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ZT (Lead zirconate titanate) • Higher strain ceramic material for larger deflections • Common choice for MEMS devices • Fabrication: deposited in layers by a sol-gel (spin on) process, thermally cured, then poled in an electric field > Piezoelectric polymers • Example: poly (vinylidene fluoride) • Very high strains Ci...
https://ocw.mit.edu/courses/6-777j-design-and-fabrication-of-microelectromechanical-devices-spring-2007/511de50fb6460b31fd47ee2f9e4958a0_07lecture05.pdf
Livermore, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. CL: 6.777J/2.372J Spring 2007, Lecture 5 - 49 Outline > Soft Lithography • Materials an...
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test device accurately enough so that the accuracy of the extracted properties are limited by geometric errors • A good procedure is to make a family of test structures with a systematic variation in geometry, and extract constitutive properties from all of the data. > Preventing systematic errors in the measurement...
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G Stable Region Pull-In Instability Unstable Region Applied Voltage Courtesy of Joel Voldman. Used with permission. VPI Image removed due to copyright restrictions. Figure 10 on page 116 in: Osterberg, P. M., and S. D. Senturia. "M-TEST: A Test Chip for MEMS Material Property Measurement Using Electrostatically Actua...
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:19)(cid:25)!(cid:17)(cid:9)(cid:18)(cid:10)!(cid:26)(cid:29)#$(cid:26)(cid:29)(cid:13)(cid:16)$(cid:25)(cid:5)(cid:9)!(cid:20) (cid:30).(cid:29)(cid:29).(cid:29)(cid:30)(cid:10) +++/(cid:18)(cid:14)(cid:17)(cid:20)(cid:8)(cid:4)(cid:5)(cid:17)(cid:3)(cid:4)(cid:5)(cid:7)(cid:8)(cid:18)(cid:9)(cid:23)(cid:5)/(cid:12)(c...
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cid:3)(cid:4) –  Christiana Figueres, former UNFCCC negotiator for Costa Rica, now Executive Secretary of the UNFCCC, 2009(cid:4) (cid:4) (cid:2)...delegates [in Bonn] complained that their heads were spinning as they were trying to understand the science and assumptions underlying the increasing number of proposals...
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5)(cid:7)(cid:8)(cid:18)(cid:9)(cid:23)(cid:5)/(cid:12)(cid:7)(cid:6)(cid:10) 11 C-ROADS Scientific Review Panel •  Dr. Robert Watson - Department for Environment, Food and Rural Affairs (DEFRA) and former chair, IPCC -- Panel Chair •  Dr. Eric Beinhocker - McKinsey Global Institute •  Dr. Klaus Hasselmann - Max-P...
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Bayesian Networks Representation and Reasoning Marco F. Ramoni Children’s Hospital Informatics Program Harvard Medical School HST 951 (2003) Harvard-MIT Division of Health Sciences and Technology HST.951J: Medical Decision Support Introduction � Bayesian network are a knowledge representation formalism for reasoning...
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The link — should be « . A B C Characters: D E Adjacent set: the nodes one step away from A: Adj(A)={B|(A,B)˛ L}. Path: The set of n nodes Xi from A to B via links: Loop: A closed path: X1 = Xn. Acyclic graph: A graph with no cycles. HST 951 Directed Graphs Parent: A is parent of B if there is a directed l...
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L 0.2 H 0.2 H A E I I A E I A E L L Y L L Y H L Y H L Y L H Y H Y L H H Y H H Y L L O L L O H L O H O L L O H O H L O H H O H H p(I|A,E) p(I|A,E) 0.9 0.9 0.1 0.1 0.5 0.5 0.5 0.5 0.7 0.7 0.3 0.3 0.2 0.2 0.8 0.8 A=A=AgeAge; E=; E=Education HST 951 Income Educa...
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but, at the same time, breaks down the whole system in separate regions: conditional independence. � This is independence used by Bayesian networks. HST 951 Conditional Independence � When two variables are independent given a third, they are said to be conditionally independent. p(A|B � C)=p(A � B � C)/p(B � C)...
