Problem 14 - Differential Equations and Linear Algebra

Question

Determine the inverse Laplace transform of \(F.\) $$F(s)=\frac{e^{-3 s}}{s+4}$$.

Step-by-Step Solution

Answer

The inverse Laplace transform of \(F(s) = \frac{e^{-3s}}{s + 4}\) is \(f(t) = \frac{1}{4}(1 - e^{12}e^{-4t}).\)

Step 1: Recognize the Laplace Transform Property

We identify a property of the inverse Laplace transform: the convolution theorem states that if \(F(s) = G(s)H(s)\), then the inverse Laplace transform of F(s) (denoted as \(f(t)\)) is the convolution of the inverse Laplace transforms of G(s) and H(s), i.e., \[f(t) = \mathcal{L^{-1}}\{G(s)H(s)\} = (\mathcal{L^{-1}}\{G(s)\}) * (\mathcal{L^{-1}}\{H(s)\}) = g(t) * h(t).\] In this exercise, the given Laplace transform is \(F(s) = e^{-3s} \cdot \frac{1}{s + 4}\), which is the product of two functions.

Step 2: Determine the Inverse Laplace Transforms of the Functions

Using the Laplace transform table, find the inverse Laplace transforms of the two individual functions: \[\mathcal{L}^{-1}\{e^{-3s}\} = u(t - 3),\] \[\mathcal{L}^{-1}\left\{\frac{1}{s + 4}\right\} = e^{-4t}.\] Here, \(u(t - 3)\) is the unit step function.

Step 3: Apply the Convolution Theorem

We can denote the inverse Laplace transforms found in the previous step as \(g(t) = u(t - 3)\) and \(h(t) = e^{-4t}\), respectively. We will now use the convolution theorem to find the inverse Laplace transform of the given function \(F(s)\): \[f(t) = g(t) * h(t) = \int_0^t g(\tau)h(t - \tau) \, d\tau = \int_0^t u(\tau - 3)e^{-4(t - \tau)} \, d\tau.\]

Step 4: Solve the Integral

To determine the inverse Laplace transform \(f(t)\), we solve the integral: \[f(t) = \int_0^t u(\tau - 3)e^{-4(t - \tau)} \, d\tau.\] For \(\tau - 3 < 0\), the unit step function is equal to zero, so the integral only needs to be evaluated for \(\tau - 3 > 0\), which implies \(\tau > 3\). Then, we have: \[f(t) = \int_3^t e^{-4(t - \tau)} \, d\tau = \frac{1}{4}e^{-4t}\int_3^t e^{4\tau} \, d\tau = \frac{1}{4}e^{-4t}\left[e^{4\tau}\right]_3^t\] \[f(t) = \frac{1}{4}e^{-4t}(e^{4t} - e^{12}) = \frac{1}{4}(1 - e^{12}e^{-4t}).\]

Step 5: Final Answer

The inverse Laplace transform of \(F(s) = \frac{e^{-3s}}{s + 4}\) is: \[f(t) = \frac{1}{4}(1 - e^{12}e^{-4t}).\]

Key Concepts

Convolution Theorem

The Convolution Theorem is a useful tool when working with inverse Laplace transforms. It relates the inverse transform of a product of two Laplace-transformed functions to a convolution operation in the time domain. Here’s how it works: if you have two functions in the Laplace domain, say \(G(s)\) and \(H(s)\), their product \(F(s) = G(s)H(s)\) corresponds to the convolution of their inverse transforms in the time domain. Mathematically, this is expressed as:\[ f(t) = g(t) * h(t) = \int_0^t g(\tau)h(t - \tau) \, d\tau\]Where *\(g(t)\)* is the inverse transformation of *\(G(s)\)* and *\(h(t)\)* that of *\(H(s)\)*. This theorem is incredibly useful because it breaks down a complex product into a manageable integration, giving you a way to find the inverse of troublesome Laplace functions by tackling them piece by piece. The goal is to simplify complex operations through integration instead of direct transformation.

Unit Step Function

The Unit Step Function, often denoted as \(u(t)\), is a function that "turns on" at a certain point in time. It is defined as 0 for \(t < 0\) and 1 for \(t \geq 0\). When the shift \(u(t-a)\) is used, it represents a step function that turns on at time \(t = a\). It is particularly useful in control systems and signal processing when dealing with sudden changes or start-ups of signals.

• \(u(t-a) = 0\) when \(t < a\)
• \(u(t-a) = 1\) when \(t \ge a\)

In the realm of Laplace transforms, the unit step function often appears when a function is delayed by a particular amount of time. Thus, it is particularly used in the inverse transformation process to determine the time-domain response of systems that includes delays. Its presence simplifies calculations by removing contributions before a certain time point, effectively filtering the signals.

Laplace Transform Table

A Laplace Transform Table is an essential reference tool for anyone working with Laplace transforms, as it provides the transforms of common functions and their inverses. For students and engineers, this table is a huge time-saver as it avoids the heavy lifting of computing transforms and inverse transforms directly from their definitions.Some common entries in this table might include:

• \(\mathcal{L}\{1\} = \frac{1}{s}\)
• \(\mathcal{L}\{e^{at}\} = \frac{1}{s-a}\)
• \(\mathcal{L}^{-1}\{\frac{1}{s-a}\} = e^{at}\)
• \(\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}\)

This comprehensive list allows users to quickly match the form of the function they have with the transform or inverse needed, enabling efficient problem solving. The table bridges the gap between the time and frequency domain, making it an indispensable tool for solving differential equations and analyzing systems.