Solution Manual for Medical Imaging Signals and Systems, 2nd Edition

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Medical ImagingSignals and Systems (2e):Solutions ManualVersion 1.4 (December 21, 2016)Jerry L. PrinceElectrical and Computer EngineeringWhiting School of EngineeringJohns Hopkins UniversityJonathan M. LinksEnvironmental Health SciencesBloomberg School of Public HealthJohns Hopkins University

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ContentsPreface1Signals and Systems2Image Quality37Physics of Radiography61Projection Radiography70Computed Tomography96The Physics of Nuclear Medicine134Planar Scintigraphy141Emission Computed Tomography162The Physics of Ultrasound176Ultrasound Imaging Systems188Physics of Magnetic Resonance208Magnetic Resonance Imaging216iii

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2Signals and SystemsSIGNALS AND THEIR PROPERTIESSolution 2.1(a)δs(x, y) =∑∞m=−∞∑∞n=−∞δ(x−m, y−n) =∑∞m=−∞δ(x−m)·∑∞n=−∞δ(y−n), therefore it is aseparable signal.(b)δl(x, y)is separable ifsin(2θ) = 0. In this case, eithersinθ= 0orcosθ= 0,δl(x, y)is a product of aconstant function in one axis and a 1-Ddeltafunction in another. But in general,δl(x, y)is not separable.(c)e(x, y) = exp[j2π(u0x+v0y)] = exp(j2πu0x)·exp(j2πv0y) =e1D(x;u0)·e1D(y;v0), wheree1D(t;ω) =exp(j2πωt). Therefore,e(x, y)is a separable signal.(d)s(x, y)is a separable signal whenu0v0= 0. For example, ifu0= 0,s(x, y) = sin(2πv0y)is the productof a constant signal inxand a 1-D sinusoidal signal iny. But in general, when bothu0andv0are nonzero,s(x, y)is not separable.Solution 2.2(a)Not periodic.δ(x, y)is non-zero only whenx=y= 0.(b)Periodic. By definitioncomb(x, y) =∞∑m=−∞∞∑n=−∞δ(x−m, y−n).For arbitrary integersMandN, we havecomb(x+M, y+N)=∞∑m=−∞∞∑n=−∞δ(x−m+M, y−n+N)=∞∑p=−∞∞∑q=−∞δ(x−p, y−q) [letp=m−M, q=n−N]=comb(x, y).2

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3So the smallest period is 1 in bothxandydirections.(c)Periodic. Letf(x+Tx, y) =f(x, y), we havesin(2πx) cos(4πy) = sin(2π(x+Tx)) cos(4πy).Solving the above equation, we have2πTx= 2kπfor arbitrary integerk. So the smallest period forxisTx0= 1. Similarly, we find that the smallest period foryisTy0= 1/2.(d)Periodic. Letf(x+Tx, y) =f(x, y), we havesin(2π(x+y)) = sin(2π(x+Tx+y)).So the smallest period forxisTx0= 1and the smallest period foryisTy0= 1.(e)Not periodic. We can see this by contradiction. Supposef(x, y) = sin(2π(x2+y2))is periodic; then thereexists someTxsuch thatf(x+Tx, y) =f(x, y), andsin(2π(x2+y2))=sin(2π((x+Tx)2+y2))=sin(2π(x2+y2+ 2xTx+T2x)).In order for the above equation to hold, we must have that2xTx+T2x=kfor some integerk. The solutionforTxdepends onx. Sof(x, y) = sin(2π(x2+y2))is not periodic.(f)Periodic. Letfd(m+M, n) =fd(m, n). Thensin(π5m)cos(π5n)= sin(π5 (m+M))cos(π5n).Solving forM, we find thatM= 10kfor any integerk. The smallest period for bothmandnis therefore10.(g)Not periodic. Following the same strategy as in (f), we letfd(m+M, n) =fd(m, n), and thensin(15m)cos(15n)= sin(15 (m+M))cos(15n).The solution forMisM= 10kπ.Sincefd(m, n)is a discrete signal, its period must be an integerif it is to be periodic.There is no integerkthat solves the equality forM= 10kπfor someM.So,fd(m, n) = sin(15m)cos(15n)is not periodic.Solution 2.3(a)We haveE∞(δs)=∫∞−∞∫∞−∞δ2s(x, y)dx dy=limX→∞limY→∞∫X−X∫Y−Y∞∑m=−∞∞∑n=−∞δ(x−m, y−n)dx dy=limX→∞limY→∞(2bXc+ 1)(2bYc+ 1)=∞,

