Key Properties and Methods of Inverse Laplace Transform
The Laplace Transform and Inverse Laplace Transform is a powerful tool for solving non-homogeneous linear differential equations (the solution to the derivative is not zero).
The Laplace Transform finds the output Y(s) in terms of the input X(s) for a given transfer function H(s), where s = jω. The inverse Laplace Transform finds the input X(s) in terms of the output Y(s) for a given transfer function H(s), where s = jω.
What is Laplace Transform:
The Laplace Transform is a linear operator on continuous functions. It maps the function's domain onto the complex plane and transforms the function's variables from time-domain to frequency-domain. The inverse of this transformation, also taking place in the complex plane, consists of rotating counterclockwise around a point on the unit circle by 90 degrees and then scaling down by a factor of -1 in the vertical direction.
What is Inverse Laplace Transform:
The derivative of s is jω; when finding what X(s) equals when Y(s) equals zero, we find the inverse function (X(-s)) and remember that multiplying both sides by j gives complex conjugate. We can multiply both sides by j to get rid of the (-s).
The Inverse Laplace Transform takes the output Y(s) and finds what X(s) it is in terms of, for a given transfer function H(s).
Transfer Functions: The transfer function is simply s divided by jω. Since Laplace transforms are linear, the transfer function can be factored into a product of simpler functions.
Laplace Transform and Inverse Laplace Transform
The Inverse Laplace Transform can be described as the transformation into a function of time. In the Laplace inverse formula, F(s) is the Transform of F(t), while in Inverse Transform F(t) is the Inverse Laplace Transform of F(s). Therefore, we can write this Inverse Laplace transform formula as follows:
f(t) = L⁻¹{F}(t) = \[\frac{1}{2\pi i} \lim_{T\rightarrow \infty} \oint_{\gamma - iT}^{\gamma + iT} e^{st} F(s) ds\]
If the integrable functions differ on the Lebesgue measure, then the integrable functions can have the same Laplace transform. Therefore, there is an inverse transform on the very range of transform. The inverse of a complex function F(s) to generate a real-valued function f(t) is an inverse Laplace transformation of the function. If a unique function is continuous on 0 to ∞ limit and also has the property of Laplace Transform. This function is, therefore an exponentially restricted real function.
Inverse Laplace Transform Table
This inverse laplace table will help you in every way possible.
Inverse Laplace Transform Theorems
Theorem 1: When a and b are constant,
L⁻¹ {a f(s) + b g(s)} = a L⁻¹ {f(s)} + b L⁻¹{g(s)}
Theorem 2: L⁻¹ {f(s)} = \[e^{-at} L^{-1}\] {f(s - a)}
Inverse Laplace Transform Examples
Example 1 Compute the inverse Laplace transform of Y (s) = \[\frac{2}{3−5s}\].
Solution Adjust it as follows:
Y(s) = \[\frac{2}{3 - 5s} = \frac{-2}{5}. \frac{1}{s - \frac{3}{5}}\]
Thus, by linearity,
Y(t) = \[L^{-1}[\frac{-2}{5}. \frac{1}{s - \frac{3}{5}}]\]
= \[\frac{-2}{5} L^{-1}[\frac{1}{s - \frac{3}{5}}]\]
= \[\frac{-2}{5} e^{(\frac{3}{5})t}\]
Example 2 Compute the inverse Laplace transform of Y (s) = \[\frac{5s}{s^{2} + 9}\]
Solution Adjust it as follows:
Y (s) = \[\frac{5s}{s^{2} + 9} = 5. \frac{s}{s^{2} + 9}\]
Thus, by linearity,
y(t) = \[L^{-1} [5. \frac{s}{s^{2} + 9}]\]
= \[5 L^{-1} [\frac{s}{s^{2} + 9}]\]
= 5 Cos 3t
Example 3 Compute the inverse Laplace transform of Y (s) = \[\frac{2}{3s^{4}}\].
Solution Adjust it as follows:
Y (s) = \[\frac{2}{3s^{4}} = \frac{1}{9} . \frac{3!}{s^{4}}\]
Thus, by linearity,
y(t) = \[L^{-1} [ \frac{1}{9}. \frac{3!}{s^{4}}]\]
= \[\frac{1}{9} L^{-1} [\frac{3!}{s^{4}}]\]
= \[\frac{1}{9}t^{3}\]
Example 4 Compute the inverse Laplace transform of Y (s) = \[\frac{3s + 2}{s^{2} + 25}\].
Solution Adjust it as follows:
Y (s) = \[\frac{3s + 2}{s^{2} + 25}\]
= \[\frac{3s}{s^{2} + 25}\] + \[\frac{2}{s^{2} + 25}\]
= \[3. \frac{s}{s^{2} + 25} + \frac{2}{5} . \frac{5}{s^{2} + 25}\]
Thus,
\[L^{-1}[3. \frac{s}{s^{2} + 25} + \frac{2}{5} . \frac{5}{s^{2} + 25}]\]
= \[3 L^{-1} [\frac{s}{s^{2} + 25}] + \frac{2}{5} L^{-1} [\frac{5}{s^{2} + 25}]\]
= 3 Cos 5t + \[\frac{2}{5}\] Sin 5t.