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C B HST 951 Example Background knowledge: General rules of behavior. p(Age=<5)=0.3 p(T-shirt=small| Age=<5)=0.5 p(T-shirt=small|Age=>5)=0.3 p(Literacy=yes|Age=>5)=0.6 p(Literacy=yes|Age=<5)=0.2. Evidence: Observation p(T-shirt=small). Solution: The posterior probability distribution of the unobserved nodes gi...
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1 1 1 1 0.40 0.60 0.45 0.55 0.60 0.40 0.30 0.70 0 1 0 1 0 1 0 1 0 0 1 1 0 0 1 1 D E G p(G|D,E) 0 0 0 0 1 1 1 1 0.90 0.10 0.70 0.30 0.25 0.75 0.15 0.85 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 HST 951 Brute Force � Compute the Joint Probability Distribution: p(a,b,c...
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ytrees � In a polytree, each node breaks the graph into two independent sub-graphs and evidence can be independently propagated in the two graphs: � E+: evidence coming from the parents (E+ = {c}). � E-: evidence coming from the children (E- = {g}). A B D E C D F G HST 951 Message Passing � Message passing...
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p -messages from all other parents are in, send l -message to parent. Repeat until no message is sent. � Closure: � For each X/e, compute b (x)= p (x) l (x). � For each X/e, compute p(x)= b (x)/S l (xi). HST 951 Properties Distributed: Each node does not need to know about the others when it is passing the inf...
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E=e C D E F G F G HST 951 Algorithm Input: a (multiply connected) BBN and evidence e. Output: the posterior probability p(x|e) for each X. Procedure: Identify a loop cutset C=(C1, …, Cn). 1. 2. For each member of combinations c=(c1, …, cn). � Generate a polytree BBNs for each c. � Use Pearl’s Algorithm to...
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2 0.8 0 1 0 1 A C p(C|A) 0 0 1 1 0.2 0.8 0.50 0.50 0 1 0 1 C F p(F|C) 0 0 1 1 0.1 0.9 0.4 0.6 0 1 0 1 HST 951 Example A B C D E F B C E p(E|B,C) 0 0 0 0 1 1 1 1 0.4 0.6 0.5 0.5 0.7 0.3 0.2 0.8 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 Example � Loop cutset: ...
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.219 0.781 E 0.427 0.573 0 1 F 0.277 0.723 0 1 Clustering Methods The basic strategy (Lauritzen & Spiegelhalter 1988) is: 1. Convert a BBN in a undirected graph coding the same conditional independence assumptions. 2. Ensure the resulting graph is decomposable. 3. This operation clusters nodes in loca...
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(b|a)p(a). A B C A B C Moralize 1.Marry parents 2.Drop arrows D E D E HST 951 Reading Independence � The translation method via moralization reads the conditional independence statements in BBN. � DAGs cannot encode any arbitrary set of conditional independence assumptions. I(D,A|(B,C)) I(C,B|(A,D))...
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Acceleration Structures for Ray Casting MIT EECS 6.837 Computer Graphics Wojciech Matusik, MIT EECS © ACM. All rights reserved. This content is excluded from our Creative Commons license. For more information, see http://ocw.mit.edu/help/faq-fair-use/. Hašan et al. 2007 1 Recap: Ray Tracing trace ray Intersec...
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to generate nice pictures – Intersecting every ray with every primitive becomes the bottleneck • Bounding volumes • Bounding Volume Hierarchies, Kd-trees For every pixel Construct a ray from the eye For every object in the scene Find intersection with the ray Keep if closest Shade 10 Accelerating Ra...
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Z, of course) 18 Find Intersections Per Dimension • Basic idea – Determine an interval along the ray for each dimension – The intersect these 1D intervals (remember CSG!) – Done! y=Y2 y=Y1 Ro x=X1 x=X2 19 Find Intersections Per Dimension • Basic idea – Determine an interval along the ray for each dime...
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2 - Rox) / Rdx – [t1, t2] is the X interval y=Y2 y=Y1 t2 t1 Rd Ro x=X1 x=X2 28 Then Intersect Intervals • Init tstart & tend with X interval • Update tstart & tend for each subsequent dimension y=Y2 y=Y1 tstart tend x=X1 x=X2 29 Then Intersect Intervals • Compute t1 and t2 for Y... t2 y=Y2 t1...
