The document discusses convolution and its applications in imaging. Convolution describes how one function is modified by another function. In imaging, the point spread function (PSF) describes how a point object is blurred by the imaging system. The image formation process can be described as the convolution of the object with the PSF plus noise. Fourier analysis is useful because convolution in the spatial domain is equivalent to multiplication in the frequency domain. This allows filtering techniques to be used to modify images. Examples are provided of using Fourier analysis to determine the PSF and modulation transfer function of an imaging system.

1. Convolution
Fourier Convolution
Outline
• Review linear imaging model
• Instrument response function vs Point spread function
• Convolution integrals
• Fourier Convolution
• Reciprocal space and the Modulation transfer function
• Optical transfer function
• Examples of convolutions
• Fourier filtering
• Deconvolution
• Example from imaging lab
• Optimal inverse filters and noise

2. Instrument Response Function
The Instrument Response Function is a conditional mapping, the form of the map
depends on the point that is being mapped.
This is often given the symbol h(r|r’).
Of course we want the entire output from the whole object function,
and so we need to know the IRF at all points.
IRF(x,y | x0, y0 ) = S δ(x − x0 )δ(y − y0 ){ }
E(x,y) =
∞
∫∫
−∞
∞
∫∫
−∞
I(x,y)S δ(x − x0 )δ(y − y0 ){ } dxdydx0dy0
E(x,y) =
∞
∫∫
−∞
∞
∫∫
−∞
I(x,y)IRF(x, y | x0,y0 )dxdydx0dy0

3. Space Invariance
Now in addition to every point being mapped independently onto the detector,
imaging that the form of the mapping does not vary over space (is independent of
r0). Such a mapping is called isoplantic. For this case the instrument response
function is not conditional.
The Point Spread Function (PSF) is a spatially invariant approximation of the IRF.
IRF(x,y | x0, y0 ) = PSF(x − x0,y − y0 )

4. Space Invariance
Since the Point Spread Function describes the same blurring over the entire sample,
The image may be described as a convolution,
or,
IRF(x,y | x0, y0 ) ⇒ PSF(x − x0,y − y0 )
E(x,y) =
∞
∫∫
−∞
I(x0,y0 )PSF(x − x0,y − y0 )dx0dy0
Image(x, y) =Object(x,y)⊗ PSF(x,y)+noise

5. Convolution Integrals
Let’s look at some examples of convolution integrals,
So there are four steps in calculating a convolution integral:
#1. Fold h(x’) about the line x’=0
#2. Displace h(x’) by x
#3. Multiply h(x-x’) * g(x’)
#4. Integrate
f (x) = g(x) ⊗ h(x) =
−∞
∞
∫ g(x')h(x − x')dx'

6. Convolution Integrals
Consider the following two functions:
#1. Fold h(x’) about the line x’=0
#2. Displace h(x’) by x x

7. Convolution Integrals

8. Convolution Integrals
Consider the following two functions:

9. Convolution Integrals

10. Some Properties of the Convolution
commutative:
associative:
multiple convolutions can be carried out in any order.
distributive:
f ⊗ g = g ⊗ f
f ⊗ (g ⊗ h) =( f ⊗ g) ⊗ h
f ⊗ (g +h) = f ⊗ g + f ⊗ h

11. Recall that we defined the convolution integral as,
One of the most central results of Fourier Theory is the convolution
theorem (also called the Wiener-Khitchine theorem.
where,
Convolution Integral
f ⊗ g = f (x)g(x'−x)dx
−∞
∞
∫
ℑ f ⊗ g{ } = F(k)⋅G(k)
f (x) ⇔ F(k)
g(x) ⇔ G(k)

12. Convolution Theorem
In other words, convolution in real space is equivalent to
multiplication in reciprocal space.
ℑ f ⊗ g{ } = F(k)⋅G(k)

13. Convolution Integral Example
We saw previously that the convolution of two top-hat functions
(with the same widths) is a triangle function. Given this, what is
the Fourier transform of the triangle function?
-3 -2 -1 1 2 3
5
10
15
=
-3 -2 -1 1 2 3
5
10
15
?

14. Proof of the Convolution Theorem
The inverse FT of f(x) is,
and the FT of the shifted g(x), that is g(x’-x)
f ⊗ g = f (x)g(x'−x)dx
−∞
∞
∫
f (x) =
1
2π
F(k)eikx
dk
−∞
∞
∫
g(x'−x) =
1
2π
G(k')eik'(x'−x)
dk'
−∞
∞
∫

15. Proof of the Convolution Theorem
So we can rewrite the convolution integral,
as,
change the order of integration and extract a delta function,
f ⊗ g = f (x)g(x'−x)dx
−∞
∞
∫
f ⊗ g =
1
4π2
dx F(k)eikx
dk
−∞
∞
∫
−∞
∞
∫ G(k')eik'(x'−x)
dk'
−∞
∞
∫
f ⊗ g =
1
2π
dkF(k) dk'G(k')eik'x' 1
2π
eix(k−k')
dx
−∞
∞
∫
δ(k−k')
1 24 4 34 4−∞
∞
∫
−∞
∞
∫

16. Proof of the Convolution Theorem
or,
Integration over the delta function selects out the k’=k value.
f ⊗ g =
1
2π
dkF(k) dk'G(k')eik'x' 1
2π
eix(k−k')
dx
−∞
∞
∫
δ(k−k')
1 24 4 34 4−∞
∞
∫
−∞
∞
∫
f ⊗ g =
1
2π
dkF(k) dk'G(k')eik'x'
δ(k −k')
−∞
∞
∫
−∞
∞
∫
f ⊗ g =
1
2π
dkF(k)G(k)eikx'
−∞
∞
∫

17. Proof of the Convolution Theorem
This is written as an inverse Fourier transformation. A Fourier
transform of both sides yields the desired result.
f ⊗ g =
1
2π
dkF(k)G(k)eikx'
−∞
∞
∫
ℑ f ⊗ g{ } = F(k)⋅G(k)

18. Fourier Convolution

19. Reciprocal Space
real space reciprocal space

20. Filtering
We can change the information content in the image by
manipulating the information in reciprocal space.
Weighting function in k-space.

