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2 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 Line Production Rate Assembly System Production Rate Production rate = rate of flow of material into M1 1 3 = µ1 p(n1, n2) n1=0 ...
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µ 1 µ 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 Asse...
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1, n2) = n2 p(n1, n2) n1=0 n2=0 � � � n1=0 n2=0 � � � N = 2 1 N = 3 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 �201...
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parameters; that is, µi = µi , i = 1, ..., kM and ′ Nb = Nb , b = 1, ..., kB . ′ ′ ◮ There is a subset of buffers Ω such that for j 6∈ Ω, u (j) = 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. ...
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(0) and s ′(0) are related as follows: nj ′(0) = nj (0) for j 6∈ Ω, and nj (0) = Nj − nj (0) for j ∈ Ω. ′ ◮ Then P ′ (n ′ (0)) = P(n(0)) ′ (n ′ (0)) = n¯b(n(0)), for j 6∈ 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 Equiva...
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B 1 µ 2 N 2 B 1 B 2 N 2 N 1 µ 3 µ 2 µ 3 µ 2 µ 1 N 1 N 2 µ 1 µ 3 2.852 Manufacturing Systems Analysis 38/41 Copyright �2010 Stanley B. Gershwin. c Equivalence Equivalence classes of four-machine systems Representative members 2.852 Manufacturing Systems Analysis 39/41 Copyright �2010 Stanley B...
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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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Meshness of Random Networks 2/16/2011 Basic Network Metrics © Daniel E Whitney 1997-2010 4/45 CAIDA Paper on Internet Structure • Nice review and comparison of many metrics • Follows up early 2000s papers purporting to find the structure of the internet • Shows that there are three ways to do this, each approxi...
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1 2 2 2 3 1 2 3 2 2 3 2 3 1 3 4 2 3 4 2 3 2 2 5 2 2 5 x = 2.2 y = 2.2 2/16/2011 Basic Network Metrics © Daniel E Whitney 1997-2010 7/45 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 co...
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for Pearson (symmetric) r = ∑(x − x)(y − y) ∑(x − x)2 ∑ (y − y)2 Look at numerator, ignore xbar for the moment ∑(xiy j )= xiδij y j = x Ax ' ' δij = 1 if i links to j δij = 0 if i does not link to j Essentially the calculation is a quadratic form. Pearsondir does the calculation for asymmetric networks 9/45 2/...
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11 Basic Network Metrics © Daniel E Whitney 1997-2010 12/45 Degree Distribution for V8 Engine 2/16/2011 Basic Network Metrics © Daniel E Whitney 1997-2010 13/45 K Nearest Neighbors for V8 Missing marks indicate that there are no nodes with that degree 2/16/2011 Basic Network Metrics © Daniel E Whitney 1997-201...
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© Daniel E Whitney 1997-2010 20/45 Degree-preserving Pair-wise Rewiring • Picks two pairs of nodes at random and swaps their links so that each node retains its nodal degree • Usually used to randomize a network – Rewire at random, a lot • Can also be used to change a network’s degree correlation or clustering ...
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5 2/16/2011 Basic Network Metrics © Daniel E Whitney 1997-2010 24/45 Finding Communities • Big topic in social network analysis • Many algorithms exist, based on different principles, several in UCINET • Recent one based on network flow by Newman and Girvan: M. E. J. Newman and M. Girvan, Phys. Rev. E 69, 0261...
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(the same % format as specified in UCINET). QRecord2, dendrogramRecord, and MarkCut A1=load('TEST.txt'); outputFileName1='Q_resultTEST'; outputFileName2='dendrogramTEST'; outputFileName3='CutSequenceTEST'; m=max(max(A1(:,1:2))); % This code builds the adjacency matrix from the edgelist in TEST.txt % You can change the ...
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00000e+00 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 ...
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nodes 1 - 3 are in community 1 while 4 - 6 are isolates in communities 2 - 4 respectively. In column 3 we see that nodes 1 - 3 are in community 1 while nodes 4 - 6 are in community 2. 1 1 1 6 5 2 3 6 5 2 3 6 5 2 3 4 4 4 Q = -0.18367 Q = 0.04081 Q = 0.20408 2/16/2011 Basic Network Metrics © Daniel E Whit...
