Representation of 1D Function

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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( = g(ξδ( ξdξ. To gain intuition, consider the approimation g( n= g(nδ 1 Δ Π nδ Δ. Δ g( Representation of 2D Function Similarly, we can write a 2D function as g(,y = g(ξ,ηδ( ξ,y ηdξdη. To gain intuition, consider the approimation g(, y g(nδ,mδy 1 Δ Π nδ 1 n= Δ Δy Π y mδy ΔΔy. m= Δy 1

2 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. 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( 2 ;ξ = L[ δ( 1 ξ ] 1D Impulse Response h( 2,y 2 ;ξ,η = L[ δ( 1 ξ,y 1 η ] 2D Impulse Response y 1 Impulse at ξ,η y 2 h( 2, y 2 ;ξ,η 1 2 Superposition Integral What is the response to an arbitrary function g( 1,y 1? Write g( 1,y 1 = g(ξ,ηδ( 1 ξ, y 1 ηdξdη. - - The response is given by [ ] I( 2,y 2 = L g 1 ( 1,y 1 [ - - ] = L g(ξ,ηδ( 1 ξ,y 1 ηdξdη [ ] = g(ξ,ηl δ( 1 ξ,y 1 η dξdη - - = g(ξ,ηh( 2,y 2 ;ξ,η dξdη - - 2

3 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( 2, y 2 ;ξ,η = h( 2 ξ, y 2 η I( 2, y 2 = - 2D Convolution For a space invariant linear system, the superposition integral becomes a convolution integral. - g(ξ,ηh( 2,y 2 ;ξ,η dξdη = g(ξ,ηh( 2 ξ,y 2 η dξdη - - = g( 2,y 2 **h( 2,y 2 where ** denotes 2D convolution. This will sometimes be abbreviated as *, e.g. I( 2, y 2 = g( 2, y 2 *h( 2, y 2. I( = 1D Convolution For completeness, here is the 1D version. - g(ξh(;ξdξ = g(ξh( ξ dξ - = g( h( 3

4 Convolution with Dirac delta function g( δ( = g(ξδ( ξ dξ - = g( g( δ( Δ = g(ξδ( Δ ξ dξ - = g( Δ Convolution of g( with a shifted Dirac delta function just shifts g( g(,y δ(,y = g(ξ,ηδ( ξ,y η dξdη - - = g(,y g(,y δ( 0,y y 0 = g(ξ,ηδ( 0 ξ,y y 0 η dξdη - - = g( 0,y y 0 2D Convolution Eample g(= δ(+1/2,y + δ(,y y h(=rect(,y y -1/2 12 I(,y=g(**h(,y 2D Convolution Eample 4

5 1D Convolution Review g( h( = - g(ξh( ξdξ Basic Rule: Flip one function, slide it past the other function, and integrate as you go. g(=rect( h(=rect(-1/2-1/2 1/2 1 1D Convolution Review h(-1/2-ξ g(ξ I( h(-ξ h(1/2-ξ -1/2 1/2 3/2 h(3/2-ξ Convolution/Modulation Theorem [ ] e j 2πk F{ g( h( } = g(u h( udu = g(u h( u e j 2πk = g(uh(k e j 2πk u du = G(k H(k ddu d Convolution in the spatial domain transforms into multiplication in the frequency domain. Dual is modulation F{ g(h( } = G( k H(k 5

6 2D Convolution/Multiplication Convolution F[ g(, y h(,y ] = G(k,k y H(k,k y Multiplication F[ g(, yh(,y ] = G(k,k y H(k,k y Application of Convolution Thm. 1 <1 Λ( = 0 otherwise 1 F(Λ( = 1 e j 2πk d =?? 1 ( -1 1 Application of Convolution Thm. Λ( = Π( Π( F(Λ( = sinc 2 ( k -1 1 = * 6

