SIMG Solution Set #1

Transcription

1 SIMG Solution Set #1 1. Find the zero-order Hankel transforms of the following functions and make appropriate sketches of your results IN BOTH DOMAINS (e.g., axial profiles, top views, perspective plots, etc.) (a) f (r) CY L(r) ³ r f (r) CY L(r) CY L 0.³ r by scaling theorem of Hankel transform:h 0 nf d d o F (dρ) ³ π H 0 {f (r)} F (ρ) (0.) SOMB (0. ρ) π ³ ρ ³ 16 SOMB ρ SOMB 1

2 (b) g (r) CY L(r) FCY L(r) (where F denotes -D correlation) Correlation convolution for circularly symmetric real-valued functions ³ π H 0 {CY L(r) FCY L(r)} SOMB (ρ) n ³ r o H 0 {CY L(r) FCY L(r)} H 0 {CY L(r) CY L(r)} ³H 0 CY L Ã µ1 π π SOMB (0.ρ) 6 SOMB ³ ρ g (r) π 6 H 1 0 n SOMB ³ ρ 0.! ³ π ³ ρ 16 SOMB ³ SOMB ρ π 16 CTRI (r) CTRI o µ π 6 π CTRI (r) Ã! r 1

3 (c) h (r) CY L(0.r) CY L(0.r) ³ r n ³ r H (ρ) H 0 ncy L H 0 CY L ³ o o π ³ SOMB (ρ) π SOMB (ρ) h ³ ρ ³ ρ i π SOMB SOMB

4 (d) t (r) SOMB (r) FSOMB (r) T (ρ) H 0 SOMB (r) ª H 0 {SOMB (r)} µµ π CY L(ρ) F Ã " µ1 ³ #! ρ CY L(ρ) π π CY L " µ 3 µ # 1 ³ ρ (CY L(ρ) FCY L(ρ)) CY L π " µ 3 µ # 1 (CY L(ρ) FCY L(ρ)) π 6 π (CY L(ρ) FCY L(ρ)) (CY L(ρ) FCY L(ρ)) 3 because autocorrelation is contained within large cylinder t (r) H 1 0 {T (ρ)} " µ π 3 µ 1 # ³π SOMB (r) " µ π µ # 1 SOMB (r) 16 π SOMB (r) SOMB (r) SOMB (r) This is the -D circularly symmetric analogue of SINC [x] SINC [x]

5 (e) v (r) π r GAUS (r) (πir) GAUS (r) we derived the relation : H 0 r GAUS (r) ª µ 1 π ρ e πρ V (ρ) H 0 π (r) GAUS (r) ª π H 0 (r) GAUS (r) ª µ µ 1 1 ³ ρ π π e π( ρ ) µ 1 ³ ρ π π e π ( ρ ) µ ³πρ π e π( ρ )

6 . Given an LSI system with impulse response: h [x, y] SINC [x, y] find the output g i [x, y] for the following input signals: (a) f 1 [x, y] 1+cos[6πx]+cos[1πx] H [ξ,η] F SINC [x, y] ª TRI ξ, η is the transfer function F 1 [ξ,η] δ [ξ,η]+ 1 (δ [ξ +3,η]+δ [ξ 3,η]+δ [ξ 6,η]+δ [ξ +6,η]) ( Dirac delta functions) G 1 [ξ,η] δ [ξ,η] TRI, (δ [ξ +3,η]+δ [ξ 3,η]) TRI, (δ [ξ +6,η]+δ [ξ 6,η]) δ [ξ,η]+ 1 (δ [ξ +3,η]+δ [ξ 3,η]) + 0 (δ [ξ +6,η]+δ [ξ 6,η]) δ [ξ,η]+ (δ [ξ +3,η]+δ [ξ 3,η]) 10 g 1 [x, y] F ½δ 1 [ξ,η]+ 10 ¾ (δ [ξ +3,η]+δ [ξ 3,η]) 1[x, y]+ 10 cos[6πx] 1+ cos [6πx] (b) f 3 [x, y] δ [x] 1[y] (note that g 3 [x, y] is a line response ) ξ H [ξ,η] TRI, η ξ h η TRI TRI i F 3 [ξ,η] 1[ξ] δ [η] µ ξ ³ h η G 3 [ξ,η] F 3 [ξ,η] H [ξ,η] TRI 1[ξ] TRI i µ µ ξ 0 ξ TRI TRI δ [η] TRI δ [η] ½ ¾ ξ g 3 [x, y] F 1 TRI δ [η] SINC [x] 1[y] δ [η] 6

