Laplace transform is an effective method for solving ordinary and partial differential equations, and it has been successful in many applications. These equations describe how certain quantities change over time, such as the current in an electrical circuit, the vibrations of a membrane, or the flow of heat through a conductor. The Laplace transform helps convert these differential equations into simpler algebraic equations. Both the Laplace transform and its inverse are important tools for analyzing dynamic control systems.
In this article, we will cover the Laplace transform, its definition, various properties, solved examples, and its applications in various fields such as electronic engineering for solving and analyzing electrical circuits.
Laplace transform is an integral transform used in mathematics and engineering to convert a function of time f(t) into a function of a complex variable s, denoted as F(s), where s = \sigma+\iota\omega.
Let us assume f(t) is a function, be it a real or complex function of the variable t>0, where t is time. Then, the Laplace transform F(s) of f(t) is the complex function defined for s\in \Complex, given by:
F(s) = \mathcal{L}\{f(t)\} = \int_{0}^{\infty} e^{-st} f(t) \, dt
where, s =\sigma+i\omega.
Standard Notation:
If a function of t is indicated as f(t),g(t), or y(t), then their respective Laplace transforms are represented by F(s), G(s) and Y(s). Besides the notation F(s), we can also use \mathcal{L}\{f(t)\} or \mathcal{L}\{f\}(s).
Function f(t) | Laplace Transform \mathcal{L}\{f(t)\} =F(s) |
|---|
1 | \frac{1}{s};\,s>0 |
t | \frac{1}{s^2};\,s>0 |
t^n\\n = 0,1,2,... | \frac{n!}{s^n+1};\,s>0 |
e^{at} | \frac{1}{s-a};\,s>a |
\sin at | \frac{a}{s^2+a^2};\,s>0 |
\cos at | \frac{s}{s^2+a^2};\,s>0 |
\sinh at | \frac{a}{s^2-a^2};\,s>|a| |
\cosh at | \frac{s}{s^2-a^2};\,s>|a| |
Here are some definitions before delving into the sufficient conditions for the existence of the Laplace transform:
- Sectional Continuity: A function is said to be sectionally or piecewise continuous in an interval t_1\le t\le t_2 if that interval can be subdivided into a finite number of subintervals, in each of which the function is continuous and has finite left and right-hand limits.
- Functions of Exponential Order: If real constants k>0 and \gamma exist such that for all t>N, the function f(t) satisfies the condition |f(t)|\le ke^{\gamma t}, then f(t) is said to be of exponential order \bm\gamma.
If f(t) is sectionally continuous in every finite interval 0\le t\le N and of exponential order \bm\gamma for \bm{t>N}, then its Laplace transform F(s) exists for all \bm{s>\gamma}.
In the following properties, it is assumed that all functions satisfy the conditions for the existence of the Laplace transform.
- Linearity
- Shifting
- Change of Scale
- Laplace Transforms of Derivatives
- Laplace Transforms of Integrals
- Multiplication by t^n
- Division by t
- Laplace Transform of Periodic function
- Behaviour of F(s) as s\to\infty
- Initial value theorem
- Final value theorem
- Convolution theorem for Laplace transform
Linearity
If c_1 and c_2 are two constants, while f_1(t) and f_2(t) are function, and F_1(s) and F_2(s) are their respective Laplace transforms, then:
\begin{aligned}\mathcal{L}\{c_1f_1(t) +c_2f_2(t)\} &= c_1\mathcal{L}\{f_1(t)\}+c_2\mathcal{L}\{f_2(t)\}\\&=c_1F_1(s) +c_2F_2(s)\end{aligned}
Shifting
It constitutes of two properties
- First Shifting property: If \mathcal{L}\{f(t)\}=F(s), then \mathcal{L}\{e^{at}f(t)\}=F(s-a).
- Second Shifting property: If \mathcal{L}\{f(t)\}=F(s) and g(t)=\begin{cases}
f(t-a)\quad&; {t>a}\\
0 \quad&;{t<a}
\end{cases}, then \mathcal{L}\{g(t)\}=e^{-as}F(s).
Change of Scale
If f(t) is a function and F(s) is its Laplce transfrom, and c is a constant, then \mathcal{L}\{f(at)\}=\frac{1}{a}F(\frac{s}{a}).
