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If the parameters denoted pure translation, then one way to classify y is to determine for which i , the following quantity is maximized: ( ) ( ) s i t y t d t where t = (x,y) denotes spatial coordinates. This is


  1.  If the parameters θ denoted pure translation, then one way to classify y is to determine for which i , the following quantity is maximized:  (  θ ) ( ) s i t y t d t where t = (x,y) denotes spatial coordinates.  This is called a matched filter and it is equivalent to GMLC if all the candidate signals s i had the same magnitude, and we assume additive white Gaussian noise.

  2.  Consider the following compressive acquisition model:       2 , ~ ( 0 , ), y x N  ,      m n m n , , y R x R R m n  The MLC for this case is now:   arg max ( | ) j p y s  { 1 , 2 ,..., } k P k   2   y s 1      i ( | ) exp p y s    i   2 2 2  

  3.  But the compressive measurement y could be acquired from an image which was acquired in a different pose than any of the images in the database.  The GMLC is now given as: ~   θ This has an interesting arg max ( | , ) j p y s  { 1 , 2 ,..., } k P k k name – the smashed filter   ~ 2   θ (derived from the name   ( ) y s ~ 1 i i    θ “matched filter”), taking ( | , ) exp   p y s  i i   2   2 2 into account the   compressive nature of the ~   θ θ measurements. argmax ( | , ) p y s   θ i i

  4.  These are all essentially nearest neighbour classifiers with Euclidean distance.  What is special about these classifiers?  Let us assume that the sensing matrix Φ and hence (matrix Φ U for any orthonormal U ) obey the restricted isometry property.  Then for k -sparse signals s 1 and s 2 and RIC δ 2k , we have: 2 2 2           ( 1 ) ( ) ( 1 ) s s s s s s 2 1 2 1 2 2 1 2 k k

  5. Image source: Davenport et al, “The smashed filter for compressive classification and target recognition”

  6. Image size: 128 x 128 Compressive measurements taken by a Rice single pixel camera. Though the sensing matrix of the camera does not obey RIP since it contains values that are 0 or 1, it can be converted into a matrix with entries that are either -1 or +1. This Image source: Davenport et is by taking two measurements of al, “The smashed filter for the same scene, where the second compressive classification measurement is taken by flipping and target recognition” the 0 and 1 values in the first sensing matrix.

  7.  The procedure is summarized below: Φ Randomly pick a by matrix . m n Repeat until convergenc e {       Φ Φ ΦΨ ( ( )) Pick the step-size adaptively so that you Φ actually descend on the mutual } coherence. ΦΨ ΦΨ ( ) ( )    ΦΨ j ( ) max i  i j ΦΨ ΦΨ ( ) ( ) i j 2 2

  8.  The main problem is how to find a derivative of the “max” function which is non - differentiable!  Use the softmax function which is differentiable:    1 n     n lim log exp( ) max{ } x x      1 i i i    1 i

  9.  This method does not directly target  but instead considers the Gram matrix D T D where D  ΦΨ with all columns unit - normalized .  The aim is to design Φ in such a way that the Gram matrix resembles the identity matrix as much as possible, in other words we want:  Ψ T Φ T ΦΨ I

  10. Note:  need not be  Ψ Φ ΦΨ T T orthonormal and hence   T I need not be identity.   ΨΨ Φ ΦΨΨ ΨΨ T T T T  ΨΨ ΛV T T (SVD) V     Φ ΦVΛV T T T T V V V V     Φ ΦV T T T T V V V V V     Φ ΦV T T V   ΦV      T 2       T minimize w.r.t. F

  11.   ΦV      T 2       T minimize w.r.t. F E j      Consider diag( , ,..., ) 1 2 n      Rank one matrix T ( | | ... | m ) 1 2 2 We want a rank one  2             matrix that T t t where i i j j approximates E j as F  i j closely as possible in F         the Frobenius sense. ( , ,..., ) 1 , 1 2 , 2 , i i i n i n The solution lies in SVD!

