Mathematics.

transforms

Integral Transforms Overview

Calculus II50 minDifficulty6 out of 10

You should know: integral, improper integrals

Overview

An integral transform converts a function f(t) into another function F(s) by integrating f against a kernel k(s,t): T[f](s) = integral k(s,t) f(t) dt. The two most important transforms are: the Laplace transform L{f}(s) = integral_0^inf e^{-st} f(t) dt (used for ODEs, signal processing), and the Fourier transform F{f}(xi) = integral_{-inf}^{inf} f(t) e^{-2*pi*i*xi*t} dt (used for frequency analysis). Transforms convert differential equations to algebraic equations in the transformed domain, solve them, then invert.

Intuition

The Laplace transform converts a time-domain function into a frequency/s-domain function where differential equations become algebraic. To solve y'' + y = sin(t): take Laplace transform, solve for Y(s), then invert to get y(t). It's like working in a different 'language' where calculus is easier -- similar to how logarithms turn multiplication into addition. The Fourier transform decomposes a function into sinusoidal components at different frequencies.

Formal Definition

Definition

Laplace transform: L{f}(s) = integral_0^inf e^{-st} f(t) dt, defined for s > s_0 (abscissa of convergence). Key pairs: L{1} = 1/s, L{e^{at}} = 1/(s-a), L{sin(at)} = a/(s^2+a^2), L{t^n} = n!/s^{n+1}. Differentiation: L{f'} = s*L{f} - f(0). Convolution: L{f*g} = L{f}*L{g} (f*g = integral_0^t f(tau)g(t-tau)dtau). Fourier transform: F{f}(xi) = integral_{-inf}^{inf} f(t) e^{-2*pi*i*xi*t} dt.

L{f}(s)=∫0∞e−stf(t) dt\mathcal{L}\{f\}(s) = \int_0^\infty e^{-st}f(t)\,dt
Laplace transform
L{f′}(s)=s L{f}(s)−f(0)\mathcal{L}\{f'\}(s) = s\,\mathcal{L}\{f\}(s) - f(0)
Transform of derivative
L{eat}=1s−a,L{sin⁡(at)}=as2+a2\mathcal{L}\{e^{at}\} = \frac{1}{s-a},\quad \mathcal{L}\{\sin(at)\} = \frac{a}{s^2+a^2}
Key pairs
f^(ξ)=∫−∞∞f(t) e−2πiξt dt\hat{f}(\xi) = \int_{-\infty}^\infty f(t)\,e^{-2\pi i\xi t}\,dt
Fourier transform

Notation

NotationMeaning
L{f}(s)\mathcal{L}\{f\}(s)Laplace transform of f at s
f^(ξ)\hat{f}(\xi)Fourier transform of f at frequency xi
f∗gf * gConvolution of f and g

Theorems

Theorem 1: Laplace Transform Linearity and Differentiation Rule
Lislinear:Laf+bg=a∗Lf+b∗Lg.Differentiationrule:Lf(n)(s)=sn∗Lf(s)−sn−1∗f(0)−...−f(n−1)(0).ThisconvertsalinearODEwithconstantcoefficientsintoanalgebraicequationinLy(s).L is linear: L{af+bg} = a*L{f} + b*L{g}. Differentiation rule: L{f^{(n)}}(s) = s^n*L{f}(s) - s^{n-1}*f(0) - ... - f^{(n-1)}(0). This converts a linear ODE with constant coefficients into an algebraic equation in L{y}(s).
Theorem 2: Convolution Theorem
L(f∗g)(s)=Lf(s)∗Lg(s)where(f∗g)(t)=integral0tf(tau)∗g(t−tau)dtauistheconvolution.EquivalentlyforFourier:Ff∗g=Ff∗Fg.Thisiswhytransformsconvertconvolutionequationstomultiplicationinthetransformeddomain.L{(f*g)}(s) = L{f}(s) * L{g}(s) where (f*g)(t) = integral_0^t f(tau)*g(t-tau) dtau is the convolution. Equivalently for Fourier: F{f*g} = F{f} * F{g}. This is why transforms convert convolution equations to multiplication in the transformed domain.

Worked Examples

  1. 1

    By linearity: L{3e^{2t}} - L{4sin(3t)} = 3*L{e^{2t}} - 4*L{sin(3t)}.

    3L{e2t}−4L{sin⁡(3t)}3\mathcal{L}\{e^{2t}\} - 4\mathcal{L}\{\sin(3t)\}
  2. 2

    Use pairs: L{e^{at}} = 1/(s-a) and L{sin(at)} = a/(s^2+a^2).

    =3s−2−12s2+9= \frac{3}{s-2} - \frac{12}{s^2+9}

✓ Answer

L{3e^{2t} - 4sin(3t)} = 3/(s-2) - 12/(s^2+9).

Practice Problems

Mediumapplication

Use the Laplace transform to solve y' - 2y = 0 with y(0) = 3.

Common Mistakes

Common Mistake

Assuming the inverse Laplace transform is unique without specifying the function class.

The Laplace transform is injective on functions of exponential order that are piecewise continuous: if L{f} = L{g} and both are piecewise continuous of exponential order, then f = g at all points of continuity. But without such conditions, there can be non-uniqueness. In practice, partial fraction decomposition and a table of transform pairs are used to invert; the result is unique under the piecewise continuous exponential order assumption.

Quiz

The Laplace transform converts a linear ODE with initial conditions into:

Historical Background

Pierre-Simon Laplace (1782) introduced his transform to solve probability problems. Euler had used similar ideas earlier. Oliver Heaviside (1893) developed operational calculus based on the Laplace transform for circuit analysis, introducing the 'D-operator' method. Joseph Fourier (1822) introduced the Fourier transform in his study of heat conduction. Norbert Wiener (1933) and others rigorously developed Fourier analysis in L^2. Today, the Fast Fourier Transform (Cooley-Tukey, 1965) makes the discrete Fourier transform computationally tractable.

  1. 1782

    Laplace introduces his transform for probability and differential equations

    Pierre-Simon Laplace

  2. 1822

    Fourier introduces the transform bearing his name for heat equation

    Joseph Fourier

  3. 1965

    Cooley and Tukey publish the Fast Fourier Transform (FFT) algorithm

    James Cooley, John Tukey

Summary

  • L{f}(s) = integral_0^{inf} e^{-st}f(t)dt. Converts ODEs to algebraic equations in s.
  • Key pairs: L{1}=1/s, L{e^{at}}=1/(s-a), L{sin at}=a/(s^2+a^2), L{t^n}=n!/s^{n+1}.
  • Differentiation rule: L{f'} = s*L{f} - f(0). Convolution theorem: L{f*g} = L{f}*L{g}.
  • Fourier transform: converts time domain to frequency domain; F{f*g} = F{f}*F{g}.

References

  1. BookBoyce, W.E. and DiPrima, R.C. Elementary Differential Equations. 10th ed. Wiley, 2012.
  2. WebsiteWikipedia -- Laplace transform