# 라플라스 변환

둘러보기로 가기 검색하러 가기

## 개요

• 푸리에 변환의 변형
• 어떤 미분방정식들의 해를 대수적 조작을 통해 얻을 수 있게 해주는 변환
• 라플라스 변환을 미분방정식에 응용한 사람은 Oliver Heaviside http://en.wikipedia.org/wiki/Oliver_Heaviside 이다
• operational calculus 또는 Heaviside calculus 의 도구

## 정의

• 함수 $$f(t)$$에 대한 라플라스 변환을 다음과 같이 정의함$F(s) = \mathcal{L} \left\{f(t)\right\}=\int_0^{\infty} e^{-st} f(t) \,dt$

## 성질

• 함수 $$f$$에 대한 도함수의 라플라스 변환은 다음과 같다$\mathcal{L}\left\{\frac{df}{dt}\right\} = s\cdot\mathcal{L} \left\{ f(t) \right\}-f(0)$

(정리)

$$f$$가 유계이고, $$t\geq 0$$에서 조각적 연속(piecewise continuous)라 하자.

$$\mathfrak{R}(s)\geq 0$$에서 정의된 함수 $$F(s) = \mathcal{L} \left\{f(t)\right\}=\int_0^{\infty} e^{-st} f(t) \,dt$$ 가 $$\mathfrak{R}(s)\geq 0$$에서 해석함수로 확장되면,

$$\int_0^{\infty} f(t) \,dt$$이 존재하고, $$F(0) = \int_0^{\infty} f(t) \,dt$$가 성립한다.

## 예

$$\left(\frac{t^ne^t}{n!}\right)'=\frac{t^{n-1}e^t}{(n-1)!}+\frac{t^ne^t}{n!}$$ 로부터 $$\mathcal{L}\left\{\frac{t^{n-1}e^t}{(n-1)!}\right\} = (s-1)\cdot\mathcal{L} \left\{ \frac{t^ne^t}{n!}\right\}$$

$$\mathcal{L}\left\{e^t\right\} = \frac{1}{s-1}$$

$$\mathcal{L}\left\{t e^t\right\} = \frac{1}{(s-1)^2}$$

$$\mathcal{L}\left\{\frac{t^2 e^t}{2!}\right\} = \frac{1}{(s-1)^3}$$

$$\mathcal{L}\left\{\frac{t^3 e^t}{3!}\right\} = \frac{1}{(s-1)^4}$$

...

## 상수계수 미분방정식에의 응용

• $$y''(t)-2 y'(t)+y(t)=e^t$$
• 양변에 라플라스 변환을 취하면,$s^2 Y(s)+Y(s)-2 (s Y(s)-1)-s+1=\frac{1}{s-1}$, 여기서 $$Y(s)=\mathcal{L} \left\{ f(t) \right\}$$.
• $$Y(s)=\frac{1}{s-1}-\frac{2}{(s-1)^2}+\frac{1}{(s-1)^3}$$
• $$y(t)=e^t-2t e^t+\frac{t^2}{2}e^t$$ 는 주어진 미분방정식의 해가 된다

## 멜린변환과의 관계

• 푸리에 변환 항목 참조$\hat{f}(s)= \int_{0}^{\infty} f(x) x^{s}\frac{dx}{x}$
• 멜린변환에서 $$x=e^{-t}$$로 변수를 치환하면, 라플라스 변환을 얻는다$\int_{0}^{\infty} f(e^{-t}) e^{-st}\,dt$

