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is called the scattering amplitude. We now relate fk(θ, φ) to cross section!! dσ = (cid:104)# particles scattered per unit time into solid angle dΩ about (θ, φ) (cid:104) flux of incident particles = # particles area · time (cid:105) (cid:105) (7.1.10) dσ is called differential cross section, it’s the area that removes f...
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m |fk(θ, φ)|2 dΩ dσ = (cid:19) (cid:19)(cid:126)k (cid:19) m hence Differential cross section: dσ dΩ = |fk(θ, φ)|2 Total cross section: σ = (cid:90) (cid:90) dσ = |fk(θ, φ)|2 dΩ (7.1.13) (7.1.14) 7.2. PHASE SHIFTS 7.2 Phase shifts 137 Assume V (r) = V (r), so that we are dealing with a central potential. First recall t...
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The solution to (7.2.3) is uE(cid:96)(ρ) = A(cid:96)ρj(cid:96)(ρ) + B(cid:96) ρ n(cid:96)(ρ) or uE(cid:96)(r) = A(cid:96)rj(cid:96)(kr) + B(cid:96) r n(cid:96)(kr) (7.2.4) where j(cid:96)(ρ) is the spherical Bessel function n(cid:96)(ρ) is the spherical Neumann function j(cid:96)(ρ) is non singular at the origin n(cid...
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2.9) This is an incredible relation in which a plane wave is built by a linear superposition of spherical waves with all possible values of angular momentum! Each (cid:96) contribution is a partial wave. Each partial wave is an exact solution when V = 0. We can see the spherical ingoing and outgoing waves in each par...
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2i (cid:17) eikxe2iδk − e−ikx (cid:124) (cid:123)(cid:122) (cid:125) same ingoing wave for x > a (7.2.12) (7.2.13) where the outgoing wave can only differ from the ingoing one by a phase, so that probability is conserved. Finally we defined So do a similar transformation to write a consistent ansatz for ψ(r). We have fro...
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(cid:96),0(θ) 1 2i (cid:124) (cid:16) (cid:17) e2iδ(cid:96) − 1 (cid:123)(cid:122) eiδ(cid:96) sin δ(cid:96) (cid:125) eikre− i(cid:96)π r 2 = fk(θ) eikr r √ 4π k = ∞ (cid:88) √ (cid:96)=0 2(cid:96) + 1 Y(cid:96),0(θ)eiδ(cid:96) sin δ(cid:96) eikr r , (7.2.16) (7.2.17) where we noted that e− i(cid:96)π 2 = (−i)(cid:96)...
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π k2 ∞ (cid:88) (cid:96)=0 (2(cid:96) + 1) sin2 δ(cid:96) . dΩ Y ∗ (cid:96),0(Ω)Y(cid:96)(cid:48),0(Ω) (cid:125) (cid:123)(cid:122) δ(cid:96)(cid:96)(cid:48) (7.2.20) Now let us explore the form of f (θ) in the forward direction θ = 0. Given that Y(cid:96),0(θ) = (cid:114) 2(cid:96) + 1 4π P(cid:96)(cos θ) =⇒ Y(cid:96)...
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12) (cid:12) (cid:12)(cid:96) = (A(cid:96)j(cid:96)(kr) + B(cid:96)n(cid:96)(kr)) Y(cid:96),0(θ) x > a (7.2.22) If B (cid:54)= 0 then V (cid:54)= 0. As a matter of fact, if V = 0 the solution should be valid everywhere and n(cid:96) is singular at the origin, thus B(cid:96) = 0. Now, let’s expand ψ(x)|(cid:96) for larg...
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2.25) = 1 1 kr 2i (cid:39) e−iδ 1 kr (cid:104) ei(kr− (cid:96)π 2 +δ(cid:96)) − e−i(kr− (cid:96)π 2 +δ(cid:96))(cid:105) Y(cid:96),0(θ) (cid:104) ei(kr− (cid:96)π 2 +2δ(cid:96)) − e−i(kr− (cid:96)π 2 )(cid:105) 1 2i Y(cid:96),0(θ) (7.2.26) 7.2. PHASE SHIFTS 141 This shows that our definition of δk above is indeed consi...
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(cid:12)(cid:96) ∼ e−i(kr− (cid:96)π 2 ) kr Y(cid:96),0(θ) (7.2.29) Each wave is a solution, but not a scattering one. 7.2.2 Example: hard sphere V (r) = (cid:40) ∞ 0 r ≤ a r > a Radial solution R(cid:96)(r) = u r = A(cid:96)j(cid:96)(kr) + B(cid:96)n(cid:96)(kr) ψ(a, θ) = (cid:88) R(cid:96)(a)P(cid:96)(cos θ) ≡ 0 (cid...
