1Thomas Liu, BE280A, UCSD, Fall 2005Bioengineering 280APrinciples of Biomedical ImagingFall Quarter 2005Linear Systems Lecture 2Thomas Liu, BE280A, UCSD, Fall 2005Representation 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 approximationg(x) ≈ g(nΔx)1ΔxΠx − nΔxΔx n= −∞∞∑Δx.g(x)Thomas Liu, BE280A, UCSD, Fall 2005Representation 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 approximationg(x, y) ≈ g(nΔx, mΔy)1ΔxΠx − nΔxΔx n= −∞∞∑1ΔyΠy − mΔyΔy ΔxΔym= −∞∞∑.2Thomas Liu, BE280A, UCSD, Fall 2005Intuition: the impulse response is the response ofa system to an input of infinitesimal width andunit area.Impulse ResponseSince any input can be thought of as theweighted sum of impulses, a linear system ischaracterized by its impulse response(s).Blurred ImageOriginalImageThomas Liu, BE280A, UCSD, Fall 2005Impulse Response€ The impulse response characterizes the response of a system over all space to a Dirac delta impulse function at a certain location. h( x2;ξ) = Lδx1−ξ( )[ ] 1D Impulse Response h( x2, y2;ξ,η) = Lδx1−ξ, y1−η( )[ ] 2D Impulse Responsex1y1x2y2€ h(x2, y2;ξ,η)€ Impulse at ξ,ηThomas Liu, BE280A, UCSD, Fall 2005Superposition Integral€ What is the response to an arbitrary function g(x1,y1)? Write g(x1,y1) = g(ξ,η)δ(x1-∞∞∫-∞∞∫−ξ, y1−η)dξdη.The response is given by I( x2, y2) = L g1(x1,y1)[ ] = L g(ξ,η)δ(x1-∞∞∫-∞∞∫−ξ, y1−η)dξdη[ ] = g(ξ,η)Lδ(x1−ξ, y1−η)[ ]-∞∞∫-∞∞∫dξdη = g(ξ,η)h( x2, y2;ξ,η)-∞∞∫-∞∞∫dξdη3Thomas Liu, BE280A, UCSD, Fall 2005Space Invariance€ If a system is space invariant, the impulse response depends onlyon the difference between the output coordinates and the position ofthe impulse and is given by h(x2, y2;ξ,η) = h x2−ξ, y2−η( ) Thomas Liu, BE280A, UCSD, Fall 20052D Convolution€ I( x2, y2) = g(ξ,η)h( x2, y2;ξ,η)-∞∞∫-∞∞∫dξdη= g(ξ,η)h( x2−ξ, y2−η)-∞∞∫-∞∞∫dξdη= g(x2, y2) **h(x2, y2)For a space invariant linear system, the superposition integralbecomes a convolution integral.where ** denotes 2D convolution. This will sometimes beabbreviated as *, e.g. I(x2, y2)= g(x2, y2)*h(x2, y2).Thomas Liu, BE280A, UCSD, Fall 20051D Convolution€ I( x) = g(ξ)h(x;ξ)dξ-∞∞∫= g(ξ)h(x −ξ)-∞∞∫dξ= g(x) ∗ h(x)For completeness, here is the 1D version.4Thomas Liu, BE280A, UCSD, Fall 2005Convolution 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 deltafunction 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 − x0, y − y0) = g(ξ,η)δ( x − x0−ξ, y − y0−η)-∞∞∫dξdη-∞∞∫= g(x − x0, y − y0)Thomas Liu, BE280A, UCSD, Fall 20052D Convolution Example-1/2 12xyh(x)=rect(x,y)yxg(x)= δ(x+1/2,y) + δ(x,y)xI(x,y)=g(x)**h(x,y)Thomas Liu, BE280A, UCSD, Fall 20052D Convolution Example5Thomas Liu, BE280A, UCSD, Fall 20051D 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 1/21Thomas Liu, BE280A, UCSD, Fall 20051D Convolution Reviewh(-1/2-ξ) g(ξ)h(-ξ)h(1/2-ξ)h(3/2-ξ)I(x)-1/2 1/2 3/2Thomas Liu, BE280A, UCSD, Fall 2005Convolution/Modulation Theorem€ F g(x) ∗ h(x){ }= g(u) ∗ h(x − u) du−∞∞∫[ ]e− j 2πkxx−∞∞∫dx= g(u) h( x − u)−∞∞∫e− j2πkxx−∞∞∫dxdu= g(u)H(kx)e− j2πkxu−∞∞∫du= G(kx)H(kx)Convolution in the spatial domain transforms intomultiplication in the frequency domain. Dual ismodulation€ F g(x)h(x){ }= G kx( )∗ H(kx)6Thomas Liu, BE280A, UCSD, Fall 20052D Convolution/Multiplication€ ConvolutionF g(x, y) ∗∗h(x, y)[ ]= G(kx,ky)H(kx,ky)MultiplicationF g(x, y)h(x, y)[ ]= G(kx,ky) ∗∗H(kx,ky)Thomas Liu, BE280A, UCSD, Fall 2005Application of Convolution Thm.€ Λ(x) =1− x x < 10 otherwise F(Λ(x)) = 1− x( )−11∫e− j2πkxxdx = ??-1 1 Thomas Liu, BE280A, UCSD, Fall 2005Application of Convolution Thm.€ Λ(x) = Π(x) ∗ Π(x)F (Λ(x)) = sinc2kx( )-1 1 *=7Thomas Liu, BE280A, UCSD, Fall 2005Eigenfunctions€ z(x) = g(x) ∗ ej 2πkxx= g(u)−∞∞∫ej 2πkx(x−u)du= G(kx)ej 2πkxxThe fundamental nature of the convolution theorem may bebetter understood by observing that the complex exponentialsare eigenfunctions of the convolution operator.g(x) z(x)€ ej 2πkxxThe response of a linear shift invariant system to a complexexponential is simply the exponential multiplied by the FT ofthe system’s impulse response.Thomas Liu, BE280A, UCSD, Fall 2005Eigenfunctions€ h( x) = H(kx−∞∞∫)ej 2πkxxdkxNow consider an arbitrary input h(x).h(x) g(x) z(x)Recall that we can express h(x) as the integral of weightedcomplex exponentials.Each of these exponentials is weighted by G(kx) so that theresponse may be written as€ z(x) = G(kx)H(kx−∞∞∫)ej 2πkxxdkxThomas Liu, BE280A, UCSD, Fall 2005Convolution Example8Thomas Liu, BE280A, UCSD, Fall 2005Analog vs. DigitalThe Analog World:Continuous time/space, continuous valued signals orimages, e.g. vinyl records, photographs, x-ray films.The Digital World:Discrete time/space, discrete-valued signals orimages, e.g. CD-Roms, DVDs, digital photos, digitalx-rays, CT, MRI, ultrasound.Thomas Liu, BE280A, UCSD, Fall 2005The Process of Samplingg(x) g[n]=g(n Δx)ΔxsampleThomas Liu, BE280A, UCSD, Fall 2005QuestionsHow finely do we need to sample?What happens if we don’t sample finelyenough?Can we reconstruct the original signal orimage from its samples?9Thomas Liu, BE280A, UCSD, Fall 2005Sampling in the Time DomainThomas Liu, BE280A, UCSD, Fall 2005Sampling in Image SpaceThomas Liu, BE280A, UCSD, Fall 2005Sampling in k-space10Thomas Liu, BE280A, UCSD, Fall 2005Sampling in k-spaceThomas Liu, BE280A,
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