SKKU 공학수학, Week 2

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

http://www.hanbit.co.kr/EM/sage/2_chap11.html

Week 2

11.7. Fourier Integral

Indefinite integral

Sage Coding

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

f(x) = x^2

integral(f(x), x)

1/3*x^3

Definite integral

Sage Coding

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

f(x) = sin(x)

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

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

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

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

and gives the answer

(6) .

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

Sage Coding

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

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

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

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

1/4*pi

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

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

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

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

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

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

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

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

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

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

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

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

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)