Laplace Transform and 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
1. What is meant by inverse Laplace transform?
The inverse Laplace transform is the transformation that takes a function in the frequency domain and transforms it back to a function in the time domain. This transformation is accomplished by 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 this means is that once you have found the Laplace transform of a function, you can use the inverse Laplace transform to find the original function. Additionally, if you have two functions that are integrable on the Lebesgue measure, they will have the same Laplace transform. This information can be used to find solutions to differential equations.
2. Why do we use inverse Laplace transform?
The Laplace transform is used to solve linear differential equations; the inverse Laplace transform is used to solve nonlinear differential equations. This can be understood by thinking of linear differential equations as relations between two continuous variables, x(t) and y(t). An example would be dy/dx=y, for which an inconstant solution could be given with a common substitution. A nonlinear differential equation will have relations between more than two continuous variables, x(t), y(t), and z(t). An example of this would be dx/dy=xz+y, which can also be solved using the Laplace transform. In fact, the solution is given in terms of certain constants, and the system of differential equations is converted into an algebraic equation. By using the inverse Laplace transform, one can convert this algebraic equation back into a system of differential equations. This is tremendously helpful in solving real-world problems.
3. What is the difference between the Laplace transform and the Fourier transform?
The Laplace transform and the Fourier transform is two different ways of solving linear differential equations. The Laplace transform takes the input and transforms it into the frequency domain, while the Fourier transform takes the input and transforms it into the time domain. Additionally, the Laplace transform is only valid for linear differential equations, while the Fourier transform is valid for any type of differential equation. Finally, the Laplace transform is a unilateral transformation, while the Fourier transform is a bilateral transformation. This means that the Laplace transform only affects positive frequencies, while the Fourier transform affects both positive and negative frequencies.
4. How can I use the inverse Laplace transform to solve differential equations?
The inverse Laplace transform is a tool that can be used to solve linear differential equations. To use the inverse transform, one must first find the Laplace transform of the given function and then apply the inverse Laplace transform. The result should be a function in terms of time, which will contain constants as well as an unknown function. In general, this equation can not be solved algebraically and must be solved using numerical methods. However, if the differential equation is of a particular form, then it can be solved algebraically. Additionally, there are certain properties that can be used to simplify the solution process. For more information on how to use the inverse Laplace transform to solve differential equations, our instructors at Vedantu can help you out.
5. What are some of the properties of the Laplace transform?
There are a few important properties of the Laplace transform that can be used to simplify the solution process. These include the following:
The Laplace transform is a linear operator. This means that the Laplace transform of a sum is the sum of the Laplace transforms.
If a function is multiplied by a constant, then the Laplace transform of that function is multiplied by the same constant as well.
The inverse transform of an exponential function equals one divided by that exponential function.
The properties of the Laplace transform can be used to simplify the solution process, but it is important to remember that they are not all applicable to every problem.
6. What is the Inverse Laplace Transform of 1?
First, we have to discuss the unit impulse function:-
In mechanics, the idea of a large force acting for a short time occurs frequently.
Therefore, to deal with such similar ideas, we use the unit impulse function which is also called the Dirac delta function. We can define the unit impulse function by its limiting form of it.
δₑ(t - a) = 1/e , a ≤ t ≤ a + e
Otherwise, 0.
If ϵ → 0, the height of the strip will increase indefinitely and the width will decrease in such a manner that its area is always unity.
Thus the unit impulse function δ(t - a) can be defined as
δ(t - a) = ∞ where t= a
Solving it, our end result would be L⁻¹[1] = δ(t).
7. What is the Main Purpose or Application of 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, Inverse Laplace can basically convert any variable domain back to the time domain or any basic domain for example, from the frequency domain back to the time domain. These properties allow them to be utilized for solving and analyzing linear dynamical systems and optimization purposes. Both Laplace and inverse Laplace transforms can be used to solve differential equations in an extremely easy way.