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4CHAPTER2:SIGNALS AND SYSTEMSwherebXcis the greatest integer that is smaller than or equal toX. We also haveP∞(δs)=limX→∞limY→∞14XY∫X−X∫Y−Yδ2s(x, y)dx dy=limX→∞limY→∞14XY∫X−X∫Y−Y∞∑m=−∞∞∑n=−∞δ(x−m, y−n)dx dy=limX→∞limY→∞(2bXc+ 1)(2bYc+ 1)4XY=limX→∞limY→∞{4bXcbYc4XY+ 2bXc+ 2bYc4XY+14XY}=1.(b)We haveE∞(δl)=∫∞−∞∫∞−∞|δ(xcosθ+ysinθ−l)|2dx dy=∫∞−∞∫∞−∞δ(xcosθ+ysinθ−l)dx dy©1=∫∞−∞1|sinθ|dx,sinθ6= 0∫∞−∞1|cosθ|dy,cosθ6= 0E∞(δl)=∞.Equality©1comes from the scaling property of the point impulse. The 1-D version of Eq. (2.8) in the text isδ(ax) =1|a|δ(x). Supposecosθ6= 0. Thenδ(xcosθ+ysinθ−l) =1|cosθ|δ(x+ysinθcosθ−lcosθ).Therefore,∫∞−∞δ(xcosθ+ysinθ−l)dx=1|cosθ|.We also haveP∞(δl)=limX→∞limY→∞14XY∫X−X∫Y−Y|δ(xcosθ+ysinθ−l)|2dx dy=limX→∞limY→∞14XY∫X−X∫Y−Yδ(xcosθ+ysinθ−l)dx dy .Without loss of generality, assumeθ= 0andl= 0, so that we havesinθ= 0andcosθ= 1. Then it follows

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5thatP∞(δl)=limX→∞limY→∞14XY∫X−X∫Y−Yδ(x)dx dy=limX→∞limY→∞14XY∫Y−Y{∫X−Xδ(x)dx}dy=limX→∞limY→∞14XY∫Y−Y1dx=limX→∞limY→∞2Y4XY=limX→∞12X=0.(c)We haveE∞(e)=∫∞−∞∫∞−∞|exp [j2π(u0x+v0y)]|2dx dy=∫∞−∞∫∞−∞1dx dy=∞.And alsoP∞(e)=limX→∞limY→∞14XY∫X−X∫Y−Y|exp[j2π(u0x+v0y)]|2dx dy=limX→∞limY→∞14XY∫X−X∫Y−Y1dx dy=1.(d)We haveE∞(s)=∫∞−∞∫∞−∞sin2[2π(u0x+v0y)]dx dy©2=∫∞−∞∫∞−∞1−cos[4π(u0x+v0y)]2dx dy=∫∞−∞∫∞−∞12dx dy−∫∞−∞∫∞−∞cos[4π(u0x+v0y)]2dx dy©3=∞.Equality©2comes from the trigonometric identitycos(2θ) = 1−2 sin2(θ).Equality©3holds becausethe first integral goes to infinity. The absolute value of the second integral is bounded, although it does not

