Radon Transformation - Digital Signal Processing for Medical Imaging Using Matlab - page 4

Hence,( 1.6 ) is rewritten as

π ∞

exp j 2 π( x cos θ + y sin θ) U f ( x , y ) = G ( U ,θ) | U | dUd θ (1.7)

− π

0

From ( 1.1 ), we get R ( l ,θ) = R ( − l ,θ + π) . This implies

( − ,θ) = ( ,θ + π)

G U G U (1.8)

Splitting ( 1.7 )intotwoterms,
I-term:

0 ∞

exp j 2 π( x cos θ + y sin θ) U

G ( U ,θ) | U | dUd θ (1.9)

− π −∞

II-term:

π ∞

exp j 2 π( x cos θ + y sin θ) U

( ,θ) | | θ

G U U dUd (1.10)

0 0

Change the variable φ = θ + π and U 1 =− U in the first term ( 1.9 ), we get

π ∞

exp j 2 π( x cos φ + y sin φ) U 1 G ( − U 1 ,φ − π) | U 1 | dU 1 d φ (1.11)

0 0

Using ( 1.7 )-( 1.11 ), we get

π ∞

exp j 2 π( x cos φ + y sin φ) U 1

f ( x , y ) = 2 G ( U 1 ,φ − π + π) | U 1 | dU 1 d φ.

0 0

(1.12)

Replacing the dummy variables U 1 and φ with U and θ respectively in ( 1.12 )

and writing the second limit ranging from −∞ to ∞ , we get the following.

π ∞

exp j 2 π( x cos θ + y sin θ) U f ( x , y ) = 2 ( 1 / 2 ) G ( U ,θ) | U | dUd θ (1.13)

0 −∞

Let l = x cos θ + y sin θ . Thus rewriting ( 1.12 )asfollows.

π ∞

exp j 2 π lU

f ( x , y ) = G ( U ,θ) | U | dUd θ (1.14)

−∞

0