Fourier and Laplace Transforms Made Easy for Students
In this topic we have introduced two extremely powerful methods to solving differential equations i.e the Fourier and the Laplace transforms. Besides its practical use, the Fourier transform is also used in quantum mechanics, by providing the correspondence between the position and momentum representations of the Heisenberg commutation relations. An integral transform is very useful as it allows the transformation of a complicated problem into a simpler one. The transformation we will be studying in this chapter is mostly useful to solve differential equations and, to a lesser extent, integral equations. Here we will learn about Fourier transform integral, Laplace transform of integration, and their applications along with real-life applications.
Integral transform is a mathematical operation that produces a new function f(y) after integrating the product of an existing function F(x), and it has a kernel function K(x,y) between the suitable limits. The process is called transformation, and it is symbolically represented by the equation f(y) = ∫K(x, y)F(x)dx. Several transforms are named by the name of mathematicians who introduced them such as the Laplace transform the kernel is \[e^{-xy}\] and the limits of integration are zero and plus infinity. The Fourier transform the kernel is \[(2\prod)^{\frac{-1}{2}}e^{-ixy}\] and the limits are minuses and plus infinity.
What is the Laplace Transform?
A piecewise continuous function is a function that has a finite number of breaks and it does not continue up to infinity anywhere. Let assume the function f(t) is a piecewise continuous function, then f(t) is defined using the Laplace transform. The Laplace transformation of a function is represented by the equation L{f(t)} or F(s). It helps to solve the differential equations, where it reduces the differential equation into an algebraic problem.
Fourier Integral
The representation of a function given on a finite set of intervals by a Fourier series is very important. The representation of a function f given on the whole axis by a Fourier integral is given below:
\[f(x)=\int_{0}^{\infty}\left [ A(\lambda)cos\lambda x + B(\lambda)sin\lambda x\right]d\lambda\]
Where \[A(\lambda)=\frac{1}{\prod}\int_{-\infty}^{\infty}f(\xi)cos\lambda \xi d\xi\]
\[B(\lambda)=\frac{1}{\prod}\int_{-\infty}^{\infty}f(\xi)sin\lambda \xi d\xi\]
Mellin Transform
The Mellin transform is mostly useful for various applications like solving Laplace’s equation in polar coordinates, as well as is used for estimating integrals.
Laplace Transform of Integral
If \[G(s)=\mathcal{L}\left\{g(t)\right\},then\mathcal{L}\left\{\int_{0}^{t}g(t)dt\right\}=\frac{G(s)}{s}\].
For the general integral, if
\[\left [ \int g(t)dt \right ]_{t=0}\]
when the value of the integral t=0, then:
\[\mathcal{L}\left\{\int g(t)dt\right\}=\frac{G(s)}{s}+\frac{1}{s} \left [ \int g(t)dt \right ]_{t=0}\]
Let’s understand this with an example
Find the Laplace transform of \[\int_{0}^{t}sin\,at\,cos\,at\,dt\]
Sol: We know the formula of \[sin\,2\alpha=2 sin\alpha\,cos\alpha\]
We the help of the above formula we can rearrange our integrand, and then integrate
\[sin\,at+cos\,at=\frac{1}{2}sin\,2at\]
So the Laplace Transform of the integral becomes:
\[\mathcal{L}\{\int_{0}^{t}sin\,at\,cos\,at\,dt\}=\frac{1}{2}\mathcal{L}\{\int_{0}^{t}sin\,2at\,dt\}\]
\[=\frac{1}{2}\frac{2a}{s(s^{2}+4a^{2})}\]
\[=\frac{a}{s(s^{2}+4a^{2})}\]
Hence this is the required answer.
Fourier Transform Integral
If the complex function \[g\epsilon L^{2}(R)\] then the function is given by the Fourier integral, i.e.