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Erdos-Gallai theorem tests if a degree sequence is “graphic” (routine isgraphic.m) • Generating the graph is fraught and often ends up incomplete or disconnected, or else it has some self-loops and multiple edges between nodes 2/16/2011 Basic Network Metrics © Daniel E Whitney 1997-2010 31/45 Random Graph Reali...
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7-2010 33/45 random_graph.m % Random graph construction routine with various models % Gergana Bounova, October 31, 2005 function [adj] = random_graph(N,p,E,distribution,fun,degrees) % INPUTS: % N - number of nodes % p - probability, 0<=p<=1 % E - fixed number of edges % distribution - probability distributio...
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with permission. 2/16/2011 Basic Network Metrics © Daniel E Whitney 1997-2010 35/45 Example Calls to random_graph random_graph(10) random_graph(10,0.1,20) random_graph(10,0,0,'normal') random_graph(10,0,0,'custom',@mypdf) degs = [3 1 1 1]; random_graph(10,0,0,'custom',@mypdf,degs) 2/16/2011 Basic Network Metrics © D...
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stop); Nseqabs=abs(Nseq); %protect against negative values Nseqint=int16(Nseqabs); %Volz routine requires integers dlmwrite('degdist.txt',Nseqint,'\t') %Volz routine requires tab delimited input !java -jar RandomClusteringNetwork.jar degdist.txt 100 .001 output.txt % n = 100, desired clust = % if you use 0.0 for desire...
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not %protect against disconnecting the network or isolating nodes. % % Inputs: % nv - number of nodes % p - rewiring probability % Kreg - initial node degree of for regular graph (use 1 or even numbers) % % Output: % G is a structure implemented as data structure in this as well as other % graph theory algori...
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0 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 sizes. • Here is a small example to demonstrate how to use the code. This code creates a seed ...
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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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and the velocity is randomized: net velocity� in direction � of field τc The average net 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...
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∝∝∝∝carrier concentration ∝∝∝∝carrier charge Drift current densities: drift = −qnvdn = qnµ n E Jn drift = qpvdp = qpµ p E J p Check signs: E vdn - drift Jn 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 ...
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1 Doping (cm-3) 6.012 Spring 2009 Lecture 3 9 Numerical Example: Si with Nd = 3 x 1016 cm-3 at room temperature µn ≈ 1000 cm 2 / V • s ρρρρn ≈ 0.21Ω • cm n ≈ 3X1016 cm −3 Apply E = 1 kV/cm vdn ≈ −106 cm / s << vth drift ≈ qnvdn = qnµnE = σσσσE = Jn E ρρρρ drift ≈ 4.8 ×103 A / cm 2 Jn Time to drift t...
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1] D measures the ease of carrier diffusion in response to a concentration gradient: D ↑ ⇒ Fdiff ↑ D limited by vibration of lattice atoms and ionized dopants. 6.012 Spring 2009 Lecture 3 12 Diffusion Current Diffusion current density =charge ××××carrier flux diff = qDn Jn dn dx diff = −qDp J p dp dx Che...
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dp dx J total = J n + J p 6.012 Spring 2009 Lecture 3 15 What did we learn today? Summary of Key Concepts • Electrons and holes in semiconductors are mobile and charged – ⇒⇒⇒⇒ Carriers of electrical current! • Drift current: produced by electric field drift ∝ E J drift J ∝ dφ dx • Diffusion current: p...
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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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icular displacement or stress component; is the associated wavenumber kernel. The order of the Bessel function is "#(cid:10)%$ , except for the horizontal displace- ment and shear stress, Eqs. (4.36) and (4.39), where "&(cid:10)%’ . The numerical evaluation of this integral is complicated by the following features, whi...
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WavenumberIntegration 225 of modal singularities only the relatively smooth continuous spectrum remains, being less susceptible to discretization problems. Further, with the field required at only a few receivers, the integration, and the associated sampling of the kernel, and therefore the solution of the depth-separat...
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evaluation of the acoustic field at a large number of receiver ranges. The FFT technique is also well suited to illustrate the discretization problem because of the direct analogy to periodic solution to cylindrical problems. Then the more direct numerical integrations schemes, based on either fixed or adaptively deter- ...
