Numerical Methods Fourier Transform Pair Part: Frequency and Time Domain

Numerical Methods

Fourier Transform Pair Part: Frequency and Time Domain

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Chapter 11.03: Fourier Transform Pair: Frequency and

Time Domain

Major: All Engineering Majors

Authors: Duc Nguyen


Numerical Methods for STEM undergraduateshttp://numericalmethods.eng.usf.edu 504/10/23

Lecture # 5

http://numericalmethods.eng.usf.edu6

)()()( 1101 tSinbtCosaatf

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

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

Example 1

)(2for

0for)(

Tt

tttf

)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

tikw

keCtf 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:

12

2220

Tfw

2

0

1~dtedtet

TC iktikt

k



Define:

00

0

11dte

ike

iktdtetA iktiktikt

or

22

2

11

11

ke

ke

k

i

ek

eik

A

ikik

ikik



Also,

2

2

ik

edteB iktikt

ikikikik eek

iee

ikB

22



Thus: BACk

21~

2

22

11

2

1~ ikik

k ek

i

kk

i

kk

ieC

Using the following Euler identities

)cos(

)sin()cos(

)sin()cos(

k

kik

kike ik

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



Noting that 1)2cos( k for any integerk

)2(

11)(

2

1~22

kCosk

i

kkkCosCk

1



Also,

,...)8,6,4,2(1

,...)7,5,3,1(1)cos(

numberevenkfor

numberoddkfork

Thus,

k

i

kkC

k

k

22

11

2

1~

ikk

C k

k

2

111

2

1~2



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:

iiiba

C2

11

2

)1(2

2~ 111

ii

ibaC

4

10

221

0

2~ 222



ii

ibaC

8

10

241

0

2~ 444

ii

ibaC

10

1

25

1

251

252

2~ 555

ii

ibaC

6

1

9

1

231

92

2~ 333



ii

ibaC

12

10

261

0

2~ 666

ii

ibaC

14

1

49

1

271

492

2~ 777

ii

ibaC

16

10

281

0

2~ 888



In general, one has

numberevenkfori

k

numberoddkforikk

Ck

,..8,6,4,22

1

,..7,5,3,12

11

~ 2


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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





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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 i

k )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

A

R

A

R kk 1cos)cos(

A

I

A

I kk 1sin)sin(

and



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

kC~

)13770783.2(

1 )59272353.0(2

11~ 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

A

RCos 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(

4

10

~ ii

eeiC

)77990097.1(

3 )17037798.0(6

1

9

1~ 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(

8

10

~ ii

eeiC

)69743886.1(

5 )100807311.0(10

1

25

1~ ieiC



)57079633.1(26 )08333333.0()08333333.0(

12

10

~ ii

eeiC


)66149251.1(

7 )07172336.0(14

1

49

1~ ieiC

28 )0625.0(

16

10

~ i

eiC




Acknowledgement




The End - Really

Numerical Methods

Fourier Transform Pair Part: Non-Periodic Functions





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

Recall

k

tikw

keCtf 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~

0ikwFT

Ck

And Equation (39) becomes

k

tikweikwFT

tf 00 )(ˆ

1)(

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

fti

np efiFdftf 2)2(ˆ)(

dfefiFtf fti

np 2)2(ˆ)(


Non-Periodic Functions cont.

Multiplying and dividing the right-hand-side of the equation by , one obtains2

; inverse Fourier transform

)()(ˆ2

1)( 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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Numerical Methods Fourier Transform Pair Part: Frequency and Time Domain