Representation of 1D Function

Transcription

1 Bioengineering 280A Principles of Biomedical Imaging Fall Quarter 2005 Linear Systems Lecture 2 Representation of 1D Function From the sifting property, we can write a 1D function as g(x) = g(ξ)δ(x ξ)dξ. To gain intuition, consider the approximation g(x) n= g(nδx) 1 Δx Π x nδx Δx. Δx g(x) 1

2 Representation of 2D Function Similarly, we can write a 2D function as g(x, y) = g(ξ,η)δ(x ξ,y η)dξdη. To gain intuition, consider the approximation g(x, y) g(nδx,mδy) 1 Δx Π x nδx 1 n= Δx Δy Π y mδy ΔxΔy. m= Δy Impulse Response Intuition: the impulse response is the response of a system to an input of infinitesimal width and unit area. Original Image Blurred Image Since any input can be thought of as the weighted sum of impulses, a linear system is characterized by its impulse response(s). 2

3 Impulse Response The impulse response characterizes the response of a system over all space to a Dirac delta impulse function at a certain location. h(x 2 ;ξ) = L[ δ( x 1 ξ) ] 1D Impulse Response [ ( )] 2D Impulse Response h(x 2, y 2 ;ξ,η) = L δ x 1 ξ, y 1 η y 1 Impulse at ξ,η y 2 h(x2,y 2 ;ξ,η) x 1 x 2 Superposition Integral What is the response to an arbitrary function g(x 1,y 1 )? Write g(x 1,y 1 ) = g(ξ,η)δ(x 1 ξ, y 1 η)dξdη. - - The response is given by [ ] I(x 2,y 2 ) = L g 1 (x 1,y 1 ) [ - - ] = L g(ξ,η)δ(x 1 ξ,y 1 η)dξdη [ ] = g(ξ,η)l δ(x 1 ξ, y 1 η) dξdη - - = g(ξ,η)h(x 2,y 2 ;ξ,η) dξdη - - 3

4 Space Invariance If a system is space invariant, the impulse response depends only on the difference between the output coordinates and the position of ( ) the impulse and is given by h(x 2, y 2 ;ξ,η) = h x 2 ξ,y 2 η I(x 2,y 2 ) = 2D Convolution For a space invariant linear system, the superposition integral becomes a convolution integral. - - g(ξ,η)h(x 2,y 2 ;ξ,η) dξdη = g(ξ,η)h(x 2 ξ, y 2 η) dξdη - - = g(x 2,y 2 ) **h(x 2,y 2 ) where ** denotes 2D convolution. This will sometimes be abbreviated as *, e.g. I(x 2, y 2 )= g(x 2, y 2 )*h(x 2, y 2 ). 4

5 I(x) = 1D Convolution For completeness, here is the 1D version. - g(ξ)h(x;ξ)dξ = g(ξ)h(x ξ) dξ - = g(x) h(x) Convolution with Dirac delta function g(x) δ(x) = g(ξ)δ(x ξ) dξ - = g(x) g(x) δ(x Δ) = g(ξ)δ(x Δ ξ) dξ - = g(x Δ) Convolution of g(x) with a shifted Dirac delta function just shifts g(x) g(x,y) δ(x,y) = g(ξ,η)δ(x ξ,y η) dξdη - - = g(x,y) g(x,y) δ(x x 0,y y 0 ) = g(ξ,η)δ(x x 0 ξ,y y 0 η) dξdη - - = g(x x 0,y y 0 ) 5

6 2D Convolution Example g(x)= δ(x+1/2,y) + δ(x,y) y h(x)=rect(x,y) y x -1/2 12 x I(x,y)=g(x)**h(x,y) x 2D Convolution Example 6

7 1D Convolution Review g(x) h(x) = - g(ξ)h(x ξ)dξ Basic Rule: Flip one function, slide it past the other function, and integrate as you go. g(x)=rect(x) h(x)=rect(x-1/2) -1/2 1/2 1 1D Convolution Review h(-1/2-ξ) g(ξ) I(x) h(-ξ) h(1/2-ξ) -1/2 1/2 3/2 h(3/2-ξ) 7

8 Convolution/Modulation Theorem { } = g(u) h(x u)du F g(x) h(x) [ ] e j 2πk xx = g(u) h(x u) e j 2πk xx = g(u)h(k x )e j 2πk xu du = G(k x )H(k x ) dxdu dx Convolution in the spatial domain transforms into multiplication in the frequency domain. Dual is modulation F{ g(x)h(x) } = G( k x ) H(k x ) Convolution 2D Convolution/Multiplication [ ] = G(k x,k y )H(k x,k y ) F g(x, y) h(x,y) Multiplication [ ] = G(k x,k y ) H(k x,k y ) F g(x, y)h(x, y) 8

9 Application of Convolution Thm. 1 x x <1 Λ(x) = 0 otherwise F(Λ(x)) = 1 ( 1 x ) e j 2πkxx dx =?? Application of Convolution Thm. Λ(x) = Π(x) Π(x) F(Λ(x)) = sinc 2 ( ) k x -1 1 = * 9

