What’s
Your
Function?
Convolution jct.
The Team:
Jeff Swanson
John Rivera
Craig Rykal
Justin Malaise
Ben Rougier
Brief Overview
• Functions: Basic Definitions
• Convolutions: Definition and reasons for
• Applying Convolutions to Computer Graphics
• Fourier Transforms: Uses and relationship
to Convolutions
• Applying Convolutions to the viewing of stars
Functions: Basic Definitions
f(x)
x
• Association
f(x)
Domain
Range
Input
x
f
Output
f(x)
• Logical Machine
Functions: Basic Definitions
f(x)
• Sets
(x3,f(x3))
(x2,f(x2))
of Ordered Pairs
• Graphical
(x1,f(x1))
Picture
x
• Computer Graphics
Convolutions
F (t )   f ( x)k (t  x)dx
R
Convolutions can be thought of as a
method of averaging unruly functions.
Unruly Functions include:
• Discontinuous functions
• Functions with sharp or jagged edges
Origin of Convolutions
Weighted Averages
n
 y
j 1
j
j
Substitute y  f (x) into the equation to receive
n

j 1
j
f (x j )
n
We need

j 1
j
f ( x j ) to return a function
Therefore we create a new equation
n

j 1
Which
returns
a
new
function
f
(
x

x
)
j
j
One more substitution of
 j  g(x j )
Gives us the equation:
n
 g(x ) f (x  x )
j 1
j
j
This equation will give us a new discrete function
Discrete Functions to
Continuous
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:
g
(
x
)
f
(
x

x
)
dt
j
j

R
Which is a Convolution!
Examples
Take the function
0 for x  -h

f ( x)  1 for  h  x  h
0 for h  x

Convolve the function with itself to receive the new function
0
t  2h

F (t )  
2 h  t
0
for
t  2h
for  2h  t  0
for
for
0  t  2h
2h  t
One more convolution gives use the continuous function
0

2
(
t

3
h
)
/2


G (t )   t 2  3h 2
(t  3h) 2 / 2

0
t  3h
 3h  t  h
ht  h
h  t  3h
3h  t
Convolutions of multi-dimensional functions
Convolutions are also useful in smoothing functions
of more than one dimension
The convolution of a two dimensional function is
expressed as:
F (s, t )    f ( x, y)k (s  x, t  y)dxdy
RR
When represented by pictures the effects of convolutions
on two dimensional functions can be easily seen:
Function
Convolution
Applying Convolutions to
Computer Graphics
How Convolutions Effect Graphics
• Blurring of graphics
•Positive Blurring
•Negative Blurring
Blurring Has Positive Effect
• Aliasing
• Antialiasing
Blurring Has Negative Effect
Blurred
Deconvolution
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
ˆf ( x) 



 2 itx
f (t ) e
dt
Fourier Transforms Related to
Convolutions
First define a function h as f convolved with
g:
h(t )  ( f  g )(t )
Then the Fourier Transform of h is:
ˆh( x)  ( fˆ  gˆ )( x)
Fourier Transforms on
2
R : f(s,t)
The Fourier Transform of f(s,t) on the plane
would be represented by:
fˆ ( x, y ) 
 
 2 πi(sx ty)
  f ( s, t ) e
  
dsdt
Application:
The Diameter of a Star
How can we find it
using Convolutions?
THE 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.
Viewing the star:
WE SEE:
TRUE IMAGE(shape):
THE STAR(true image):

x0
  (x inside disc)
f ( x)  
otherwise.
0
f ( x)   D(
x  x0

)
Where D(x) is the equation of the unit circle,
And is defined as follows
D( x)  1
for x  1,
D( x)  0
for x  1
Since f has radial symmetry, it follows that the fourier
transform of f has radial symmetry also!!
Center: Point source
of light.
X
Y
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.
The Photographic Plate
X-Y
O
|| Y ||
X
Y
Let
 : “Brightness” at O
 ': “Brightness” at Y
 K t ( X ) : “Brightness” at X
 ' K t ( X  Y ) : “Brightness” at X-Y
K t ( X ) : Apparent Brightness (At time t)
True Brightness
We Get:
n
A(x)  " Brightness at x"    j K t ( X  Y j )
 ,  ,...,  n 
1
Y ,Y
1
j1
2
2
“Brightness” of Yj given point sources
,...,Y n  Images(in absence of atmospheric effects)
||Y || ,||Y || ,...,||Y ||n 
1
2
Given are very small
Discrete to Continuous
Since we think of light as a continuous entity,
We can easily change
n
 K ( X
j 1

