Question

Find the inverse Laplace transform of the given function. $$ F(s)=\frac{2 e^{-2 s}}{s^{2}-4} $$

Short answer

Answer: The inverse Laplace transform of the given function is \(f(t) = \frac{1}{2}e^{-2(t+2)} - \frac{1}{2}e^{-2(t-2)}\).

Step-by-step solution

Identify the given function and find the roots

We are given the function, $$ F(s) = \frac{2e^{-2s}}{s^2 - 4} $$ First, we need to find the roots of the denominator: \(s^2 - 4 = 0\). This quadratic equation has roots at \(s=2\) and \(s=-2\). We will use these roots while performing the partial fraction decomposition.

Perform partial fraction decomposition

To perform partial fraction decomposition, we can write F(s) as the sum of two fractions with numerators A and B and denominators \((s-2)\) and \((s+2)\), respectively. $$ \frac{2e^{-2s}}{s^2 - 4} = \frac{A}{s-2} + \frac{B}{s+2} $$ Now, we need to find the values of A and B. To do this, we will manipulate the equation and solve for A and B: $$ 2e^{-2s} = A(s+2) + B(s-2) $$ To find A, let \(s = 2\). The equation becomes: $$ 2e^{-4} = A(4) $$ So, we get \(A=\frac{1}{2}e^{-4}\). Similarly, to find B, let \(s = -2\). The equation becomes: $$ 2e^{4} = B(-4) $$ So, we get \(B = -\frac{1}{2}e^{4}\). Now, we rewrite F(s) in terms of A and B: $$ F(s) = \frac{\frac{1}{2}e^{-4}}{s-2} - \frac{\frac{1}{2}e^{4}}{s+2} $$

Apply the inverse Laplace transform

Now apply the inverse Laplace transform to F(s) and use the linearity property (which means we can apply the inverse Laplace transform to each fraction separately and then subtract the results): $$ \mathcal{L^{-1}}\{F(s)\} = \mathcal{L^{-1}}\left\{\frac{\frac{1}{2}e^{-4}}{s-2}\right\} - \mathcal{L^{-1}}\left\{\frac{\frac{1}{2}e^{4}}{s+2}\right\} $$ Applying the inverse Laplace transform, we get: $$ f(t) = \frac{1}{2}e^{-4}e^{2t} - \frac{1}{2}e^{4}e^{-2t} $$ By combining these exponentials, we obtain the inverse Laplace transform of F(s): $$ f(t) = \frac{1}{2}e^{-2(t+2)} - \frac{1}{2}e^{-2(t-2)} $$ And, that's the final result. So, the inverse Laplace transform of the given function is: $$ f(t) = \frac{1}{2}e^{-2(t+2)} - \frac{1}{2}e^{-2(t-2)} $$

Key concepts

Partial Fraction Decomposition

Partial Fraction Decomposition is a technique used to break down complex rational expressions into simpler fractions, making them easier to work with. In this exercise, we needed to find the inverse Laplace transform of a function with a denominator of the quadratic form, \(s^2 - 4\).
To decompose \(F(s)\), we express it as a sum of simpler fractions:

• \(\frac{A}{s-2} + \frac{B}{s+2}\)

By using the roots from the quadratic equation, \(s = 2\) and \(s = -2\), each part of the decomposition simplifies the expression.
By strategically choosing values for \(s\) and substituting back, we find the constants \(A\) and \(B\), which allows us to reframe the original function into simpler parts. This simplification sets the stage for applying the inverse Laplace transform.

Quadratic Equation

In the realm of mathematical equations, a Quadratic Equation is any equation that takes the form \(ax^2 + bx + c = 0\). In our case, the quadratic equation \(s^2 - 4 = 0\) can be factored to find the roots essential for partial fraction decomposition.
Finding the roots involves solving this quadratic equation which is quite straightforward here since the equation is in the format\((s-2)(s+2) = 0\).

• The solutions are \(s = 2\) and \(s = -2\).

The solutions help us in splitting up the fraction correctly. By identifying these roots, partial fractions can be distributed across these linear factors, thus simplifying the inverse Laplace transform process.

Linearity Property

The Linearity Property is a key feature of the Laplace transform and its inverse. This property allows us to transform linear combinations of functions separately and then combine the results. When we look to find \(\mathcal{L}^{-1}\{F(s)\}\), the formula breaks down into smaller pieces:

• \(\mathcal{L^{-1}}\left\{\frac{\frac{1}{2}e^{-4}}{s-2}\right\} - \mathcal{L^{-1}}\left\{\frac{\frac{1}{2}e^{4}}{s+2}\right\}\)

Using linearity simplifies the process significantly, meaning you just solve each segment separately and subtract the results in the end.
This makes complex transforms much easier by handling smaller, more manageable pieces of the function separately.

Exponential Functions

Exponential Functions play a crucial role in the inverse Laplace transform process. The general form of an exponential function is \(ae^{bx}\), and it often appears in the solutions found via Laplace transforms.
In our solution for \(f(t)\), the exponential terms \(e^{2t}\) and \(e^{-2t}\) emerged directly from applying the inverse Laplace transform to each decomposed fraction.

• They represent exponential growth and decay, which are critical components in modeling dynamic systems.

Recognizing these patterns allows us to interpret and find the behavior of the function \(f(t)\), concluding the transformation from the Laplace domain back to the time domain.