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6CHAPTER2:SIGNALS AND SYSTEMSconverge asXandYgo to infinity. We also haveP∞(s)=limX→∞limY→∞14XY∫X−X∫Y−Ysin2[2π(u0x+v0y)]dx dy=limX→∞limY→∞14XY∫Y−Y{∫X−X1−cos[4π(u0x+v0y)]2dx}dy=limX→∞limY→∞14XY∫Y−Y[X+ sin[4π(u0X+v0y)]−sin[4π(−u0X+v0y)]8πu0]dy©4=limX→∞limY→∞14XY∫Y−Y[X−sin(4πu0X) cos(4πv0y)4πu0]dy=limX→∞limY→∞14XY(2XY−2 sin(4πu0X) sin(4πv0Y)(4π)2u0v0)=12.In order to get©4, we have used the trigonometric identitysin(α+β) = sinαcosβ+ cosαsinβ. The restof the steps are straightforward.Sinces(x, y)is a periodic signal with periodsX0= 1/u0andY0= 1/v0, we have an alternative way tocomputeP∞by considering only one period in each dimension. Accordingly,P∞(s)=14X0Y0∫X0−X0∫Y0−Y0sin2[2π(u0x+v0y)]dx dy=14X0Y0(2X0Y0−2 sin(4πu0X0) sin(4πv0Y0)(4π)2u0v0)=14X0Y0(2X0Y0−2 sin(4π) sin(4π)(4π)2u0v0)=12.SYSTEMS AND THEIR PROPERTIESSolution 2.4Suppose two LSI systemsS1andS2are connected in cascade. For any two input signalsf1(x, y),f2(x, y), andtwo constantsa1anda2, we have the following:S2[S1[a1f1(x, y) +a2f2(x, y)]]=S2[a1S1[f1(x, y)] +a2S1[f2(x, y)]]=a1S2[S1[f1(x, y)]] +a2S2[S1[f2(x, y)]].So the cascade of two LSI systems is also linear. Now suppose for a given signalf(x, y)we haveS1[f(x, y)] =g(x, y), andS2[g(x, y)] =h(x, y). By using the shift-invariance of the systems, we can prove that the cascade oftwo LSI systems is also shift invariant:S2[S1[f(x−ξ, y−η)]] =S2[g(x−ξ, y−η)] =h(x−ξ, y−η).

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7This proves that two LSI systems in cascade is an LSI systemTo prove Eq. (2.46) we carry out the following:g(x, y)=h2(x, y)∗[h1(x, y)∗f(x, y)]=h2(x, y)∗∫∞−∞∫∞−∞h1(ξ, η)f(x−ξ, y−η)dξ dη=∫∞−∞∫∞−∞h2(u, v)[∫∞−∞∫∞−∞h1(ξ, η)f(x−u−ξ, y−v−η)dξ dη]du dv=∫∞−∞∫∞−∞∫∞−∞∫∞−∞h2(u, v)h1(ξ, η)f(x−u−ξ, y−v−η)dξ dηdu dv=∫∞−∞∫∞−∞h1(ξ, η)[∫∞−∞∫∞−∞h2(u, v)f(x−ξ−u, y−η−v)du dv]dξ dη=h1(x, y)∗[h2(x, y)∗f(x, y)].This proves the second equality in (2.46). By lettingα=u+ξ, andβ=v+η, we haveg(x, y)=∫∞−∞∫∞−∞∫∞−∞∫∞−∞h2(u, v)h1(ξ, η)f(x−u−ξ, y−v−η)dξ dηdu dv=∫∞−∞∫∞−∞[∫∞−∞∫∞−∞h2(α−ξ, β−η)h1(ξ, η)dξ dη]f(x−α, y−β)dα dβ=[h1(x, y)∗h2(x, y)]∗f(x, y),which proves the second equality in (2.46).To prove (2.47) we start with the definition of convolutiong(x, y) =∫∞−∞∫∞−∞h2(ξ, η)h1(x−ξ, y−η)dξ dη=h1(x, y)∗h2(x, y).We then make the substitutionα=x−ξandβ=y−ηand manipulate the resultg(x, y) =∫−∞+∞∫−∞+∞h2(x−α, y−β)h1(α, β)(−dα) (−dβ)=∫+∞−∞∫+∞−∞h1(α, β)h2(x−α, y−β)dα dβ=∫+∞−∞∫+∞−∞h1(ξ, η)h2(x−ξ, y−η)dξ dη=h2(x, y)∗h1(x, y),where the next to last equality follows sinceαandβare just dummy variables in the integral.