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wavefield at very short ranges and is therefore neglected. Next we replace (cid:8) by its asymptotic form [30], (cid:3)(cid:18)(cid:17)(cid:20)(cid:19) (cid:5)(cid:8)(cid:7)(cid:10)(cid:9) (cid:19)(cid:8)(cid:20)(cid:22)(cid:21) (cid:3)(cid:18)(cid:17)(cid:20)(cid:19) )(+* ,/. (cid:5)(cid:8)(cid:7)(cid:10)(cid:9) ’& (ci...
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than the Bessel function, particularly in terms of computation time. Since the numerical im- plementations used in underwater acoustics are almost without exception based on the fast-field approximation, we will focus on the evaluation of this integral in the follow- ing. It should be pointed out, however, that the trun...
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(cid:1) (cid:27) (cid:0) (cid:28) & (cid:30) (cid:3) (cid:24) * , (cid:8) - , (cid:16) (cid:28) (cid:19) 4.5. WavenumberIntegration 227 Instead, the wavenumber axis is truncated, allowing for numerical quadrature with- out the complication of a semi-infinite integration interval. The reason for this ap- proach being ap...
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, we can take advantage of the oscillatory nature of the exponential function in Eq. (4.96). Thus, for it , even for source and receiver will ensure the convergence of the integral for at the same depth where the kernel alone is non-integrable. Therefore, the contribu- tion to the integral beyond a certain wavenumber (...
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)(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 transforms that under- sampling in one domain causes aliasing (wrap-around) in the other domain (see e.g., Ref. [31], Sec. 3....
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0) (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:9)(cid:8) represents the entire field...
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:2) (cid:10) (cid:23) (cid:10) & (cid:7) (cid:12) (cid:15) (cid:16) 2 (cid:17) (cid:28) (cid:30) (cid:19) (cid:15) " $ 5 4.5. WavenumberIntegration 229 In this form it is clear that adding (cid:0) multiple of yields a periodic solution, to the range (cid:0) simply adds an integer to the argument of the exponential fun...
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boundaries, (cid:16)(cid:0)(cid:18)(cid:17)(cid:20)(cid:19) (cid:13)(cid:10) (cid:17)(cid:9)(cid:19) (cid:1)(cid:7)(cid:0) 5(cid:31)5 . This property is the true culprit of the aliasing problem. (cid:17)(cid:20)(cid:19) (cid:1)(cid:9)(cid:8) (cid:11)(cid:0) (cid:8)(cid:1)(cid:7)(cid:0) (cid:2)(cid:4)(cid:3)*(cid:17)(ci...
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these components are ignored here, the positive wavenumbers will contribute at small negative ranges, as indicated in the figure, yielding a non-vanishing field which will also wrap into the current window at (cid:0) also will have to appear as negatively propagating waves at in the neighbor window (cid:1) . (cid:13)(cid...
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:10) (cid:25) (cid:8) 5 (cid:6) $ (cid:6) 5 5 (cid:8) 5 (cid:10) (cid:0) (cid:0) (cid:11) (cid:1) (cid:27) (cid:1) (cid:23) (cid:1) (cid:27) (cid:0) (cid:10) (cid:0) (cid:11) (cid:6) (cid:1) (cid:10) (cid:25) (cid:8) 5 (cid:8) (cid:0) (cid:10) $ (cid:4) (cid:10) (cid:10) $ $ (cid:6) (cid:0) (cid:10) (cid:10) ’ (c...
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curve near the origin. The discrete wavenumber integration yields the periodic result shown in the lower frame by a dashed curve, approximating the correct con- tinuous result shown as a solid curve. The discrete result is a superposition of the ’true’ field produced by the mirror sources in all the range windows. (cid:...
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which is attenuated enough to reduce the periodic multiples to insignificance. from both sides of the actual interval and therefore also from ranges smaller than (cid:0) If Because of the two-sided nature of the discrete Fourier transform, aliasing occurs (cid:16)(cid:0)(cid:18)(cid:17)(cid:20)(cid:19) . . are wrapped i...
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) (cid:6) (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. WavenumberInteg...
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) (cid:25)(cid:17) (cid:4)(cid:19) (cid:1)(cid:15) (cid:5)(cid:14)(cid:13) (cid:9)(cid:18)(cid:9) "%$ "%$ "%$ (4.109) (cid:17)(cid: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 ha...
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��nite contribution from the small wavenumber components, and more inportantly the numerical artifact of the dis- continuity of the kernel at factor introduced by the FFP approximation forces the kernel to vanish, the derivatives remain discontinuous, and an artificial backward propagating field will result, which will b...