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intersection at tstart • Else → closest intersection at tend – Eye is inside box y=Y2 y=Y1 tend tstart x=X1 x=X2 36 Ray-Box Intersection Summary • For each dimension, – If Rdx = 0 (ray is parallel) AND Rox < X1 or Rox > X2 → no intersection • For each dimension, calculate intersection distances t1 and t...
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the bounding box! (xmax, ymax, zmax) (x'max, y'max, z'max) = (max(x0,x1,x2,x3,x4,x5,x6,x7), max(y0,y1,y2,y3,y4,x5,x6,x7), max(z0,z1,z2,z3,z4,x5,x6,x7)) M (x3,y3,z3) = M (xmax,ymax,zmin) (x2,y2,z2) = M (xmin,ymax,zmin) (x1,y1,z1) = M (xmax,ymin,zmin) (x0,y0,z0) = M (xmin,ymin,zmin) (x'min, y'min, z'min)...
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Split objects/primitives into two, compute child BVs • Recurse, build a binary tree 49 Bounding Volume Hierarchy (BVH) • Find bounding box of objects/primitives • Split objects/primitives into two, compute child BVs • Recurse, build a binary tree 50 Bounding Volume Hierarchy (BVH) • Find bounding box of obje...
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plane • A space partition: The nodes do not overlap! – This is in contrast to BVHs 63 Data Structure KdTreeNode: KdTreeNode* backNode, frontNode //children int dimSplit // either x, y or z float splitDistance // from origin along split axis boolean isLeaf List of triangles //only for leaves backNode front...
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• If t<tstart => intersect only back Note: “Back” and “Front” depend on ray direction! 73 Kd-tree Traversal Pseudocode travers(orig, dir, t_start, t_end): #adapted from Ingo Wald’s thesis #assumes that dir[self.dimSplit] >0 if self.isLeaf: return intersect(self.listOfTriangles, orig, dir, t_start, t_end) t ...
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averse(orig, dir, t, t_end) 75 Early termination is powerful! travers(orig, dir, t_start, t_end): #adapted from Ingo Wald’s thesis #assumes that dir[self.dimSplit] >0 if self.isLeaf: return intersect(self.listOfTriangles, orig, dir, t_start, t_end) t = (self.splitDist - orig[self.dimSplit]) / dir[self.dimSpl...
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) else: # case three: traverse both sides in turn t_hit = self.frontSideNode.traverse(orig, dir, t_start, t) stop at first intersection if t_hit <= t: return t_hit; # early ray termination return self.backSideNode.traverse(orig, dir, t, t_end) 79 Important Details • For leaves, do NOT report intersection i...
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traversal of that child – number of primitives (simplistic heuristic) • This heuristic likes to put big densities of primitives in small-area nodes 86 Is it Important to Optimize Splits? • Given the same traversal code, the quality of Kd-tree construction can have a big impact on performance, e.g. a factor of ...
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r 1:  P 1 8.3 1 4 re   u Lect inciples  of  Applied  Mathematics   Rodolfo  Rosales Spring  2014 an  1  now  th e r Mo Con  is Flu  q x ect v equal  sources  &  sink -­‐D.  a   w a  l ion t serva or.  Us s u  Ga e s s).  2-­‐D  or  3-­‐ s  in e  th orem  t...
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a t i tart  w s m: ma f  o b  blo ll  s  ink,  R  = f b  o  blo e  th f  o s  the  radius at  i  α (t) s:  R y  sa s si sional  analy n a  co ink,  a  R(  √(nu*t) nd  a t),  as  th :   sk e  blo ffee  cup  without  stirring?  Idealized   b  ex...
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essel.   1                                                                     r 1:  P 1 8.3 1 4 re   u Lect inciples  of  Applied  Mathematics   Rodolfo  Rosales Spring  2014 T  h in red   su ...
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6.826—Principles of Computer Systems 2002 6.826—Principles of Computer Systems 2002 18. Consensus Consensus (sometimes called ‘reliable broadcast’ or ‘atomic broadcast’) is a fundamental building block for distributed systems. Informally, we say that several processes achieve consensus if they all agree on some ...
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for transactions; there the replication takes the form of redoing a sequence of actions that is remembered in a log. Suppose, for example, that we want to build a highly available file system. The transitions are read and write operations on the files (and rename, list, … as well). We make several copies of the fil...
