Content Curator
Laplace transform is a transformation that can change one signal into another using a series of rules or equations. It was first formulated by Pierre Simon De Laplace.
The best way to turn differential equations into algebraic equations is by the Laplace transformation.
- The Laplace transformation is crucial in electronics engineering for resolving issues with signals and systems, digital signal processing, and control systems.
- Both the inverse Laplace transformations and Laplace transform properties are used to investigate the dynamic control system.
A function with a finite number of breaks that maintains consistency at infinity is said to be piecewise continuous. The Laplace transform of the function f(t) will be L f(t) or F if the function is piecewise continuous (s). Any signal can be converted using this transform into the frequency domain "s," where the problem's complexity is reduced.
Key Terms: Piecewise Continuous, Differential Equation, Complex Function, Variables, Functions
What is Laplace Transform?
[Click Here for Sample Questions]
If a function has a finite number of breaks and does not blow up to infinity anywhere, it is said to be piecewise continuous.
- Assuming that f(t) is a piecewise continuous function, the Laplace transform is used to define f(t). Lf(t) or F is the symbol for a function's Laplace transform (s).
- When a differential equation is reduced to an algebraic issue, the Laplace transform aids in its solution.
Laplace Transform Formula
[Click Here for Previous Year Questions]
To translate a given derivative function with a real variable t into a complex function with a variable s, the Laplace transform is used. Let f(t) be given for t 0 and suppose that the function meets some later-explained requirements.
The formula for the Laplace transform of f(t), represented by Lf(t) or F(s) is as follows:
Standard notation: Where the notation is clear, we will use an uppercase letter to indicate the Laplace transform, e.g, L(f; s) = F(s).
The one-sided Laplace transform is another name for the Laplace transform that we defined. There is a two-sided version where the integral goes from −∞ to ∞.
Read More:
Laplace Equation
Laplace’s equation, a second-order partial differential equation, claims that the sum of the second-order partial derivatives of f (i.e. the unknown function) is equivalent to zero for the Cartesian coordinates. The two-dimensional Laplace equation for the function f can be expressed as:
The Laplace equation for three-dimensional coordinates is represented as:
Properties of Laplace Transform
[Click Here for Sample Questions]
Assuming that f1 (t) ⟷ F1 (s) and [wherein, the sign “⟷” signifies Laplace Transform]
f2 (t) ⟷ F2 (s), then:
| Type | Property |
|---|---|
| Linearity Property | A f1(t) + B f2(t) ⟷ A F1(s) + B F2(s) |
| Frequency Shifting Property | es0t f(t)) ⟷ F(s – s0) |
| Integration | t∫0 f(λ) dλ ⟷ 1⁄s F(s) |
| Multiplication by Time | T f(t) ⟷ (−d F(s)⁄ds) |
| Complex Shift Property | f(t) e−at ⟷ F(s + a) |
| Time Reversal Property | f (-t) ⟷ F(-s) |
| Time Scaling Property | f (t⁄a) ⟷ a F(as) |
Laplace Transform Table
[Click Here for Previous Year Questions]
The differential equations for various functions can be solved using the Laplace transform table shown below:
| f(t) | L(f(t)) = F(s) | Sl No. | f(t) | L(f(t)) = F(s) |
|---|---|---|---|---|
| 1 | 1/s | 11 | e(at) | 1/(s − a) |
| tn at t = 1,2,3,… | n!/s(n+1) | 12 | tp, at p>-1 | Γ(p+1)/s(p+1) |
| √(t) | √π/2s(3/2) | 13 | t(n-1/2) at n = 1,2,.. | (1.3.5…(2n-1)√π)/(2n s(n+1/2) |
| sin(at) | a/(s2+a2) | 14 | cos(at) | s/(s2+a2) |
| t sin(at) | 2as/(s2+a2)2 | 15 | t cos(at) | (s2-a2)/(s2+a2)2 |
| sin(at+b) | (s sin(b)+ a cos(b)/(s2+a2) | 16 | cos(at+b) | (s cos(b)-a sin(b)/(s2+a2) |
| sinh(at) | a/(s2-a2) | 17 | cosh(at) | s/(s2-a2) |
| e(at)sin(bt) | b/((s-a)2+b2) | 18 | e(at)cos(bt) | (s-a)/((s-a)2+b2) |
| e(ct)f(t) | F(s-c) | 19 | tnf(t) at n = 1,2,3.. | (-1)n Fn s |
| f'(t) | sF(s) – f(0) | 20 | f”(t) | s2F(s) − sf(0) − f'(0) |
Laplace Transform of Differential Equation
[Click Here for Sample Questions]
One of the crucial methods for solving a differential equation is the Laplace transform. Transformations are used to solve a variety of mathematical issues. The inverse transform, on the other hand, is useful to determine the answer to the given problem.
Let's use the Laplace transformation to solve a first-order differential equation:
Consider y’- 2y = e³x and y(0) = -5. Find the value of L(y).
First step of the equation can be solved with the help of the linearity equation:
L(y’ – 2y] = L(e³x)
L(y’) – L(2y) = 1/(s-3)
(because L(eax) = 1/(s-a))
L(y’) – 2s(y) = 1/(s-3)
sL(y) – y(0) – 2L(y) = 1/(s-3)
(Using Linearity property of the Laplace transform)
L(y)(s-2) + 5 = 1/(s-3) (Use value of y(0) ie -5 (given))
L(y)(s-2) = 1/(s-3) – 5
L(y) = (-5s+16)/(s-2)(s-3) …..(1)
here (-5s+16)/(s-2)(s-3) can be written as -6/s-2 + 1/(s-3) using partial fraction method
(1) implies L(y) = -6/(s-2) + 1/(s-3)
L(y) = -6e²x + e³x
Unilateral Laplace Transform
[Click Here for Previous Year Questions]
Unilateral Laplace Transform of an arbitrary signal can be defined as:
| \(F(p)= L |f| (p) = \int_0^\infty f(t) e^{-pt}dt\) |
The Unilateral Laplace Transform of every signal is discovered to be identical to its Bilateral Transform even if it differs from the Bilateral Laplace Transform.