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4. WavenumberIntegrationTechniques peak in Fig. 4.6(a) clearly introduces errors of up to 2 dB in the predicted transmission loss, with the largest errors at longer ranges. However, even at short ranges the aliasing introduces errors in the modal interference pattern. 4.5.5 Complex Contour Integration (cid:14)(cid:23) ...
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particularly in cases where the field is needed only at relatively short ranges. The aliasing problem can, however, be eliminated by moving the integration con- tour out into the complex plane. According to Cauchy’s theorem the integral in the complex plane between two points is invariant to a change in the integration ...
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(cid:4) (cid:17) 0 (cid:17) (cid:25) (cid:17) (cid:8) (cid:17) (cid:10) (cid:17) (cid:25) (cid:8) (cid:17) 0 (cid:19) (cid:8) 4.5. WavenumberIntegration 235 B r a n c h c u t f o r k z = 2 2 mk -kr -k m * C 1 m P o l e s k * C 2 k r C 3 k m i n k m a x Fig. 4.7. Complex integration contours for evaluation of wavenumbe...
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:3) (cid:19) 0 (cid:19) (cid:8) (cid:6) (cid:10) (cid:23) (cid:28) (cid:5) (cid:3) ( (cid:9) (cid:9) (cid:19) (cid:8) (cid:19) (cid:8) (cid:10) & (cid:7) (cid:12) (cid:15) (cid:16) (cid:15) (cid:25) (cid:9) (cid:5) (cid:6) (cid:28) (cid:30) (cid:19) (cid:9) (cid:15) " $ 2 (cid:17) (cid:15) (cid:25) (cid:9) (cid:5) (cid...
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) . On the other hand, signals wrapped around (cid:8) . As was the case also for the (cid:0)(cid:3)(cid:2)(cid:5)(cid:4) The explanation for this is as follows. The contour offset moves the integration path away from singularities such as branch points and modes resulting in a similar but smoother integration kernel. I...
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vertical contours. Figure 4.8 illustrates the effects of using the complex integration contour for eval- uation of the field in the Pekeris waveguide treated above. Figure 4.8(a) shows the magnitude of the kernel of the summation in Eq. (4.114) for two contour offsets de- fined by Eq. (4.115). The solid curve is the kern...
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) (cid:0) (cid:28) (cid:3) (cid:19) (cid:5) (cid:10) (cid:11) (cid:0) (cid:19) (cid:28) (cid:10) (cid:11) (cid:1) (cid:27) (cid:3) (cid:24) (cid:25) ’ (cid:8) (cid:19) (cid:28) (cid:25) (cid:17) (cid:8) (cid:6) (cid:6) (cid:17) (cid:25) (cid:17) (cid:10) ’ $ (cid:1) (cid:10) ’ (cid:1) (cid:10) ’ $ (cid:1) (cid:10) ’ (c...
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Chapter4. WavenumberIntegrationTechniques (cid:17)(cid:20)(cid:19) out to a range of approximately 10 km, after which the result shows increasing errors. These errors appear in spite of the fact that the maximum range for this sampling is (cid:14) km. This is due to the fact that even though we have removed (cid:8) , t...
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Field Program (FFP) approach described above has gained popularity because of its efficiency in producing field estimates at a large number of ranges. However, being based on the large argument asymptotic of the Hankel functions it is associated with errors for small arguments (cid:17)(cid:9)(cid:19) (cid:0) of the Besse...
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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 Fast Hankel Transform [25] is very efficient for relatively smooth kernels, but not well-suited for the rapidly vary- ing kernels of waveg...
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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 where to incoming waves, bo...
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:9) $ (cid:8) 2 " $ (cid:13) (cid:19) (cid:28) & (cid:30) (cid:5) (cid:24) , , (cid:9) $ 240 Chapter4. WavenumberIntegrationTechniques 0 10 −1 10 −2 10 −3 10 | ) r k ( J | m −4 10 −5 10 −6 10 −7 10 0 10 20 30 40 50 kr 60 70 80 90 100 Fig. 4.9. Error of far-field approximation of (cid:0) approximation, Dashed: Error. (c...
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3) it is only necessary to cor- rect the contributions corresponding tovalues of . This is performed in a numerically stable manner by a Hanning weighted average of the contributions of the exact Bessel function and the approximate FFP kernel, (cid:9)(cid:7) KR (cid:10) for (cid:3)(cid:18)(cid:17)(cid:20)(cid:19) (cid:...