  12.  SVD gives us the best possible rank- r approximation to any matrix (it may or may not be a natural image matrix).  In other words, the solution to the following optimization problem: 2 ˆ ˆ A A A    min where rank( ) min( , ) r,r m n ˆ A F is given using the SVD of A as follows:   r Note: We are using the singular vectors ˆ A S u v A USV  t T , where corresponding to the r largest singular ii i i  i 1 values. This property of the SVD is called the EckartYoung Theorem . 12

  13. E j Rank one matrix 2  2             T t t where i i j j F  i j F         ( , ,..., ) 1 , 1 2 , 2 , i i i n i n We want a rank one matrix that approximates Ej as closely as possible in the Frobenius sense. The solution lies in SVD.    T t E USU S u u j kk k k Assuming that S 11 is the k largest singular value   t S u 11 1 j

  14.  Initialize Φ to a random matrix.  By SVD, we have  T = V  V T  We have Γ = Φ V .  For each j = 1 to m :  Compute E j  Find the largest singular value and corresponding singular vector of E j  Use these to update Γ (via τ j ).  Compute the optimal Φ = Γ V T . Ajit Rajwade

  15.  Leading to smaller values of column-column dot products  Better reconstruction errors  For more details refer to figure 1 and table 1 of “ Learning to Sense Sparse Signals: Simultaneous Sensing Matrix and Sparsifying Dictionary Optimization ” by Duarte - Carvajalino and Sapiro. Ajit Rajwade

  16.  It is the task of reconstructing a 2D image (object) from its 1D projections, or a 3D image (object) from its 2D projections.  What is a projection (also called tomographic projection)?  It is defined as the Radon transform of the image in a particular direction (see next slide).

  17. Imagine a line was drawn through the 2D image in a certain direction α , and you integrated the intensity values along that line. Now you repeat this for lines parallel to the original one but at different offsets. Each such summation produces a bin of the tomographic projection. The collection of bins form a 1D array which is called the tomographic projection or the Radon transform of the object in the https://en.wikipedia.org/wiki/Rado direction α . n_transform

  18. The degree of absorption of detector Xray-beam X-Rays at each point is measured by an X-Ray absorption detector. This detector produces a 1D signal whose amplitude/intensity is directly proportional to the extent of absorption. Any point in the signal = sum of the absorptivity values across the path of a single ray in the X-Ray beam that spatially maps onto that point. The image is a simplification of a set of real biological tissues: example, an organ/tumor surrounded by a background consisting of soft, uniform tissue. The set of tissues is bombarded with an X-Ray beam. The tumor has higher rate of absorption as compared to the surrounding tissue.

  19. detector Xray-beam Given the 1D signal (called a projection signal ), we try to reconstruct the original 2D image by smearing backwards along the direction of projection. This is called as back-projection . The 1D signal that was measured is duplicated along the columns of the image to be estimated (see the directions marked in yellow). Sum-total of the two back-projections

  20.  Given projections in K different directions, we can hope to reconstruct the original image by performing back-projection along all these directions, and adding up the results.  Back-projection refers to smearing the 1D projection back across the 2D image, i.e. duplicating the 1D signal across the image in a direction perpendicular to the direction of projection.  The shape of the object will be approximated better and better as K increases.

  21. Even with many (32) back-projections, there is a blur artifact in the reconstruction. This is called as a “halo effect”.

  22.  A 3D object is illuminated with a large cone-shaped X- Ray beam. This will produce a projection which is a 2D- image.  Changing the direction of the X-ray beam will produce another image. This set of images when back-projected will yield the 3D volume/object.  However in conventional Computed Tomography (CT), each slice of the volume is measured at a time. A slice is a 2D entity obtained by cutting the 3D volume transversely through a plane parallel to the XY plane.  This allows for the employment of a smaller number of detectors at a time, for the same resolution of the measurement.

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