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8CHAPTER2:SIGNALS AND SYSTEMSSolution 2.51. Suppose the PSF of an LSI system is absolutely integrable.∫∞−∞∫∞−∞|h(x, y)|dx dy≤C <∞(S2.1)whereCis a finite constant. For a bounded input signalf(x, y)|f(x, y)| ≤B <∞,for every(x, y),(S2.2)for some finiteB, we have|g(x, y)|=|h(x, y)∗f(x, y)|=∣∣∣∣∫∞−∞∫∞−∞h(x−ξ, y−η)f(ξ, η)dξdη∣∣∣∣≤∫∞−∞∫∞−∞|h(x−ξ, y−η)| · |f(ξ, η)|dξdη≤B∫∞−∞∫∞−∞|h(x, y)|dx dy≤BC <∞,for every(x, y)(S2.3)Sog(x, y)is also bounded. The system is BIBO stable.2. We use contradiction to show that if the LSI system is BIBO stable, its PSF must be absolutely integrable.Suppose the PSF of a BIBO stable LSI system ish(x, y), which is not absolutely integrable, that is,∫∞−∞∫∞−∞|h(x, y)|dx dyis not bounded. Then for a bounded input signalf(x, y) = 1, the output is|g(x, y)|=|h(x, y)∗f(x, y)|=∫∞−∞∫∞−∞|h(x, y)|dx dy,which is also not bounded. So the system can not be BIBO stable. This shows that if the LSI system is BIBO stable,its PSF must be absolutely integrable.Solution 2.6(a)Ifg′(x, y)is the response of the system to input∑Kk=1wkfk(x, y), theng′(x, y)=K∑k=1wkfk(x,−1) +K∑k=1wkfk(0, y)=K∑k=1wk[fk(x,−1) +fk(0, y)]=K∑k=1wkgk(x, y)

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9wheregk(x, y)is the response of the system to inputfk(x, y). Therefore, the system is linear.(b)Ifg′(x, y)is the response of the system to inputf(x−x0, y−y0), theng′(x, y) =f(x−x0,−1−y0) +f(−x0, y−y0);whileg(x−x0, y−y0) =f(x−x0,−1) +f(0, y−y0).Sinceg′(x, y)6=g(x−x0, y−y0), the system is not shift-invariant.Solution 2.7(a)Ifg′(x, y)is the response of the system to input∑Kk=1wkfk(x, y), theng′(x, y)=(K∑k=1wkfk(x, y)) (K∑k=1wkfk(x−x0, y−y0))=K∑i=1K∑j=1wiwjfi(x, y)fj(x−x0, y−y0),whileK∑k=1wkgk(x, y) =K∑k=1wkfk(x, y)fk(x−x0, y−y0).Sinceg′(x, y)6=∑Kk=1gk(x, y), the system is nonlinear.On the other hand, ifg′(x, y)is the response of the system to inputf(x−a, y−b), theng′(x, y)=f(x−a, y−b)f(x−a−x0, y−b−y0)=g(x−a, y−b)and the system is thus shift-invariant.(b)Ifg′(x, y)is the response of the system to input∑Kk=1wkfk(x, y), theng′(x, y)=∫∞−∞K∑k=1wkfk(x, η)dη=K∑k=1wk(∫∞−∞fk(x, η)dη)=K∑k=1wkgk(x, y),wheregk(x, y)is the response of the system to inputfk(x, y). Therefore, the system is linear.