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.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 FFT evaluation of Eq. (4.118). Secondly, the Bessel functions will only be needed for a limited number of discrete values of the argument,...
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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 transform (cid:3)(cid:30)(cid:0)(cid:31)(cid:6) (cid:17)(cid:20)(cid:19) (cid:3)(cid...
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5 (cid:1) (cid:3) (cid:23) (cid:23) (cid:0) (cid:8) (cid:6) (cid:4) (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) (ci...
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.121) the array, often requiring a wavenumber sampling which is fixed for all frequencies, which is not optimal. As a consequence broad-band and discrete array computations are in general opti- mally being performed by direct numerical quadrature schemes such as the trapezoidal rule integration. This scheme approximates...
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, for example they do not have to use an integration contour parallel to the real axis such as the contour C in Fig. 4.7, but can instead apply an ’exact’ Cauchy contour such as the one shown by the dashed line in Fig. 4.7, totally eliminating the contributions from the two vertical contour sections. This is particular...
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rule integration, for the omni- directional field components, " $ : (cid:2)(cid:4)(cid:3)(cid:30)(cid:0) (cid:2)(cid:4)(cid:3)*(cid:17)(cid:25)(cid:19) (cid:3)*(cid:17)(cid:20)(cid:19)(cid:26)(cid:0) (cid:17)(cid:20)(cid:19) (cid:3)(cid:18)(cid:17)(cid:20)(cid:19)(cid:26)(cid:0) (cid:10)(cid:13)(cid:12) (cid:17)(cid:25)...
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:27) (cid:8) (cid:1) (cid:3) ’ $ (cid:27) (cid:8) (cid:8) . (cid:1) (cid:1) ’ $ (cid:27) (cid:15) (cid:0) (cid:15) (cid:1) $ (cid:27) $ (cid:0) (cid:0) (cid:1) $ (cid:27) (cid:0) (cid:10) (cid:1) $ (cid:27) (cid:27) (cid:1) (cid:1) $ 4.5. WavenumberIntegration 245 4.5.8 Filon Integration While the trapezoidal rule int...
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23) (cid:8) . scheme is exact for linear variations of the kernel (cid:21) In the present case (cid:24) (cid:17)(cid:20)(cid:19) , i.e., the exponent is inherently a linear function of (cid:17)(cid:25)(cid:19) . For the equidistant sampling given in Eq. (4.97), it is easily shown that the Filon integration scheme leads...
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)(cid:2) (cid:17)(cid:20)(cid:19) (cid:4)(cid:2) (4.131) (cid:17)(cid:9)(cid:19) Here it is interesting to note that Eq. (4.130) is identical to Eq. (4.108) except for the simple change in integration weight from (cid:23) , basically applying a sinc-function squared to the field amplitude vs range. The summation can aga...
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13) 5 (cid:2) 246 Chapter4. WavenumberIntegrationTechniques (a) (b) k1 k 2 k1 k 2 Fig. 4.12. Adaptive evaluation of wavenumber integral. , the Filon scheme requires a sampling which is approxi- inversely proportional to mately inversely proportional to . However, they considered seismic reflectivity problems characteri...
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These integration problems may be overcome by an adaptive selection of the wave- number sampling. Shown in Fig. 4.12 is an example of such an adaptive scheme, de- veloped by Krenk et al. [34]. Here, the kernel is first sampled on a coarse wavenumber grid, which is then subsequently subdivided by bisection, until a stabl...
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the same issues of windowing and sampling must be properly addressed. Since the evaluation of the frequency integral is common to all numerical approaches solving the Helmholtz equation, the associated numerical issues will be deferred to Chap. 8. However, we present time-domain calculations also here (Sec. 4.8.4), due...
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Lecture 1 Overview of some probability distributions. In this lecture we will review several common distributions that will be used often throughtout the class. Each distribution is usually described by its probability function (p.f.) in the case of discrete distributions or probability density function (p.d.f.) i...
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in this case for any a expectation of �(X) is defined by ∞ R we have P(X = a) = 0. Given a function � : R, the X � E�(X) = � �(x)p(x)dx. � −� Notation. The fact that a random variable X has distribution P will be denoted by P. Normal (Gaussian) Distribution N(�, π2). Normal distribution is a continuous dis­ X...