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2 ∞ (cid:88) (cid:96)=0 (2(cid:96) + 1) j2 (cid:96) (ka) (cid:96) (ka) + n2 j2 (cid:96) (ka) Griffiths gives you the low energy ka (cid:28) 1 expansion σ (cid:39) 4π k2 ∞ (cid:88) (cid:96)=0 1 2(cid:96) + 1 (cid:20) 2(cid:96)(cid:96)! (2(cid:96))! (cid:21)4 (ka)4(cid:96)+2 (7.2.36) (7.2.37) (7.2.38) At low energy the dom...
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= 0). This must be matched to the general solution that holds for V = 0, as it holds for r > a: At r = a must match the function and its derivative R(cid:96)(a) = A(cid:96)j(cid:96)(ka) + B(cid:96)n(cid:96)(ka) (7.2.43) 7.2. PHASE SHIFTS aR(cid:48) (cid:96)(a) = ka (cid:16) A(cid:96)j(cid:48) (cid:96)(ka) + B(cid:96)...
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:16) j(cid:96)−iη(cid:48) (cid:96) j(cid:96)−in(cid:96) (cid:16) j(cid:96)+iη(cid:48) (cid:96) j(cid:96)+in(cid:96) which we can also write as e2iδ(cid:96) = e2iξ(cid:96) (cid:124)(cid:123)(cid:122)(cid:125) Hard sphere phase shift   β(cid:96) − ka β(cid:96) − ka (cid:17)   (cid:17) (cid:16) j(cid:96)−iη(cid:48) (c...
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0 radial equation with angular momentum (cid:96) and energy (cid:126)2k2/(2m). Setting the effective potential equal to the energy: (cid:126)2(cid:96)((cid:96) + 1) 2mr2 = (cid:126)2k2 2m we find that the turning point for the solution is at k2r2 = (cid:96)((cid:96) + 1) . Thus we expect j(cid:96)(kr) to be exponentially...
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U (r)(cid:3) ψ(r) = k2ψ(r) (cid:0)∇2 + k2(cid:1) ψ(r) = U (r)ψ(r) This is the equation we want to solve. Let us introduce G(r − r(cid:48)), a Green function for the operator ∇2 + k2, i.e. (cid:0)∇2 + k2(cid:1) G(r − r(cid:48)) = δ(3)(r − r(cid:48)) Then we claim that any solution of the integral equation ψ(r) = ψ0(r) +...
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as a matter of fact ∇2 (cid:0)− 1 4πr (cid:0)∇2 r + k2(cid:1) e±ikr r = 0 ∀ r (cid:54)= 0 (cid:1) = δ3(r) (7.3.9) (7.3.10) G±(r) = − 1 4π e±ikr r 146 CHAPTER 7. SCATTERING and we are going to refer to G+ as the outgoing wave Green function and to G− as the ingoing wave Green function. Let’s verify that this works, wri...
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(cid:18) 1 4πr − e±ikr (cid:8)(cid:8)(cid:8)(cid:8) e±ikr + δ3(r) + e±ikrδ3(r) + 2 (cid:8)(cid:8)(cid:8)(cid:8) e±ikr 2ik ± (cid:8)(cid:8) 4πr 2ik = −k2G±(r) ∓ (cid:8)(cid:8) 4πr = −k2G±(r) + δ3(r) (cid:88) We will use ψ0(r) = eikz and G = G+ Thus, with these choices, (7.3.6) takes the form ψ(r) = eikz + (cid:90) d3r(c...
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n·r(cid:48) U (r(cid:48))ψ(r(cid:48)) (cid:21) eikr r 147 (7.3.20) (7.3.21) The object in brackets is a function of the unit vector n in the direction of r. This shows that the integral equation, through the choice of G, incorporates both the Schr¨odinger equation and the asymptotic conditions. By definition, the object...
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)) (7.3.27) 148 CHAPTER 7. SCATTERING By iterating this procedure we can form an infinite series which schematically looks like ψ = eiki·r + (cid:90) G U eikir + (cid:90) (cid:90) G U G U eikir + (cid:90) (cid:90) G U (cid:90) G U G U eikir + . . . (7.3.28) The approximation in which we keep the first integral in this ...
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further by performing the radial integration. We have f Born k (θ) = − (cid:90) 1 4π 2m (cid:126)2 d3re−iK·rV (r) . (7.3.33) By spherical symmetry this integral just depends on the norm K of the vector K. This is why we have a result that only depends on θ: while K is a vector that depends on both θ and φ, its magnitud...
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e−µr sin(Kr) = − 2mβ (cid:126)2(µ2 + K2) . (7.3.37) We can give a graphical representation of the Born series. Two waves reach the desired point r. The first is the direct incident wave. The second is a secondary wave originating at the scattering “material” at a point r(cid:48). The amplitude of a secondary source at r...
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MIT OpenCourseWare http://ocw.mit.edu 18.917 Topics in Algebraic Topology: The Sullivan Conjecture Fall 2007 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. Introduction (Lecture 1) Let X be an algebraic variety defined over a field k. If k is the field C of complex...
