Question

Find the inverse Laplace transform of (a) \(\frac{3}{s+2}\) (b) \(\frac{-2}{s-1}\) (c) \(\frac{2}{(s+2)^{3}}\) (d) \(\frac{1}{s}+\frac{1}{s+1}\) (e) \(\frac{1}{(s-3)^{2}}+\frac{1}{(s-3)^{3}}\)

Short answer

Question: Determine the inverse Laplace transforms of the following functions: (a) \(\frac{3}{s+2}\) (b) \(\frac{-2}{s-1}\) (c) \(\frac{2}{(s+2)^{3}}\) (d) \(\frac{1}{s}+\frac{1}{s+1}\) (e) \(\frac{1}{(s-3)^{2}}+\frac{1}{(s-3)^{3}}\) Answer: (a) \(3e^{2t}\) (b) \(-2e^{-t}\) (c) \(\frac{t^2}{2}e^{2t}\) (d) \(1 + e^t\) (e) \(te^{3t} + \frac{t^2}{2}e^{3t}\)

Step-by-step solution

Identify the inverse Laplace transform formula

We can recognize this function as the Laplace transform of \(e^{at}\). That is, the inverse Laplace transform formula is: \(g(t) = e^{-at} \mathcal{L}^{-1}\{F(s)\}(t) = e^{-at}\) In our case, \(a = -2\).

Calculate the inverse Laplace transform

Using the inverse Laplace transform formula, we have: \(g(t) = 3 e^{(-(-2))t} = 3e^{2t}\) So the inverse Laplace transform of \(\frac{3}{s+2}\) is \(3e^{2t}\). (b) \(\frac{-2}{s-1}\)

Identify the inverse Laplace transform formula

Similar to part (a), this function is of the form \(e^{at}\). The inverse Laplace transform formula is: \(g(t) = e^{-at}\) In this case, \(a = 1\).

Calculate the inverse Laplace transform

Substituting \(a = 1\) in the formula, we get: \(g(t) = -2e^{-(1)t} = -2e^{-t}\) So the inverse Laplace transform of \(\frac{-2}{s-1}\) is \(-2e^{-t}\). (c) \(\frac{2}{(s+2)^{3}}\)

Identify the inverse Laplace transform formula

This function is of the form \(\frac{1}{(s-a)^n}\), where \(a = -2\) and \(n = 3\). The inverse Laplace transform formula for this function is: \(g(t) = \frac{t^{n-1}}{(n-1)!}e^{at}\)

Calculate the inverse Laplace transform

Substituting \(a = -2\) and \(n = 3\) in the formula, we get: \(g(t) = \frac{2 t^{3-1}}{(3-1)!}e^{-(-2)t} = \frac{t^2}{2}e^{2t}\) So the inverse Laplace transform of \(\frac{2}{(s+2)^{3}}\) is \(\frac{t^2}{2}e^{2t}\). (d) \(\frac{1}{s}+\frac{1}{s+1}\)

Identify the inverse Laplace transform formula for each term

Both terms in this function are of the form \(e^{at}\). The inverse Laplace transform formula for this function is: \(g(t) = e^{-at}\) For the first term, \(a = 0\) and the second term, \(a = -1\).

Calculate the inverse Laplace transform for each term

For the first term, we have: \(g_1(t) = e^{-0t} = 1\) For the second term, we have: \(g_2(t) = e^{-(-1)t} = e^t\)

Combine the inverse Laplace transforms

Adding \(g_1(t)\) and \(g_2(t)\) together, we get the overall inverse Laplace transform: \(g(t) = 1 + e^t\) So the inverse Laplace transform of \(\frac{1}{s}+\frac{1}{s+1}\) is \(1 + e^t\). (e) \(\frac{1}{(s-3)^{2}}+\frac{1}{(s-3)^{3}}\)

Identify the inverse Laplace transform formula for each term

Both terms in this function are of the form \(\frac{1}{(s-a)^n}\). The inverse Laplace formula for this function is: \(g(t) = \frac{t^{n-1}}{(n-1)!}e^{at}\) For the first term, \(a = 3\) and \(n = 2\). For the second term, \(a = 3\) and \(n = 3\).

Calculate the inverse Laplace transform for each term

For the first term, we have: \(g_1(t) = \frac{t^{2-1}}{(2-1)!}e^{3t} = te^{3t}\) For the second term, we have: \(g_2(t) = \frac{t^{3-1}}{(3-1)!}e^{3t} = \frac{t^2}{2}e^{3t}\)

Combine the inverse Laplace transforms

Adding \(g_1(t)\) and \(g_2(t)\) together, we get the overall inverse Laplace transform: \(g(t) = te^{3t} + \frac{t^2}{2}e^{3t}\) So the inverse Laplace transform of \(\frac{1}{(s-3)^{2}}+\frac{1}{(s-3)^{3}}\) is \(te^{3t} + \frac{t^2}{2}e^{3t}\).

Key concepts

Laplace Transform

The Laplace transform is a mathematical operation that converts a function of a real variable, such as time (often denoted as t), into a function of a complex variable (s). This technique is widely used in engineering and physics to simplify differential equations and study systems' behavior without solving the entire equation explicitly.

The process involves an integral transform, particularly beneficial because it transforms differential equations into algebraic equations which are easier to manipulate and solve. The Laplace transform of a function f(t) is generally represented as F(s), and the transformation itself is defined by the integral:
\[\begin{equation}F(s) = \[5pt] \[5pt] \[5pt] \[5pt] \[5pt] \[5pt] \[5pt] \/mathscr{L}\{f(t)\} = \int_0^\infty e^{-st}f(t) dt.\end{equation}\]
When it comes to finding the inverse Laplace transform, it is a process of determining the original function gain from its transformed counterpart. In educational terms, it's like 'decoding' the solution back to the problem's language.

Exponential Function

At the heart of many inverse Laplace transforms lies the exponential function, often expressed as e^{at}, where e is the base of the natural logarithm, approximately equal to 2.71828, and a is a constant. This function is fundamental because it describes processes involving growth or decay, such as radioactive decay, population growth, and even compound interest in finance.

The exponential function's presence in an inverse Laplace transform often signals a system returning to equilibrium after some perturbation. For instance, in the solution for the provided exercises, the inverse Laplace transform results in functions like 3e^{2t} or -2e^{-t}, indicating processes that grow or decay exponentially over time.

Understanding the behavior of exponential functions is essential for interpreting these solutions within the context of the real-world phenomena they represent.

Transfer Function

The transfer function is a fundamental concept in control theory and systems engineering, which relates the output of a system to its input using Laplace transform. In mathematical terms, it is the Laplace transform of the system's response divided by the Laplace transform of the input under the assumption that all initial conditions are zero.

A transfer function, usually denoted as G(s), provides a compact representation of the system's behavior in the frequency domain. It allows engineers to analyze stability, control, and the frequency response of systems without directly tackling the time domain differential equations.

Within the context of the exercise solutions, one can think of the given Laplace transform functions as simplified versions of transfer functions. The process of finding the inverse Laplace transforms of these functions is somewhat akin to understanding how various system inputs (expressed in the 's' domain) would affect the system's time-domain behavior, which is crucial for designing and analyzing control systems.