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)/π. This, of course, − EY = � � −� y 1 ≤2ϕ 2 y 2 dy = 0 e− since the integrand is an odd function. To compute the second moment EY 2 , let us first note 1 is a probability density function, it integrates to 1, i.e. that since � 2 y 2 2� e− If we integrate this by parts, we get, � −� � 1 = 1 ≤2ϕ 2 y d...
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π2 is a variance of a normal distribution. Let us recall (without giving a proof) that if we have several, say n, independent random variables N (�i, π2) then their sum will also have a normal distribution Xi, 1 i n, such that Xi � ≈ ≈ X1 + . . . + Xn � N(�1 + . . . + �n, π1 2 + . . . + π2 ).n Normal distribut...
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. Bernoulli Distribution B(p). This distribution describes a random variable that can . The distribution is described by a probability 0, 1 } { take only two possible values, i.e. function = X p(1) = P(X = 1) = p, p(0) = P(X = 0) = 1 − p for some p [0, 1]. ∞ It is easy to check that EX = p, Var(X) = p(1 −...
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0 is � E(�) then P(X ∼ t) = P(X [t, → ∞ )) = t � � �e− �xdx = e− �t . Given another s > 0, the conditional probability that X will exceed level t + s given that it will exceed level t can be computed as follows: P(X t + s X | ∼ ∼ t) = P(X t + s, X ∼ P(X t) ∼ �t = e− �(t+s)/e− ∼ = e− t) = P(X ...
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. Poisson distribution could be used to describe the following random objects: the number of stars in a random area of the space; number of misprints in a typed page; number of wrong connections to your phone number; distribution of bacteria on some surface or weed in the field. All these examples share some common ...
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this interval into a large number n of small equal subintervals of length T /n and denote by Xi the number of random objects in the ith subinterval, i = 1, . . . , n. By the first property above, EXi = �T . n On the other hand, by definition of expectation EXi = kP(Xi = k) = 0 + P(Xi = 1) + αn, 0 k � � k 2 kP(...
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λ]. This distribution has probability density function 1 λ , x [0, λ], ∞ 0, otherwise. p(x) = � Matlab review of probability distributions. Matlab Help/Statistics Toolbox/Probability Distributions. Each distribution in Matlab has a name, for example, normal distribution has a name ’norm’. Adding a suffix defines ...
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MIT 6.972 Algebraic techniques and semidefinite optimization February 28, 2006 Lecturer: Pablo A. Parrilo Scribe: ??? Lecture 6 Last week we learned about explicit conditions to determine the number of real roots of a univariate polynomial. Today we will expand on these themes, and study two mathematical objects o...
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• Sylvester matrix: If p(x0) = q(x0) = 0, then we can write the following (n + m) × (n + m) linear system: ⎡ pn pn−1 p n . . . qm qm−1 qm ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ . . . p1 . . . p0 . . . . . . . . . q0 . . . . . . . . . ⎤ ⎡ n+m−1 x ⎥ 0 ⎥ n+m−2 x ⎥ 0 ⎢ ⎥ ⎢ . . . ⎥ ⎢ ⎥ ⎢ p1 p0 ⎥ ⎢ ...
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⎥ ⎥ ⎥ ⎥ ⎥ ⎦ = 0. This implies that the matrix on the left­hand side, called the Sylvester matrix Sylx(p, q) associated to p and q, is singular and thus its determinant must vanish. It is not too difficult to show that the converse is also true; if det Sylx(p, q) = 0, then there exists a vector in the kernel of Sylx(p,...
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be shown that all these constructions are equivalent. They define exactly the same polynomial, called the resultant of p and q, denoted as Resx(p, q): Resx(p, q) = det Sylx(p, q) n = p m det q(Cp) n = (−1)nm q det p(Cq ) m m= pn q n det(Cp ⊗ Im − I m n ⊗ Cq ). The resultant is a homogeneous multivariate polynomia...
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. Definition 2. The discriminant of a univariate polynomial p(x) is defined as Disx(p) := (−1)n(n−1)/2 Resx p(x), 1 pn � � . dp(x) dx Similar to what we did in the resultant case, the discriminant can also be obtained by writing a natural condition in terms of the roots αi of p(x): Disx(p) = p 2n−2 n � (αj ...