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prophecy variables. The idea is that the outcome of consensus should be one and only one of the allowed values. In the spec there is an outcome variable initialized to nil, and an action Allow(v) that can be invoked any number of times. There is also an action Outcome to read the outcome variable; it must return ei...
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-> (V + Null) = << RET outcome [] RET nil >> END Consensus Note that Outcome is allowed to return nil even after the choice has been made. This reflects the fact that in code with several replicas, Outcome is often coded by talking to just one of the replicas, and that replica may not yet have learned about the cho...
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We would also like to have the property that eventually Outcome stops returning nil. In the code, this happens when every process’ outcome variable is non-nil. However, this could take a long time if some process is very slow (or down for a long time). We can change Consensus to express this with an internal action ...
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transition, and a link can take an arbitrary amount of time to deliver a message. In general a process can send a message only to certain other processes; this “can send message” relation defines a graph whose edges are the links. The graph may be directed (it’s possible that A can talk to B but B can’t talk to A), ...
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non-faulty processes or to none of them. Real systems get around it by using timeout to make the system synchronous. • Even in a synchronous system with perfect processes there is no consensus algorithm that is guaranteed to terminate if an unbounded number of messages can be lost (that is, if communication is effe...
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thing when the number of faults exceeds f. It’s often more important to do either the right thing or nothing. The simplest consensus algorithms There are two simple and popular algorithms for consensus. Both have the problem that they are not very fault-tolerant. • A fixed ‘leader’, ‘master’, or ‘coordinator’ proc...
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1989, finally published in ACM Transactions on Computer Systems 16, 2 (May 1998), pp 133-169. Unfortunately, the terminology of this paper is confusing. B. Liskov and B. Oki, Viewstamped replication: A new primary copy method to support highly available distributed systems, Proc. 7th ACM Conference on Principles of ...
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, trying to get a majority to accept v, and if successful, distributes v as the outcome to everyone. The outcome is the value accepted by a majority in some round. The tricky part of the algorithm is to ensure that there is only one such value, even though there may be lots of rounds. Most descriptions of Paxos cal...
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Z % implements Consensus % data Value to agree on % Leader; <= a total order % Agent; majorities must intersect % round Number; <= is total % one agent’s state in one round % Agents’ states VAR s : S := {*->{*->neutral} outcome : A -> (V+Null) := {*->nil} allowed : L -> SET V := {*->{}} % agent working St...
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the N = [i, l] pairs. This means that we must assume a total ordering on the leaders L. With these preliminaries, we can give the abstraction function from Paxos to LateConsensus. For allowed it is just the union of the leaders’ allowed sets. For outcome it is the value of a successful round. ABSTRACTION FUNCTION ...
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.rng)) Initially all the vn are nil so that (I3) and (I4) hold trivially. The Paxos algorithm maintains (I3) by choosing the value of a round so that the round is safe. To accomplish this, the leader chooses a new n and queries all the agents to learn their state in all rounds with numbers less than n. Before an ag...
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no no no y leader’s choices for round 4 x, y, or w y w no no no no no w w y w w w Note that only the latest V state from each agent is of interest, so only that state actually has to be transmitted. Now in a second round trip the leader commands everyone for round n. Each agent that is still neut...
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os terminate? If no leader starts another round until after an existing one is successful, then the algorithm definitely terminates as soon as the leader succeeds in both querying and commanding a majority. It doesn’t have to be the same majority for both, and the agents don’t all have to be up at the same time. The...
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practice this is usually coded by individual messages to each agent. We describe continuous retransmission; in practice agents retransmit only in response to the leader’s retransmission. A process acting as a leader uses messages to communicate with the same process acting as an agent, so we describe the two roles o...
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RET outcome(a) >> = << VAR l | allowed(l) := allowed(l) \/ {v} >> THREAD LeaderActions(l) = VAR n phase reports v: (V+Null) := nil | := N{i := 1, l := l}, := idle, := S{}, % leader state (volatile except n) % last round started % leader’s phase % info about agents so far % used iff phase = commanding D...
