Question

Find the inverse Laplace transform of each of the following expressions: (a) \(\frac{3}{s}-\frac{2}{s^{2}}\) (b) \(\frac{16}{s^{4}}\) (c) \(\frac{3 s}{s^{2}+9}\) (d) \(\frac{s+3}{(s+3)^{2}+1}\) (e) \(\frac{10}{(s+2)^{3}}\)

Short answer

#Answer# (a) \(3 + 2t\) (b) \(\frac{8t^3}{3}\) (c) \(3\cos(3t)\) (d) \(e^{-3t}\cos(3t)\) (e) \(5t^{2}e^{-2t}\)

Step-by-step solution

(a) Inverse Laplace transform of \(\frac{3}{s}-\frac{2}{s^{2}}\)

To find the inverse Laplace transform of this expression, break it down into two simpler expressions: 1. Inverse Laplace transform of \(\frac{3}{s}\) Using the Laplace transform table, we look for the expression corresponding to \(\frac{1}{s}\). We find that the inverse Laplace transform of \(\frac{1}{s}\) is \(u(t)=1\). Now, we multiply by 3 to get the inverse Laplace transform of \(\frac{3}{s}\), which is: \(3u(t)=3\). 2. Inverse Laplace transform of \(\frac{2}{s^{2}}\) Using the Laplace transform table, we look for the expression corresponding to \(\frac{1}{s^{2}}\). We find that the inverse Laplace transform of \(\frac{1}{s^{2}}\) is \(u(t)=t\). Now, we multiply by 2 to get the inverse Laplace transform of \(\frac{2}{s^{2}}\), which is: \(2u(t)=2t\). Finally, add the results from steps 1 and 2 to get the inverse Laplace transform of the whole expression: \(3 + 2t\).

(b) Inverse Laplace transform of \(\frac{16}{s^{4}}\)

Using the Laplace transform table, we look for the expression corresponding to \(\frac{1}{s^{4}}\). We find that the inverse Laplace transform of \(\frac{1}{s^{4}}\) is \(u(t)=\frac{t^3}{6}\). Now, we multiply by 16 to get the inverse Laplace transform of \(\frac{16}{s^{4}}\), which is: \(16u(t)=\frac{8t^3}{3}\).

(c) Inverse Laplace transform of \(\frac{3s}{s^{2}+9}\)

Using the Laplace transform table, we look for the expression corresponding to \(\frac{s}{s^{2}+a^{2}}\). We find that the inverse Laplace transform of this expression is \(u(t)=\cos(at)\). In our case, \(a=3\), so the inverse Laplace transform of \(\frac{s}{s^{2}+9}\) is \(u(t)=\cos(3t)\). Now, we multiply by 3 to get the inverse Laplace transform of \(\frac{3s}{s^{2}+9}\), which is: \(3u(t)=3\cos(3t)\).

(d) Inverse Laplace transform of \(\frac{s+3}{(s+3)^{2}+1}\)

Through the Laplace transform table, we find that the inverse Laplace transform of \(\frac{s-a}{(s-a)^{2}+a^{2}}\) is \(u(t)=e^{at}\cos(at)\). In our case, \(a=3\), so the inverse Laplace transform of \(\frac{s+3}{(s+3)^{2}+1}\) is \(u(t)=e^{-3t}\cos(3t)\).

(e) Inverse Laplace transform of \(\frac{10}{(s+2)^{3}}\)

Using the Laplace transform table, we find that the inverse Laplace transform of \(\frac{1}{(s-a)^{3}}\) is \(u(t)=\frac{t^{n-1}e^{at}}{(n-1)!}\). In our case, \(a=-2\) and \(n=3\), so the inverse Laplace transform of \(\frac{1}{(s+2)^{3}}\) is \(u(t)=\frac{t^{2}e^{-2t}}{2}\). Now, we multiply by 10 to get the inverse Laplace transform of \(\frac{10}{(s+2)^{3}}\), which is: \(10u(t)=5t^{2}e^{-2t}\).

Key concepts

Laplace Transform Table

The Laplace transform table is a crucial tool for engineers and mathematicians working with differential equations and systems control. It is essentially a collection of commonly encountered functions transformed into the Laplace domain, paired alongside their corresponding inverse transforms.

For instance, the function \( u(t) = 1 \), which represents a step function in the time domain, has a Laplace transform equivalent to \( \frac{1}{s} \). Conversely, the inverse Laplace transform of \( \frac{1}{s} \) is the step function \( u(t) = 1 \). The Laplace transform of functions involving \( t^n \) (where \( n \) is a positive integer) translates to factorial expressions in the Laplace domain.

Understanding this table and how to use it is pivotal for exercising problems related to control systems, signal processing, and other areas where complex time domain behaviors need to be analyzed. By matching the given Laplace function with its corresponding inverse in the table, we can effectively move between the time and complex frequency domains—a method highlighted in the provided exercise solutions.

Partial Fraction Decomposition

Partial fraction decomposition is a mathematical technique used to break down complex rational expressions into simpler ones that can be more easily transformed or integrated. It's often employed when working with Laplace transforms to simplify the inverse process.

This technique is essential when dealing with expressions where the denominator consists of a polynomial with a degree higher than the numerator. The decomposition allows for the separation of the complex expression into a sum of simpler fractions, where each part corresponds to a component of the denominator's polynomial factors.

For example, a fraction with a denominator that can be factored into linear terms would decompose into a form where each term's denominator is one of these linear factors. Then, by comparing the coefficients, we solve for the constants that complete the decomposition. In the context of Laplace transforms, after decomposition, each simple fraction can individually be inverted using the Laplace transform table, often resulting directly in exponential or trigonometric time-domain functions.

Exponential Functions

Exponential functions are a fundamental class of mathematical functions, represented as \( f(t) = e^{at} \), where \( e \) is the base of the natural logarithm and \( a \) is a constant. These functions describe processes that change exponentially over time and are particularly common in areas of natural science and finance.

In the realm of Laplace transforms, the appearance of exponential functions is quite typical when dealing with inverse transformations. They often arise from terms like \( \frac{1}{(s-a)} \) where 'a' is a negative real number, indicating a decay process in the time domain. When the Laplace transform involves a negative exponent, it suggests a function that decays over time, as seen with the term \( e^{-at} \) in the time domain.

The ability to relate exponential decay or growth in the time domain to their Laplace counterparts is especially useful in solving differential equations that model physical systems where the concept of decay or growth rate is significant—for example, radioactive decay, population growth, or the discharge of a capacitor in electrical circuits.