Problem 15 - 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. $$F(s)=\frac{1}{s}, \quad G(s)=\frac{1}{s-2}$$

Step-by-Step Solution

Answer

\(h(t) = \frac{1}{2}(e^{2t} - 1)\)

Step 1: Identify F(s) and G(s)

In the given exercise, we are provided with the following Laplace transforms: $$F(s)=\frac{1}{s}, \quad G(s)=\frac{1}{s-2}$$ (a) Convolution Theorem Method:

Step 2: Find inverse Laplace transforms of F(s) and G(s)

Before we can calculate the convolution of f(t) and g(t), we must first find their inverse Laplace transforms: $$f(t) = \mathcal{L}^{-1}\left[ \frac{1}{s}\right], \quad g(t) = \mathcal{L}^{-1}\left[ \frac{1}{s-2}\right]$$ Using standard Laplace transform tables, $$f(t) =1, \quad g(t) = e^{2t}$$

Step 3: Evaluate the convolution of f(t) and g(t)

Now that we have the inverse Laplace transforms of F(s) and G(s), we can express the convolution of f(t) and g(t) as: $$h(t) =f(t) * g(t) = \int_0^t f(\tau)g(t-\tau)d\tau$$ Substitute f(t) and g(t): $$h(t) = \int_0^t (1)(e^{2(t-\tau)})d\tau$$

Step 4: Evaluate the integral

Evaluate the integral to get the final result for the inverse Laplace transform of F(s)G(s): $$h(t) = \int_0^t e^{2(t-\tau)}d\tau = \left[-\frac{1}{2}e^{2(t-\tau)}\right]_0^t = -\frac{1}{2}e^{2(t-t)} + \frac{1}{2}e^{2t} = \frac{1}{2}\left(e^{2t} - 1\right)$$ (b) Partial Fractions Method:

Step 5: Find the product of F(s) and G(s), H(s)

To proceed with the partial fractions method, we first calculate the product of F(s) and G(s): $$H(s)=F(s)G(s) = \frac{1}{s} \cdot \frac{1}{s-2} = \frac{1}{s(s-2)}$$

Step 6: Perform partial fraction decomposition on H(s)

Next, decompose H(s) using partial fractions: $$\frac{1}{s(s-2)}= \frac{A}{s} + \frac{B}{s-2}$$ To find A and B, multiply both sides by the common denominator (s)(s-2), and then set the numerators equal to each other: $$1=A(s-2)+Bs$$ At this point, we can use the method of equating coefficients to find the values of A and B: $$A = \lim_{s\to 0} \frac{1-A(s-2)}{s} = 1$$ $$B = \lim_{s\to 2} \frac{1-B(s-2)}{s-2} = -\frac{1}{2}$$ So the partial fraction decomposition is: $$H(s)=\frac{1}{s} - \frac{1}{2}\frac{1}{s-2}$$

Step 7: Find the inverse Laplace transform of the partial fractions

Now, compute the inverse Laplace transform of the decomposed expression: $$h(t)=\mathcal{L}^{-1}\left[\frac{1}{s} - \frac{1}{2}\frac{1}{s-2}\right]$$ Using standard Laplace transform tables, we obtain the same result as in the convolution method: $$h(t) = 1 - \frac{1}{2}e^{2t} = \frac{1}{2}(e^{2t} - 1)$$ Both methods give us the same result for the inverse Laplace transform of the product F(s)G(s): $$\boxed{h(t) = \frac{1}{2}(e^{2t} - 1)}$$

Key Concepts

Convolution Theorem

The Convolution Theorem is a powerful tool in the realm of Laplace Transforms, allowing us to find the inverse transform of a product of two Laplace-transformed functions. In simple terms, it states that if you have two functions, say \( F(s) \) and \( G(s) \), and you want to find the inverse Laplace Transform of their product \( F(s)G(s) \), you can do that by convolving their inverse transforms in the time domain.

Here’s how you apply it:

• First, identify the functions \( F(s) \) and \( G(s) \) that you're working with.
• Next, find their respective inverse Laplace Transforms, \( f(t) \) and \( g(t) \).
• Then, by integrating from 0 to \( t \), calculate the convolution \( h(t) = f(t) * g(t) \), which is written as \( h(t) = \int_0^t f(\tau) g(t-\tau) d\tau \).

For the given problem, \( F(s)=\frac{1}{s} \) and \( G(s)=\frac{1}{s-2} \). The inverse of \( F(s) \) is \( f(t)=1 \) and for \( G(s) \), \( g(t)=e^{2t} \). Using these, the convolution integrates to yield the final solution: \( h(t) = \frac{1}{2}(e^{2t} - 1) \). Through this method, we can solve many complex product integrals that might otherwise be taxing to handle!

Partial Fraction Decomposition

Partial Fraction Decomposition is a method used to break down complex fractions into simpler, more manageable pieces. It’s especially useful when dealing with rational functions, which are fractions where both the numerator and the denominator are polynomials.

When applying this method to Laplace Transforms:

• First, you express your function as a sum of simpler fractions, where each denominator is a part of the original fraction's denominator. This is known as decomposition.
• Secondly, solve for any unknown constants by equating coefficients or substituting specific values.

In our case, we were given \( H(s) = \frac{1}{s(s-2)} \). We decomposed it into \( \frac{A}{s} + \frac{B}{s-2} \). By solving for \( A \) and \( B \), we found \( A = 1 \) and \( B = -\frac{1}{2} \). This allowed us to write \( H(s) \) as a combination of two simpler fractions. Finally, taking the inverse Laplace Transform of each term gave us the overall solution as \( h(t) = \frac{1}{2}(e^{2t} - 1) \). This method simplifies the process of finding inverse transforms by making the expressions easier to handle.

Inverse Laplace Transform

The inverse Laplace Transform is a technique used to revert back from the frequency domain to the time domain, yielding the original function from its Laplace Transform. This process is fundamental in analyzing systems and solving differential equations in engineering and physics.

How does it work?

• Begin with a Laplace-transformed function \( F(s) \) and carefully choose or calculate functions whose transforms match your equation.
• To find \( f(t) \), consult standard Laplace tables or apply known techniques like partial fractions or convolution.

For example, from the original exercise, finding \( \mathcal{L}^{-1} \left[ \frac{1}{s} \right] \) gives \( f(t)=1 \) and \( \mathcal{L}^{-1} \left[ \frac{1}{s-2} \right] \) results in \( g(t)=e^{2t} \). By combining these techniques—whether through convolution or partial fractions—we returned to the time domain and obtained the inverse solution \( h(t) = \frac{1}{2}(e^{2t} - 1) \).

The inverse Laplace Transform is essential for translating complex frequency-domain problems into more intuitive time-domain solutions, representing systems' actual behavior.