\[f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}g(k)e^{ikx}dk\]
Exists and \[f\epsilon L^{2}(R)\] Furthermore, we have the equality
\[\int_{-\infty}^{\infty}\left|f(x)\right|^{2}dx=\int_{-\infty}^{\infty}\left|g(k)\right|^{2}dx\],
The function g(k) is called Fourier transform of f(x) and it can be recovered from the
following inverse Fourier integral
\[g(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)e^{-ikx}dx\]
Laplace Stieltjes Transform
When we have to investigate the growth and approximation of entire functions we use Laplace Stieltjes transform. It is convergent on the whole complex plane. For Laplace-Stieltjes transforms,
\[G(s)=\int_{0}^{+\infty}e^{-sx}d\alpha(x),\,s=\sigma +it\]
Where \alpha(x) is a bounded variation on any finite interval [0, Y]\[(0<Y<+\infty)\] and σ and t are real variables. In fact, if \[\alpha(t)\] is absolutely continuous, then G(s) becomes the classical Laplace integral form:
\[G(s)=\int_{0}^{\infty}e^{-st}\varphi(t)dt\]
Fourier Sine Integral
The Fourier integral of an odd function in the interval \[(-\infty,\infty)\] is the sine integral of the function
\[f(x)=\frac{2}{\prod}\int_{0}^{\infty}B(\alpha)sin\alpha x\,d\alpha\]
Where \[B(\alpha)=\int_{0}^{\infty}f(x)sin\alpha x\,dx\]
Laplace Transform Formula
Laplace transform is the integral transform of the given derivative function with respect to real variable t. It is also used to convert into a complex function with variable s. For t ≥ 0, let f(t) be given and assume the function satisfies certain conditions to be stated later on.
The Laplace transform of f(t), that is denoted by L{f(t)} or F(s) is defined by the Laplace transform formula:
\[F(s)=\int_{0}^{\infty}f(t).e^{-s.t}dt\]
whenever the improper integral converges.
Standard notation: Where the notation is clearly known, we use an uppercase letter to indicate the Laplace transform, for e.g, L(f; s) = F(s).
Sometimes we define Laplace transform as the one-sided Laplace transform. Two-sided version Laplace transform is also available where the integral goes from \[(-\infty\,to\,\infty)\].
Integration of Fourier Series
Consider g(x) to be a \[2\prod\] periodic piecewise continuous function on the interval \[\left[-\prod,\prod\right]\]. Then this function can be integrated term by term on this interval. The Fourier series for g(x) is given by
\[g(x)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty}(a_{n}cosnx+b_{n}sinnx)\]
Consider the function
\[G(x)\int_{0}^{x}g(t)dt\sim\frac{A_{0}}{2}+\sum_{n=1}^{\infty}(A_{n}cosnx+B_{n}sinnx)\]
Where \[A_{n}=-\frac{b_{n}}{n},B_{n}=\frac{a_{n}}{n}\]
By setting x=0, we see that
\[G(0)=0=\frac{A_{0}}{2}+\sum_{n=1}^{\infty}A_{n}=\frac{A_{0}}{2}-\sum_{n=1}^{\infty}\frac{b_{n}}{n}\]
or \[\frac{A_{0}}{2}=\sum_{n=1}^{\infty}\frac{b_{n}}{n}\]
Therefore, the Fourier series expansion of the function G(x) is given by
\[G(x)\int_{0}^{x}g(t)dt=\int_{0}^{x}\frac{a_{0}}{2}dx+\sum_{n=1}^{\infty}(a_{n}cosnx+b_{n}sinnx)dx\]
\[=\frac{a_{0}x}{2}+\sum_{n=1}^{\infty}\frac{a_{n}sinnx+b_{n}(1-cosnx)}{n}\]
where the series on the right-hand side is obtained by the formal step-by-step integration of the Fourier series for g(x).
As the presence of the term depending on x, on the right-hand side, this is not clearly a Fourier series expansion of the integral of g(x). The result we can rearrange in the form of Fourier series expansion of the function
\[\Phi(x)=\int_{0}^{x}g(t)dt-\frac{a_{0}x}{2}\]
The Fourier series of the function (x)is given by the expression
\[\Phi(x)=\int_{0}^{x}g(t)dt-\frac{a_{0}x}{2}\]
\[=\frac{A_{0}}{2}+\sum_{n=1}^{\infty}{A_{n}cos\,nx+B_{n}sin\,nx}\]
where the Fourier coefficients are defined by using the following relationships:
\[\frac{A_{0}}{2}=\sum_{n=1}^{\infty}\frac{b_{n}}{n},A_{n}=-\frac{b_{n}}{n},B_{n}=\frac{a_{n}}{n}\]
Inverse Laplace Transform Integral
In the inverse Laplace transform, the transformation of F(s) is given and we have to find what function we have initially. The inverse transform of the function F(s) is given as:
\[f(t)=L^{-1}\left\{{F(s)}\right\}\]
Let’s understand the two Laplace transform. Suppose the two functions are F(s) and G(s), the inverse Laplace transform is defined by:
\[L^{-1}\left\{aF(s)+bG(s)\right\}=aL^{-1}\left\{F(s)+bL^{-1}\{G(s)\right\}\]
Where a and b are constants.
In the above case, we can take the inverse transform for the individual transforms, and then add their constant values in their respective places, that perform the operation to get the result.
