2018학년도 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

 주차 주교재 부교재 2 11.7: Fourier Integral 11.8: Fourier Cosine and Sine Transforms 11.4 푸리에 급수의 수렴 11.5 스투름-리우빌 정리와 직교함수 web

Week 2

11.7. Fourier Integral

Indefinite integral

Sage Coding

 f(x) = x^2 integral(f(x), x)

1/3*x^3

Definite integral

Sage Coding

 f(x) = sin(x) integral(f(x), x, 0, pi/2)

1

Fourier series are powerful tools for problems involving functions that are periodic or are of interest on a finite interval only.

Many problems involve functions that are nonperiodic and are of interest on the whole axis.

 Definition Fourier integrals Consider a special function of period and see what happens to its Fourier series if we let . Do the same for an arbitrary function of period .

 Example 1 Rectangular Wave Consider the period rectangular wave of period given by

The left part of Fig. shows this function for as well as the nonperiodic function , which we obtain from if we let ,

.

We now explore what happens to the Fourier coefficients of as increases.

Since is even, for all . For the Euler formulas , Sec. 11.2, give

.

Sage Coding

 var('L, n') assume(n, 'integer') a0 = 1/(2*L)*integral(1, x, -1, 1) print "a0 =", a0 an = 1/L*integral(cos(n*pi*x/L), x, -1, 1) print "an =", an

a0 = 1/L

an = 2*sin(pi*n/L)/(pi*n)

This sequence of Fourier coefficients is called the amplitude spectrum of because is the maximum amplitude of the wave . Figure shows this spectrum for the periods . We see that for increasing these amplitude because more and more dense on the positive axis, where . Indeed, for we have amplitude per “half-wave” of the function (dashed in the figure). Hence for we have amplitude per half-wave, so that these will eventually be everywhere dense on the positive axis (and will decrease to zero).

From Fourier Series to Fourier Integral

We now consider any periodic function of period that can be represented by a Fourier series

,

.

imply

and find out what happens if we let .

Together with Example the present calculation will suggest that we should expect an integral(series) involving and with no longer restricted to integer multiples of but taking all values. We shall also see what from such an integral might have.

If we insert and from the Euler formulas , Sec. 11.2, and denote the variable of integration , the Fourier series of becomes

.

We now set

The Fourier series in the form

(1)

This representation is valid for any fixed , arbitrarily large, but finite.

We now let and assume that the resulting nonperiodic function

is absolutely integrable on axis; that is, the following(finite) limits exist :

(2)               .

Then , and the value of the first term on the right side of approaches zero.

Also and it seems plausible that the infinite series in becomes an integral from to , which represents, namely,

(3)       .

If we introduce the notations

(4)          ,

we can write this in the form

(5)                    .

This is a representation of by a Fourier integral.

 Theorem 1 Fourier Integral is piecewise continuous in every finite interval. has a right-hand derivative and left-hand derivative at every point. If the integral exist, then can be represented by a Fourier integral with and . At a point where is discontinuous the value of the Fourier integral equals the average of the left-hand and right-hand limits of at the point.

Applications of Fourier Integrals

 Example 2 Single Pulse, Sine Integral. Dirichlet’s Discontinuous Factor. Gibbs Phenomenon Find the Fourier integral representation of the function                                .

Solution

From we obtain

(6)                     .

The average of the left-hand and right-hand limits of at is equal to .

Sage Coding

 var('w, v') A(w) = 1/pi*integral(cos(w*v), v, -1, 1) print "A(w) =", A(w) B(w) = 1/pi*integral(sin(w*v), v, -1, 1) print "B(w) =", B(w)

A(w) = 2*sin(w)/(pi*w)

B(w) = 0

From and Theorem we obtain(multiply by )

(7)               .

Sage Coding

 var('w') g(x)=integral(cos(w*x)*sin(w)/w, w, 0, +oo) print g(0) print g(1/2) print g(1) print g(2)

1/2*pi

1/2*pi

1/4*pi

0

We mention that this integral is Dirichlet’s discontinuous factor.

