EECS490: Digital Image Processing. Lecture #16

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1 Lecture #16 Wiener Filters Constrained Least Squares Filter Computed Tomography Basics Reconstruction and the Radon Transform Fourier Slice Theorem Filtered Backprojections Fan Beams

2 Motion Blurring Model motion in x- and y-directions over a period T for an integrating detector such as a camera. g( x, y)= f x x t T Gu,v = g x, y (), y y () t e j 2 ux+vy T = f x x t T Gu,v = f x x t dt Fourier transform g(x,y) and reverse order of integration dxdy (), y y () t (), y y () t dt e e dxdy j 2 ux+vy dxdydt j 2 ux+vy

3 Motion Blurring Replace innter term by F(u,v), the Fourier transform of f(x,y) T Gu,v = Fu,v e j 2 ( ux ()+vy t () t ) dxdy dt H( u,v)= e j 2 ux t T ( ()+vy () t ) dt Gu,v = H( u,v)fu,v e j 2 ux t = Fu,v Identify the motion blurring transfer function as We can then model motion degradation as Where, for x (t)=at/t, y (t)= T ( ()+vy () t ) dt T H( u,v)= e j 2ux () at t j 2u dt = e T dt = T ua T jua sin( ua)e

4 Modeling Image Degradation Original image (1st edition cover) Motion blurring with a=b=.1 and T=1 Motion blurring transfer function H( u,v)= T ua + vb sin ua + vb j ( ua+vb ) e

5 Inverse Filtering If degraded image is given by degradation + noise Gu,v = H( u,v)fu,v + N( u,v) Estimate the image by dividing by the degradation function H(u,v) F ( u,v)= Gu,v Fu,v + N ( u,v ) = Fu,v + N( u,v) H u,v H u,v H u,v = H u,v We can never recover F(u,v) exactly: 1. N(u,v) is not known since (x,y) is a r.v. estimated 2. If H(u,v) -> then noise term will dominate. Helped by restricting analysis to (u,v) near origin.

6 Modeling of Degradation 48x48 No radial limiting of H(u,v) H(u,v) cut off at R=4 H(u,v) cut off at R=7 H(u,v) cut off at R=85 F ( u,v)= Fu,v + N u,v H u,v whereh ( k u u,v )= e M v N

7 Wiener Filter Minimize e 2 = E {( f ˆf ) 2 } Assuming: 1. f and n are uncorrelated 2. f and/or n is zero mean 3. gray levels in f are a linear function of the gray levels in f The best estimate ˆF ( u,v) is then given by ˆF ( u,v)= ˆF ( u,v)= S f H *( u,v)s f u,v ( u,v) H u,v 1 H u,v 2 + S ( u,v) H( u,v) S ( u,v) ( u,v) H u,v S f Gu,v ( )= H * u,v H u,v S f Gu,v 2 + S ( u,v) ( u,v) Gu,v H(u,v) = degradation function H*(u,v) = complex conjugate of H H(u,v) = H*(u,v) H(u,v) S (u,v)= N(u,v) 2 =power spectrum of noise (estimated) S f (u,v)= F(u,v) 2 =power spectrum of original image (not known)

8 Modeling of Degradation Inverse filtering N u,v H u,v Radially limit at D o =75 ˆF ( u,v)= Wiener filtering 1 H u,v H( u,v) K H u,v Gu,v In practice we don t know the power spectrum S f (u,v)= F(u,v) 2 of the original image so we replace the S /S f term with a constant K which we vary

9 Modeling of Degradation Inverse filtering w/o noise modeling Wiener filtering w/ noise modeling Image + motion blur + Gaussian noise ( 2 =65) Reduce 2 to 65 Reduce 2 to.65

10 Constrained Least Squares Filter g = Hf + g = [g row1 (x,y) g row2 (x,y) g roww (x,y)... g rown (x,y) ] f, have the same form and dimensions as g, i.e., MNx1 H has dimensions MNxMN which is VERY large Mininize smoothness constraint C = M 1 N 1 x= y= 2 f x, y g Hˆf = 2 = T 2 Solution (Castleman, 1996) ˆF ( u,v)= H *( u,v) 2 + Pu,v 2 H u,v Pu,v = F p x, y p( x, y)= Gu,v

11 Constrained Least Square Filter Image + motion blur + Gaussian noise ( 2 =65) Reduce 2 to 65 Reduce 2 to.65 Does a nice job of removing motion degradation and noise

12 Constrained Least Square Filter Correct noise parameters Wrong noise parameters can be determined interactively but the optimum value of can be determined from the mean and variance of the noise. Knowing the correct noise parameters is important to the process.