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´etale cohomology provides a purely algebraic recipe for extracting the cohomology groups H∗(X(C); Z/pZ). This raises the question: to what extent can we recover the topological space X(C) itself in purely algebraic terms? Of course, algebro-topological invariants like cohomology do not contain enough information to...
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a space M is, in some sense, determined up to homotopy equivalence by the mod-p cohomology H∗(M ; Z/pZ). Of course, for this to be true one must consider H∗(M ; Z/pZ) as endowed with more structure than just a graded vector space. The precise statement is that M ∨ can be reconstructed from the cochain complex C ∗(M ...
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one has an underlying topological space of real points X(R). To what extent can this topological space be reconstructed in purely algebraic terms? We note that there is a canonical inclusion X(R) ⊆ X(C). The group Z/2Z acts on X(C), via complex conjugation, and we can identify X(R) with the fixed set X(C)Z/2Z with r...
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above, the functor M �→ M G need not preserve homotopy equivalences. One explanation for this is that the G-space ∗ is badly behaved. To get a better functor, we need to replace ∗ by a better G-space. Definition 2. Let G be a group. We let EG denote a contractible space on which G acts freely, and BG the quotient sp...
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sequence of maps X(R) � X(C)Z/2Z → X(C)hZ/2Z → (X(C)∨)hZ/2Z . Here X(C)∨ denotes the p-adic completion of X(C). The left hand side of the above diagram is what we are interested in: the space of R-valued points of the algebraic variety X. The right hand side is what we can understand in purely algebraic terms, usi...
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ure of Sullivan: Conjecture 6 (Sullivan). Let p be a prime number. Let M be a topological space with an action of a finite p-group G. Assume that M is sufficiently nice (for simplicity, a simply connected finite G-CW complex). Then the canonical map M G → (M ∨)hG induces an isomorphism on mod-p cohomology. This con...
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To prove that the mapping space Map(BG, M ) is equivalent to M , it suffices to prove the result after completing M at each prime number q. The essential case is that in which p = q. In this case, the essence of the problem is to compute the cohomology ring and to show that it agrees with the cohomology ring H∗(Map(B...
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of stable cohomology operations forms a graded ring, where multiplication is given by composition. This ring is called the mod-p Steenrod algebra and is usually denoted by A. If Y is any topological space, then the cohomology H∗(Y ; Z/pZ) has the structure of a module over A. In fact, a bit more is true: H∗(Y ; Z/pZ...
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elegant proofs of Conjecture 6 and Theorem 7. We now sketch how to use Theorem 9 to prove THeorem 7 in a special case. Assume that M is the p-adic completion of a simply connected finite CW -complex. We wish to show that the canonical map M Map(BV, M ) → induces an equivalence on mod-p cohomology. In view of Theorem...
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opy groups of any E -algebra over the field Z/pZ. Once we understand the Steenrod algebra well enough, we will proceed to study its category U of unstable modules, and introduce the functor TV . To establish the basic properties of TV (such as exactness), we will need to have a good understanding of the structure of ...
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MIT 2.852 Manufacturing Systems Analysis Lectures 15–16: Assembly/Disassembly Systems Stanley B. Gershwin http://web.mit.edu/manuf-sys Massachusetts Institute of Technology Spring, 2010 2.852 Manufacturing Systems Analysis 1/41 Copyright �2010 Stanley B. Gershwin. c Assembly-Disassembly Systems Assembly System ...
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a machine does an operation, it removes one part from each upstream buffer and inserts one part into each downstream buffer. ◮ Continuous material systems: when machine Mi operates during [t, t + δt], it removes µi δt from each upstream buffer and inserts µi δt into each downstream buffer. ◮ A machine is starved if a...
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steady-state distribution is a function of the initial conditions. ◮ Example: if the system below has K pallets at time 0, it will have K pallets for all t ≥ 0. Therefore, the probability distribution is a function of K . Raw Part Input Empty Pallet Buffer Finished Part Output ◮ This applies to more general sys...
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(i, m) Mi B (n, i) B (i, q) Mj • • • Mn Mm • • • Mq Part of Original Network 2.852 Manufacturing Systems Analysis 10/41 Copyright c�2010 Stanley B. Gershwin. Assembly-Disassembly Systems Decomposition Mu (j, i) B (j, i) Md (j, i) • • • Mu (n, i) B (n, i) Md (n, i) Mu (i, m) B (i, m) Md (i, ...
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2 2.852 Manufacturing Systems Analysis 14/41 Copyright c�2010 Stanley B. Gershwin. Numerical examples Eight-Machine Systems 4.5640 7.7077 7.7077 7.7077 4.1089 6.5276 2.2923 Case 3: Same as Case 1 except p1 = .2 2.852 Manufacturing Systems Analysis 15/41 Copyright c�2010 Stanley B. Gershwin. Numeric...