- If \mathcal{L}\{f(t)\}=F(s), then \mathcal{L}\{f'(t)\}=sF(s)-f(0), where f(t) is continous for 0\le t\le N and of exponential order \gamma, and its derivative f'(t) is sectionally continous for 0\le t\le N.
- If f(t) fails to be continous at t=0 but \lim\limits_{t\to0}f(t)=f(0^+), then \mathcal{L}\{f'(t)\}=sF(s)-f(0^+).
- If f(t) fails to be continous at t=a, then \mathcal{L}\{f'(t)\}=sF(s)-f(0)-e^{-as}\{f(a^+)-f(a^-)\}
- If \mathcal{L}\{f(t)\}=F(s), then \mathcal{L}\{f^{(n)}(t)\}=s^nF(s)-s^{n-1}f(0)-s^{n-2}f'(0)-...-f^{(n-1)}(0), where f(t),\,f'(t),\,f''(t),\,...\,,f^{(n-1)}(t) are continous for 0\le t\le N and of exponential order for t>N while f^{(n)}(t) is sectionally continous for 0\le t\le N.
If \mathcal{L}\{f(t)\} =F(s), then \mathcal{L}\{\int\limits_0^t f(v)dv\}=\frac{F(s)}{s}.
Multiplication by t^n
If \mathcal{L}\{f(t)\} = F(s), then \mathcal{L}\{t^nf(t)\} = (-1)^n \frac{d^n}{ds^n}F(s), where \frac{d^n}{ds^n} denotes the n-th derivative.
Division by t
If \mathcal{L}\{f(t)\} = F(s), then \mathcal{L}\{\frac{f(t)}{t}\} = \int\limits_{s}^{\infty}f(v)dv.
Let a function f(t) is periodic with period T>0, such that f(t+T)=f(t). Then \mathcal{L}\{f(t)\}=\frac{\int\limits_0^T e^{-st}f(t)dt}{1-e^{-sT}}.
Behaviour of F(s) as s\to\infty
If \mathcal{L}\{f(t)\}=F(s), then \lim\limits_{s\to\infty}F(s)=0.
Initial Value Theorem
Let f(t) be a sectional continous with Laplace transform F(s). Then \lim\limits_{s\to\infty}sF(s) = f(0^+), where the limit s\to\infty has to be taken in such a way that real part of s, Re s \to\infty as well.
Final Value Theorem
Let f(t) be a sectional continous with Laplace transform F(s). When f(\infty) =\lim\limits_{t\to\infty}f(t) exists, then \lim\limits_{s\to0}sF(s) = f(\infty), where the limit s\to0 has to be taken in such a way that real part of s, Re(s) \to0 as well.
Let f(t) and g(t) are piecewise continous and of exponential order \gamma with their Laplace transforms F(s)=\mathcal{L}\{f(t)\} and G(s)=\mathcal{L}\{g(t)\}, respectively. Then \mathcal{L}\{f*g\} exists for Re(s)>\gamma and \mathcal{L}\{f*g\}=F(s)\cdot G(s).
If the Laplace transform of a function f(t) is F(s), i.e., if \mathcal{L}\{f(t)\}=F(s), then f(t) is called the inverse Laplace transform of F(s), and we write it symbolically f(t)=\mathcal{L}^{-1}\{F(s)\} where \mathcal{L^{-1}} is called the inverse Laplace transformation operator.
- Example: Find the inverse Laplace transform of F(s)=\frac{1}{s+3}.
We know that \mathcal{L}\{e^{at}\}=\frac{1}{s-a}, s>a. Then, \mathcal{L^{-1}}\{\frac{1}{s-a}\}=e^{at}. Similarly, \mathcal{L^{-1}}\{\frac{1}{s+3}\}=e^{-3t}.
Some special Functions:
- Gamma function
- Bessel's function
- Unit Step function
- Unit Impulse function
- Sine and Cosine integrals
- Step up function
Gamma function
If n>0, then the gamma function is defined by \Gamma(n)=\int_{0}^{\infty} t^{n-1} e^{-t}dt. Here are some properties of the gamma function are:
- \Gamma(n+1)=\Gamma(n)
- \Gamma(\frac{1}{2})=\sqrt{\pi}
- For large values of n, \Gamma(n+1)\sim\sqrt{2\pi n}\,n^{n}e^{-n}.