13 Computed Tomography 1. Forward projection from a source through an absorber to a linear detector array 2. Backproject the signal back to the source to reconstruct the absorber 3. Rotate the source and detector array 9 and measure a new forward projection through the same absorber 4. Sum the two back projections to reconstruct a better image of the absorber

14 Reconstruction of a Single Object Absorber Reconstruction using 1 backprojection Reconstruction using 2 backprojections 45 apart Reconstruction using 3 backprojections 45 apart Reconstruction using 32 backprojections apart Reconstruction using 4 backprojections 45 apart

15 Reconstruction of Multiple Objects Darker because forward projection was only through smaller less highly absorbing object Brightest because most absorption projection was through both absorbing objects Large highly absorbing object Reconstruction using 2 backprojections 9 apart Smaller less highly absorbing object Reconstruction using 3 backprojections 45 apart Reconstruction using 32 backprojections apart Reconstruction using 64 backprojections apart

16 Computed Tomography G1: Pencil beam, single source and detector move detector source pair G2: Narrow fan beam, single source, small linear detector array move detector source pair (not as much movement required) G5: G4 with electromagnetically aimed sources to eliminate mechanical movement G6: patient moves linearly through rotation scanner describing a helical scan G7: multi-slice scanners with thick fan beams and 2-D detector arrays G3: Wide fan beam, single source, large linear detector array move detector source pair (not as much movement required) G4: Wide fan beam, rotating source, circular detector array move source around circle

17 Basics of the Radon Transform Describe a line in (x,y) coordinates as: ( x cos + ysin )

18 Basics of the Radon Transform The Radon transform is the line integral (projection) of f(x,y) along a line at an angle + + R { f( x, y) }= g(,)= f( x, y) ( x cos + ysin ) dxdy

19 Basics of the Radon Transform f( x, y)= A x2 + y 2 r 2 otherwise Since the absorber is circularly symmetric we only need to do the projection for = + + g(, )= f( x, y) ( x ) dxdy + = f, y dy = Ady + r 2 2 r 2 2 = 2A r2 2 r otherwise g(,)= 2A r2 2 r otherwise This is the Radon transform g() for =

20 Sinograms The plot of g(, ) as an intensity with on the horizontal and on the vertical is called a sinogram. The sinogram of a bar. The Shepp-Logan phantom is designed to simulate the absorption of a brain with small tumors The sinogram of the Shepp-Logan phantom.

21 Backprojection of Sinograms There is significant blurring using this approach The backprojection for a specific angle is the line The complete reconstruction is the integral over all f ( x, y)= g xcos + ysin, f( x, y)= f ( x, y) d

22 Fundamental CT Reconstruction Physically measure the projection Back project each projection Sum all the projections to generate one image Results in blurred images

23 Fourier Slice Theorem + G(,)= g(,)e j 2 d G(,)= f( x, y) ( x cos + ysin )dxdye j 2 d + + G(,)= f x, y + + G(,)= f( x, y)e + + G(,)= f( x, y)e + e j 2 ddxdy x cos + ysin dxdy j 2 x cos + y sin dxdy j 2 ux+vy Compute the 1-D Fourier transform of g(,) Substitute the expression for the projection g(,) Reverse the order of integration and evaluate This is the Fourier transform of the absorption f(x,y) evaluated at u=cos; v=sin u = cos;v= sin

24 Fourier Slice Theorem G(,)= Fu,v u = cos;v= sin = F cos, sin The Fourier transform of a projection is a slice of the 2-D Fourier transform of the density f(x,y)

