Fourier Reconstruction. Polar Version of Inverse FT. Filtered Backprojection [ ] Bioengineering 280A Principles of Biomedical Imaging

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1 Fourier Reconstruction Bioengineering 280A Principles of Biomedical Imaging Fall Quarter 2014 CT/Fourier Lecture 5 F Interpolate onto Cartesian grid with appropriate density correction then take inverse transform Polar Version of Inverse FT µ(, y) = 2π G(k,k y )e j 2π (k +k y y ) dk dk y = G(k,θ)e j 2π (k cosθ +yk sinθ ) kdkdθ 0 π 0 = G(k,θ)e j 2πk( cosθ +y sinθ ) k dkdθ 0 Note : g(l,θ + π) = g( l,θ) So G(k,θ + π) = G( k,θ) µ(, y) = Filtered Backprojection π 0 π G(k,θ)e j 2π (k cosθ +yk sinθ ) k dkdθ = k G(k,θ)e j 2πkl dkdθ 0 π = g (l,θ)dθ 0 where l = cosθ + y sinθ g (l,θ) = k G(k,θ)e j 2πkl dk = g(l,θ) F 1 k = g(l,θ) q(l) [ ] Backproject a filtered projection 1

2 Fourier Interpretation Ram-Lak Filter Density N circumference N 2π k Low frequencies are oversampled. So to compensate for this, multiply the k-space data by k before inverse transforming. Typically will pick k ma =1/Δs where Δs is the beamwidth Kak and Slaney; Reconstruction Path Reconstruction Path * Filtered F F F -1 Back- Project F -1 Filtered Back- Project F -1 Filtered 2

3 Eample Eample Prince and Links 2005 Kak and Slaney Eample Filters Filtered s Prince and Links

4 CT Sampling Requirements View Aliasing What should the size of the detectors be? How many detectors (or lines) do we need? How many views do we need? Kak and Slaney 4

5 Aliasing Artifacts Object Effect of Noise Aliasing due to insufficient number of detectors Aliasing due to insufficient number of views Kak and Slaney Analog vs. Digital The Analog World: Continuous time/space, continuous valued signals or images, e.g. vinyl records, photographs, -ray films. The Digital World: Discrete time/space, discrete-valued signals or images, e.g. CD-Roms, DVDs, digital photos, digital -rays, CT, MRI, ultrasound. The Process of Sampling g() Δ sample g[n]=g(n Δ) 5

6 Questions Sampling in the Time Domain How finely do we need to sample? What happens if we don t sample finely enough? Can we reconstruct the original signal or image from its samples? Comb Function comb() = δ( n) n= Scaled Comb Function comb # & % ( = δ( $ Δ ' Δ n) n= = δ( nδ ) Δ n= = Δ δ( nδ) n= Other names: Impulse train, bed of nails, shah function. Δ 6

7 1D spatial sampling g S () = g() 1 Δ comb # & % ( $ Δ ' Recall the sifting property = g() δ( nδ) n= = g(nδ)δ( nδ) n= g()δ( a) = g(a) But we can also write g(a)δ( a) = g(a) δ( a) = g(a) So, g()δ( a) = g(a)δ( a) 1D spatial sampling g() Δ comb(/δ)/ Δ g S () Fourier Transform of comb() F[ comb() ] = comb(k ) = δ(k n) n= Fourier Transform of comb(/ Δ) comb(/ Δ)/ Δ # F 1 Δ comb( $ % Δ ) & ' ( = 1 Δ Δcomb(k Δ) = δ(k Δ n) n= = 1 Δ n= δ(k n Δ ) 1/Δ Δ 1/Δ F comb(k Δ) k 7

