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SEOUL NATIONAL UNIVERSITY

School of Mechanical & Aerospace Engineering

400.002

Eng Math II

13 Fourier Transform

13.1 Fourier Cosine and Sine Transforms

- An "integral transform" is a transformation that produces from given functions new func- tions that depend on a di®erent variable and appear in the form of an integral. !Laplace transform, Fourier transform

Fourier Cosine Transforms

- For an even functionf(x), (a)f(x) =Z 1 0

A(!)cos!xd!where (b)A(!) =2

Z 1 0 f(v)cos!vdv(1) - We now setA(!) =p

2=¼¢^fc(!).

- Fourier cosine transform off(x): fc(!) =r 2 Z 1 0 f(x)cos!xdx(2) - Inverse Fourier cosine transform of ^fc(!): f(x) =r 2 Z 1

0^fc(!)cos!xd!(3)

- Fourier cosine transform : the process of obtaining the transform ^fc(!) from a givenf

Fourier Sine Transforms

- For an odd functionf(x), (a)f(x) =Z 1 0

B(!)sin!xd!where (b)B(!) =2

Z 1 0 f(v)sin!vdv(4) - We now setB(!) =p

2=¼¢^fs(!).

- Fourier sine transform off(x): fs(!) =r 2 Z 1 0 f(x)sin!xdx(5) - Inverse Fourier sine transform of ^fs(!): f(x) =r 2 Z 1

0^fs(!)sin!xd!(6)

1

Example 1.

Fourier cosine and Fourier sine transforms

f(x) =½kif 0< x < a

0 ifx > a:

fc(!) =r 2 kZ a 0 cos!xdx=r 2 kµsina! fs(!) =r 2 kZ a 0 sin!xdx=r 2 ¡k cos!xja0=r 2 kµ1¡cosa!

Example 2.

fc(e¡x). fc(e¡x) =r 2 Z 1 0 e¡xcos!xdx=r 2

¢e¡x

1 +w2(¡cos!x+!sin!x)j10

p

2=¼

1 +!2

13.2 Properties of Fourier Cosine/Sine Transforms

- Linear operations fc(af+bg) =r 2 Z 1 0 [af(x) +bg(x)]cos!xdx =ar 2 Z 1 0 f(x)cos!xdx+br 2 Z 1 0 g(x)cos!xdx =a^fc(f) +b^fc(g) (a) ^fc(af+bg) =a^fc(f) +b^fc(g) (b) ^fs(af+bg) =a^fs(f) +b^fs(g)(7)

Theorem 1

[Cosine and sine transforms of derivatives] Letf(x) be continuous and absolutely integrable on thex-axis, letf0(x) be piecewise con- tinuous on each ¯nite interval, and letf(x)!0 asx! 1. Then (a) ^fc[f0(x)] =!^fs[f(x)]¡q 2 f(0) (b) ^fs[f0(x)] =¡!^fc[f(x)](8) 2

Proof.

fc[f0(x)] =r 2 Z 1 0 f0(x)cos!xdx r 2 f(x)cos!xj10+!Z 1 0 f(x)¢sin!xdx¸ =!^fs[f(x)]¡r 2 f(0) fs[f0(x)] =r 2 Z 1 0 f0(x)sin!xdx r 2 f(x)sin!xj10¡!Z 1 0 f(x)cos!xdx¸ =¡!^fc[f(x)] (a) ^fc[f00(x)] =!^fs[f0(x)]¡q 2 f0(0) =¡!2^fc[f(x)]¡q 2 f0(0) (b) ^fs[f00(x)] =¡!^fs[f0(x)] =¡!2^fs[f(x)] +q 2 !f(0)(9)

Example 3.

An application of the operational formula (9)

Find the Fourier cosine transform off(x) =e¡ax, wherea >0. solution) (e¡ax)00=a2¢e¡ax=)a2f(x) =f00(x) a

2^fc(f) =^fc(f00) =¡!2^fc(f)¡r

2 f0(0) =¡!2^fc(f) +ar 2 (a2+!2)^fc(f) =ar 2 ^fc(e¡ax) =r 2 2 a (a >0)

13.3 Fourier Transform

Complex Form of the Fourier Integral

- The real Fourier integral is f(x) =Z 1 0 [A(!)cos!x+B(!)sin!x]d! where

A(!) =1

Z 1 ¡1 f(v)cos!vdv; B(!) =1 Z 1 ¡1 f(v)sin!vdv: 3 - SubstitutingA(!) andB(!) into the integral forf, we have f(x) =1 Z 1 0Z 1 ¡1 f(v)[cos!vcos!x+ sin!vsin!x]dvd! 1 Z 1 0· Z1 ¡1 f(v)cos(!x¡!v)dv¸ d!

F(!) =Z

1 ¡1 f(v)cos(!x¡!v)dv -F(!) is an even function, since cos!(x¡v) is an even function of!. For example, f(x) =Rc

¡ccosxtdt=1

x sinxtjc¡c=2sincx x f(¡x) =2sin(¡cx)

¡x=2sincx

x =f(x) )f(x) is an even function. -f(v) does not depend on!, and we integrate with respect tov. f(x) =1 Z 1 0

F(!)d!=1

2¼Z

1 ¡1

F(!)dx

f(x) =1

2¼Z

1

¡1·

Z1 ¡1 f(v)cos(!x¡!v)dv¸ d! (10) f(x) =1

2¼Z

1

¡1·

Z1 ¡1 f(v)sin(!x¡!v)dv¸ d!= 0 (11)

G(!) =Z

1 ¡1 f(v)sin(!x¡!v)dv -G(!) is an odd function since sin(!x¡!v) is an odd function of!. - (1) +i(2) witheix= cosx+isinx(Euler formula) f(v)cos(!x¡!v) +if(v)sin(!x¡!v) =f(v)ei(!x¡!v) -Complex Fourier Integral )f(x) =1quotesdbs_dbs4.pdfusesText_8