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19/41 Copyright c�2010 Stanley B. Gershwin. Numerical Examples Alternate Assembly Line Designs 2.852 Manufacturing Systems Analysis 20/41 Copyright c�2010 Stanley B. Gershwin. Equivalence Simple models Consider a three-machine transfer line and a three-machine assembly system. Both are perfectly reliable (p...
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µ 3 µ 2 µ 1 11 µ 1 µ 2 µ 3 µ 2 22 21 µ 3 µ 3 03 µ 3 02 µ 3 01 µ 3 µ 1 00 µ 1 10 20 2.852 Manufacturing Systems Analysis 23/41 Copyright �2010 Stanley B. Gershwin. c Equivalence Unlabeled State Space ◮ The transition graphs of the two systems are the same except for the labels of the sta...
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� � N = 2 1 µ 1 µ 3 µ 2 N = 3 2 µ 1 µ 1 20 10 µ 2 µ 3 µ 2 µ 3 µ 1 µ 1 11 µ 2 µ 3 µ 2 µ 1 12 µ 1 µ 3 21 22 µ 2 µ 3 µ 2 µ 3 00 µ 3 01 µ 3 02 µ 3 µ 1 03 µ 1 13 23 2.852 Manufacturing Systems Analysis 25/41 Copyright �2010 Stanley B. Gershwin. c Equivalence Transfer L...
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�2010 Stanley B. Gershwin. c Equivalence Assembly System ¯n1 2 3 2 3 n¯1 = n1p(n1, n2) = n1 p(n1, n2) n1=0 n2=0 � � N = 2 1 � n2=0 n1=0 � � � µ 1 µ 3 µ 2 N = 3 2 µ 1 µ 1 20 10 µ 2 µ 3 µ 2 µ 3 µ 1 µ 1 11 µ 2 µ 3 µ 2 µ 1 12 µ 1 µ 3 21 22 µ 2 µ 3 µ 2 µ 3 00 µ 3 01 µ 3...
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1 10 20 03 3 02 3 01 3 00 2.852 Manufacturing Systems Analysis 29/41 Copyright �2010 Stanley B. Gershwin. c Equivalence Equal n¯1 Assembly System Production Rate Therefore T A = n¯1 n¯1 2.852 Manufacturing Systems Analysis 30/41 Copyright �2010 Stanley B. Gershwin. c Equivalence Assembly Sy...
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2 µ 1 µ 2 µ 3 µ 1 µ 1 23 13 µ 2 µ 3 µ 2 µ 3 µ 1 µ 1 12 µ 2 µ 3 µ 2 µ 1 11 µ 1 µ 3 22 21 µ 2 µ 3 µ 2 µ 3 03 µ 3 02 µ 3 01 µ 3 µ 1 00 µ 1 10 20 2.852 Manufacturing Systems Analysis 32/41 Copyright �2010 Stanley B. Gershwin. c Equivalence Complementary n¯1 Assembly System P...
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u(j) and d (j) = d(j); and for j ∈ Ω, u (j) = d(j) and d (j) = u(j). That is, there is a set of buffers such that the direction of flow is reversed in the two networks. ′ ′ ′ ◮ Then, the transition equations for network Z are the same as those of Z , except that the buffer levels in Ω are replaced by the amounts of...
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Ω n¯b ′ (n ′ (0)) = Nb − n¯b(n(0)), for j ∈ Ω n¯b 2.852 Manufacturing Systems Analysis 36/41 Copyright �2010 Stanley B. Gershwin. c Equivalence Theorem Corollary: That is, ◮ the production rates of the two systems are the same, ◮ the average levels of all the buffers in the systems whose direction of flow has ...
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Equivalence Example of equivalent loops 1 1 4 2 4 2 3 Ω = (3, 4) 3 1 1 4 2 4 2 3 3 (a) A Fork/ Join Network (b) A Closed Network 2.852 Manufacturing Systems Analysis 40/41 Copyright �2010 Stanley B. Gershwin. c Equivalence To come ◮ Loops and invariants ◮ Two-machine loops ◮ Instability of ...
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Basic Network Metrics and Operations • Meshness ratio • Degree correlation – Joint degree distribution – K-nearest neighbors – Pearson degree correlation • Rich club metric • Degree-preserving rewiring • Generating a graph that has a specified degree sequence • Finding Pearson degree correlation • Finding commu...
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s up early 2000s papers purporting to find the structure of the internet • Shows that there are three ways to do this, each approximate, using different methods, each with a bias • Shows that each way gives different results, providing caution about artifacts inherent in data collection • Joint Degree Distributio...
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/45 node x y 1 1 2 2 2 3 4 4 5 5 average 2 2 3 3 3 1 2 2 2 2 2.2 3 2 1 2 2 3 3 3 2 2 x = 2.2 y = 2.2 pearson -0.676753 Calculating x-bar x = sum of column values number of column values each node of degree k creates k rows with k in each row number of rows = sum of entries in kvec(A) = sum...
