Problem 20 - Differential Equations and Linear Algebra

Question

Determine \(L^{-1}[F(s) G(s)]\) in the following two ways: (a) using the Convolution Theorem, (b) using partial fractions. $$F(s)=\frac{1}{s^{2}}, \quad G(s)=\frac{e^{-\pi s}}{s^{2}+1}$$

Step-by-Step Solution

Answer

The inverse Laplace transform of the product F(s)G(s) is given by: $$L^{-1}[F(s)G(s)] = \sin t - \pi t \cos t$$

Step 1: (a) Convolution Theorem

The Convolution Theorem states that if \(F(s) = L[f(t)]\) and \(G(s) = L[g(t)]\), then the inverse Laplace transform of the product is given by: $$L^{-1}[F(s)G(s)] = f(t)* g(t) = \int_{0}^{t} f(\tau)g(t-\tau) d\tau$$ where * denotes convolution. Let's find the inverse Laplace transforms of F(s) and G(s): For F(s): $$L^{-1}\left[\frac{1}{s^2}\right] = f(t) = t$$ For G(s) (using the first translation theorem): $$L^{-1}\left[\frac{e^{-\pi s}}{s^2 + 1}\right] = g(t) = e^{-\pi t}\sin(t)$$ Now, we need to compute the convolution: $$L^{-1}[F(s)G(s)] = \int_{0}^{t} \tau e^{-(t-\tau)}\sin(t-\tau) d\tau$$

Step 2: Solve the Convolution Integral

To solve the convolution integral, we perform integration by parts: Let \(u = \tau\) and \(dv = e^{-(t-\tau)}\sin(t-\tau) d\tau\). Then \(du = d\tau\) and \(v = - e^{-(t-\tau)}\cos(t-\tau)\). Using integration by parts formula (\(\int u dv = uv - \int v du\)), we get: $$L^{-1}[F(s)G(s)] = -\tau e^{-(t-\tau)}\cos(t-\tau)\bigg|_0^t + \int_0^t e^{-(t-\tau)}\cos(t-\tau)d\tau$$ Calculate the bounds: $$L^{-1}[F(s)G(s)] = -t e^{-t}\cos(0) + \int_0^t e^{-(t-\tau)}\cos(t-\tau)d\tau$$ To solve the remaining integral, use substitution: Let \(z = t - \tau\), then \(dz = -d\tau\) and \(\tau = t - z\) Our integral becomes: $$\int_0^t e^{-z}\cos z (-dz) = \int_t^0 e^{-z}\cos z dz$$ Now, we will integrate by parts again: Let \(u = e^{-z}\) and \(dv = \cos z dz\). Then \(du = -e^{-z}dz\) and \(v = \sin z\). Applying the integration by parts formula, we have: $$\int_t^0 e^{-z}\cos z dz = e^{-z}\sin z - \int_t^0 -e^{-z}\sin z dz$$ Now, we will integrate by parts one more time: Let \(u = -e^{-z}\) and \(dv = \sin z dz\). Then \(du = e^{-z}dz\) and \(v = -\cos z\). Applying the integration by parts formula, we get: $$\int_t^0 -e^{-z}\sin z dz = -e^{-z}\cos z - \int_t^0 -e^{-z}\cos z dz$$ Now, we have all the components required to find the inverse Laplace transform of F(s)G(s):

Step 3: Combine the Results

Combine the results to find the inverse Laplace transform: $$L^{-1}[F(s)G(s)] = -t e^{-t} + \left(e^{-t}\sin t - e^{-t}\cos t + C\right)$$ Where \(C\) is the constant of integration. Now, we will continue with the second method.

Step 4: (b) Using Partial Fractions

To use partial fractions, first, we need to find the product of F(s) and G(s): $$H(s) = F(s)G(s) = \frac{1}{s^2} \cdot \frac{e^{-\pi s}}{s^2 + 1} = \frac{e^{-\pi s}}{s^2(s^2 + 1)}$$ Now, apply partial fraction decomposition to H(s): $$\frac{e^{-\pi s}}{s^2(s^2 + 1)} = \frac{A}{s}+\frac{B}{s^2}+\frac{Cs+D}{s^2+1}$$ Our goal is to find the values of A, B, C, and D. Next, we will clear the denominator: $$e^{-\pi s} = A(s^2+1)(s^2)+B(s)(s^2+1)+s(Cs+D)(s^2)$$ Now, solve for A, B, C, and D by comparing the coefficients. We obtain: $$A = 0, B = 0, C = 1, D = -\pi$$ So, our decomposition becomes: $$\frac{1}{s^2(s^2 + 1)} = \frac{1}{s^2+1} - \frac{\pi s}{s^2+1}$$ Now, we can find the inverse Laplace transform L^{-1}[H(s)] for both terms: $$L^{-1}\left[\frac{1}{s^2+1}\right] = \sin t$$ $$L^{-1}\left[-\frac{\pi s}{s^2+1}\right] = -\pi t \cos t$$ Finally, combining the results using linearity property of the Laplace transform, we get: $$L^{-1}[H(s)] = \sin t - \pi t \cos t$$ Both methods yield the same result, and the given exercise is solved.

Key Concepts

Convolution Theorem

The Convolution Theorem is a powerful tool in solving problems involving Laplace transforms. It relates the product of two Laplace transforms to the convolution of their corresponding time functions. This allows us to find inverse Laplace transforms in a straightforward way.

Here's how it works:

• If you have two Laplace transforms, say \( F(s) = L[f(t)] \) and \( G(s) = L[g(t)] \), their product \( F(s)G(s) \) in the frequency domain corresponds to the convolution \( f(t) * g(t) \) in the time domain.
• The convolution is given by an integral: \[ L^{-1}[F(s)G(s)] = \int_{0}^{t} f(\tau)g(t-\tau) \, d\tau \]

In the exercise, by finding the inverse Laplace transforms of \( F(s) \) and \( G(s) \), and then performing the convolution integral, you can determine the solution without reverting to partial fractions or other decompositions. This method is particularly useful when the direct computation of inverse transforms is complex and cumbersome.

Partial Fractions

Partial fraction decomposition is a method to break down complex rational expressions into simpler, more manageable parts. This is especially helpful in inverse Laplace transforms, where simplifying the expression allows for more straightforward calculations.

Here's the general approach:

• Express the function as a sum of simpler fractions. For example, \( \frac{e^{-\pi s}}{s^2(s^2 + 1)} \) can be decomposed into \( \frac{A}{s} + \frac{B}{s^2} + \frac{Cs+D}{s^2+1} \).
• Solve for the coefficients by clearing the denominator and comparing coefficients of like powers of \( s \).
• Once the expression is decomposed, each term's inverse Laplace transform can be easily calculated.

In the exercise, the decomposition facilitated the calculation of the inverse by making each term align with known Laplace transforms such as \( L^{-1}\left[\frac{1}{s^2+1}\right] = \sin t \). This method is crucial when dealing with polynomials or factors that appear in the Laplace domain.

Inverse Laplace Transform

The inverse Laplace transform is a technique to revert from the frequency domain back to the time domain. It is akin to finding the original function given its Laplace transform.

Here's the step-by-step process:

• Identify the given Laplace transform and if possible, match it with standard Laplace pairs that are often found in tables.
• For complex expressions, use methods like partial fraction decomposition to simplify it into recognisable components.
• Calculate each component's inverse Laplace transform separately and use linearity to combine the results.

In this exercise, both the Convolution Theorem and partial fraction decomposition are methods that aid in finding the inverse Laplace transform effectively. Knowing how to use these techniques makes it easier to tackle problems involving differential equations and control systems.