Week 32018학년도 2학기
반도체 공학과 공학수학2 (GEDB005) 강의교안 (2018학년도용)
주교재: Erwin Kreyszig, Engineering Mathematics, 10th Edition
부교재: 최신공학수학 I 과 II, 한빛출판사 및 (영문) 다변량 Calculus (이상구, 김응기 et al) (http://www.hanbit.co.kr/EM/sage/)
강의시간: BD1615(화 16:30-17:45/ 목 15:00-16:15), 반도체관 400102호 담당교수: 김응기 박사
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주차 |
주교재 |
부교재 |
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3 |
11.9: Fourier Transform. Discrete and Fast Fourier Transforms 12.1: Basic Concepts of PDEs 12.2: Modeling: Vibrating String, Wave Equation |
11.6 푸리에 변환 12.1 편미분방정식이란 |
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web |
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Week 3
11.9. Fourier Transform. Discrete and Fast Fourier Transforms:
Complex Form of the Fourier Integral
The (real) Fourier integral is
where
, 
.
Substitution 
and 
into the integral for 
, we have
(1) 
is an even function of 
.
.
is an odd function of 
Euler formula
complex Fourier integral
is complex Fourier integral.
Fourier Transform and Its Inverse
(5) 
.
Writing 
, we have
is the Fourier transform of 
is the inverse Fourier transform of 
.
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Remark |
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Other notations for the Fourier transform is
so that
|
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The process of obtaining the Fourier transform 
from a given 
is also called the Fourier transform or the Fourier transform method.
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Theorem 1 |
Existence of the Fourier Transform |
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If |
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Example 1 |
Fourier Transform |
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Find the Fourier transform of |
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Solution
Using Fourier transform and integrating,
.
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Example 2 |
Fourier Transform |
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Find the Fourier transform |
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Solution
From the definition Fourier transform we obtain by integration
.
Linearity. Fourier Transform of Derivatives
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Theorem 2 |
Linearity of the Fourier Transform |
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The Fourier transform is a linear operation ; that is, for any functions (8) |
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Proof
This is true because integration is a linear operation, so that 
gives
.
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Theorem 3 |
Fourier Transform of the Derivative of |
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Let (9) |
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Proof
From the definition of the Fourier transform we have
.
Two successive application of 
give
.
We have for the transform of the second derivative of 
(10) 
.
Similarly for higher derivatives.
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Example 3 |
Application of the Operational Formula (9) |
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Find the Fourier transform of |
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Solution
We use 
, By formula 
in Table III.
.
Convolution
The convolution 
*
of function 
and 
is defined by
(11) 
*
.
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Theorem 4 |
Convolution Theorem |
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Suppose that (12) |
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Proof
By the definition,
*
.
Taking 
.
Hence 
*
This double integral can be written as a product of two integrals and gives the desired result
*
.
By taking the inverse Fourier transform on both sides of 
, writing 
and 
as before, and nothing that in 
and 
cancel each other, we obtain
(13) 
*
,
a formula that will help us in solving partial differential equation.
Chapter 12. Partial Differential Equations (PDEs)
12.1 Basic Concepts of PDEs
Partial Differential Equation(PDE)
An equation involving one or more partial derivatives of an (unknown) function that depends on two or more variables.
Order of the PDE is the order of the highest derivative.
PDEs
Linear : The first degree in the unknown function 
and its partial derivatives
Homogeneous : each of its terms contains either 
or one of its partial derivatives
Nonhomogeneous
Nonlinear
Example 1 Important Second-Order PDEs
(1) 
One-dimension wave equation
(2) 
One-dimension heat equation
(3) 
Two-dimension Laplace equation
(4) 
Two-dimension Poisson equation
(5) 
Two-dimension Wave equation
(6) 
Three-dimension Laplace equation
Here 
is positive constant, 
is time, 
, 
, 
are Cartesian coordinates, and dimension is the number of these coordinates in the equation.
A solution of a PDE in some region 
of the space of the independent variables is a function that has all the partial derivatives appearing in the PDE in some domain 
containing 
, and satisfies the PDE in some everywhere in 
.
In general, the totality of solution of a PDE is very large.
The unique solution of a PDE corresponding to a given physical problem will be obtained by the use of additional conditions arising from the problem.
Addition Conditions
Boundary Conditions
Initial Conditions
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Theorem 1 |
Fundamental Theorem on Superposition |
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If
with any constants |
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Example 2 Solving 
like an ODE
Find solution 
of the PDE 
depending on 
and 
.
Solution
Since no 
derivatives.
The PDE 
, where 
and 
are constant.
Here 
and 
are function of 
.
Answer is
with arbitrary function 
and 
.