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Implementation function prs = pearson(A) %calculates pearson degree correlation of A [rows,colms]=size(A); won=ones(rows,1); k=won'*A; ksum=won'*k'; ksqsum=k*k'; xbar=ksqsum/ksum; num=(k-won'*xbar)*A*(k'-xbar*won); kkk=(k'-xbar*won).*(k'.^.5); denom=kkk'*kkk; prs=num/denom; 2/16/2011 Basic Network Metrics © Daniel E W...
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2011 Basic Network Metrics © Daniel E Whitney 1997-2010 17/45 JDD for Random Matrix 2/16/2011 Basic Network Metrics © Daniel E Whitney 1997-2010 18/45 Rewiring • A way to deliberately transform a graph • Several ways this is done – Unhooking one end of an edge and hooking it in somewhere else – Adding a new e...
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rgrowdgoal (grows r to a desired value called goal, ignores connectedness) • • • You can easily write your own to do what you want 2/16/2011 Basic Network Metrics © Daniel E Whitney 1997-2010 22/45 Rewired V8 Engine Maslov-Sneppen randomizing Volz clust reduction 2/16/2011 Basic Network Metrics © Daniel E Whit...
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, Q], where mainNum is the # of % components that have at least two nodes as members, SingletonNum is the # % of singletons, and Q is the Q defined by Newman-Girvan. % dendrogramRecord: First row is mainNum, second row is singletonNum, and % the third row is Q, and the rest rows is the partition of nodes (the same % fo...
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1.0000000e+00 1.0000000e+00 2.0000000e+00 1.0000000e+00 1.0000000e+00 3.0000000e+00 1.0000000e+00 1.0000000e+00 4.0000000e+00 2.0000000e+00 2.0000000e+00 5.0000000e+00 3.0000000e+00 2.0000000e+00 6.0000000e+00 4.0000000e+00 2.0000000e+00 Q is based on density of Links inside groups compared To links between groups The...
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0.20408 2/16/2011 Basic Network Metrics © Daniel E Whitney 1997-2010 29/45 Rich Club Metric • Measures the extent to which the high degree nodes link to each other • A subset of Pearson degree correlation since it focuses on the high degree nodes • Large RCM indicates that hi...
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of random_graph Graph generation is slow for n > 100 - 200 Only one type of graph sfng in folder Barabasi-Albert Power law with 2 < k < 3 typically Folder Volz No As above As above Generates a symmetric edge list Can choose the clustering coeff No buildSmax Builds graph with max positive degree ...
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% only N and the kind of distribution would be used. Others, like E, % will be ignored Courtesy of Gergana Bounova. Used with permission. 2/16/2011 Basic Network Metrics © Daniel E Whitney 1997-2010 34/45 degree_dist.m function [Nseq] = degree_dist(N,p,distribution) % Random graph degree sequence constructi...
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network_generator_script % script to generate random networks with given degree sequence % Java executable RandomClusteringNetwork.jar must be in your matlab % directory N=100 p=0.1 E=10 distribution='normal' fun=1 degrees=1 stop=1 Random networks with tunable degree distribution and clustering Erik Volz Cornell...
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Generator function [G]=erdosRenyi(nv,p,Kreg) %Function [G]=edosRenyi(nv,p,Kreg) generates a random graph based on %the Erdos and Renyi algoritm where all possible pairs of 'nv' nodes are %connected with probability 'p'. It does this by creating a connected %regular grid with k = Kreg at every node and then rewires....
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= 3(z /2 −1) 2(z −1) (1− p)3 2/16/2011 Basic Network Metrics © Daniel E Whitney 1997-2010 42/45 SFNG • Text from the “read me:” • B-A Scale-Free Network Generation and Visualization • By Mathew Neil George • The *SFNG* m-file is used to simulate the B-A algorithm and returns scale- free networks of given size...
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Lecture 3 Semiconductor Physics (II) Carrier Transport Outline • Thermal Motion • Carrier Drift • Carrier Diffusion Reading Assignment: Howe and Sodini; Chapter 2, Sect. 2.4-2.6 6.012 Spring 2009 Lecture 3 1 1. Thermal Motion In thermal equilibrium, carriers are not sitting still: • Undergo collisions wi...
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velocity in direction of the field: time v = vd = ± τ c = ± qE 2m n,p qτ c 2m n,p E This is called drift velocity [cm s-1] Define: µn, p = qτ c 2m n,p Then, for electrons: and for holes: ≡ mobility [cm 2 V−1 s −1] v dn = −µn E v dp = µp E 6.012 Spring 2009 Lecture 3 5 Mobility - is a measure of e...
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E vdp + drift Jp x x 6.012 Spring 2009 Lecture 3 7 Total Drift Current Density : drift J drift = J n + Jp drift = q n( µ n + pµ p )E Has the form of Ohm’s Law J = σσσσE = E ρρρρ Where: Then: σ ≡ conductivity [Ω-1 • cm-1] ρ ≡ resistivity [Ω • cm] σσσσ= 1 ρρρρ = q n( µn + pµ p ) 6.012 Spring 2009...