Bessel functions
The Bessel functions of order n is given by,
J_n(t) = \frac{t^n}{2^n \Gamma(n+1)} \left\{ 1 - \frac{t^2}{2(2n+2)} + \frac{t^4}{2 \cdot 4(2n+2)(2n+4)} - \cdots \right\}
Some important properties of Bessel function are
- If n is a positive integer, then J_{-n}(t)=(-1)^{n}J_n(t).
- J_{n+1}(t)=\frac{2n}{t}J_n(t)-J_{n-1}(t)
- e^{\frac{1}{2} t \left( u - \frac{1}{u} \right)} = \sum_{n=-\infty}^{\infty} J_n(t) u^n
Unit Step function
Unit Step function is defined as, u(t) = \begin{cases} 0 & \text{for } t < 0 \\1 & \text{for } t \ge 0 \end{cases}.
Unit Impulse function
Unit impulse function, also known as the Dirac delta function, is defined as:
\delta(t) = \begin{cases} +\infty & \text{for } t = 0 \\0 & \text{for } t \neq 0 \end{cases}
Sine and Cosine integrals
- The Sine integral, denoted as Si(t), is given by:
\text{Si}(t) = \int_0^t \frac{\sin u}{u}du. - The Cosine integral, denoted as Ci(t), is given by:
\text{Ci}(t) = \int_t^\infty \frac{\cos u}{u}du
Step Up funtion
A step-up function can be represented as a piecewise function that "steps up" by a fixed amount at certain points. For example, a function f(x) is defined as:
f(x) =\begin{cases}0 & \text{for } x < 1, \\1 & \text{for } 1 \leq x < 2, \\2 & \text{for } x \geq 2.\end{cases}
Some laplace transformation of special function are:
f(t) | F(s)=\mathcal{L}\{f(t)\} |
|---|
J_0(at) | \frac{1}{\sqrt{s^2 + a^2}} |
J_n(at) | \frac{(\sqrt{s^2 + a^2} - s)^n}{a^n \sqrt{s^2 + a^2}} |
\sin \sqrt{t} | \frac{\sqrt\pi}{2s^{3/2}}e^{-1/4s} |
\frac{\cos \sqrt{t}}{\sqrt{t}} | \sqrt{\frac{\pi}{s}} e^{-1/4s} |
\mathrm{Si}(t) | \frac{1}{s} \tan^{-1} \frac{1}{s} |
\mathrm{Ci}(t) | \frac{\ln(s^2 + 1)}{2s} |
u(t - a) | \frac{e^{-as}}{s} |
\delta(t) | 1 |
\delta(t-a) | e^{-as} |
The bilateral Laplace transform involves the values of a function for both t<0 and t\ge0. This means that the bilateral Laplace transform is well-suited for non-causal signals and functions. If f(t) is the function and F(s) is its bilateral Laplace transform, then F(s) given by:
F(s) = \int_{-\infty}^{\infty} f(t) e^{-st}dt
Various relation of laplace transform with other transform are:
Let us assume f(t) is a function of the variable t>0, where t is time. Then, the bilateral Laplace transform F(s) of f(t) is the complex function defined for s=\sigma+i\omega, s\in \Complex, given by:
\begin{aligned}F(s) &= \int_{-\infty}^{\infty} f(t) e^{-st}dt\\&=\int_{-\infty}^{\infty} f(t) e^{-\sigma t}e^{-\omega t}dt\end{aligned}
If the Re(s)=0, the Laplace transform becomes similar to the Fourier transform, given by
F(i\omega) = \int_{-\infty}^{\infty} f(t) e^{-i\omega t}dt
where s=i\omega and \omega is a real number.
Read More: Fourier Transform
The Laplace–Stieltjes transform of a function g(t) is defined as:
\mathcal{L}\{g(t)\}=\int_{0}^{\infty}e^{-st}dg(t)
where s is a complex number.
Mellin transform and its inverse are related to the bilateral Laplace transform by a simple change of variables. If, in the Mellin transform
G(s)=\mathcal{M}\{g(\theta)\}=\int\limits_{0}^{\infty}\theta^sg(\theta)\frac{d\theta}{\theta}
\theta becomes equal to e^{-t},
then we get the bilateral Laplace transform.
Z-transform can be seen as a discrete counterpart of the Laplace transform. This means that Z-transform can be converted into Laplace transform of an ideally sampled signal by the substitution of z=e^{sT}. The Z-transform of a discrete-time signal \( x[n] \) is defined as:
X(z) = \sum_{n=-\infty}^{\infty} x[n] z^{-n}
where z is a complex number.