Example 5 Compute the inverse Laplace transform of Y (s) = \[\frac{1}{3 - 4s} + \frac{3 - 2s}{s^{2} + 49}\]
Solution Adjust it as follows:
Y (s) = \[\frac{1}{3 - 4s} + \frac{3 - 2s}{s^{2} + 49}\]
= \[\frac{1}{-4} . \frac{1}{s - \frac{3}{4}} + \frac{3}{s^{2} + 49} - \frac{2s}{s^{2} + 49}\]
= \[\frac{1}{-4} . \frac{1}{s - \frac{3}{4}} + \frac{3}{7} . \frac{7}{s^{2} + 49} -2. \frac{s}{s^{2} + 49}\]
Thus,
y(t) = \[L^{-1} [\frac{-1}{4}. \frac{1}{s - \frac{3}{4}} + \frac{3}{7} . \frac{7}{s^{2} + 49} -2. \frac{s}{s^{2} + 49}]\]
= \[-\frac{1}{4} L^{-1} [\frac{1}{s - \frac{3}{4}}] + \frac{3}{7} L^{-1}[\frac{7}{s^{2} + 49}] -2 L^{-1} [\frac{s}{s^{2} + 49}]\]
= \[-\frac{1}{4} e^{(\frac{3}{4})t} + \frac{3}{7} sin 7t - 2 cos 7t\]
Example 6 Compute the inverse Laplace transform of Y (s) = \[\frac{5}{(s + 2)^{3}}\]
Solution The transform pair is:
\[t \Leftrightarrow \frac{2}{s^{3}}\]
According to the proposition,
\[e^{-2t}t^{2} \Leftrightarrow \frac{2}{(s + 2)^{3}}\]
Therefore,
y(t) = \[L^{-1} [\frac{5}{(s + 2)^{3}}]\]
= \[L^{-1} [\frac{5}{2} . \frac{2}{(s + 2)^{3}}]\]
= \[\frac{5}{2} L^{-1} [\frac{2}{(s + 2)^{3}}]\]
= \[\frac{5}{2} e^{-2t}t^{2}\]
Example 7 Compute the inverse Laplace transform of Y (s) = \[\frac{4(s - 1)}{(s - 1)^{2} + 4}\]
Solution The transform pair is:
\[cos 2t \Leftrightarrow \frac{s}{s^{2} + 4}\]
According to the proposition,
\[e^{t} cos 2t \Leftrightarrow \frac{s - 1}{(s - 1)^{2} + 4}\]
Hence,
y(t) = \[L^{-1} [\frac{4(s - 1)}{(s - 1)^{2} + 4}]\]
= \[4 L^{-1} [\frac{s - 1}{(s - 1)^{2} + 4}]\]
= \[4e^{t} cos 2t\]
Conclusion:
The Laplace Transform is an operator that maps the input to the output of a linear differential equation involving derivatives of functions. It transforms variables from time-domain to frequency domain. The inverse of this transformation, also taking place in the complex plane, consists of rotating counterclockwise around a point on the unit circle by 90 degrees and then scaling down by a factor of -1 in the vertical direction. The Inverse Laplace Transform takes the output and finds what X(s) it is in terms of, for a given transfer function H(s). If the integrable functions differ on the Lebesgue measure, then the integrable functions can have the same Laplace transform. Students can use this information to learn more about a topic they are studying.
FAQs on Inverse Laplace Transform Explained
1. What is the inverse Laplace transform?
The inverse Laplace transform is a mathematical process that converts a function from the frequency (s) domain back to the time (t) domain. It helps solve differential equations in engineering and science by moving from transformed solutions back to real-world time-dependent answers.
2. How do you calculate the inverse Laplace transform?
To calculate the inverse Laplace transform, you usually use Laplace transform tables, partial fraction decomposition, or complex integration. Each method finds the original function of time, $f(t)$, from its Laplace counterpart, $F(s)$. Tables are the most common and practical for many problems.
3. Why is the inverse Laplace transform important in engineering?
Inverse Laplace transforms are important in engineering because they allow professionals to analyze and solve linear time-invariant systems. These transforms turn results in the Laplace domain into solutions in the time domain, which describe real-world signals, stresses, and responses.
4. What are the steps for finding the inverse Laplace transform using partial fractions?
To find the inverse Laplace transform using partial fractions:
- Express the Laplace function as a sum of simpler fractions.
- Match each simple fraction with its known inverse from a Laplace table.
- Sum these inverses for the complete time-domain solution.
5. What is the Laplace inverse of $\frac{1}{s+a}$?
The inverse Laplace transform of $ \frac{1}{s+a} $ is $ e^{-a t} $, where $a$ is a constant. This result is useful for solving simple exponential decay or growth problems in time-dependent engineering or physics equations.
6. Can the inverse Laplace transform be used for non-rational functions?
Yes, the inverse Laplace transform can be used on non-rational functions, like those involving exponentials or trigonometric terms. These often require table lookup, integral methods, or advanced techniques, especially when the function does not factor easily into simple fractions.
7. What is the difference between Laplace and inverse Laplace transforms?
Laplace transforms convert functions from the time domain to the complex frequency domain, making problems easier to solve. Inverse Laplace transforms return these solutions back to the time domain, providing answers that describe real-world system behavior.
8. Are there any conditions required for the inverse Laplace transform to exist?
For the inverse Laplace transform to exist, the original function must be piecewise continuous and of exponential order as $t \to \infty$. This ensures the Laplace transform converges, allowing a valid inverse to be found mathematically.
9. What are some common applications of the inverse Laplace transform?
The inverse Laplace transform is widely applied in:
- Electrical circuit analysis
- Mechanical vibrations
- Control systems
- Signal processing
- Solving differential equations
10. How does the inverse Laplace transform help solve differential equations?
The inverse Laplace transform helps solve differential equations by:
- Transforming them into algebraic equations in the Laplace domain
- Solving for $F(s)$
- Converting $F(s)$ back to the time domain using the inverse transform
11. What is the mathematical formula for the inverse Laplace transform?
The inverse Laplace transform is given by the Bromwich integral:
$$f(t) = \frac{1}{2\pi i} \int_{\gamma-i\infty}^{\gamma+i\infty} F(s) e^{st} ds$$
where $F(s)$ is the Laplace transform and $\gamma$ is a real constant to the right of all singularities.
