What’s Your Function? Convolution jct.. Jeff Swanson John Rivera Ben Rougier Justin Malaise Craig...
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Transcript of What’s Your Function? Convolution jct.. Jeff Swanson John Rivera Ben Rougier Justin Malaise Craig...
Brief Overview
• Functions: Basic Definitions
• Fourier Transforms: Uses and relationship to Convolutions
• Convolutions: Definition and reasons for
• Applying Convolutions to Computer Graphics
• Applying Convolutions to the viewing of stars
x
Input
Domain Range
f(x)
x f(x) • Association
f • Logical Machine
f(x)
Output
Functions: Basic Definitions
(x1,f(x1))
(x2,f(x2))
(x3,f(x3))
• Graphical Picture
• Sets of Ordered Pairs
f(x)
x
Functions: Basic Definitions
• Computer Graphics
Convolutions can be thought of as a method of averaging unruly functions.
Unruly Functions include:
• Discontinuous functions• Functions with sharp or jagged edges
Weighted Averages
n
jjj y
1
Origin of ConvolutionsOrigin of Convolutions
Substitute into the equation to receive)(xfy
n
jjj xf
1
)(
We need
n
jjj xf
1
)(
Therefore we create a new equation
to return a function
Which returns a new function
n
jjj xxf
1
)(
One more substitution of )( jj xg
Gives us the equation:
n
jjj xxfxg
1
)()(
This equation will give us a new discrete function
Discrete Functions to Discrete Functions to ContinuousContinuous
For the new function to be useful to us it must be in a continuous form
To accomplish this, change the function to an integration
This gives us:
R
jj dtxxfxg )()(
Which is a Convolution!
Convolve the function with itself to receive the new function
th
htth
thht
ht
tF
2for 0
20for 2
02for 2
2 for 0
)(
One more convolution gives use the continuous function
th
hthht
hthht
hthht
ht
tG
3 0
3 2/)3(
3
3 2/)3(
3 0
)(2
22
2
Convolutions of multi-dimensional functions
R R
dxdyytxskyxftsF ),(),(),(
Convolutions are also useful in smoothing functions of more than one dimension
The convolution of a two dimensional function is expressed as:
When represented by pictures the effects of convolutions on two dimensional functions can be easily seen:
Function Convolution
Fourier Transforms
The Fourier Transform can be used as an Analog for functions that are not
Periodic, like the Taylor Series.
It also has great importance in its relation to Convolutions.
Definition of a Fourier Transform
Define a function f(t) then the Fourier Transform of f on R is
dttfxf eitx2
)()(ˆ
Fourier Transforms Related to Convolutions
First define a function h as f convolved with g:
Then the Fourier Transform of h is:
))(ˆˆ()(ˆ xgfxh
))(()( tgfth
Fourier Transforms on : f(s,t)
The Fourier Transform of f(s,t) on the plane would be represented by:
2R
dsdttsfyxf e
ty)πi(sx2),(),(ˆ
THE PROBLEMTHE PROBLEM
• Can we find an accurate calculation of the diameter using a photograph of the star?
• Atmospheric effects cause stars to appear to us as ‘twinkling’ patches of light.
• A true photograph should show them as tiny discs.
Where D(x) is the equation of the unit circle,
And is defined as follows
10)(
,11)(
xforxD
xforxD
Since f has radial symmetry, it follows that the fourier transform of f has radial symmetry also!!
X
The further from the point source you get, the less luminous intensity their is.
This makes point X less “bright” than Y because it’s affected by less light.
Center: Point source of light.
Y
The Photographic Plate
O
XY
X-Y
Let: “Brightness” at O
:' “Brightness” at Y
:)(XK t “Brightness” at X
|||| Y
:)(' YXK t “Brightness” at X-Y
:)(XK tApparent Brightness
True Brightness
(At time t)
n ,..., ,2 1
YYY n,...,,
21
|||||||||||| ,...,,21
YYY n
“Brightness” of Yj given point sources
Images(in absence of atmospheric effects)
Given are very small
)( at x" Brightness" A(x)n
1jj YX jtK
We Get:
Discrete to ContinuousDiscrete to ContinuousSince we think of light as a continuous entity,
We can easily change
)(1
YX jt
n
jj K
)()()( YdAYXYf K t
To
This replaces a ‘TRUE IMAGE’ with an ‘ACTUAL IMAGE’.
TRUE IMAGE(brightness) = f(x) ACTUAL IMAGE = f*Kt
Continuous form in Detail
)()()( YdAYXYf K t
:)(Yf
:)( YXK t
:)(YdA
Brightness at point Y
Blurring effect(kernel) of the atmosphere at time t.
Integrate with respect to Y
The Convolution:
K tf * =
Where
THE PROBLEM
i. Atmospheric Effects cause Kt to be RANDOM.
ii. How can we calculate an accurate representation of this?
iii. How does this Blurring Kernel tell us
about ?
THE SOLUTION(S)
1. Average the image received at various times. This will give us the Average Blur.
2. Use the Convolution Theorem for Fourier Transforms
1. Averaging Kt:
Average the image received at various times,
)(),...,3(),2(),1( nttttThis will give a fixed Kt or the “Average Blur”:
KKjt
n
j n )(1
1~
: AVERAGE BLUR
Averaging the Convolutions,
n
jjtKf
n 1)(
*1
K jt
n
j nf
)(1
1*
~
* Kf
2. Using the Convolution Theorem for Fourier Transforms.
Take the FOURIER TRANSFORM of our image(convolution):
K ttf *
The Above convolution becomes a multiplication under the Convolution Theorem for Fourier Transforms:
K ttf
Laybeyrie’s idea is as follows:
Furthermore, if we take the sequence
)()3()2()1(
,...,,,ntttt
fAre the zeros of
K ttfby
We see is RANDOM, due to Kt being RANDOM.
Therefore, the only zeros(roots) the ‘s
)( jt
)( jt
Will have in common with all other ‘s
)( jt
Superimposing
)( jt
forms
K jt
n
j
n
j jtf
)(11 )(
)ζ()ζ()ζ(
Clearly the zeros of )ζ(f
Will stand out as the zeros of )ζ(This is our “average” picture!!!
***Using the ABSOLUTE VALUE signs guarantees that our values are real and NOT complex.
- Where is an ordered pair in the s,t plane (s,t)
The ZERO’s of f^
)(ε
)ζ(2
xiζπ20)( xdAxx
Df e
Substitute y = x – x0
)(ε
)ζ(2
)xy(π2 0)( ydAy
Df e i
Move out the constants
)()ε
()ζ(2
yiζπ2xζiπ2 0
ydAy
Df ee
)()()ζ(2
wεiζπ2xζiπ22 0
ε wdAwDf ee
Substitute w = y/ y = w
You’ll notice, by the definition of Fourier Transforms
)εζ()()(2
wεiζπ2
DwdAwD e
So
)εζ()ζ(0xζiπ22
ε
Df e
We see that the zeros of )ζ(f
Are the same as the zeros of )εζ(D
TO COMPUTE , we need to know the location of the rings of zeros of
)εζ(D
HOW?
)εζ(D
The zeros occur on the circle of radius = r0,
Which should be visually evident from our picture.
i.e. 0)εζ(
D When = r0
It’s well known,
ζ
)ζ(π2)ζ( 1JD
- Where J1 is the first order Bessel function
BUT! The zeros of )ζ(D
Occur on the circle of radius = r1
i.e. 0)ζ(
D When = r1
r1 can be found analytically.