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12 6.826—Principles of Computer Systems 2002 6.826—Principles of Computer Systems 2002 [] % This round is dead if a majority has no state. Try another round. Dead(reports, n) => phase := idle >> OD THREAD AgentActions(a) = % State is in sa and outcomea , which are stable. DO << % Pick an enabled action and do...
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formulas more readable. FUNC IsTotal(le: (L, L) -> Bool) -> Bool = % Is le a total order? RET ( ALL l1, l2, l3 | (le(l1, l2) \/ le(l2, l1)) /\ (le(l1, l2) /\ le(l2, l3) ==> le(l1, l3)) ) \/ {Value(n)} \/ (phasen.l = commanding => {vn.l} [*] {}) FUNC SafeVals(n) -> SET V = % The v’s that would make n a safe rou...
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IN UnreliableCh.q /\ m.x IS S | m.x} \/ {l | reportsl} | (ALL a, n | s1!a /\ s1(a)!n /\ s1n a # neutral ==> s1n a = sn a)) % (5) Every round value is allowed ( ALL n | Value(n) IN (Allowed()\/ {nil}) ) % (6) If anyone thinks v is the value of a round, it is a good round value. ( ALL n | ValAnywhere(n) <= v IN Good...
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1(a)!n | s1n a } Handout 18. Consensus 13 Handout 18. Consensus 14 6.826—Principles of Computer Systems 2002 6.826—Principles of Computer Systems 2002 Reducing state and message sizes Combining rounds It’s not necessary to store or transmit the complete agent state sa. Instead, everything can be encoded i...
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of A’s, lastmax, v). If phase = commanding, reports consists of a set of vn or no in round n, so it can be encoded as the set of agents responding. We only care about a majority, so we only need to count the number of agents responding. The leaders need not be the same processes as the agents. A leader doesn’t rea...
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private state. A new leader needs to learn the first unused K. If it tries to get consensus using a number that’s too small, it will discover that there’s already an outcome for that action. If it uses a number k that’s too big, however, it can get consensus. This is tricky, since it leads to a gap in the action nu...
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set of agents you have to get a majority of them in order to make any further changes. Handout 18. Consensus 15 Handout 18. Consensus 16 6.826—Principles of Computer Systems 2002 6.826—Principles of Computer Systems 2002 Leases In a synchronous system, if you want to avoid running a full-blown consensus alg...
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might get an old result from some replica that isn’t up to date. (Of course, you could settle for a result as of action k, rather than a current one. Then you would need neither a lease nor a state machine action, but the client has to interpret the k that it gets back along with the result it wanted. Usually this i...
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changes to the disks. Compare-and-swap agents An alternative to using simple memory and leases is to use memory that implements a compare- and-swap or conditional store operation. The spec for compare-and-swap is APROC CAS(a, old: V, new: V) -> V = << IF m(a) = old => m(a) := new; RET old [*] RET m(a) FI >> Many m...
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action of the state machine. The agents must reach consensus on the complete sequence of actions that makes up the transaction. In practice, this means that each agent logs all the updates, and then they reach consensus on committing the transaction. When an agent recovers from a failure, it runs redo recovery in th...
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Massachusetts Institute of Technology Department of Materials Science and Engineering 77 Massachusetts Avenue, Cambridge MA 02139-4307 3.205 Thermodynamics and Kinetics of Materials—Fall 2006 October 26, 2006 Lecture 1: Fields and gradients; fluxes; continuity equation; entropy production; driving forces and fluxes 1. ...
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jugate “force” (see KOM Eq. 2.15). • Familiar empirical laws are linear relationships between fluxes and their conjugate forces: Fourier’s law of heat conduction, Fick’s law for diffusion, and Ohm’s law for electrical conduction (see KOM Table 2.1). • The basic postulate of irreversible thermodynamics is that, near e...
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18.700 JORDAN NORMAL FORM NOTES These are some supplementary notes on how to find the Jordan normal form of a small matrix. First we recall some of the facts from lecture, next we give the general algorithm for finding the Jordan normal form of a linear operator, and then we will see how this works for small matrices...
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, it cannot happen that each dim(E(X−λi)k ) < dim(E(X−λi)k+1 ), for each k = 1, . . . , n. Therefore there is some least integer ei ≤ n such that E(X−λi )ei = E(X−λi)ei+1 . As was proved in class, for each k ≥ ei we have E(X −λi)k = E(X−λi)ei , and we defined the generalized eigenspace Egen to be E(X−λi)ei . λi It w...