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cm / s << vth drift ≈ qnvdn = qnµnE = σσσσE = Jn E ρρρρ drift ≈ 4.8 ×103 A / cm 2 Jn Time to drift through L = 0.1 µm td = L vdn = 10 ps fast! 6.012 Spring 2009 Lecture 3 10 3. Carrier Diffusion Diffusion = particle movement (flux) in response to concentration gradient n x Elements of diffusion: • A ...
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At the core of drift and diffusion is same physics: collisions among particles and medium atoms ⇒⇒⇒⇒ there should be a relationship between D and µ Einstein relation [will not derive in 6.012] D µ = kT q In semiconductors: D n µn = kT q = D p µ p kT/q ≡ thermal voltage At room temperature: kT q ≈ ...
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16 MIT OpenCourseWare http://ocw.mit.edu 6.012 Microelectronic Devices and Circuits Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
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224 Chapter4. WavenumberIntegrationTechniques perimentally [23], which should be kept in mind when comparing synthetic and exper- imental reflection data. 4.5 Wavenumber Integration Integral transform techniques such as wavenumber integration is an important mod- eling tool in all disciplines dealing with wave propagati...
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39), where "&(cid:10)%’ . The numerical evaluation of this integral is complicated by the following features, which must be considered when choosing an integration technique: (cid:2)(cid:4)(cid:3)(cid:18)(cid:17) (cid:19)!(cid:6) (cid:1)(cid:9)(cid:8) ( The infinite integration interval. ( The wavenumber discretization ...
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refinement, or by modern adaptive integration techniques, described later. In contrast, the traditional application of modeling in underwater acoustics has been aimed at predicting the transmission loss over very large horizontal distances, typically 2-3 orders of magnitude larger than the vertical scales of the ocean a...
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evaluation of the inverse Hankel transform, Eq. (4.93), can be obtained by the so-called FFP (Fast Field Program) integration technique intro- 226 Chapter4. WavenumberIntegrationTechniques duced by DiNapoli and Deavenport [1]. First the Bessel function is expressed in terms of Hankel functions, (cid:3)(cid:18)(cid:17)...
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25) (cid:17)(cid:20)(cid:19) (cid:29)(cid:28)(cid:31)(cid:30)! #"%$ to arrive at the following expression for the inverse Hankel transform, (cid:2)(cid:4)(cid:3)(cid:5)(cid:0)(cid:7)(cid:6) (cid:2))(cid:3)*(cid:17)(cid:20)(cid:19)(cid:21)(cid:6) (cid:17)(cid:20)(cid:19) (cid:27)(cid:29)(cid:17)(cid:20)(cid:19) 65 )( (c...
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.96), which is different for the various ranges considered. It would therefore lead to differ- ent quadrature points for each range, with a significant additional computational effort as a result. Secondly, the variation of the kernel is strongly dependent on the environ- mental model and frequency, and moreover charact...
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depth, the integration kernel will be rapidly decaying with increasing (cid:17)(cid:20)(cid:19) . For small separations the decay is slower, and in the extreme situation of source and receiver at the same depth, the kernel only decays as (cid:17) . Based on this information, it is usually straightforward to truncate th...
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��eld have propagation wavenumbers far out in the evanescent regime. (cid:14)(cid:0)(cid:3)(cid:2)(cid:5)(cid:4) 4.5.3 Wavenumber Discretization - Aliasing To numerically evalute the wavenumber integral in Eq. (4.96) the integration kernel must be evaluated at a discrete number of wavenumbers. Even though dedicated qua...
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:8)(cid:16) (cid:16)(cid:15) (cid:16)(cid:15) )(+* ,/. (cid:11)(cid:10) (cid:17)(cid:25)(cid:19) ’& (cid:9)(cid:15)- (cid:18)(cid:17) (cid:1)(cid:25)(cid:8) (cid:1)(cid:25)(cid:8) (cid:28)(cid:31)(cid:30)! (cid:6)(cid:5)(cid:8)(cid:7) (cid:14)(cid:13) (4.98) It is well known from the discretization of time–frequency t...
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the wavenumber integral (4.99) (4.100) (cid:3)(cid:30)(cid:0)(cid:31)(cid:6) (cid:2))(cid:3)*(cid:17)(cid:20)(cid:19) (cid:17)(cid:20)(cid:19) (cid:27)(cid:29)(cid:17)(cid:20)(cid:19) 15 (cid:1)(cid:9)(cid:8)(cid:11)(cid:10) (cid:1)(cid:25)(cid:8) 32 (cid:30)4" (4.101) (cid:3)(cid:30)(cid:0)(cid:31)(cid:6) (cid:1)(cid:...