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two univariate polynomials p(x, y0), q(x, y0). If y0 corresponds to the y­component of a root, then these two univariate polynomials clearly have a common root, hence their resultant vanishes. Therefore, to solve (2), we can compute Resx(p, q), which is a univariate polynomial in y. Solving this univariate polynomia...
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One of the main reasons why nonnegativity conditions about polynomials are difficult is because these sets can have a quite complicated structure, even though they are always convex. Recall from last lecture that we have defined Pn ⊂ Rn+1 as the set of nonnegative polynomials of degree n. It is easy to see that if p(x)...
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+ b. For what values of a, b does it hold that p(x) ≥ 0 ∀x ∈ R? Since the leading term x4 has even degree and is strictly positive, p(x) is strictly positive if and only if it has no real roots. The discriminant of p(x) is equal to 256 b (a − b)2 . 2 Here is a slightly different example, showing the same phenomenon. ...
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511.52a-1-0.50.511.52b-3-2-1123a-1123b0220444-3-2-1123a-1123b Figure 3: A three­dimensional convex set, described by one quadratic and one linear inequality, whose projection on the (a, b) plane is equal to the set in Figure 1. One has to do with its algebraic structure, and the other one with convexity, and in parti...
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projection on the plane (a, b) is exactly the one discussed in Example 5 and Figure 1. The presence of “extraneous” components of the discriminant inside the feasible set is an important roadblock for the availability of “easily computable” barrier functions. Indeed, every polynomial that vanishes on the boundary o...
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how useful this approach is in practical optimization problems. 1A face of a convex set S is a convex subset F ⊆ S, with the property that x, y ∈ S, 1 F is exposed if it can be written as F = S ∩ H, where H is a supporting hyperplane of S. 2 (x + y) ∈ F ⇒ x, y ∈ F . A face 6­5 −20246024012345abt Figure 4: The dis...
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Self­concordant barriers for cones generated by Chebyshev systems. SIAM J. Optim., 12(3):770–781 (electronic), 2002. [KM02] C. Y. Kao and A. Megretski. A new barrier function for IQC optimization problems. In Proceedings of the American Control Conference, 2002. [RG95] M. Ramana and A. J. Goldman. Some geometric re...
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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. Steenrod Operations (Lecture 2) The objective of today’s lecture is to introduce the Steenrod operations a...
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�n, by permuting the tensor factors. One of the most important examples of an F2-module spectrum is the cochain complex C ∗(X; F2) of a topological space X. The cohomology groups of this F2-module spectrum are simply the cohomology groups of X. The cohomology H∗(X; F2) has the structure of a graded commutative rin...
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nth extended power of V is given by the homotopy coinvariants ⊗ Vh Σ n n . This is a complex which we will denote by Dn(V ). Remark 3. In concrete terms, Dn(V ) may be computed in the following way. Let M denote the vector space F2, with the trivial action of Σn. Choose a resolution . . . → P −1 → → P 0 M b...
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over the field F2. Then A has an underlying F2-module spectrum, which is equipped with a symmetric multiplication. ∞ Our goal in this lecture is to study the consequences of the existence of a symmetric multiplication on a complex V . Notation 8. Let n be an integer. We let F2[−n] denote the complex which consists o...
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(unique) isomorphism H∗(RP ∞; F2) � F2[t], where the polynomial generator t lies in H1(RP ∞; F2). We have a dual description of the homology H (RP ∞; F2): this is just a one-dimensional vector space in each degree m, with a unique generator which we will denote by xm. ∗ Definition 9. Let V be a complex, and let v ∈...
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respect to the multiplication on V . This is why the operations Sqi are called “Steenrod squares”. Example 11. Let X be a topological space, and let V = C ∗(X; F2) be the cochain complex of X, equipped with its usual symmetric multiplication. Then Definition 9 yields operations Sqi : Hn(X; F2) → Hn+i(X; F2). These ...
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(v + v�) = Sq k (v) + Sq (v�) ∈ Hn+k D2(V ). In particular, if V is equipped with a symmetric multiplication, we have Sqk(v + v�) = Sqk(v) + Sqk(v�) ∈ Hn+k V. Proof. If k > n, then both sides are zero and there is nothing to prove. If k = n, then k Sq k (v + v�) = (v + v�)2 = Sq k (v) + Sq (v�) + (vv� + v�v). Sinc...
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