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λiIV )e(v) = 0}, λi Date: Fall 2001. 1 (4) 2 18.700 JORDAN NORMAL FORM NOTES give a direct sum decomposition of V . Moreover, we have dim(Egen) equals the algebraic multiplicity of λi, mi. λi (B) The semisimple part S of T and the nilpotent part N of T defined to be the unique C­linear operators V → V such t...
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� Egen . λi λi (6) If (S�, N �) is any pair of a diagonalizable operator S� and a nilpotent operator N � such that T = S� + N � and S�N � = N �S�, then S� = S and N � = N . We call the unique pair (S, N ) the semisimple­nilpotent decomposition of T . (C) For each i = 1, . . . , r, choose an ordered basis B(i) = ...
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⎜ ⎝ C (1) 0m2×m1 . . . 0m1×m2 C (2) . . . 0mr ×m1 0mr ×m2 [N ] B,B = . . . 0m1×mr . . . 0m2×mr . . . . . λr Imr . . . . . . . 0m1×mr . . . 0m2×mr . . . . . C (r) . . . . ⎞ ⎟ ⎟ ⎠ , ⎞ ⎟ ⎟ ⎟ ⎠ . (8) (9) Notice that D(i) has a nice form with respect to ANY basis B(i) for Egen . But we might hope to im...
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other words, (11) Jae1 = e2, Jae2 = e3, . . . , Jaea−1 = ea, Jaea = 0. Notice that the powers of Ja are very easy to compute. In fact J a = 0a,a, and for d = 1, . . . , a − 1, we have a d e1 = ed+1, Ja Ja Notice that we have ker(Ja d e2 = ed+2, . . . , Ja d) = span(ea+1−d, ea+2−d, . . . , ea). a d ea−d = ea, J d...
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in Jordan normal form. , . . . , B(r) puts T in Jordan normal form if each B(i) puts T (i) � B(1) � We say that a basis B = in Jordan normal form. WARNING: Usually such a basis is not unique. For example, if T is diagonalizable, then ANY basis B(i) puts T (i) in Jordan normal form. In this section we present th...
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−1). If a2 = a1, then this is also a distinguishing feature of ea1+1. But if a2 < a1, this doesn’t work. In this case it turns out that the distinguishing feature is that ea1+1 is in ker(J a2 ) but is not in ker(J a2−1) + J (ker(J a2+1)). This motivates the following definition: Definition 1. Suppose that B ∈ Mn×n(C)...
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a nontrivial primitive 1 ≤ k1 for Gkj . Then the subspace Gkj . For each j = 1, . . . , u, choose a basis v[j]1, . . . , v[j]pj desired basis is simply � � � B(i) = v[u]1, Bv[u]1, . . . , Bu−1 v[u]1, v[u]2, Bv[u]2, . . . , Bku−1 v[u]2, . . . , v[u] v[j]i, Bv[j]i, . . . , Bkj −1 v[j]i, . . . , v[1]1, . . . , Bk...
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I2. By performing Gauss­Jordan elimination we may find a basis for ker(Bi). In fact each kernel will be one­dimensional, so let v1 be a basis � (16) (17) (18) (19) (20) (21) 18.700 JORDAN NORMAL FORM NOTES 5 for ker(B1) and let v2 be a basis for ker(B2). Then with respect to the basis B = (v1, v2), we will h...
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), so we are in the case discussed above. The two eigenvalues are −4 and 3. First we consider the eigenvalue λ1 = −4. Then we have B1 = A + 4I2 = � 42 21 − 35 −70 � . (22) Performing Gauss­Jordan elimination on this matrix gives a basis of the kernel: v1 = (5, 3)†. Next we consider the eigenvalue λ2 = 3. The...
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case is readily identified, so if A is not already in diagonal form at the beginning of the problem, we are in the second case. In the second case Eλ1 has dimension 1. According to our algorithm, we must find a 2) = C2 . Such a subspace necessarily has dimension 1, i.e. primitive subspace G2 ⊂ ker(B1 it is of the for...
