1 dream plan idea implementation. 2 3 introduction to image processing dr. kourosh kiani email:...
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DREAMDREAM
PLANPLANIDEAIDEA
IMPLEMENTATIONIMPLEMENTATION
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Introduction to Image ProcessingIntroduction to Image Processing
Dr. Kourosh KianiEmail: [email protected]: [email protected]: [email protected]: www.kouroshkiani.com
Present to:Amirkabir University of Technology (Tehran
Polytechnic) & Semnan University
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Lecture 08Fourier Transform
Lecture 08Fourier Transform
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Development of the Fourier TransformRepresentation of an Aperiodic Signal
In the last lecture we saw how a periodic signal could be represented as a linear combination of cos(nω) and sin(nω). In fact, these results can be extended to develop a representation of aperiodic signals as a linear combination of cos(nω) and sin(nω).
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Continuous Fourier transform
T1 T1
)(~ tx
T0 2T0T1 T1-T0-2T0-3T0
T0T1 T1T0
)(tx
T1 T1
)(~ tx
)(~ tx
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)2()(~1
)1()(~
0
0
0
0
2
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dtetxT
a
eatx
tjk
T
Tk
tjk
kk
2)()(~ 0Ttfortxtx Since
And also since x(t)=0 outside this interval, equation (2) can be rewritten as:
)3()(1
)(~100
0
0 0
2
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dtetxT
dtetxT
a tjktjk
T
Tk
Therefore, defining the envelope X(ω) of T0ak as: )4()()( dtetx tj
We have that the coefficients ak can be expressed as: )(1
00
kT
ak
Combining (1) and (4), can be expressed in the term X(ω) as:)(~ tx
)5()(1
)(~ 00
0
tjk
k
ekT
tx
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)6()(1
)(~ 00
0
tjk
k
ekT
tx
Or equivalently, since 002 T
)7()(2
1)(~
000
tjk
k
ektx
As , approaches x(t), and consequenetly eq. (7) becomes a representation of x(t). Furthermore, as and the right-hand side of eq (7) passes to an integral.
0T )(~ tx0T00
tje )(
tjkek 0)( 0
0k00 k
00 .)( 0 tjkekArea
Each term in the summation on the right-hand side of eq. (7) is the area of a rectangle of height and width (here t is regarded as fixed). As this by definition converges to the integral of . Therefore, using the fact that as eq. (7) and (4) become
tjkek 0)( 0 0
00
tje )()()(~ txtx 0T
dtetx
detx
tj
tj
)()(
)(2
1)(
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Fourier Transform
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Comments
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Example
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Samples of Fourier Transforms of Aperiodic Signals
Spectrum
0 f 3f 5f
0 f 3f 5f
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CTFT Properties
x 0 X f df
X 0 x t dt
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Example: the Fourier Transform of arectangle function: rect(t)
1/ 21/ 2
1/ 2
1/ 2
1( ) exp( ) [exp( )]
1[exp( / 2) exp(
exp( / 2) exp(
2
sin(
F i t dt i ti
i ii
i i
i
( sinc(F Imaginary Component = 0
F(w)
w
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Example: the Fourier Transform of adecaying exponential: exp(-at) (t > 0)
0
0 0
0
( exp( )exp( )
exp( ) exp( [ )
1 1exp( [ ) [exp( ) exp(0)]
1[0 1]
1
F at i t dt
at i t dt a i t dt
a i ta i a i
a i
a i
1(F i
ia
A complex Lorentzian!
Questions? Discussion? Suggestions?
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