Laplace transform of the sampled signal of x_a(t) is
\begin{aligned}X_a(s) &= \int_{0}^{\infty} x_a(t) e^{-st} dt\\&= \int_{0}^{\infty} \sum_{n=0}^{\infty} x[n] \delta(t - nT) e^{-st} dt\\&= \sum_{n=0}^{\infty} x[n] \int_{0}^{\infty} \delta(t - nT) e^{-st} dt\\&= \sum_{n=0}^{\infty} x[n] e^{-nsT}\end{aligned}
substituting e^{sT}=z we get,
X_a(s)=X(z)|_{\small z=e^{sT}}.
Various application of Laplace Transform includes:
- Laplace transform can be used to find the transfer function of linear time-invariant continuous-time systems: In linear time-invariant continuous-time systems (LTI systems), h(t) is the impulse response. The system function or transfer function H(s) of the LTI system is the Laplace transform of h(t).
- Laplace transform can be used to solve differential equation problems, including initial value problems. In an initial value problem, the solution to a differential equation is determined by the initial conditions of the system, such as the initial values of the function and its derivatives.
- Let H(s) be the transfer function of a causal LTI system described by the following differential equation,
a_n \frac{d^n y(t)}{dt^n} + a_{n-1} \frac{d^{n-1} y(t)}{dt^{n-1}} + \cdots + a_1 \frac{dy(t)}{dt} + a_0 y(t) = b_m \frac{d^m x(t)}{dt^m} + b_{m-1} \frac{d^{m-1} x(t)}{dt^{m-1}} + \cdots + b_1 \frac{dx(t)}{dt} + b_0 x(t)
satisfying the following condition of initial rest,
y(0)=y'(0)=y''(0)=...=y^{m-2}(0)=y^{m-1}(0)=0 Then the system is stable if and only if the poles of H(s) lie in the left half-plane Re(s)<0.
- Laplace transform can be applied to analyze electrical circuits, simplifying the process of solving circuits with capacitors, inductors, and resistors by converting the time-domain equations into s-domain equations.
- Laplace transform is used in probability theory to find the distribution of sums of random variables and to solve problems related to stochastic processes. For example: Transforming Probability Density Functions (PDFs). Laplace transform can be used to transform the probability density function (PDF) of a random variable. For a non-negative random variable X with PDF f_X(x), the Laplace transform is:
\mathcal{L}\{f_X(x)\} = F(s) = \int_0^\infty e^{-sx} f_X(x) \, dx.
Find the Laplace transform of the following functions.
e^{-2t}.
\mathcal{L}\{e^{-2t}\} = \int_0^\infty e^{-st} e^{-2t} \, dt = \int_0^\infty e^{-(s+2)t} \, dt = \frac{1}{s+2}, for s>-2.
e^{(2+3\iota)t}.
\mathcal{L}\{e^{(2+3i)t}\} = \int_0^\infty e^{-st} e^{(2+3i)t} \, dt = \int_0^\infty e^{-(s-(2+3i))t} \, dt = \frac{1}{s-(2+3i)}, for Real part of s>2
\sin 3t.
\begin{aligned}F(s) = \mathcal{L}\{f(t)\} &= \int_{0}^{\infty} e^{-st} f(t) dt\\\mathcal{L}\{\sin(3t)\} &= \int_{0}^{\infty} e^{-st} \sin(3t) dt\\\end{aligned}
Using Laplace transform property for sine functions:
\begin{aligned}\mathcal{L}\{\sin(at)\} &= \frac{a}{s^2 + a^2}, \quad \text{where } s > 0\\\mathcal{L}\{\sin(3t)\} &= \frac{3}{s^2 + 9}\end{aligned}
Therefore, the Laplace transform of f(t) = \sin(3t) is \frac{3}{s^2 + 9}.
Conclusion
In this article, we have discussed the Laplace transform, the conditions for its existence, and its various properties. We have also solved some examples based on these properties. Additionally, we have examined the Laplace transform of elementary functions, such as sine and cosine functions.
Furthermore, we have explored its applications, such as analyzing electrical circuits by converting them to the s-domain, solving differential equations and initial value problems, and finding the distribution of sums of random variables.
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