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10CHAPTER2:SIGNALS AND SYSTEMSOn the other hand, ifg′(x, y)is the response of the system to inputf(x−x0, y−y0), theng′(x, y)=∫∞−∞f(x−x0, η−y0)dη=∫∞−∞f(x−x0, η−y0)d(η−y0)=∫∞−∞f(x−x0, η)dη.Sinceg(x−x0, y−y0) =∫∞−∞f(x−x0, η)dη, the system is shift-invariant.Solution 2.8From the results in Problem 2.5, we know that an LSI system is BIBO stable if and only if its PSF is absolutelyintegrable.(a)Not stable. The PSFh(x, y)goes to infinite whenxand/orygo to infinity.∫∞−∞∫∞−∞|h(x, y)|dx dy=∫∞−∞∫∞−∞(x2+y2)dx dy=∫∞−∞[∫∞−∞x2dx]dy+∫∞−∞[∫∞−∞y2dy]dx. Since∫∞−∞x2dx=∫∞−∞y2dyis not bounded, then∫∞−∞∫∞−∞(x2+y2)dx dyis not bounded.(b)Stable.∫∞−∞∫∞−∞|h(x, y)|dx dy=∫∞−∞∫∞−∞(exp{−(x2+y2)})dx dy=[∫∞−∞e−x2dx]2=π, which isbounded. So the system is stable.(c)Not stable. The absolute integral∫∞−∞∫∞−∞x2e−y2dx dy=∫∞−∞x2[∫∞−∞e−y2dy]dx=∫∞−∞√πx2dxisunbounded. So the system is not stable.Solution 2.9(a)g(x) =∫∞−∞f(x−t)f(t)dt.(b)Given an input asaf1(x) +bf2(x), wherea, bare some constant, the output isg′(x)=[af1(x) +bf2(x)]∗[af1(x) +bf2(x)]=a2f1(x)∗f1(x) + 2abf1(x)∗f2(x) +b2f2(x)∗f2(x)6=ag1(x) +bg2(x),whereg1(x)andg2(x)are the output corresponding to an input off1(x)andf2(x)respectively.Hence, the system is nonlinear.(c)Given a shifted inputf1(x) =f(x−x0), the corresponding output isg1(x)=f1(x)∗f1(x)=∫∞−∞f1(x−t)f1(t)dt=∫∞−∞f(x−t−x0)f1(t−x0)dt.

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11Changing variablet′=t−x0in the above integration, we getg1(x)=∫∞−∞f(x−2x0−t′)f1(t′)dt′=g(x−2x0).Thus, if the input is shifted byx0, the output is shifted by2x0. Hence, the system is not shift-invariant.CONVOLUTION OF SIGNALSSolution 2.10(a)f(x, y)δ(x−1, y−2)=f(1,2)δ(x−1, y−2)=(1 + 22)δ(x−1, y−2)=5δ(x−1, y−2)(b)f(x, y)∗δ(x−1, y−2)=∫∞−∞∫∞−∞f(ξ, η)δ(x−ξ−1, y−η−2)dξ dη=∫∞−∞∫∞−∞f(x−1, y−2)δ(x−ξ−1, y−η−2)dξ dη=f(x−1, y−2)∫∞−∞∫∞−∞δ(x−ξ−1, y−η−2)dξ dη=f(x−1, y−2)=(x−1) + (y−2)2(c)∫∞−∞∫∞−∞δ(x−1, y−2)f(x,3)dx dy©1=∫∞−∞∫∞−∞δ(x−1, y−2)f(1,3)dx dy=∫∞−∞∫∞−∞δ(x−1, y−2)(1 + 32)dx dy=10∫∞−∞∫∞−∞δ(x−1, y−2)dx dy©2=10Equality©1comes from the Eq. (2.7) in the text. Equality©2comes from the fact:∫∞−∞∫∞−∞δ(x−1, y−2)dx dy=∫∞−∞∫∞−∞δ(x, y)dx dy= 1.

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12CHAPTER2:SIGNALS AND SYSTEMS(d)δ(x−1, y−2)∗f(x+ 1, y+ 2)©3=∫∞−∞∫∞−∞δ(x−ξ−1, y−η−2)f(ξ+ 1, η+ 2)dξ dη©4=∫∞−∞∫∞−∞δ(x−ξ−1, y−η−2)f((x−1) + 1,(y−2) + 2)dξ dη=∫∞−∞∫∞−∞δ(x−ξ−1, y−η−2)f(x, y)dξ dη©5=f(x, y) =x+y2©3comes from the definition of convolution;©4comes from the Eq. (2.7) in text;©5is the same as©2in part(c). Alternatively, by using the sifting property ofδ(x, y)and definingg(x, y) =f(x+ 1, y+ 2), we haveδ(x−1, y−2)∗g(x, y)=g(x−1, y−2)=f(x−1 + 1, y−2 + 2)=f(x, y)=x+y2.Solution 2.11(a)f(x, y)∗g(x, y)=∫∞−∞∫∞−∞f(ξ, η)g(x−ξ, y−η)dξ dη=∫∞−∞∫∞−∞f1(ξ)f2(η)g1(x−ξ)g2(y−η)dξ dηf(x, y)∗g(x, y)=(∫∞−∞f1(ξ)g1(x−ξ)dξ) (∫∞−∞f2(η)g2(y−η)dη).Hence, their convolution is also separable.(b)f(x, y)∗g(x, y) = (f1(x)∗g1(x)) (f2(y)∗g2(y)).

