inverse transformations in Laplace analysis.

Laplace Transform Properties

The Laplace transform is governed by several properties that streamline complex mathematical operations involving differential equations. These properties include linearity, time-shifting, frequency-shifting, scaling, and the convolution property. In our problem, these principles were pivotal:

• **Linearity**: This property allowed us to decompose \( F(s) \) into simpler parts that could each be inversely transformed independently. Each part was scaled appropriately.
• **Time-Shifting**: The exponential term \( e^{-2s} \) translated into a delay in time of 2 units in the transform inverse. This introduced the \( u(t-2) \) function, effectively shifting the response signal.
• **Convolution**: Enabled the inverse transform of the product, given that each part's inverse is simpler. It bridges the transform property with practical signal response.

Understanding these properties is essential for efficiently solving complex differential equations and finding inverse Laplace transforms. They reduce the complexity, making it manageable to solve using straightforward steps.