sol1 - Digital Image Processing Solution 1 Spring 2002 1. F...

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Unformatted text preview: Digital Image Processing Solution 1 Spring 2002 1. F (u; v) = F = = (1 X 1 X m= 1 n= 1 1 1 X X Z1Z1 (x m x; y n y) 1 ) m= 1 n= 1 1 1 XX m= 1 n= 1 1 (x m x; y n y)e j 2 (ux+vy)dxdy e j 2 (um x+vn y) The last term is the Fourier series exapansion (for proof see question 4) of 1 X 1 X m= 1 n= 1 u m ;v x n y 1 xy We found out that the Fourier transform of an impulse train in the space domain is an impulse train in the frequency domain with an inverse period. Note also the scaling factor added 2. F (u; v) = = = = = Z1Z1 11 Z2 Z1 Z2 Z1 Z1 Z0 f (x; y)e j 2 (ux+vy) dxdy 0 0 f^(r)e j 2 (ur cos( )+vr sin( )) rdrd f^(r)e j 2 r cos( Z2 0 1 f^(r)r 0 )rdrd )d dr 0 ^ f^(r)rB (r; )dr = F ( ) 0 e j 2 r cos( Where we used the following variable transformation:x = r cos( ); y = r sin( ); u = cos( ); v = sin( ). 1 3. T f (x; y)] = f (x; y) h(x; y) = = = = = = Z1Z1 Z1Z1 1 Z1 1 Z1 1 Z1 r1 1 (x m )2 (y )2 d e ( 11 2 2 m )e (x ) (y ) d d f ( ; )h(x ;y )d d e x +2xm m 2 2 2 2 2 y2 +2y 2 2 d 2 e (m +1) +(2xm+2y) (x +y ) d 4( (1) (2) m2 + 1 e r = m2 + 1 e Z1 xm+y)2 4( 2 +1)( 2 + 2 ) 4( 2 +1) m m xy ( 1+ x my)2 m2 The last integral (1) was computed by transformingp to a known Gaussian it R1 x integral. First we will show that I = 1 e dx = : 2 I2 = = Z 1Z 2 1 Z1 Z1Z1 e x dx e y dy = e x y dxdy 2 2 2 2 1 By variable transformation from (3) it can be shown that a To use this result on (1) we need to bring (1) to the form of (4): 1 Z1 Z1 0 0 e r rdrd = 2 2 Z1 1 1 0 e r rdr = 2 (3) (4) pm2 +1 p m+y 2 x m2 +1 d e e (ax+b)2 dx = p The result (2) is then produced from applying (4) to (5). 4. (a) 1 (x2 +y2 ) 2 2 2 2 e (m +1) +(2xm+2y) (x +y ) d = e ( xm+y)2 m2 +1 Z1 1 (5) F (u; v) = = = Z1Z1 11 Z1Z1 1 Z1 1 f (x; y)e j 2 (ux+vy) dxdy (y g(x))e j 2 (ux+vy) dxdy 1 o n e j 2 uxe j 2 vg(x) dx = F e j 2 vg(x) Since this is a periodic function with one variable it can be described by a 1-D Fourier series expansion. Then we will calculate the Fourier transform of the Fourier series. Note that F 1 f (f Therfore, f0 )g = Z1 1 f0 x (f f0 )ej 2 fx df = e j 2 f x 0 F ej 2 2 = (f f0 ) And speci cally F ejkx = f k= 1 k 2 1 X e j 2 vg(x) = ak ejkx ak u 2k k= 1 Where ak are the Fourier series coe cients of e j 2 vg(x) : 1 Z e j 2 vg(x)e j kxdx ak = 2 1 Z 0 e j 2 vg(x)e j kxdx + 1 Z e j 2 vg(x)e j kxdx =2 20 Z0 Z = 21 e j 2 v e j kxdx + 21 ej 2 v e j kxdx 0 1 1 ( 1)k e j v2 ejv2 =2 jk (k = 2 1 1 jk 1) ( 2j sin(2v )) = k1 sin(2v )(1 ( 1)k ) The treatment for k = 0 should be done sepeartely: 1 e j 2 v Z 0 e j kxdx + 1 ej 2 v Z e j kxdx = cos(2 v) 2 2 0 To conclude: 1 X 1 ( 1)k F (u; v) = cos(2 v) (u) + sin(2 v) u 2k k k= 1;k6=0 F (u; v) = (b 1 X ) The non-zero support of F(u,v) is: 5. After the sampling, in the frequency domain there will be spectral islands (areas of nonzero value) in the shape of rectangles. The islands will be periodic with inverse period of the space sampling intervals: 1;1 xy We want to keep only the central island with the reconstruction lter which is a circle. In order to avoid aliasing we require the lter's circle will bound the central rectangle: 1 > u + pv2 + u2 max max x max 1 > v + pv2 + u2 max max y max 3 u = 2k ; v 6= n ; n 2 Z k 2 2Z+ 1 ; 2 u = 0; v 6= 1 + n ; n 2 Z 42 mx ny 1 6. (a) Let's denote pAB e 2 j ( A + B ) by m;n . h m;n ; p;q i = Z A=2 Z B=2 2 m px n qy e 2 j ( A ) dx = AB e 2 j ( B ) dy A=2 B =2 n q y B =2 m p x ) A=2 e 2 j( B ) 1 e 2 j( A = AB 2 j (mA p) A=2 2 j (nB q) B=2 1 = 4 2(m p)(n q) e j (p m) e j (p m) e j (q n) e j (q n) 1 = 4 2(m p)(n q) (2j sin( (p m)))(2j sin( (q n))) = sinc( (p m))sinc( (q n)) = p;m q;n ( ) ( ) ( ) ( ) A=2 B=2 1 Z A=2 1 Z A=2 Z B=2 e 2 j ( m Ap x + n Bq x ) dxdy dxdy = AB m;n p;q ( ) ( ) Z B=2 A=2 B =2 (b) The condition jjf mn mn < f; mn > jj2 = 0 implies that the orthonormal set is closed. This means that every function can be approximated by its projection on the subspace spanned by the orthonormal set to arbitrary accuracy (The projection of f converge uniformly to f as basis members are added). This condition is obviously not true in the case of non-periodic functions since the orthonormal set is periodic. The condition is as follows: Any 2-D piecewise continouos fucntion f : R2 ! R which is de ned on R2 and has a 2-D period A=n and B=m for some positive integers m and n, satis es the condition. P 4 ...
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