25 + + f( x, y)= Fu,v e 2 EECS49: Digital Image Processing Reconstruction Using the Fourier Slice Theorem dudv j 2 ux+vy j 2 x cos + y sin f( x, y)= F( cos, sin )e dd 2 j 2 x cos + y sin f( x, y)= G(,)e dd j 2 x cos + y sin f( x, y)= G(,)e dd j 2 x cos + y sin + G(,)e dd j 2 x cos + y sin f( x, y)= G(,)e dd j 2 x cos + y sin + G(, )e dd j 2 x cos + y sin f( x, y)= G(,)e dd f( x, y)= G(,)e j 2 d The inverse 2-D Fourier transform of F(u,v) = x cos + y sin 2 d Rewrite in polar coordinates Use the Fourier Slice Theorem to recognize G(,) Recombine Rearrange Separate Use symmetry

26 Reconstruction Using the Fourier Slice Theorem f( x, y)= G(,)e j 2 d = x cos + y sin d The inner term is a 1-D inverse Fourier transform. The is a ramp filter (whose inverse Fourier transform does not exist since it is not bounded) modifying the transform. In practice we multiply by another filter, a windowing function, which limits its high frequency response.

27 Filtered Backprojections multiplied by a box function in the frequency domain. Hamming window function Hamming filtered and its inverse Fourier transform

28 Reconstruction Using Filtered Backprojections Physically measure the projection for angle k Compute the 1-D Fourier Transform of each projection Multiply each Fourier Transform by the filter function and an appropriate window, e.g., Hamming Obtain the inverse 1-D Fourier Transform of the windowed, filtered transform Integrate (sum) all the 1-D transforms to generate one image

29 Filtered Backprojection Backprojection filtered by a ramp filter. Backprojection filtered by a Hamming-windowed ramp filter.

30 Filtered Backprojection Backprojection filtered by a ramp filter. Backprojection filtered by a Hamming-windowed ramp filter.

31 Computed Tomography To analyze the fan-beam geometry for G4 and later machines we switch from frequency domain analysis to convolution. f( x, y)= G(,)e j 2 d = x cos + y sin f( x, y)= sg (,) = x cos + y sin d f( x, y)= g(,) s( xcos + ysin )d d = x cos + y sin d This is a convolution of the ramp filter and the corresponding projection

32 Computed Tomography f( x, y)= g(,) s( xcos + ysin )d f( x, y)= T T g(,) s( xcos + ysin )dd = x cos + y sin We recognize that the projections are mirror images and then convert to polar coordinates d

33 Computed Tomography We recognize that the projections are mirror images and then convert to polar coordinates Letx = r cosandy = r sin Then x cos + ysin = r cos cos + r sin sin = r cos f( x, y)= T T g, s rcos( ) dd For this beam geometry = + = sin Using this information we then transform the variables of integration to and f( x, y)= 1 2 sin 1 T D 2 sin 1 T D g( Dsin, + ) s rcos( + ) S sin D cosdd

34 Computed Tomography Since - m <<+ m and we can rotate with no loss of generality f( x, y)= m m p(,) s rcos( + ) Dsin D cosdd This is the fundamental fan beam reconstruction formula where p(,)= g( Dsin, + )

35 Computed Tomography From problem 5.33 r cos( + ) Dsin = Rsin ' f( r,)= 1 2 where 2 m m From problem 5.34 s( Rsin )= f( r,)= 2 1 R 2 p, s Rsin '+ D cosdd Rsin m m 2 s( ) q(,) h[ ' ]d d h( )= 1 2 sin 2 and q, s = p(,)d cos

36 Computed Tomography This is computationally hard to evaluate f( r,)= 2 1 R 2 m m A common simplification is q(,) h[ ' ]d d p(,)= g(,)= g( Dcos, + ) For a discrete system let == pn (,m )= g Dsin n,( m + n)

37 Computed Tomography =1 =.5 =.25 =.125

38 Computed Tomography =1 =.5 =.25 =.125

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