2019학년도 2학기
반도체 공학과 공학수학2
주교재: Erwin Kreyszig, Engineering Mathematics, 10th Edition
부교재: 이상구 외 4인, 최신공학수학 Ⅱ 1st Edition
실습실: http://www.hanbit.co.kr/EM/sage/
http://matrix.skku.ac.kr/2018-album/R-Sage-Stat-Lab-1.html
http://matrix.skku.ac.kr/2018-album/R-Sage-Stat-Lab-2.html
강의시간: 공학수학2, (화16:30-17:45) (목15:00-16:15)
담당교수: 김응기 박사
Week 3
3주차
11.8: Fourier Cosine and Sine Transforms
11.4 푸리에 급수의 수렴
*11.5 스투룹-리우빌 정리와 직교함수
11.9: Fourier Transform.
11.6 푸리에 변환
http://www.hanbit.co.kr/EM/sage/2_chap11.html
http://matrix.skku.ac.kr/2018-EM/EM-2-W3-lab.html
11.8. Fourier Cosine and Sine Transforms
An integral transformation is a transformation in the form of an integral that produces from given functions new functions depending on a different variable.
The Laplace transform is of this kind and is by far the most important integral transform in engineering.
Fourier Integral
Let us assume that following conditions on
1. and are piecewise continuous in every finite interval.
2. converges i.e. is absolutely integrable in .
The Fourier integral theorem states that
(*)
is a Fourier integral expansion of .
(**) where
Fourier Cosine Transforms
For an even function , the Fourier cosine integral
(a) , where (b) . ( in (**))
We now set , where suggests “cosine.”
Then from , writing , we have
(1)
This is the Fourier cosine transform of . Similarly, from we have
(2) .
This is the inverse Fourier cosine transform of .
Attention!
In we integrate with respect to and in with respect to . Formula gives from a new function , called the Fourier cosine transform of .
The Formula gives us back from , and we therefore call the inverse Fourier cosine transform of .
The process of obtaining the transform from a given is called the Fourier cosine transform or the Fourier cosine transform method.
Fourier Sine Transforms
For an odd function , the Fourier integral is the Fourier sine integral
(a) , where (b). ( in (**))
, where suggests “sine.”
Hence
Then from , writing , we have
(2) . Fourier sine transform of ,
Similarly, from we have
(2)
inverse Fourier sine transform of .
The process of obtain from is the Fourier sine transform or the Fourier sine transform method.
Remark
Other notations are
,
and and for the inverse of and , respectively.
Example 1 Fourier Cosine and Sine Transforms
Find the Fourier cosine and Fourier sine transformation of the function(Fig. 282)
Solution
From the definitions and ,
we have the following Fourier cosine transform
Sage Coding
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We also will have the following Fourier sine transform
.
Sage Coding
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This agrees with formulas in the first two tables in Sec. 11.10 (where ).
Note that , these transform do not exist.(Why?)
Example 2 Fourier Cosine Transform of the Exponential Function
Find .
Solution
By integration by parts and recursion,
.
This agrees with formula in Table I, Sec. 11. 10, with .
Sage Coding
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Linearity, Transforms of Derivatives
If is absolutely integrable on the positive axis and piecewise continuous on every finite interval, then the Fourier cosine and sine transforms of exists.
Theorem Fourier Cosine Transforms
If and have Fourier cosine transforms, then for any constants and has Fourier cosine transforms.
Proof
Theorem Fourier Sine Transforms
If and have Fourier sine transforms, then for any constants and has Fourier sine transforms.
Proof
Shows that the Fourier cosine and sine transforms are linear operation,
(3) .
Theorem 1 Cosine and Sine Transforms of Derivatives
Let be continuous and absolutely integrable on the axis, let be piecewise continuous on every finite interval, and let as . Then
(4)
Proof
,
.
The Formula , say , with instead of
(when , satisfy the respective assumptions for , in Theorem)
instead of
(5a) .
Similarly
(5b)
Example 3 An Application of the Operational Formula (5)
Find the Fourier cosine transform of , where .
Solution
, thus
.
From this, , and the linearity ,
.
Hence
.
Answer is (see. Table I, Sec.11.10)
.
Sage Coding
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11.9. Fourier Transform
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 .
Remark
Other notations for the Fourier transform is
,
so that
.
The process of obtaining the Fourier transform from a given is also called the Fourier transform or the Fourier transform method.
Theorem 1 Existence of the Fourier Transform
If is absolutely integrable on the axis and piecewise continuous on every finite interval, then the Fourier transformation of given by exists.
Example 1 Fourier Transform
Find the Fourier transform of if and otherwise.
Solution
Using Fourier transform and integrating,
.
Sage Coding
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Example 2 Fourier Transform
Find the Fourier transform of if and if ; here .
Solution
From the definition Fourier transform we obtain by integration
.
Sage Coding
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Linearity. Fourier Transform of Derivatives
Theorem 2 Linearity of the Fourier Transform
The Fourier transform is a linear operation ; that is, for any functions and whose Fourier transforms exist and any constant and , the Fourier transform of exists, and
(8) .
Proof
This is true because integration is a linear operation, so that gives
.
Theorem 3 Fourier Transform of the Derivative of
Let be continuous on the axis and as . Furthermore, let be absolutely integrable on the axis. Then
(9) .
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.
Example 3 Application of the Operational Formula (9)
Find the Fourier transform of from Table III, Sec. 11. 10.
Solution
We use , By formula in Table III.
.
Sage Coding
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Convolution
The convolution * of function and is defined by
(11) *.
Theorem 4 Convolution Theorem
Suppose that and are piecewise continuous, bounded, and absolutely integrable on the axis. Then
(12) *.
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.
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
Made by Prof. Sang-Gu LEE sglee at skku.edu
http://matrix.skku.ac.kr/sglee/ with Dr. Jae Hwa LEE
Copyright @ 2019 SKKU Matrix Lab. All rights reserved.
Made by Manager: Prof. Sang-Gu Lee and Dr. Jae Hwa Lee
*This research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (2017R1D1A1B03035865).