7 3. Evaluate the results of the following operations, where the symbols and F denote -D convolution and correlation, respectively: (a) J 0 (πρ 0 r) J 0 (πρ 1 r), whereρ 0 6 ρ 1 1 H 0 {J 0 (πρ 0 r)} δ (ρ ρ πρ 0 ) 0 F {J 0 (πρ 0 r)} 1 δ (ρ ρ πρ 0 )1[φ] ½ 0 1 J 0 (πρ 0 r) J 0 (πρ 1 r) F 1 δ (ρ ρ πρ 0 ) 1 ¾ δ (ρ ρ 0 πρ 1 ) 1 F 1 {0[ρ]} 0[x, y] 7

8 (b) CY L(r) (δ [x] 1[y]) the result is a line response of a system with impulse response: ³ r h (r) CY L d ³ Z r + Ãp! x + y (δ [x] 1[y]) CY L CY L d dy d h x RECT di 1 ( x d) Z + d d 1 ( xd) 1 dy à h x RECT + di d r à ³ x 1 d r 1 d à h r x RECT d 1 di!! ³ x d! ³ x h x RECT d di d x where the rectangle constrains the domain to eliminate imaginary values 8

9 (c) CY L(r) δ r ˆp 1 r ˆp 3 where ˆp Because the input function CY L(r) is circularly symmetric, the line delta can be rotated, the integral performed, and the result rotated back.. ³ s µ δ r ˆp 1 r ˆp r r ˆp r ˆp CY L RECT d 1 d d d h p RECT di pd p 1 wherepisthe1-dcoordinate. 9

10 (d) δ (r r 0 ) (δ [x] 1[y]) δ (r r 0 ) (δ [x] 1[y]) ZZ ( " + Ãp! Ãp!#) 1 x + y lim CY L x + y CY L δ (p x) dx dy 0 r 0 + r 0 ( Z " 1 + Ãp! Ãp!# ) p + y lim CY L p + y CY L dy 0 r 0 + r 0 This is about as far as we can go without consulting the picture. The figure shows that the length of the line segment within the annulus is and there obviously are of them sin [φ] ½ µ ¾ 1 p δ (r r 0 ) (δ [x] 1[y]) lim RECT δ (r r 0 ) (δ [x] 1[y]) 0 sin [φ] r 0 v Ã! u cancel the 0 s and note that sin [φ] t p 1 h p h p δ (r r 0 ) (δ [x] 1[y]) r ³ RECT r di ³ RECT di p p 1 r 0 1 r 0 r h 0 p i p r 0 p RECT d r0 10

11 . Evaluate the following -D convolutions and make appropriate sketches of the results: (a) cos π x 1[y] (δ [x] 1[y]) h F n³cos π x i 1[y] o n (δ [x] 1[y]) F µ 1 δ [ξ +,η]+1 δ [ξ,η] µ 1 δ [ξ +] δ [η]+1 δ [ξ ] δ [η] h cos π x i o 1[y] F {δ [x] 1[y]} (1 [ξ] δ [η]) (1 [ξ] δ [η]) 1 δ [ξ +] δ [η]+1 δ [ξ ] δ [η] ½ ¾ 1 h F 1 δ [ξ +] δ [η]+1 δ [ξ ] δ [η] cos π x i 1[y] (b) cos π x 1[y] CROSS [x, y] (where CROSS [x, y] δ [x] 1[y]+1[x] δ [y]) ³ h cos π x i ³ cos 1[y] h π x (δ [x] 1[y]+1[x] δ [y]) i 1[y] (δ [x] 1[y]) + ³ cos h π x i 1[y] (1 [x] δ [y]) ³ h cos π x i ³ h 1[y] (δ [x] 1[y]) cos π x i δ [x] (1 [y] 1[y]) ³ h cos π x i ³ h δ [x] cos π x i ³ h cos π x i ³ h 1[y] (1 [x] δ [y]) cos π x i 1[x] (1 [y] δ [y]) 0[x] 1[y] 0[x, y] δ [x] 11

12 (c) e +iπr (δ [x] 1[y]), wherer p x + y e +iπr (δ [x] 1[y]) e +iπ(x +y ) (δ [x] 1[y]) ³ ³ e +iπx δ [x] e +iπy 1[y] ³ e +iπx e +iπy 1[y] Evaluate the y dependence via the filter theorem: o F 1 ne +iπy 1[y] F 1 ne +iπyo F 1 {1[y]} e +i π e iπη δ [η] e +i π e iπ0 δ [η] e +i π δ [η] e +iπy 1[y] e +i π 1[y] Hence the convolution is a -D separable function that is a chirp with an additional constant phase along x and constant along y: ³ e +iπr (δ [x] 1[y]) e +i π e +iπx 1[y] e +iπ (x + 1 ) 1[y] 1

13 13

14 (d) e +iπr CROSS [x, y] e +iπr CROSS [x, y] e +iπ (x +y ) (δ [x] 1[y]+1[x] δ [y]) ³ ³ e +i π e +iπx 1[y] + e +i π ³1[x] +iπy e e +i π ³e +iπy +iπx 1[y]+1[x] e e +i π ³e +iπy +iπx + e 1