Laplace Transform
Laplace Transform is used to simplify the process of solving differential equations in applications like engineering and physics.
1.0Laplace Transform: Definition and Formula
The technique of the Laplace Transform represents a function of time, f(t), as a function of complex frequency, s, which simplifies many operations like differentiation, integration, and solving differential equations. This is the formula for Laplace Transform.
L{f(t)}=F(s)=∫0∞f(t)e−stdt
Here,
- f(t) is the function of time.
- s is a complex number; here, j is an imaginary unit, and e-st is an exponential factor.
2.0Differentiation in Laplace Transform
Derivative Laplace Transform:
If f(t) has a Laplace F(s), then the derivative of the Transform will be given as:
L{f′(t)}=sF(s)−f(0)
You can easily convert the operation of derivatives in the time domain to simple algebraic operations in the frequency with the help of the above formula.
3.0Applications of Laplace Transform
Application of Laplace Transform in Real Life
- Electrical Engineering: Laplace transforms are considered for modeling circuits, especially in analyzing transient responses of RLC circuits (Resistor, Inductor, and Capacitor).
- Control Systems: In control theory, Laplace Transforms aid in designing and analyzing systems in terms of stability and response time.
- Mechanical Systems: Used in solving differential equations related to vibrations, motion, and forces in mechanical systems.
Application of Laplace Transform in Engineering
- Signal Processing: Laplace Transform is applied when analysing signals and systems, as well as filtering and stability analysis.
- Structural Engineering: Structural models are designed based on the behaviour of structures under loads and forces.
- Fluid Mechanics: Laplace transforms help to solve fluid flow equations and heat conduction in different types of systems.
4.0Differential Equation and Laplace Transform
The Laplace Transforms have an important application in solving differential equations. The differential equation relates any function with its derivatives, but Laplace Transforms simplify this solution process.
How to solve the differential equation with Laplace Transform?
These are the steps that need to be followed for solving linear ordinary differential equations using Laplace Transforms:
- Take the Laplace transform of both sides of the equation.
- Solve the algebraic equation for F(s), the transformed function.
- Take the inverse Laplace to revert to the time domain and obtain the solution f(t).
Example: Find the first-order differential equation:
dtdf(t)+af(t)=0
Solution: In the given equation, apply the Laplace to both sides:
L{dtdf(t)}+L{af(t)}=L{0}
Using the property of Laplace for derivatives, we get:
sF(s)−f(0)+aF(s)=0
Here, F(s) is the Laplace of f(t), and f(0) is the initial.
Now, Rearrange the equation above:
(s+a)F(s)=f(0)orF( s)=s+af(0)
Now, take the Inverse Laplace transform:
F(s)=L−1{s+af(0)}
Here 1/s+a = e-at
Thus,
f(t) = f(0)e-at
5.0The Formulas for Laplace Transform
Laplace Transform of Common Functions
Inverse Laplace Transform
The inverse Laplace is used to get back to the time domain. It is denoted as:
L−1{F(s)}=f(t)
6.0Solved Examples
Problem 1: Find the Laplace transform of f(t) = t3sin(5t).
Solution: The Laplace transform of f(t) is given by:
L{t3sin(5t)} = 0t3sin(5t)e-stdt
From the Laplace transform formula table, we know:
Ltn.sin(at)=n! a(s2+a2)(n+1)
Here, n = 3, and a = 5:
L{t3.sin(5t)}= 3! . 5(s2+52)(3+1)
=3215(s2+25)4
=30(s2+25)4
Problem 2: Find the Laplace transform of f(t)=e2tcos(3t).
Solution: Let’s convert the following equation into Laplace transform standard form.
L{e2tcos(3t)}=0e2tcos(3t)e-stdt
Comparing the equation with the Laplace transform table of formulas:
L{cos(at)ebt}=s-b(s-b)2+a2
Here, a = 3 and b = 2:
L{e2tcos(3t)}=s-2(s-2)2+9
=s-2s2+4+4s+9
=s-2s2+13+4s
Problem 3: Find the Laplace transform of f(t) = t2e3tsin(3t).
Solution: The Laplace transform of f(t) is:
L{t2e3tsin(3t)}=0t2e3tsin(3t)e-stdt
= 0t2sin(3t)e-(s-3)tdt
Comparing the following equation with the Laplace transform table,
Ltn.sin(at) = n! a(s2+a2)(n+1)
Here, n = 2 and a = 3
Lt2.sin(3t) =2!3(s2+9)(2+1)
=213(s2+9)3
=6(s2+9)3