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cA(X) = X 2 + 6X + 9 = (X + 3)2 . So the characteristic polynomial has a repeated root of λ1 = −3. We form the matrix B1 = A + 3I2, B1 = A + 3I2 = � � . −2 −4 2 1 Performing Gauss­Jordan elimination (or just by inspection) a basis for the kernel is given by (2, −1)†. So for v1 we choose ANY vector which is no...
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(XI3 − A). But a faster way of calculating this is as follows. We know that the characteristic polynomial has the form cA(X) = X 3 − trace(A)X 2 + tX − det(A), (34) for some complex number t ∈ C. Usually trace(A) and det(A) are not hard to find. So it only remains to determine t. This can be done by choosing any co...
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since det(2I3 − A) = 0). In fact it is easy to see that cA(X) = (X − 2)3 . Now that we know how to compute cA(X) in a more efficient way, we can begin our analysis. There are three cases depending on whether cA(X) has three distinct roots, two distinct roots, or only one root. Three roots: Suppose that cA(X) = (X − λ...
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It is easy to see that trace(A) = 0 and also det(A) = 0. Finally we consider the determinant of I3 − A. Using cofactor expansion along the third column, this is: det −8 9 −2 ⎠ = −2((−8)4 − 9(−4)) − (−2)((−6)4 − 7(−4)) = −2(4) + 2(4) = 0. (41) −6 7 −2 ⎞ ⎛ ⎝ −4 4 0 So we have the linear equation 13 − 0 ∗ 12 + t ...
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1 0 2 ⎠ , ⎝ 2 1 3 B = ⎝⎝ 4 2 ⎠⎠ , P = ⎝ 4 2 ⎛ 3 1 2 2 1 1 0 ⎛ [A]B,B = ⎝ −1 0 0 0 0 0 0 0 1 ⎞ ⎛ ⎠ , A = P ⎝ ⎞ ⎠ P −1 . −1 0 0 0 0 0 0 0 1 Two roots: Suppose that cA(X) has two distinct roots, say cA(X) = (X − λ1)2(X − λ2). Then we form B1 = A − λ1I3 and B2 = A − λ2I3. By perfor...
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λ2 0 0 λ2 In this case S = A and N = 0. (43) (44) (45) 18.700 JORDAN NORMAL FORM NOTES 9 The second case is when Eλ1 has dimension 2. Using Gauss­Jordan elimination we find a basis for Egen = ker(B1 which is not in Eλ1 and define λ1 v2 = B1v1. Also using Gauss­Jordan elimination we may find a vector v3 which f...
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1 0 0 ⎠ , A = P ⎝ 1 0 0 ⎠ P −1 . Let’s see how this works in an example. Consider the matrix ⎛ 3 A = ⎝ 0 −1 ⎞ 0 0 3 −1 ⎠ . 2 0 It isn’t hard to show that cA(X) = (X − 3)2(X − 2). So the two eigenvalues are λ1 = 3 and λ2 = 2. We define the two matrices 0 1 0 −1 ⎠ , B2 = A − 2I3 = ⎝ 0 −1 0 −1 0 B1 = A − 3I3 = ⎝ ...
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) which isn’t in E3 is v1 = (1, 0, −1)†. We define v2 = B1v1 = 3 1 ⎛⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎞ ⎞ 0 B = ⎝⎝ 0 ⎠ ⎝ 1 ⎠ , ⎝ 1 ⎠⎠ , P = ⎝ 0 1 1 ⎠ . 1 −1 0 1 0 0 ⎛ −1 , 0 0 1 we have ⎛ 3 0 0 ⎞ ⎛ ⎞ 3 0 0 [A]B,B = ⎝ 1 3 0 ⎠ , A = P ⎝ 1 3 0 ⎠ P −1 . 0 0 2 0 0 2 (46) (47) (48) (49) (51) (52) (53) 10 18.700 JORDAN...
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0 0 0 0 0 0 0 0 ⎛ ⎞ ⎠ P −1 = ⎝ 1 0 0 0 0 1 0 0 0 ⎞ ⎠ . [N ]B,B = ⎝ 1 0 (54) (55) One root: The final case is when there is only a single root of cA(X), say cA(X) = (X − λ1)3 . Again we form B1 = A1 − λ1I3. This case breaks up further depending on the dimension of Eλ1 = ker(B1). The simplest case i...
