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Numerical Methods

Fourier Transform Pair Part: Frequency and Time Domain

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Chapter 11.03: Fourier Transform Pair: Frequency and

Time DomainMajor: All Engineering Majors

Authors: Duc Nguyen


Numerical Methods for STEM undergraduateshttp://numericalmethods.eng.usf.edu 504/22/23

Lecture # 5

http://numericalmethods.eng.usf.edu6

)()()( 1101 tSinbtCosaatf

)2()2()()()( 221102 tSinbtCosatSinbtCosaatf

)2()2()()()( 221104 tSinbtCosatSinbtCosaatf

Example 1

)(2for

0for)(

Tttt

tf

)4()4()3()3( 4433 tSinbtCosatSinbtCosa


Frequency and Time DomainThe amplitude (vertical axis) of a given periodic function can be plotted versus time (horizontal axis), but it can also be plotted in the frequency domain as shown in Figure 2.

Figure 2 Periodic function (see Example 1 in Chapter 11.02 Continuous Fourier Series) in frequency domain.

Frequency and Time Domain cont.

Figures 2(a) and 2(b) can be described with the following equations from chapter 11.02,

k

tikwkeCtf 0~)( (39,

repeated)where

Ttikw

k dtetfT

C0

0)(1~ (41, repeated)



For the periodic function shown in Example 1 of Chapter 11.02 (Figure 1), one has:

122220

Tfw

2

0

1~ dtedtetT

C iktiktk



Define:

000

11 dteik

eik

tdtetA iktiktikt

or

22

2

11

11

ke

ke

ki

ek

eik

A

ikik

ikik



Also,

2

2

ik

edteB iktikt

ikikikik eekiee

ikB

22



Thus: BACk

21~

2

22

1121~ ikik

k eki

kki

kkieC

Using the following Euler identities

)cos()sin()cos()sin()cos(

kkikkike ik

)2cos()2sin()2cos(2 kkike ik



Noting that 1)2cos( k for any integerk

)2(11)(21~

22

kCoski

kkkCosCk

1



Also,

,...)8,6,4,2(1,...)7,5,3,1(1

)cos(numberevenkfornumberoddkfor

k

Thus,

ki

kkC

k

k

22

1121~

ikk

C kk

2111

21~

2



From Equation (36, Ch. 11.02), one has

2~ kkk

ibaC (36, repeated)

Hence; upon comparing the previous 2 equations, one concludes:

1)1(12

k

k ka

k

bk1



For

;8...4,3,2,1k the values for ka kband(based on the previous 2 formulas) are exactly

identical as the ones presented earlier in Example1 of Chapter 11.02.



Thus:

iiibaC

211

2

)1(2

2~ 111

ii

ibaC410

2210

2~ 222



ii

ibaC810

2410

2~ 444

ii

ibaC101

251

251

252

2~ 555

ii

ibaC61

91

231

92

2~ 333



ii

ibaC1210

2610

2~ 666

ii

ibaC141

491

271

492

2~ 777

ii

ibaC1610

2810

2~ 888



In general, one has

numberevenkfori

k

numberoddkforikk

Ck

,..8,6,4,221

,..7,5,3,1211

~ 2


THE ENDhttp://numericalmethods.eng.usf.edu

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Acknowledgement

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This material is based upon work supported by the National Science Foundation under Grant # 0717624. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.

The End - Really

Numerical Methods

Fourier Transform Pair Part: Complex Number in Polar Coordinates


Chapter 11.03: Complex number in polar coordinates

(Contd.)

In Cartesian (Rectangular) Coordinates, a complex number can be expressed as: kC

~

iIRC kkk ~

In Polar Coordinates, a complex number can be expressed as: kC

~

iAAiAAeC ik )sin()cos()sin()cos(~


Lecture # 6

Complex number in polar coordinates cont.

Thus, one obtains the following relations between the Cartesian and polar coordinate systems: cosARk sinAI k This is represented graphically in Figure 3.

Figure 3. Graphical representation of the complex number system in polar coordinates.



Hence

)(sin)(cossincos 222222222 AAAIR kk

AR

AR kk 1cos)cos(

AI

AI kk 1sin)sin(

and



Based on the above 3 formulas, the complex numberscan be expressed as:

kC~

)13770783.2(1 )59272353.0(

211~ ieiC



eR

mI

Notes:(a)The amplitude and angle are 0.59 and 2.14 respectively (also see Figures 2a, and 2b in chapter 11.03).

1

~C

(b) The angle (in radian) obtained from

ARCos k)( will be 2.138 radians (=122.48o).

However based on A

ISin k)(

Then = 1.004 radians (=57.52o).



)57079633.1(22 )25.0()25.0(

410~ i

ieeiC

)77990097.1(3 )17037798.0(

61

91~ ieiC

Since the Real and Imaginary components of are negative and positive, respectively, the proper selectionfor should be 2.1377 radians.


)57079633.1(24 )125.0()125.0(

810~ i

ieeiC

)69743886.1(5 )100807311.0(

101

251~ ieiC



)57079633.1(26 )08333333.0()08333333.0(

1210~ i

ieeiC


)66149251.1(7 )07172336.0(

141

491~ ieiC

28 )0625.0(

1610~ i

eiC



Acknowledgement

The End - Really

Numerical Methods

Fourier Transform Pair Part: Non-Periodic Functions


Chapter 11. 03: Non-Periodic Functions (Contd.)

Recall

k

tikwkeCtf 0~)( (39, repeated)

Ttikw

k dtetfT

C0

0)(1~ (41, repeated)

Define

2

2

00 )()(ˆ

T

T

tikw dtetfikwF


Lecture # 7

(1)


Then, Equation (41) can be written as)(ˆ1~0ikwF

TCk

And Equation (39) becomes

k

tikweikwFT

tf 00 )(ˆ1)(

From above equation

k

tikw

ffor

Tnp eikwFftftf 000

0

)(ˆ)(lim)(lim)(

k

ftik

fnp efikFftf 2

0)2(ˆ)(lim)(

or

Non-Periodic Functions

Non-Periodic Functions cont.

Figure 4. Frequency are discretized.

From Figure 4,ffk

ftinp efiFdftf 2)2(ˆ)(

dfefiFtf ftinp 2)2(ˆ)(


Non-Periodic Functions cont.Multiplying and dividing the right-hand-side of the equation by , one obtains2

; inverse Fourier transform

)()(ˆ21)( 0

00 wdeiwFtf tiw

np

Also, using the definition stated in Equation (1), one gets

)()()(ˆ 00 tdetfiwF tiw

np ; Fourier transform



Acknowledgement

The End - Really

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