## 노트

### 말뭉치

1. This limit emphasizes that any point mass located at 0 is entirely captured by the Laplace transform.[1]
2. When one says "the Laplace transform" without qualification, the unilateral or one-sided transform is usually intended.[1]
3. The Laplace transform can be alternatively defined as the bilateral Laplace transform, or two-sided Laplace transform, by extending the limits of integration to be the entire real axis.[1]
4. Two integrable functions have the same Laplace transform only if they differ on a set of Lebesgue measure zero.[1]
5. Laplace transforms and Fourier transforms are probably the main two kinds of transforms that are used.[2]
6. As we will see in later sections we can use Laplace transforms to reduce a differential equation to an algebra problem.[2]
7. Laplace transforms can also be used to solve IVP’s that we can’t use any previous method on.[2]
8. For “simple” differential equations such as those in the first few sections of the last chapter Laplace transforms will be more complicated than we need.[2]
9. The Laplace transform is particularly useful in solving linear ordinary differential equations such as those arising in the analysis of electronic circuits.[3]
10. The inverse Laplace transform is known as the Bromwich integral, sometimes known as the Fourier-Mellin integral (see also the related Duhamel's convolution principle).[3]
11. A table of several important one-sided Laplace transforms is given below.[3]
12. The Laplace transform has many important properties.[3]
13. The Laplace transform is used to quickly find solutions for differential equations and integrals.[4]
14. Like all transforms, the Laplace transform changes one signal into another according to some fixed set of rules or equations.[5]
15. In general, the Laplace transform is used for applications in the time-domain for t ≥ 0.[6]
16. The Laplace transform generates nonperiodic solutions.[6]
17. We will illustrate the usability of the Laplace transform in section 8.2.5 where we discuss an example using the Laplace transform to solve an ODE.[6]
18. In section 8.3.7 we will use the Laplace transform for solving a PDE.[6]
19. As mentioned in another answer, the Laplace transform is defined for a larger class of functions than the related Fourier transform.[7]
20. Fourier transforms are often used to solve boundary value problems, Laplace transforms are often used to solve initial condition problems.[7]
21. Also, the Laplace transform succinctly captures input/output behavior or systems described by linear ODEs.[7]
22. The Laplace transform 3{6 sinusoid: rst express f (t) = cos ![8]
23. The Laplace transform f(p), also denoted by L{F(t)} or Lap F(t), is defined by the integral involving the exponential parameter p in the kernel K = e−pt.[9]
24. Therefore, sin kx and cos kx each have a Laplace transform, since they are continuous and bounded functions.[10]
25. Furthermore, any function of the form e kx , as well as any polynomial, is continuous and, although unbounded, is of exponential order and therefore has a Laplace transform.[10]
26. Therefore, it has a Laplace transform.[10]
27. This example introduces the idea of the inverse Laplace transform operator,, L −1.[10]
28. For a function defined on , its Laplace transform is denoted as obtained by the following integral: where is real and is called the Laplace Transform Operator.[11]
29. In this chapter we will discuss the Laplace transform.[12]
30. The Laplace transform turns out to be a very efficient method to solve certain ODE problems.[12]
31. The Laplace transform also has applications in the analysis of electrical circuits, NMR spectroscopy, signal processing, and elsewhere.[12]
32. Finally, understanding the Laplace transform will also help with understanding the related Fourier transform, which, however, requires more understanding of complex numbers.[12]
33. In this article, we will be discussing Laplace transforms and how they are used to solve differential equations.[13]
34. Many kinds of transformations already exist but Laplace transforms and Fourier transforms are the most well known.[13]
35. The Laplace transforms is usually used to simplify a differential equation into a simple and solvable algebra problem.[13]
36. There is always a table that is available to the engineer that contains information on the Laplace transforms.[13]
37. Compute the Laplace transform of exp(-a*t) .[14]
38. Problems Now that we know how to find a Laplace transform, it is time to use it to solve differential equations.[15]
39. Proof To prove this theorem we just use the definition of the Laplace transform and integration by parts.[15]
40. What this tells us is that if we have a differential equation, then the Laplace transform will turn it into an algebraic equation.[15]
41. Instead we will see that the method of Laplace Transforms tackles the entire problem with one fell swoop.[15]
42. Laplace transforms convert a function f(t) in the time domain into function in the Laplace domain F(s).[16]
43. The domain of its Laplace transform depends on f and can vary from a function to a function.[17]
44. However, the Laplace transform , an integral transform, allows us to change a differential equation to an algebraic equation.[18]
45. We shall define the Laplace transform of a function $$f(t)$$ by \begin{equation*} {\mathcal L}(f)(s)= F(s) = \int_0^\infty e^{-st} f(t) \, dt, \end{equation*} provided the integral converges.[18]
46. However, before we investigate these properties, let us compute several Laplace transforms.[18]
47. The Laplace transform of a function does not always exist, even for functions that are infinitely differentiable.[18]
48. Two weeks ago I did that with the Laplace transform, and now I feel like it finally makes sense to me.[19]
49. Wikipedia sketches this briefly, but I would never have discovered it there because I was looking for that generalization rather than looking for a way to understand the Laplace transform.[19]
50. In the first part, we give an explicit formula for the Laplace transform and verify that this formula satisfies properties (??).[20]
51. In the second part, we compute a table of Laplace transforms for a number of special functions including step functions and impulse functions.[20]
52. Then the function is defined by and is called the Laplace transform of .[20]
53. Note that the Laplace transform is an improper integral — which implies that some care must be taken when discussing its properties.[20]
54. The Laplace transform of a function f (t) is dened for those values of s at which the integral converges.[21]
55. Note that the Laplace transform of f (t) is a function of s. Hence the transform is sometimes denoted L{f (t)}(s), L{f }(s), or simply F (s).[21]
56. You can integrate by parts obtain the Laplace transform of f (t) = t: Integrate by parts n times to get L{tn} = tnest dt (cid:90) 0 = n! sn+1 , for s > 0, and n = 0, 1, 2, . . .[21]
57. As (x) generalizes the factorial, the Laplace transform (8) generalizes (5).[21]
58. CHAPTER 1 Laplace Transform Methods Laplace transform is a method frequently employed by engineers.[22]
59. By applying the Laplace transform, one can change an ordinary dif- ferential equation into an algebraic equation, as algebraic equation is generally easier to deal with.[22]
60. Our rst theorem states when Laplace transform can be performed, Theorem 1.1.[22]
61. The next result shows that Laplace transform is unique in the sense that dierent continuous functions will have dierent Laplace trans- form.[22]
62. The convolution theorem for Laplace transform is a useful tool for solving certain Laplace transforms.[23]
63. An extremely powerful tool that helps us to solve this kind of real world problems are the Laplace transforms.[24]
64. But first let us become familiar with the Laplace transform itself.[24]
65. You can say that we "Laplace transform f from the t-space into F inside of the s-space.[24]
66. We can actually use the linearity in order to find even more new Laplace transforms.[24]

## 메타데이터

### Spacy 패턴 목록

• [{'LOWER': 'laplace'}, {'LEMMA': 'transform'}]