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13Solution 2.12g(x, y)=f(x, y)∗h(x, y)=∫∞−∞∫∞−∞f(x−ξ, y−η)h(ξ, η)dξdη=∫∞−∞∫∞−∞(x−ξ+y−η)exp{−(ξ2+η2)}dξdη=(x+y)∫∞−∞∫∞−∞e−ξ2−η2dξdη−∫∞−∞∫∞−∞ξe−ξ2−η2dξdη−∫∞−∞∫∞−∞ηe−ξ2−η2dξdη=(x+y)[∫∞−∞e−ξ2dξ]2−∫∞−∞e−η2[∫∞−∞ξe−ξ2dξ]dη−∫∞−∞e−ξ2[∫∞−∞ηe−η2dη]dξ=π(x+y)(S2.4)We get (S2.4) by noticing that sinceξis an odd function ande−ξ2is an even function, we must have∫∞−∞ξe−ξ2dξ= 0.Also,∫∞−∞e−ξ2dξ=√π .FOURIER TRANSFORMS AND THEIR PROPERTIESSolution 2.13(a)See the solution to part(b)below. The Fourier transform isF2{δs(x, y)}=δs(u, v)(b)F2{δs(x, y; ∆x,∆y)}=∫∞−∞∫∞−∞δs(x, y; ∆x,∆y)e−j2π(ux+vy)dx dyδs(x, y; ∆x,∆y)is a periodic signal with periods∆xand∆yinxandyaxes. Therefore it can be writtenas a Fourier series expansion. (Please review Oppenheim, Willsky, and Nawad,Signals and Systemsfor thedefinition ofFourier series expansionof periodic signals.)δs(x, y; ∆x,∆y) =∞∑m=−∞∞∑n=−∞Cmnej2π(mx∆x+ny∆y),

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14CHAPTER2:SIGNALS AND SYSTEMSwhereCmn=1∆x∆y∫∆x2−∆x2∫∆y2−∆y2δs(x, y; ∆x,∆y)e−j2π(mx∆x+ny∆y)dx dy=1∆x∆y∫∆x2−∆x2∫∆y2−∆y2∞∑m=−∞∞∑n=−∞δ(x−m∆x, y−n∆y)e−j2π(mx∆x+ny∆y)dx dy.In the integration region−∆x2< x <∆x2and−∆y2< y <∆y2there is only one impulse corresponding tom= 0,n= 0. Therefore, we haveCmn=1∆x∆y∫∆x2−∆x2∫∆y2−∆y2δ(x, y)e−j2π(0·x∆x+0·y∆y)dx dy=1∆x∆y .We have:δs(x, y; ∆x,∆y) =1∆x∆y∞∑m=−∞∞∑n=−∞ej2π(mx∆x+ny∆y).Therefore,F2{δs}=∫∞−∞∫∞−∞δs(x, y; ∆x,∆y)e−j2π(ux+vy)dx dy=∫∞−∞∫∞−∞1∆x∆y∞∑m=−∞∞∑n=−∞ej2π(mx∆x+ny∆y)e−j2π(ux+vy)dx dy=∞∑m=−∞∞∑n=−∞1∆x∆y∫∞−∞∫∞−∞ej2π(mx∆x+ny∆y)e−j2π(ux+vy)dx dy=∞∑m=−∞∞∑n=−∞1∆x∆yF2{ej2π(mx∆x+ny∆y)}=∞∑m=−∞∞∑n=−∞1∆x∆y δ(u−m∆x , v−n∆y)©5=∞∑m=−∞∞∑n=−∞1∆x∆y·∆x∆yδ(u∆x−m, v∆y−n)F2{δs}=δs(u∆x, v∆y)Equality©5comes from the propertyδ(ax) =1|a|δ(x).

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