22.058 - lecture 4, convolution and fourier convolution convolution fourier convolution outline ...

22.058 - lecture 4, Convolution and Fourier Convolution

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


Instrument Response FunctionThe 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){ }

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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


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.

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IRF(x,y|x0,y0)=PSF(x−x0,y−y0)


Space Invariance

Since the Point Spread Function describes the same blurring over the entire sample,

The image may be described as a convolution,

or,

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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


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'



Consider the following two functions:

#1. Fold h(x’) about the line x’=0

#2. Displace h(x’) by x x



Consider the following two functions:


Some Properties of the Convolution

commutative:

associative:

multiple convolutions can be carried out in any order.

distributive:

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f ⊗g=g⊗ f

€

f ⊗(g⊗h)=( f ⊗g)⊗h

€

f ⊗(g+h)=f ⊗g+f ⊗h


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)


Convolution Theorem

In other words, convolution in real space is equivalent to multiplication in reciprocal space.

€

ℑ f ⊗g{ }=F(k)⋅G(k)


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 ?


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(x)=12π

F(k)eikxdk−∞

∞∫

€

g(x'−x)=12π

G(k')eik'(x'−x)dk'−∞

∞∫



So we can rewrite the convolution integral,

as,

change the order of integration and extract a delta function,

€


∞∫

€

f ⊗g=1

4π2 dx F(k)eikxdk−∞

∞∫

−∞

∞∫ G(k')eik'(x'−x)dk'

−∞

∞∫

€

f ⊗g=12π

dkF(k) dk'G(k')eik'x'1

2πeix(k−k')dx

−∞

∞∫

δ(k−k')1 2 4 4 3 4 4 −∞

∞∫

−∞

∞∫



or,

Integration over the delta function selects out the k’=k value.

€

f ⊗g=12π

dkF(k) dk'G(k')eik'x'1

2πeix(k−k')dx

−∞

∞∫

δ(k−k')1 2 4 4 3 4 4 −∞

∞∫

−∞

∞∫

€

f ⊗g=12π

dkF(k) dk'G(k')eik'x'δ(k−k')−∞

∞∫

−∞

∞∫

€

f ⊗g=12π

dkF(k)G(k)eikx'

−∞

∞∫



This is written as an inverse Fourier transformation. A Fourier transform of both sides yields the desired result.

€

f ⊗g=12π

dkF(k)G(k)eikx'

−∞

∞∫

€

ℑ f ⊗g{ }=F(k)⋅G(k)


Fourier Convolution


Reciprocal Space

real space reciprocal space


Filtering

We can change the information content in the image by manipulating the information in reciprocal space.

Weighting function in k-space.


Filtering

We can also emphasis the high frequency components.

Weighting function in k-space.


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


Optics with lens‡ Input bit mapped image

sharp =Import @"sharp. bmp"D;

Shallow @InputForm@sharpDD

Graphics@Raster@<< 4>>D, Rule@<< 2>>DD

s = sharp @@1, 1DD;

Dimensions @sD

8480, 640<

ListDensityPlot @s, 8PlotRange ÆAll , MeshÆFalse <D


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

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üFourier transform of vertical line to show modulation

ftline =Fourier @line D;

ListPlot @RotateLeft @Abs@ftline D, 64D, 8PlotJoined ÆTrue <D

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ListPlot @RotateLeft @Abs@ftline D, 64D, 8PlotRange ÆAll , PlotJoined ÆTrue <D

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2D FT

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ftcrop =Fourier @crop D;

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xprojection = Fourier @RotateLeft @rotft @@64DD, 64DD;

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2D FT

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Deconvolution to determine MTF of Pinhole

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MTF=Abs@pinhole DêAbs@imageD;

ListPlot @MTF, 8PlotRange ÆAll , PlotJoined ÆTrue <D

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FT to determine PSF of PinholeMTF=Abs@pinhole DêAbs@imageD;


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Ö Graphics Ö PSF =Fourier @RotateLeft @MTF, 64DD;

ListPlot @RotateLeft @Abs@PSFD, 64D, 8PlotRange ÆAll , PlotJoined ÆTrue <D

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Filtered FT to determine PSF of PinholeMTF=Abs@pinhole DêAbs@imageD;


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Ö Graphics ÖFilter = Table @Exp@-Abs@x - 64Dê15D, 8x, 0, 128<DêêN;

ListPlot @Filter , 8PlotRange ÆAll , PlotJoined ÆTrue <D

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22.058 - lecture 4, convolution and fourier convolution convolution fourier convolution outline ...

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