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)(cid:19) (cid:3)(cid:30)(cid:0) (cid:3)(cid:30)(cid:0)(cid:31)(cid:6) (cid:13)(cid:6) 5(cid:31)5 (cid:31)5 (cid:2)(cid:1)(cid:3)(cid:0) (cid:5)(cid:4) (cid:1)(cid:25)(cid:8) (cid:1)(cid:9)(cid:8) (4.104) Therefore, whereas the continuous integral represents a solution over an infinite range interval, the discrete summa...
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guide. The integration kernel schematically in the upper frame in Fig. 4.5, with the dashed portion near the origin factor. Because indicating the squareroot singularity introduced by the geometric of the continuity condition at range window boundaries (cid:0) , all wave components . , propagating in produced by the ph...
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13) & (cid:14) (cid:21) (cid:11) (cid:6) (cid:6) 230 Chapter4. WavenumberIntegrationTechniques g(k ,z) r K 2K kr * f (r,z) f(r,z) −R 0 R 2R r Fig. 4.5. Aliasing associated with discrete wavenumber integration for typical Pekeris wave- guide problem. The wavenumber kernel showing the presence of a two attenuated modes ...
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wise it will be wrapped into the neighboring windows, and vice versa for the sources in the neighbor windows. This is the fundamental aliasing or wrap-around problem associated with discrete Fourier Transforms, requiring the range window to be large enough for the periodic components to be insignificant, in turn requiri...
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) (cid:2) (cid:6) (cid:10) (cid:23) (cid:4) (cid:1) (cid:27) (cid:0) " (cid:9) (cid:19) & (cid:5) (cid:24) , , . & (cid:7) (cid:12) (cid:15) (cid:16) (cid:15) (cid:6) (cid:28) (cid:30) (cid:19) (cid:9) (cid:15) (cid:17) " $ 2 (cid:17) (cid:28) (cid:30) , - (cid:8) (cid:12) (cid:6) 232 Chapter4. WavenumberIntegrationTe...
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20)(cid:19) (cid:17)(cid:20)(cid:19) (cid:3)(cid:2) (cid:8)(cid:0) (cid:16)(cid:0)(cid:18)(cid:17)(cid:20)(cid:19) (cid:1)(cid:0)(cid:18)(cid:17)(cid:20)(cid:19) Therefore, for the upper half of the wavenumber components in Eq. 4.103 is indistinguishable from the negative wavenumbers in regard to the value of the expo-...
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method for extending the range beyond the Nyquist limit (cid:0) it is not in general recommendable. Instead, the range should be extended by simply reducing the wavenumber sampling . . Even though the (cid:17)(cid:25)(cid:19) (cid:17)(cid:25)(cid:19) (cid:3)(cid:2) (cid:17)(cid:20)(cid:19) . We will illustrate the effe...
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) (cid:5) (cid:2) (cid:9) (cid:10) (cid:28) (cid:30) (cid:2) & (cid:5) (cid:2) (cid:10) (cid:28) (cid:30) (cid:5) (cid:10) & (cid:9) (cid:17) (cid:5) (cid:2) (cid:9) 5 (cid:17) (cid:10) $ (cid:10) $ (cid:0) (cid:10) (cid:1) (cid:1) (cid:0) (cid:24) (cid:23) (cid:1) (cid:1) $ (cid:6) (cid:1) (cid:1) (cid:1) (cid:10) $...
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that the field decays more rapidly with range than the geometrical spreading decay . For a perfectly lossless waveguide, the modal field decays only due to geometrical spreading. Consequently, the range window cannot in this case be made large enough to eliminate the wrap-around, which is consistent with the fact that th...
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)(cid:28)(cid:12) (cid:1)(cid:9)(cid:8) 60 (cid:1)(cid:0) (cid:30)4" (4.110) (cid:2)(cid:4)(cid:3) , and where , is the contour shown in Fig. 4.7. The contour consists of three linear sections , where the vertical sections of length (cid:5) are chosen at the points where the wavenumber axis would in any case be truncat...
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:25) (cid:9) (cid:5) (cid:16) (cid:28) (cid:30) , - (cid:8) (cid:12) 236 Chapter4. WavenumberIntegrationTechniques (cid:3)(cid:30)(cid:0) (cid:4)(cid:2) (cid:3)(cid:5)(cid:0) (cid:4)(cid:2) (cid:2)(cid:1)(cid:3)(cid:0) (cid:1)(cid:9)(cid:8) (cid:1)(cid:0) (4.114) (cid:16)(cid:0)(cid:18)(cid:17)(cid:20)(cid:19) In this...
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the contour may become significant. On the other hand, a too small value will require a very large number of sampling points. For most practical purposes an attenuation of the wrap-around by 60 dB is more than sufficient [13]. The corresponding value of the contour offset is (cid:8)(cid:7)(cid:10)(cid:9) (cid:8)(cid:7)(c...
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(4.114) for the two contours. On the dB-scale used here, the loss computed with (cid:24) (solid curve) is identical in the whole range window to the exact loss shown as the solid curve in (dashed curve) is correct only Fig. 4.4(b). However, the loss computed with (cid:24) (cid:16)(cid:14) (cid:13)(cid:12) (cid:25) (cid...
