1.Prove that a two-dimensional Gaussian filter can be extended by two one –dimensional Gaussian.SOLUTION:We know that the 2D Gaussian filter is22(2222),(δδyxeA y x h +-⋅= （1.1）accordingly ，we can get its another form)]()([),(21y h x h A y x h ⋅⋅= (1.2)where2221)(δxex h -= and2222)(δyey h -= (1.3)Obviously, h 1(x) and h 2(y) are one-dimensional Gaussian filter. Thus, it can express x or y direction filter of two-dimensional image.Thus, a two-dimensional Gaussian filter can be extended by two one –dimensional Gaussian.2. Find a formula to project ground maps onto an image of bird view in Cartesian coordinates,assuming that the camera is located at height of h above the origin of ground coordinate systems with the optical axis in downward direction. SOLUTION:Fig. 1See Fig.1. Origin of image coordinate system is located at (0, 0, f) of scene coordinate system (f is focus length of the camera), the X- and Y-axes of image coordinates are parallel with x- and y- axes of scene coordinates respectively.If the image point is(X, Y, Z), and Z=f, it’s scene is (x 0,y 0,z 0), and z=h.From above figure, the length of arc 'O A(C) is approximate to y 0.Thus00fy fy Y z h==. If f=1, then 0y Y h =Similarly, we can get00fx fx X zh==If f=1, then 0x X h=3D arc 'O A is approximated by line l , is projected onto image plane through projecting plane to getimage line L :Let we use a vector (,,)n A B C =to represent the image line L of l . Since0x X h=, 0y Y h=We have000x y A BC h h++= or 000Ax By Ch ++=It is the equation of projecting plane, the normal is (A , B , Ch ).Thus, we can use the normal of the projecting plane of 3D line to denote image plane byn= (A , B , C )3.A 8×8 image f[i, j] has gray levels given by the equation f[i, j] = |i – j |; i, j = 0, 1, 2, … 7. Find out the output images obtained by 3×3 mid-value filter and median filter over the image, respectively. SOLUTION:From the given equation, we can get 8×8 imageSetting image pixel with intensity2102N I I I +=where 21N I and I are the smallest pixel value and the biggest pixel value under themask ,respectively. N is the order of mask. Note that the edge pixels of the image were dealt with only by the part of pixels under the mask because of the absence of other pixels. Thus, the output image by mid-value filter isSetting the image pixel with intensity2012(,,,)N I m edian I I I …Thus, the output image by median filter is4.Find a method to remove the underlines in text documents; assuming that the documents had been well aligned with the image coordinates and thinned to skeletons of a single pixel width.The Techniques of Image ProcessingSOLUTION:By observing the figure ’s feature, I propose following algorithm:(1) we find the row that has the most pixels. For the sake of expressing convenience, set the rownumber is P.(2) Detect this row’s pixels (P, j) with a 8-neighbors template whose centre pixel is P row ’s pixels. (3) If the non-level neighbor pixel’s value of Pth row ’s is 1, go to the next pixel. Otherwise, signthis pixel with PD.(4) After detecting the whole row, go back to the fist pixel signed by PD and replace them with“0”pixels.(5) Perform opening operation to the image had been dealt with, then perform closing operation to the resulting image. (This step aims to delete the pixels which were reserved wrongly and smooth the image. )5.Suggest an algorithm to compute the skew angle of texts in a document image if the skew angle is small.⎪⎪⎪⎭⎫ ⎝⎛=321ˆn n n nSOLUTION:Fig. 2By looking at Fig.2, a normal width of a row is W 1, length is L. Because of the skew, the projection width W 2 is wider than the normal width W 1. Thus, when the skew angle is small, we can have21arctanW W Lθ-≈.6.Some remote sensing cameras have image border effect that makes the average pixel intensity darker when the pixels are neat to the image border. Write an algorithm to correct the image border effect.SOLUTION:Dividing the whole image into 16×16 subimage and making the median value as the subimage value. Build conicoid22(,)I x y ax bxy cy dx ey g =+++++ General mean square error is222,[()(,)]x y sax bxy cy dx ey g f x y ε∈=+++++-∑Minimize the mean square error by0aε∂=∂,0bε∂=∂,0cε∂=∂,0dε∂=∂,0eε∂=∂,0gε∂=∂Thus, we can get the coefficients: a, b, c, d, e, g. Then, let'(,)(,)1(,)kf x y f x y uI X Y =+Choose the suitable coefficients k and u, we can get new image which has got rid of the darker border.7. Given a rotation Matrix in the form find the corresponding rotation axis nˆ and rotation angle θ. SOLUTION:We know the rotation axis nˆ isMoreover: 2221231n n n ++=The element of R, rotation axis nˆ and rotation angle θ has a relationship as following:2221111(1)cos r n n θ=+- (1)12123(1cos )sin r n n n θθ=-- (2) 13132(1cos )sin r n n n θθ=-+ (3) 21123(1cos )sin r n n n θθ=-+ (4)222222(1)cos r n n θ=+- (5)23231(1cos )sin r n n n θθ=-- (6) 31132(1cos )sin r n n n θθ=-- (7) 32231(1cos )sin r n n n θθ=-+ (8)223333(1)cos r n n θ=+- (9)Thus, (4)-(2) 211232sin r r n θ-= (10) and (3)-(7) 133122sin r r n θ-= (11) and (8)-(6) 322312sin r r n θ-= (12) square and plus to both sides of above 3 equations. We can get222222211213313223321()()()4()s i n r r r r r r n n n θ-+-+-=++ where: 2221231n n n ++=Thus, we have2323s i n )a r c s )θθ==From (10)~(12) get1r r n -=2n =3n =8. Given a set of 3D point correspondences (P i , P’i ), i = 1, 2, …n; n ≥3, on an object which isuniformly changing size while it moving, described by ()P s RP T '=+, find the scale factor s andmotion parameters R and T.SOLUTION:We know ()i i P s RP T '=+Consider the initial image centroid is 11nii C P n==∑ And the moved image centroid is 11''nii C P n==∑ Thus, expansion coefficient S is11('')()ni i ni i P C S P C ==-=-∑∑Furthermore 2121()P P sR P P ''-=-3131()P P sR P P ''-=-R is the rotation matrix, and 1R =, ()RU RV R U V ⨯=⨯31213121312121()()()()[()()]P P P P R P P R P P R P P P P s''''-⨯-=-⨯-=-⨯-Thus,312123121()()[()()]P P P P R S P P P P ''''-⨯-=-⨯-。