Bilateral Laplace Transform
[Click Here for Sample Questions]
Laplace transforms come in one- and two-sided varieties. A two-sided Laplace transforms with extending limits of integrations is known as a bilateral transform.
By expanding its boundaries, a typical unilateral transform can also be converted to a bilateral Laplace transform. Below is the bilateral transform formula:
| \(F(p)= L |f| (p) = \int_0^\infty f(t) e^{-pt}dt\) |
Inverse Laplace Transform
[Click Here for Previous Year Questions]
The conversion into a function of time is known as the Inverse Laplace Transform. In contrast to Inverse Transform F(t), which is the Inverse Laplace Transform of F, the Laplace Inverse Formula F(s) is the Transform of F(t). This Laplace transform formula looks like this:
Integrable functions may have the same Laplace transform if they have different Lebesgue measures. As a result, the range of transforms has an inverse transform.
Inverse Laplace transformation of the function is the process of taking a complex function F(s) and transforming it into a real-valued function f(t).
If a unique function is continuous on 0 to ∞ limit and has the property of Laplace Transform. This function is a real function with exponential restrictions.
Applications of Laplace Transform
[Click Here for Sample Questions]
Some of the applications of Laplace Transform are:
- It helps to convert complex differential equations to a simpler form with polynomials.
- It helps convert derivatives into multiple domain variables, further helping to convert the polynomials into differential equations by means of the Inverse Laplace transform.
- One of the major applications of Laplace Transform is in the telecommunication field, helping to send signals to both sides of the medium.
- It is used in Electrical Circuit Analysis, System Modeling, Digital Signal Processing, etc.
Laplace Transform in Probability Theory
Laplace transform, in pure and applied probability theory, is expressed as the expected value.
Assuming that X is a random variable that has probability density function, f, then the Laplace transform of f can be expressed in terms of the expectation of,
L{f}(S) = E[e-sX], (also called the Laplace transform of random variable X itself).
Things to Remember
- Laplace Transform is used to simplify complicated differential equations by adding polynomials.
- It is used to translate derivatives into several domain variables and then, using the Inverse Laplace transform, to transform the polynomials back into the differential equation.
- To transfer signals to both sides of the medium, it is utilized in the field of telecommunication. For instance, when signals are transmitted over a phone line, they are first transformed into a time-varying wave before being superimposed on the medium.
- It is also utilized for a variety of technical activities, including system modeling, digital signal processing, and electrical circuit analysis.
Previous Year Questions
- If y(t) is a solution of (1+t)dydt−ty=1 and …[JEE Advanced 2003]
- The equation of the curve satisfying the differential equation ….[AMUEEE 2015]
- If xdy=y(dx+ydy),y(1)=1 and y(x)>0. Then, y(−3) is equal to...[JEE Advanced 2005]
- Which of the following is a correct solution of…..[AMUEEE 2014]
- Let f(x) be differentiable on the interval (0,∞) such that f(1)=1...[AMUEEE 2014]
- If 8√x(√9+√x)dy=(√4+√9+√x)−1dx,x>0 and $….[JEE Advanced 2017]
- A particular solution of dydx=(x+9y)2 when x=0,y=127 is..[COMEDK UGET 2007]
- If the firm employees 25 more workers, then the new level of production of items is…..[COMEDK UGET 2013]
- Let y(x) be a solution of the differential equation ….[JEE Advanced 2015]
- When y=vx , y and x are variables, the differential equation...[JKCET 2016]
- The differential equation which represents the family of curves y=c1ec2xy=c1ec2x, where c1c1 and c2c2 are arbitrary constants is… [AIEEE 2009]
Sample Questions
Ques. What is the property of “Frequency Shifting Property”? (1 mark)
Ans. The property of “Frequency Shifting Property” is es0t f(t)) ⟷ F(s – s0).
Ques. Find the Laplace transform of f(t)=1 − 2e−2t. (2 marks)
Ans. As per the given question,
⇒ F(s) = L{f(t)} = f(t)
Ques. List two applications of Laplace Transform. (2 marks)
Ans. The two applications of Laplace Transform are:
- It helps in the conversion of complex differential equations to a simpler form with polynomials.
- It is used in Electrical Circuit Analysis and System Modeling, among many others.
Ques. Solve the inverse Laplace transform of Y (s) = 2/3 - 5s (3 marks)
Ans. After readjusting the statement,
Y(s) = 23 - 5s = -25. 1s - 35
By the help of linearity,
Ques. Solve the following: f (t) = 6e-5t + e3t + 5t3 - 9. (2 marks)
Ans: 
Ques. Find the Laplace transform of f(t)=t2e−2x cos(3t) (3 marks)
Ans. The function is g(t) = cos(3t)
Thus, h(t) = e-2xcos(3t) = e-2xg(t0
Then, we can say, f(t) = t2h(t)
Hence,
Let G(s) = L{g(t)}, H(s) = L{h(t)}, F(s) = L{f(t)},
Therefore,
Ques. Solve the given differential equation: d2xdt2 + dxdt+x = 0, Given x(0+) = x’ (0+) = 1 (5 marks)
Ans. As we are aware,
Hence,
Check out:
Comments