21. Filtering
We can also emphasis the high frequency components.
Weighting function in k-space.

22. Modulation transfer function
i(x, y) = o(x,y)⊗ PSF(x,y) + noise
c c c c
I(kx,ky ) =O(kx,ky )⋅ MTF(kx ,ky )+ℑ noise{ }
E(x,y) =
∞
∫∫
−∞
∞
∫∫
−∞
I(x,y)S δ(x − x0 )δ(y − y0 ){ } dxdydx0dy0
E(x,y) =
∞
∫∫
−∞
∞
∫∫
−∞
I(x,y)IRF(x, y | x0,y0)dxdydx0dy0

23. Optics with lens
‡ Input bit mapped image
sharp =Import @" sharp . bmp" D;
Shallow @InputForm @sharp DD
Graphics@Raster@<< 4>>D, Rule@<< 2>>DD
s = sharp @@1, 1DD;
Dimensions @sD
8480, 640<
ListDensityPlot @s , 8PlotRange ÆAll , Mesh ÆFalse <D

24. Optics with lens
crop =Take @s , 8220 , 347 <, 864, 572 , 4<D;
‡ look at artifact in vertical dimension
rot =Transpose @crop D;
line =rot @@20DD;
ListPlot @line , 8PlotRange ÆAll , PlotJoined ÆTrue <D
20 40 60 80 100 120
40
60
80
100
120
Ö GraphicsÖ0 20 40 60 80 100 120
0
20
40
60
80
100
120

25. Optics with lens
20 40 60 80 100 120
40
60
80
100
120
ü Fourier transform of vertical line to show modulation
ftline =Fourier @line D;
ListPlot @RotateLeft @Abs@ftline D, 64 D, 8PlotJoined Æ
20 40 60 80 100 120
2.5
5
7.5
10
12.5
15
Ö Graphics Ö
ListPlot @RotateLeft @Abs@ftline D, 64 D, 8PlotRange ÆAll , PlotJoined ÆTrue <D
20 40 60 80 100 120
200
400
600
800
1000
Ö Graphics Ö

26. Optics with lens
2D FT
0 20 40 60 80 100 120
0
20
40
60
80
100
120
ftcrop =Fourier @crop D;
0 20 40 60 80 100 120
0
20
40
60
80
100
120

27. Optics with lens
Projections
0 20 40 60 80 100 120
0
20
40
60
80
100
120
20 40 60 80 100 120
200
400
600
800
1000
20 40 60 80 100 120
200
400
600
800
1000

28. Optics with lens
Projections
0 20 40 60 80 100 120
0
20
40
60
80
100
120
xprojection = Fourier @RotateLeft @rotft @@64 DD, 64 DD;
20 40 60 80 100 120
50
100
150
200
250
300
20 40 60 80 100 120
50
100
150
200

29. Optics with pinhole

30. 0 20 40 60 80 100 120
0
20
40
60
80
100
120
0 20 40 60 80 100 120
0
20
40
60
80
100
120

31. 2D FT
0 20 40 60 80 100 120
0
20
40
60
80
100
120
0 20 40 60 80 100 120
0
20
40
60
80
100
120

32. Optics with pinhole
Projections
0 20 40 60 80 100 120
0
20
40
60
80
100
120
20 40 60 80 100 120
200
400
600
800
1000
1200
20 40 60 80 100 120
200
400
600
800
1000
1200
1400

33. Projections
20 40 60 80 100 120
50
100
150
200
250
300
20 40 60 80 100 120
50
100
150
200
20 40 60 80 100 120
100
150
200
250
20 40 60 80 100 120
100
150
200

34. Deconvolution to determine MTF of Pinhole
20 40 60 80 100 120
200
400
600
800
1000
20 40 60 80 100 120
200
400
600
800
1000
1200
MTF =Abs@pinhole Dê Abs@image D;
ListPlot @MTF, 8PlotRange ÆAll , PlotJoined ÆTrue <
20 40 60 80 100 120
2
4
6
8
10
Ö GraphicsÖ

35. FT to determine PSF of Pinhole
MTF =Abs@pinhole Dê Abs@image D;
ListPlot @MTF, 8PlotRange ÆAll , PlotJoined ÆTrue <D
20 40 60 80 100 120
2
4
6
8
10
Ö GraphicsÖ PSF =Fourier @RotateLeft @MTF, 64DD;
ListPlot @RotateLeft @Abs@PSFD, 64D, 8PlotRange ÆAll , PlotJoined ÆTrue <D
20 40 60 80 100 120
1
2
3
4
5
6
7
Ö GraphicsÖ

36. Filtered FT to determine PSF of Pinhole
MTF =Abs@pinhole Dê Abs@image D;
ListPlot @MTF, 8PlotRange ÆAll , PlotJoined ÆTrue <D
20 40 60 80 100 120
2
4
6
8
10
Ö GraphicsÖFilter = Table @Exp@- Abs@x - 64Dê 15D, 8x , 0, 128 <D êê N;
ListPlot @Filter , 8PlotRange ÆAll , PlotJoined ÆTrue <D
20 40 60 80 100 120
0.2
0.4
0.6
0.8
1
Ö GraphicsÖ
20 40 60 80 100 120
0.25
0.5
0.75
1
1.25
1.5