j
t
 Y j)
To
f (Y ) K t ( X  Y )dA(Y )
This replaces a ‘TRUE IMAGE’ with an ‘ACTUAL IMAGE’.
TRUE IMAGE(brightness) = f(x)
ACTUAL IMAGE = f*Kt
Continuous form in Detail
The Convolution:
f *Kt
=

f (Y ) K t ( X  Y )dA(Y )
Where
f (Y ) :
Kt (X  Y) :
dA(Y ) :
Brightness at point Y
Blurring effect(kernel) of the
atmosphere at time t.
Integrate with respect to Y
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,
t (1), t (2), t (3),..., t (n)
This will give a fixed Kt or the “Average Blur”:
~
1
K  K
n
j 1 n
: AVERAGE BLUR
t( j)
Averaging the Convolutions,
n
1 n
~
1
f * K t( j)  f *  K t( j)  f * K

n j 1
j 1 n
2. Using the Convolution
Theorem for Fourier Transforms.
Laybeyrie’s idea is as follows:
Take the FOURIER TRANSFORM of our
image(convolution):

t
 f * Kt
The Above convolution becomes a multiplication
under the Convolution Theorem for Fourier
Transforms:



t
 f

K
t
Furthermore, if we take the sequence



t (1)
,

t ( 2)
,

t ( 3)
,...,
t ( n)

We see

t( j)
is RANDOM, due to Kt being RANDOM.

Therefore, the only zeros(roots) the

t ( j )‘s

Will have in common with all other

‘s
t( j)

Are the zeros of
f

by


t
 f

K
t


Superimposing
t( j)
forms
n

(ζ)   
j 1
n
t( j)

(ζ)   f (ζ)
j 1

K
t( j)
- Where z is an ordered pair in the s,t plane (s,t)
Clearly the zeros of

f (ζ )
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.
^
The ZERO’s of
f

x  x0  2 π iζ x
f (ζ )    D(
)e
dA( x)
ε
2

Substitute y = x – x0

y  2π i( y  x )
f (ζ )    D( ) e
dA( y )
ε
2
0

Move out the constants

f (ζ )  
 2 π i ζ x0
e
y 2 π iζ y
2 D( ε ) e dA( y)

w = y/
Substitute
y = w

f (ζ )  
2 π i ζ x0
εe
2
2 π iζ ε w
 D(w) e
dA( w)
2

You’ll notice, by the definition of Fourier Transforms
2 π iζ ε w
 D(w) e

dA( w)  D(εζ)
2

So

f (ζ )  
2 π i ζ x0 
εe
2
D(εζ)

We see that the zeros of
f (ζ )

Are the same as the zeros of
D(εζ)
TO COMPUTE , we need to know the location of
the rings of zeros of

D(εζ)
HOW?

D(εζ)
The zeros occur on the circle of radius z =
r0, i.e. 
When z = r0
D (εζ )  0
Which should be visually evident
from our
picture.

It’s well known,

D (ζ ) 
2π J 1 ( ζ )
ζ
- Where J1 is the first order Bessel function

BUT! The zeros of D(ζ)
Occur on the circle of radius z = r1
i.e.

D (ζ )  0
r1 can be found analytically.
When z = r1
Finally finding 
So If

D(ε r0)  0
and

D(r1)  0
then
ε r r
0
And thus
r
ε
r
1
0
1
Thank you for your cooperation.
References
• “Fourier Analysis” by T.W. Korner,
Cambridge University Press, 1988
• “Convolutions and Computer Graphics” by
Anne M. Burns,
• Dr. Steve Deckelman