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:5)(cid:4) (cid:9)(cid:2) (cid:5)(cid:4) (solid curve) and (cid:1) (solid curve) and (cid:1) (cid:7)(cid:6)(cid:8)(cid:2) (cid:10)(cid:6)(cid:8)(cid:2) (dashed curve). (b) Transmission loss (dashed curve). (cid:0) & (cid:7) (cid:0) (cid:2) (cid:6) (cid:0) (cid:6) (cid:0) (cid:2) (cid:6) (cid:0) (cid:6) 238 Chapter4. W...
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than a single range, the wavenumber sampling (cid:17)(cid:25)(cid:19) would have to be frequency independent in order to satisfy Eq. (4.107). distance (cid:23) Furthermore, since the pulse response is usually required only for a relatively small number of ranges, one of the direct numerical quadrature schemes described...
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(cid:3) (cid:24) (cid:0) (cid:0) (cid:0) (cid:5) (cid:3) (cid:5) (cid:0) (cid:0) (cid:10) (cid:1) (cid:1) (cid:10) (cid:27) (cid:1) (cid:23) 4.5. WavenumberIntegration 239 an FFT for every receiver range, but, more importantly, it requires a numerical separa- tion parameter which is not easily selected. The so-called ...
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17)(cid:25)(cid:19)(cid:26)(cid:0) (cid:3)*(cid:17)(cid:20)(cid:19) (cid:27)(cid:29)(cid:17)(cid:20)(cid:19) 3(cid:11) (4.116) (cid:5)(cid:18)(cid:7)(cid:10)(cid:9) (cid:5)(cid:14)(cid:13)(cid:15)(cid:9) (cid:1)(cid:0)(cid:3)(cid:2) (cid:3)(cid:6)(cid:5) for (cid:2) time-dependence corresponds to outgoing waves and whe...
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ficiently evaluated using a stan- dard Fast Fourier Transform if the range (cid:0) and wavenumber (cid:17) are discretized equidis- tantly, with the sampling intervals being constrained by Eq. 4.107. (cid:17)(cid:20)(cid:19) (cid:8)(cid:7) The error associated with using Eq. (4.119) is clearly associated with the approx...
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8) (cid:1) (cid:3) (cid:1) (cid:1) . (cid:1) (cid:1) (cid:1) (cid:1) (cid:15) (cid:0) (cid:15) 4.5. WavenumberIntegration 241 Equation (4.121) can be evaluated very efficiently. First of all, with the wavenumber and range sampling constrained by Eq. (4.107), all values of the exponentials are com- puted as part of the ...
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course that (cid:0) is choosen large enough to make the wrap-around insignificant in this part of the range window. even, antisymmetric for " (cid:0)(cid:3)(cid:2)(cid:5)(cid:4) (cid:17)(cid:20)(cid:19) The performance of this ’Fast Hankel Transform’ is illustrated by Fig. 4.11, which shows the evaluation of the Hankel ...
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) (cid:10) (cid:0) (cid:1) (cid:1) (cid:10) (cid:27) (cid:1) (cid:23) (cid:0) (cid:12) (cid:14) (cid:16) (cid:28) (cid:9) (cid:17) (cid:4) (cid:22) (cid:16) (cid:0) (cid:8) (cid:4) (cid:10) 2 (cid:3) (cid:0) (cid:1) (cid:8) (cid:13) (cid:25) (cid:17) (cid:13) (cid:19) (cid:0) 242 Chapter4. WavenumberIntegrationTechniq...
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, and hence is applicable only out to ranges where the product of the kernel and the exponential function is well represented by a linear function. The kernel can be smoothed by moving the contour out into the complex plane as described above, but the exponential function varies rapidly for long ranges. To ensure that ...
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rule integration. The following contour is found to work well for most ocean acoustic and seismic problems, (cid:17)(cid:20)(cid:19)(cid:31)(cid:3) (cid:17)(cid:20)(cid:19) (cid:17)(cid:25)(cid:19) (cid:6)%(cid:17)(cid:20)(cid:19) (cid:17)(cid:20)(cid:19) (cid:8)(cid:7) (cid:5)(cid:7) (cid:17) (cid:19) (cid:1)(cid:0)(c...
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)(cid:0) (cid:27)(cid:29)(cid:17)(cid:25)(cid:19) )( (cid:8)(cid:26)(cid:8) (4.127) with the weight function (cid:17)(cid:20)(cid:19) (cid:12)(cid:7) (cid:3)(cid:18)(cid:17)(cid:20)(cid:19)(cid:26)(cid:0) (cid:7)(cid:4)(cid:15) (cid:3)(cid:18)(cid:17)(cid:29)(cid:0) (cid:17)(cid:20)(cid:19) (cid:8)(cid:11)(cid:10) (cid...
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