Solve 
.
Solution
Characteristic equation is
Characteristic roots are 
and 
General solution is 
.
Example 3 Solving 
like an ODE
Find solution 
of this PDE.
Solution
Setting 
.
separable equation
integration with respect to 
integration with respect to 
.
where 
,
here, 
and 
are arbitrary.
12.2 Modeling: Vibrating String, Wave Equation
Derive the equation modeling small transverse vibrations of an elastic string.
Place the string along the 
-axis, stretch it to length 
, and fasten it at the ends 
and 
.
Distort the string, and at some instant, call it 
, we release it and allow it to vibrate.
Problem :
To determine the vibrations of the string, that is, to find its deflection 
at any point 
and at any time 
.
is the solution of a PDE.
is the model of our physical system to be derived.
Physical Assumptions
1. The mass of the string per unit length is constant. The string is perfectly elastic and
does not offer any resistance to bending.
2. The tension caused by stretching the string before fastening it at the ends is so large that the action of the gravitational force on the string can be neglected.
3. The string perform small transverse motions in a vertical plane : that is, every particle of the string moves strictly vertically and so that the deflection and the slope at every point of the string always remain small in absolute value.
Derivation of the PDE of the Model (“Wave Equation”) from Forces
Deflected string at fixed time 
.
The forces acting on a small portion of the string.
Since the string offers no resistance to bending, the tension is tangential to the curve
of the string at each point.
is the tension at the endpoints 
of that portion.
is the tension at the endpoints 
of that portion.
Since the points of the string move vertically, there is no motion in the horizontal direction.
The horizontal components of the tension
Horizontal components of the tension must be constant.
Horizontal components 
of 
Horizontal components 
of 
(1) 
.
The vertical components of the tension
Vertical components 
of 
Vertical components 
of 
The minus sign appears because the component at 
is directed downward.
By Newton’s second law
two forces = (the mass 
of the portion) 
(the acceleration 
),
evaluated at some point between 
and 
here, 
is the mass of the undeflected string per unit length
is the length of the portion of the undeflected string
(Newton’s second law).
Since the points of the string move vertically
In the vertical direction
Newton’s second law : 
.
Using (1), we can divide this by 
, obtaining
(2) 
is the slopes of the string at 
is the slopes of the string at 
Here we have to write partial derivatives because 
depends also on time 
.
Dividing (2) by 
, we thus have
.
(3) 
is the one-dimension wave equation.
is homogeneous and of the second order.
The physical constant 
is denoted by 
(instead of 
) to indicate that this constant is positive, a fact that will be essential to the form of the solutions.
“One-dimensional” means that the equation involves only one space variable, 
.
In the vertical direction
Newton’s second law : 
One-dimensional wave equation : 
http://matrix.skku.ac.kr/sglee/
[한빛 아카데미] Engineering Mathematics with Sage:
[저자] 이상 구, 김영 록, 박준 현, 김응 기, 이재 화
Contents
A. 공학수학 1 – 선형대수, 상미분방정식+ Lab
Chapter 01 벡터와 선형대수 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-1.html
Chapter 02 미분방정식의 이해 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-2.html
Chapter 03 1계 상미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-3.html
Chapter 04 2계 상미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-4.html
Chapter 05 고계 상미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-5.html
Chapter 06 연립미분방정식, 비선형미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-6.html
Chapter 07 상미분방정식의 급수해법, 특수함수 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-7.html
Chapter 08 라플라스 변환 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-8.html
B. 공학수학 2 - 벡터미적분, 복소해석 + Lab
Chapter 09 벡터미분, 기울기, 발산, 회전 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-9.html
Chapter 10 벡터적분, 적분정리 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-10.html
Chapter 11 푸리에 급수, 적분 및 변환 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-11.html
Chapter 12 편미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-12.html
Chapter 13 복소수와 복소함수, 복소미분 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-13.html
Chapter 14 복소적분 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-14.html
Chapter 15 급수, 유수 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-15.html
Chapter 16 등각사상 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-16.html
,
.
is absolutely integrable on the 
axis and piecewise continuous on every finite interval, then the Fourier transformation 
of 
given by 
exists.
if 
and 
otherwise.
of 
if 
and 
if 
; here 
.
and 
whose Fourier transforms exist and any constant 
and 
, the Fourier transform of 
exists, and
.
be continuous on the 
axis and 
as 
. Furthermore, let 
be absolutely integrable on the 
axis. Then
.
from Table III, Sec. 11. 10.
and 
are piecewise continuous, bounded, and absolutely integrable on the 
axis. Then
*
.
and 
are solutions of a 
, then
and 
is also a solution of that PDE in the region 
.