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56) Notice that there is a Jordan block of size 2 and a Jordan block of size 1. Also, S = λ1I3 and we have N = B1. Let’s see how this works in an example. Consider the matrix ⎛ ⎝ A = ⎞ 0 0 ⎠ . −1 −1 1 −3 0 0 −2 (57) It is easy to compute cA(X) = (X + 2)3 . So the only eigenvalue of A is λ1 = −2. We define B1...
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�� . 1 1 1 0 0 0 1 (59) ⎛ 0 1 0 18.700 JORDAN NORMAL FORM NOTES We have ⎛ [A]B,B = ⎝ 2 − 1 0 ⎞ ⎛ 0 0 ⎠ , A = P ⎝ 0 −2 0 −2 ⎞ 0 0 ⎠ P −1 . 0 −2 0 −2 2 − 1 0 We also have S = −2I3 and N = B1. 11 (60) Dimension One In the final case for three by three matrices, we could have that cA(X) = 2) is (X ...
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We also have S = λ1I3 and N = B1. Let’s see how this works in an example. Consider the matrix ⎛ A = ⎝ 1 ⎞ 0 ⎠ . 5 −4 0 1 2 −3 3 (61) (62) The trace is visibly 9. Using cofactor expansion along the third column, the determinant is +3(5 ∗ 1 − 1(−4)) = 27. Finally, we compute det(3I3 − A) = 0 since 3I3 − A has th...
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by inspection, we see that ker(B1 So for v1 we choose any vector not in the span of these vectors, say v1 = (1, 0, 0)†. Then we 2v1 = (0, 0, 1)†. So with respect to define v2 = B1v1 = (2, 1, 2)† and we define v3 = B1v2 = B1 2) has basis ((2, 1, 0)†, (0, 0, 1)†)). 12 18.700 JORDAN NORMAL FORM NOTES the basis and tra...
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Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.341: Discrete-Time Signal Processing OpenCourseWare 2006 Lecture 8 DT Filter Design: IIR Filters Reading: Section 7.1 in Oppenheim, Schafer & Buck (OSB). In the last lecture we studied various forms of filter realizat...
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) . | | Nonetheless, specifications can involve both magnitude and phase (or group delay). Such gen­ eralized approximation is a harder problem, but may be desired in specific applications. In particular, integer or fractional delays can only be achieved with FIR filters. Most of our following discussions will be phras...
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sample, per output sample, or per unit time (clock cycle). It is possible that an IIR of lower order actually requires more #MAD than an FIR of higher order, because FIR filters may be implemented using polyphase structures. • Different conventions exist for specifying magnitude responses for IIR and FIR filters. In ...
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: s → 1 − z−1 1 + z−1 ⇒ jΩ = j 2 T tan ω 2 , Ω Ωc → tan ω/2 tan ωc/2 π in the digital frequency domain corresponds to infinity in the analog frequency domain. Note that the bilinear transformation is really only appropriate in mapping filters which approximate piecewise constant filters. 2 Butterworth...
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that a lower order filter exists such that it satisfies the given specifications, but does not exceed them as greatly as the Butterworth design. • The following sets of figures are examples of Butterworth filter design. For an additional example, see OSB Example 7.4. 3 4 • Direct form implementation of a Butterwort...
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x > 1. | | 5 Some examples of lower order Chebyshev polynomials are: V0(x) = 1 V2(x) = cos(2 cos−1 x) = 2 cos2(cos−1 x) − 1 = 2x2 − 1 V1(x) = x See Appendix B.2 of OSB for recurrence formulas for deriving Chebyshev polynomials. • Equiripple in the passband, but decreases monotonically in the stopband. • Similar...
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iptic Filters • Four degrees of freedom: PB ripple, SB ripple, order, passband edge. • Order of the system controls the transition bandwidth. • Equiripple in both the passband and the stopband. • Elliptic filters are the lowest order rational function approximation to a given set of magnitude specifications. All...
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I.C Phase Transitions The most spectacular consequence of interactions among particles is the appearance of new phases of matter whose collective behavior bears little resemblance to that of a few particles. How do the particles then transform from one macroscopic state to a completely different one. From a formal p...
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