Laplace Transform of Integral Function
According to the theorem
If \[L\left\{f(t)\right\}=F(s)\], then
\[L\left[\int_{0}^{t}f(u)du\right]=\frac{F(s)}{s}\]
Conclusion
Integral transforms are very useful in mathematical analysis. It is used to solve differential and integral equations, and to study special functions and compute integrals. Applications of integral transform fractional, including fractional methods.
Laplace transform is used to convert complex differential equations into a simpler form that has polynomials. It is used to convert derivatives into multiple domain variables and finally convert the polynomials back to the differential equation using Inverse Laplace transformation. When we apply Fourier transform to a partial differential equation it reduces the number of independent variables by one. We also use Fourier Transform in signal and image processing. It is also useful in cell phones, LTI systems & circuit analysis.
FAQs on Integral Transform: Definition, Types & Applications
1. What do you mean by an integral transform?
An integral transform is a mathematical operator that converts a function from its original domain (like time) to a new domain (like frequency). It achieves this by integrating the function with a specific, predefined function known as a kernel. This transformation often simplifies complex operations, such as differentiation and integration, into more manageable algebraic operations, making difficult problems easier to solve.
2. What are the most common types of integral transforms used in mathematics and engineering?
The most common and important types of integral transforms have distinct applications and include:
The Laplace Transform, which is widely used in engineering to solve linear ordinary differential equations, particularly for initial value problems.
The Fourier Transform, which is essential for signal processing and image analysis as it decomposes a function into its constituent frequencies.
The Mellin Transform, which is often applied in computer science for the analysis of algorithms and in number theory.
The Hankel Transform, which is used to solve problems that exhibit cylindrical symmetry, such as wave propagation in a drumhead.
3. What are the main applications of integral transforms in the real world?
Integral transforms are fundamental tools across various scientific and engineering fields. Key real-world applications include:
Signal Processing: Analysing and filtering audio, radio, and image signals using the Fourier transform.
Control Systems: Designing and analysing feedback systems in robotics and automation with the Laplace transform to ensure stability and performance.
Electrical Circuit Analysis: Solving for voltage and current in complex RLC circuits by transforming differential equations into simpler algebraic ones.
Image Compression: Algorithms like JPEG use a type of Fourier transform to efficiently compress and store image data.
4. What is the “kernel” of an integral transform and why is it important?
The kernel of an integral transform is the specific function of two variables that is multiplied by the original function inside the integral. It is the core component that defines the transform and its unique properties. For instance, the kernel for the Laplace transform is e-st, while for the Fourier transform, it is e-iωt. The choice of kernel is crucial as it dictates the transform's domain and what kind of problems it is best suited to solve.
5. How does an integral transform actually simplify a complex problem like a differential equation?
An integral transform simplifies a differential equation by converting it into an algebraic equation. This is possible because of a key property where the transform of a derivative becomes a simple algebraic multiplication. For example, the Laplace transform converts the calculus operation of differentiation, d/dt, into an algebraic operation of multiplication by 's'. This turns a complex differential equation into a polynomial equation, which can be solved using standard algebra. Once the solution is found in the transformed domain, an inverse transform is applied to retrieve the solution in its original domain.
6. What is the key difference between the Laplace transform and the Fourier transform?
The primary difference lies in their application and the types of functions they can handle. The Fourier transform is used to analyse the frequency content of continuous, stable signals and is defined along the imaginary axis (jω). In contrast, the Laplace transform is a more general tool defined on a complex plane (s = σ + jω). This allows it to analyse not just the frequency content but also the stability and transient behaviour of a system, making it suitable for a wider range of functions, including those that grow over time.
7. Why do we need integral transforms at all? Can't problems just be solved directly?
While many problems can be solved directly, integral transforms offer a powerful alternative for problems that are extremely difficult or tedious in their original domain. The strategy is to change the problem's domain to a new one where the mathematical operations are simpler. For example, a convolution operation in the time domain becomes a simple multiplication in the frequency domain. This allows us to find the solution easily in the 'simpler world' and then transform it back. It's a strategic detour that makes solving the overall problem much more efficient.
8. Why is the Laplace Transform so effective for solving initial value problems?
The Laplace Transform is exceptionally effective for initial value problems because it automatically incorporates the initial conditions into the transformation process. When you transform a derivative like f'(t), the result is sF(s) - f(0), where f(0) is the initial value. This integrates the starting conditions of the system directly into the algebraic equation from the very beginning, eliminating the need to find a general solution first and then solve for arbitrary constants later.