10 Eigenfunctions The fundamental nature of the convolution theorem may be better understood by observing that the complex exponentials are eigenfunctions of the convolution operator. e j 2πk xx g(x) z(x) z(x) = g(x) e j 2πk x x = g(u) e j 2πk x (x u) du = G(k x )e j 2πk x x The response of a linear shift invariant system to a complex exponential is simply the exponential multiplied by the FT of the system s impulse response. Eigenfunctions Now consider an arbitrary input h(x). h(x) g(x) z(x) Recall that we can express h(x) as the integral of weighted complex exponentials. h(x) = H(k x )e j 2πk x x dk x Each of these exponentials is weighted by G(k x ) so that the response may be written as z(x) = G(k x )H(k x )e j 2πk x x dk x 10

11 Convolution Example Analog vs. Digital The Analog World: Continuous time/space, continuous valued signals or images, e.g. vinyl records, photographs, x-ray films. The Digital World: Discrete time/space, discrete-valued signals or images, e.g. CD-Roms, DVDs, digital photos, digital x-rays, CT, MRI, ultrasound. 11

12 The Process of Sampling g(x) Δx sample g[n]=g(n Δx) Questions How finely do we need to sample? What happens if we don t sample finely enough? Can we reconstruct the original signal or image from its samples? 12

13 Sampling in the Time Domain Sampling in Image Space 13

14 Sampling in k-space Sampling in k-space 14

15 Comb Function comb(x) = δ(x n) n= x Other names: Impulse train, bed of nails, shah function. Scaled Comb Function comb x = Δx n= δ( x Δx n) = δ( x nδx ) Δx n= = Δx δ(x nδx) n= x Δx 15

16 1D spatial sampling g S (x) = g(x) 1 Δx comb x Δx Recall the sifting property But we can also write = g(x) δ(x nδx) n= = g(nδx)δ(x nδx) n= So, g(x)δ(x a) = g(a)δ(x a) g(x)δ(x a) = g(a) g(a)δ(x a) = g(a) δ(x a) = g(a) 1D spatial sampling g(x) comb(x/δx)/ Δx x Δx g S (x) x 16

17 F comb(x) Fourier Transform of comb(x) [ ] = comb(k x ) = δ(k x n) n= F 1 Δx comb( x Δx ) = 1 Δx Δxcomb(k xδx) = δ(k x Δx n) n= = 1 Δx n= δ(k x n Δx ) Fourier Transform of comb(x/ Δx) comb(x/ Δx)/ Δx Δx F x 1/Δx comb(k x Δx) 1/Δx k x 17

18 F g S (x) Fourier Transform of g S (x) Δx [ ] = F g(x) 1 Δx comb x = G(k x ) F 1 Δx comb x Δx = G(k x ) 1 Δx = 1 Δx = 1 Δx n= n= n= δ k x n Δx G(k x ) δ k x n Δx G k x n Δx Fourier Transform of g S (x) G(k x ) k x 1/Δx G S (k x ) k x 1/Δx 18

19 Nyquist Condition G(k x ) K S -B B k x G S (k x ) k x K S =1/Δx To avoid overlap, we require that 1/Δx>2B or K S > 2B where K S =1/ Δx is the sampling frequency Example Assume that the highest spatial frequency in an object is B = 2 cm -1. Thus, smallest spatial period is 0.5 cm.. Nyquist theorem says we need to sample with Δx < 1/2B = 0.25 cm This corresponds to 2 samples per spatial period. 19

20 Reconstruction from Samples K S G S (k x ) K S =1/Δx Multiply by (1/K S )rect(k x /K S ) (1/K S ) G S (k x )rect(k x /K S ) =G(k x ) Example Cosine Reconstruction cos(2πk 0 x) -k 0 k 0 K S K S >2k 0 -k 0 k 0 KS K S =2k 0 -k 0 k 0 KS 20

21 Reconstruction from Samples If the Nyquist condition is met, then G ˆ S (k x ) = 1 G S (k x )rect(k x /K S ) = G(k x ) K S And the signal can be reconstructed by convolving the sample with a sinc function g ˆ S (x) = g S (x) sinc(k s x) = g(nδx)δ( x nδx) sinc(k s x) n= = g(nδx) sinc(k s (x nδx)) n= g(x) Reconstruction from Samples g S (x) Sample at Δx x ˆ g S (x) sinc(k s x) = sinc(x / Δx) 21

22 Cosine Example with K S =2k 0 Example with K s =4k 0 22

23 Example with K s =8k 0 Aliasing Example 23

24 Aliasing -B B G(k x ) k x K S Aliasing occurs when the Nyquist condition is not satisfied. This occurs for K S 2B Aliasing Example cos(2πk 0 x) -k 0 k 0 K S =k 0 -k 0 k 0 K S 24

25 Aliasing Example cos(2πk 0 x) -k 0 k 0 2k 0 >K S >k 0 -k 0 k 0 K S Practical Considerations Why sample higher than the Nyquist frequency? -- true sinc interpolation is not practical since the sinc function goes from -infinity to infinity -- the requirements on the low-pass filter are reduced. Hard -k 0 k 0 Easier -k 0 k 0 K S 25