7 Eigenfunctions The fundamental nature of the convolution theorem may be better understood by observing that the comple eponentials are eigenfunctions of the convolution operator. e j 2πk g( z( z( = g( e j 2πk = g(u e j 2πk ( u du = G(k e j 2πk The response of a linear shift invariant system to a comple eponential is simply the eponential multiplied by the FT of the system s impulse response. Eigenfunctions Now consider an arbitrary input h(. h( g( z( Recall that we can epress h( as the integral of weighted comple eponentials. h( = H(k e j 2πk dk Each of these eponentials is weighted by G(k so that the response may be written as z( = G(k H(k e j 2πk dk Convolution Eample 7

8 Analog vs. Digital The Analog World: Continuous time/space, continuous valued signals or images, e.g. vinyl records, photographs, -ray films. The Digital World: Discrete time/space, discrete-valued signals or images, e.g. CD-Roms, DVDs, digital photos, digital -rays, CT, MRI, ultrasound. The Process of Sampling g( Δ sample g[n]=g(n Δ 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? 8

9 Sampling in the Time Domain Sampling in Image Space Sampling in k-space 9

10 Sampling in k-space Comb Function comb( = δ( n n= Other names: Impulse train, bed of nails, shah function. Scaled Comb Function comb = Δ n= δ( Δ n = δ( nδ Δ n= = Δ δ( nδ n= Δ 10

11 1D spatial sampling g S ( = g( 1 Δ comb Δ Recall the sifting property = g( δ( nδ n= But we can also write g(aδ( a = g(a δ( a = g(a So, g(δ( a = g(aδ( a = g(nδδ( nδ n= g(δ( a = g(a 1D spatial sampling g( comb(/δ/ Δ Δ g S ( Fourier Transform of comb( [ ] = comb(k F comb( = δ(k n n= F 1 Δ comb( Δ = 1 Δ Δcomb(k Δ = δ(k Δ n n= = 1 Δ n= δ(k n Δ 11

12 Fourier Transform of comb(/ Δ comb(/ Δ/ Δ Δ F 1/Δ comb(k Δ 1/Δ k Fourier Transform of g S ( [ ] = F g( 1 Δ comb F g S ( Δ = G(k F 1 Δ comb Δ = G(k 1 Δ = 1 Δ = 1 Δ n= n= n= δ k n Δ G(k δ k n Δ G k n Δ Fourier Transform of g S ( G(k k 1/Δ G S (k k 1/Δ 12

13 Nyquist Condition G(k -B B k G S (k k =1/Δ To avoid overlap, we require that 1/Δ>2B or > 2B where =1/ Δ is the sampling frequency Eample 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 Δ < 1/2B = 0.25 cm This corresponds to 2 samples per spatial period. Reconstruction from Samples G S (k =1/Δ Multiply by (1/ rect(k / (1/ G S (k rect(k / =G(k 13

14 Eample Cosine Reconstruction cos(2πk 0 -k 0 k 0 >2k 0 -k 0 k 0 =2k 0 -k 0 k 0 Reconstruction from Samples If the Nyquist condition is met, then G ˆ S (k = 1 G S (k rect(k / = G(k And the signal can be reconstructed by convolving the sample with a sinc function g ˆ S ( = g S ( sinc(k s = g(nδxδ( nδx sinc(k s n= = g(nδx sinc(k s ( nδ n= g( Reconstruction from Samples g S ( Sample at Δ ˆ g S ( sinc(k s = sinc( / Δ 14

15 Cosine Eample with =2k 0 Eample with K s =4k 0 Eample with K s =8k 0 15

16 Aliasing Eample Aliasing -B B G(k k Aliasing occurs when the Nyquist condition is not satisfied. This occurs for 2B Aliasing Eample cos(2πk 0 -k 0 k 0 =k 0 -k 0 k 0 16

17 Aliasing Eample cos(2πk 0 -k 0 k 0 2k 0 > >k 0 -k 0 k 0 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 17

Representation of 1D Function

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.