The case is of particular interest. If , then gives

(8*)                          .

We see that this integral is the limit of the so-called sine integral

(8)

as . The graphs of and of the integrand are shown in Fig. .

In case of a Fourier series the graphs of the partial sums are approximation curves of the curve of the periodic function represented by the series. Similarly, in the case the Fourier integral , approximations are obtained by replacing by numbers .

Hence the integral

(9)

Setting .

, ,

Setting .

, ,

.

approximates the right side in and therefore .

The Figure shows oscillations near the points of discontinuous of .

We might expect that these oscillations disappear approaches infinity.

But this is not true: with increasing , they are shifted closer to the point .

This unexpected behavior, which also occurs in connection with Fourier series, is known as the Gibbs phenomenon

We can explain it by representing in terms of sine integrals as follows.

Using in App. A3. 1, we have

.

In the first integral on the right we set .

Then , and , corresponds to .

In the last integral we set .

Then , and , corresponds to .

Since , we thus obtain

.

From this and we see that our integral equals

and the oscillations in Fig. 283 result from those in Fig. 282.

The increase of amounts to a transformation of the scale on the axis and cause

the shift of the oscillation toward the point of discontinuity .

Fourier Cosine Integral and Fourier Sine Integral

If is an even function, then in and

(10)                          .

The Fourier integral then reduces to the Fourier Cosine Integral

(10)                        ( even).

If is an odd function, then in we have and

(11)                          .

The Fourier integral then reduces to the Fourier Sine Integral

(11)                       ( odd).

 Example 3 Laplace integrals We shall derive the Fourier cosine and Fourier sine integrals of , where and . (Fig 281) The result will be used to evaluate the so-called Laplace integrals.

Solution

From we have

(12)  .

Sage Coding

 var('k, w, v') assume(k > 0) A(w) = 2/pi*integral(exp(-k*v)*cos(w*v), v, 0, +oo) print "A(w) =", A(w)

A(w) = 2*k/((k^2 + w^2)*pi)

By substituting this into the first integral in

we thus obtain the Fourier cosine integral representation

.

From this representation we see that

.

(13)

.

(14)  .

Sage Coding

 var('k, w, v') assume(k > 0) B(w) = 2/pi*integral(exp(-k*v)*sin(w*v), v, 0, +oo) print "B(w) =", B(w)

B(w) = 2*w/((k^2 + w^2)*pi)

From we thus obtain the Fourier sine integral representation

.

From this we see that

(15)                  .

.

Sage Coding

 var('k, w') assume(k > 0) assume(x > 0) integral(w*sin(w*x)/(k^2+w^2), w, 0, +oo)

1/2*pi*e^(-k*x)

The integrals and are called the Laplace integrals.

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.

 Definition 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

 var('k, w, a') assume(k > 0) assume(a > 0) sqrt(2/pi)*integral(k*cos(w*x), x, 0, a)

sqrt(2)*k*sin(a*w)/(sqrt(pi)*w)

We also will have the following Fourier sine transform

.

Sage Coding

 var('k, w, a') assume(k > 0) assume(a > 0) sqrt(2/pi)*integral(k*sin(w*x), x, 0, a)

-(cos(a*w)/w - 1/w)*sqrt(2)*k/sqrt(pi)

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

 var('w') sqrt(2/pi)*integral(exp(-x)*cos(w*x), x, 0, +oo)

sqrt(2)/((w^2 + 1)*sqrt(pi))

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)

(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

 var('w, a') assume(a > 0) f(x) = exp(-a*x) sqrt(2/pi)*integral(f(x)*cos(w*x), x, 0, +oo)

sqrt(2)*a/((a^2 + w^2)*sqrt(pi))

[한빛 아카데미] Engineering Mathematics with Sage:

[저자] 이상구, 김영록, 박준현, 김응기, 이재화

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