8 Fourier Transform of g S () [ ] = F g() 1 Δ comb % F g S () ) + * # &, (. $ Δ '- ) = G(k ) F 1 Δ comb # &, + % (. * $ Δ '- = G(k ) 1 Δ = 1 Δ = 1 Δ n= n= n= # δ k n & % ( $ Δ ' # G(k ) δ k n & % ( $ Δ ' # G k n & % ( $ Δ ' 1/Δ Fourier Transform of g S () G(k ) k G S (k ) 1/Δ k Nyquist Condition Eample -B B G(k ) k G S (k ) Assume that the highest spatial frequency in an object is B = 2 cm -1. Thus, smallest spatial period is 0.5 cm.. Nyquist theorem says we need to sample with Δ < 1/2B = 0.25 cm =1/Δ k This corresponds to 2 samples per spatial period. To avoid overlap, we require that 1/Δ>2B or > 2B where =1/ Δ is the sampling frequency 8

9 Reconstruction from Samples G S (k ) Eample Cosine Reconstruction cos(2πk 0 ) -k 0 k 0 >2k 0 =1/Δ Multiply by (1/ )rect(k / ) -k 0 k 0 (1/ ) G S (k )rect(k / ) =2k 0 =G(k ) -k 0 k 0 Reconstruction from Samples If the Nyquist condition G ˆ S (k ) = 1 is met, then G S (k )rect(k / ) = G(k ) And the signal can be reconstructed by convolving the sample with a sinc function g() Reconstruction from Samples Sample at Δ g S () g ˆ S () = g S () sinc(k s ) ( + = * g(nδx)δ( nδx) - sinc(k s ) ), n= = g(nδx) sinc(k s ( nδ)) n= ˆ g S () sinc(k s ) = sinc( / Δ) 9

10 Cosine Eample with =2k 0 Eample with K s =4k 0 Eample with K s =8k 0 Aliasing -B B G(k ) k Aliasing occurs when the Nyquist condition is not satisfied. This occurs for 2B 10

11 Aliasing Eample Aliasing Eample cos(2πk 0 ) -k 0 k 0 =k 0 -k 0 k 0 Aliasing Eample cos(2πk 0 ) -k 0 k 0 2k 0 > >k 0 -k 0 k 0 11

12 Eample Eample 1. Consider the function g() = cos 2 ( 2πk 0 ). Sketch this function. You sample this signal in the spatial domain with a sampling rate =1/Δ (e.g. samples spaced at intervals of Δ ). What is the minimum sampling rate that you can use without aliasing? Give an intuitive eplanation for your answer. Assume that the Nyquist sampling periods of f () and g() are Δf and Δg, respectively. Determine the Nyquist sampling periods for a) f ( 0 ) b) f ()+ g() c) f ()* f () PollEv.com/be280a PollEv.com/be280a from Prince and Links 2006 Detector Sampling Requirements Smoothing of Beam Width 2/(Δs) Sampling interval Δr Beamwidth Δs Smoothed 12

13 Smoothing of g s (l,θ) = rect(l /Δs) g( l,θ) G s (k,θ) = Δssinc(k Δs)G(k,θ) Smoothed Sampling Requirements Detectors Δr Δs/2 Sampled Smooth View Aliasing View Sampling Requirements View Sampling -- how many views? Basic idea is that to make the maimum angular sampling the same as the projection sampling. πfov N views = Δr N views,360 = πfov Δr = πn proj (for 360 degrees) N views,180 = πn proj 2 (for 180 degrees) Kak and Slaney 13

14 beamwidth Δs = 1 mm Eample Field of View (FOV) = 50 cm Δr = Δs/2 = 0.5 mm 500 mm/ 0.5 mm = N = 1000 detector samples π * N = 3146 views per 360 degrees 1500 views per 180 degrees CT "Rule of Thumb" N view = N det ectors = N piels Eample Consider a rectangular object of width 20mm and height 40mm centered at (-10mm, -10mm). The attenuation coefficient of the object is 1 mm -1. The object is imaged with a 1 st generation CT scanner with a beamwidth of 1mm. The desired FOV is 100 mm. Determine the appropriate detector size Δr and the number of radial samples needed to span the FOV. Assume that the middle two samples are acquired at coordinates of -Δr/2 and Δr/2. Determine the number of angular samples required. PollEv.com/be280a 14