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

http://math3.skku.ac.kr/  http://sage.skku.edu/  http://mathlab.knou.ac.kr:8080/


We also will have the following Fourier sine transform

    

          

          

          .


Sage Coding

http://math3.skku.ac.kr/  http://sage.skku.edu/  http://mathlab.knou.ac.kr:8080/


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

http://math3.skku.ac.kr/  http://sage.skku.edu/  http://mathlab.knou.ac.kr:8080/



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

http://math3.skku.ac.kr/  http://sage.skku.edu/  http://mathlab.knou.ac.kr:8080/



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

http://math3.skku.ac.kr/  http://sage.skku.edu/  http://mathlab.knou.ac.kr:8080/


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

http://math3.skku.ac.kr/  http://sage.skku.edu/  http://mathlab.knou.ac.kr:8080/



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

http://math3.skku.ac.kr/  http://sage.skku.edu/  http://mathlab.knou.ac.kr:8080/



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).