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j j j ^ ^ sinh(m(k , N /U )(h z )) � � sinh(m(k , N /U )h) j ± ± where e is given by the equation: j 2 (j /h ) sin [j (h z )/h] e j ; for j , : : : , L (A.191) 1 The poles in the upper part of the complex k plane give the residue which follows: cos(j ) (N /U ) (j /h ) 2 2 ; p Res ( (k) exp(ikx)dk ( (ia )(i1 ), (A.192)...
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MIT OpenCourseWare https://ocw.mit.edu 2.062J / 1.138J / 18.376J Wave Propagation Spring 2017 For information about citing these materials or our Terms of Use, visit: https://ocw.mit.edu/terms.
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r 1:  P 1 8.3 1 6 re   u Lect inciples  of  Applied  Mathematics   Rodolfo  Rosales Spring  2014 acteristics  of  u +c *u  =  0  [linearized  traffic  flow]  and  u +c *u   x 0 t t x 0 Recap  solution  by  char =  a*u.   t  or  simple  variable  coefficients,  whe...
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th E • e  the  solution  is  defined. Show  wher le:   re  a ll   l l x x e e a a ro p P   V <   ∞ -­‐ n o em   ,  t  >  0. ∞ x  <   :  u +c *u  =  a*u.  I 1 mp E x t 0 1)   ,   (x u y,   +  y* -­‐∞ r o  f x) g( ∞ x  <    <   2 mp E u * x : x w 2*u x x  + ...
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MIT OpenCourseWare http://ocw.mit.edu 18.311 Principles of Applied Mathematics Spring 2014 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
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3.044 MATERIALS PROCESSING LECTURE 6 Ex. 1: glass fiber (ceramic) Ex. 2: plasma spray (ceramic and metal) Ex. 3: hot rolling steel slabs (metal) look at iron-carbon (steel) phase diagram, red hot is about 900 − 1000◦C, need to heat into gamma field to make it soft and eliminate ceramic carbide phase Problem Statement: Ho...
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) 7700 Solution: t = 22, 000s ≈ 6 hours T Ti − − Tf Tf How to decrease time?: = f (k, c 1. thinner L → constrained by casting 2. higher h (fluid) → molten metal, salt 3. hotter Tf → high energy, doesn’t drastically change time p, ρ, t, Lx, h) 4. preheat Ti ⇒ vertical integration, combine casting and rolling temperatures...
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t) = Θ(x, t)Θ(y, t) F0,x = F0,y Θ(x, t) = Θ(y, t) = √ 0.1 = 0.32 F0 = 4 t ≈ 3hrs MIT OpenCourseWare http://ocw.mit.edu 3.044 Materials Processing Spring 2013 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
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18.156 Differential Analysis II Lectures 1-2 Instructor: Larry Guth Trans.: Kevin Sackel Lecture 1: 4 February 2015 nR , and we will use coordinates x1, . . . , xn when Throughout these notes, we will be working typically over necessary. The convention will be that the Laplacian on nR with the standard Euclidean metric ...
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x)+2n(cid:3) = 2n(cid:3) > 0. Hence the previous lemma applies to show u(cid:3) attains its maximum on the boundary, and taking the limit as (cid:3) → 0 yields the result. Corollary 1.3. If Δu = Δv with u, v ∈ C 2(Ω) ∩ C 0(Ω) and u|∂Ω = v|∂Ω, thenu = v on all of Ω. Proof. We have that Δ(u − v) = 0 with u − v = 0 on ∂Ω,...
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a bit mysterious, and doesn’t really tell us what’s going on under the hood. We present a second proof which takes into account the role of symmetries. We will only do this for the case of n = 2 and with x = 0, but this can be easily generalized. Alternate proof of MVP, n = 2. For θ ∈ SO(2) (cid:8) /2π we have a corres...
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∂i(cid:11)u(y)dy, |B1/2| S1/2(x) whereby the notation (cid:10)nor, ∂i(cid:11) means the dot product of the vector ∂i with the normal vector at the point y ∈ S(x, 1/2). These coefficients don’t actually depend on y, and we see immediately that for some constant Cn depending only on n. |∂iu(x)| ≤ Cn max |u| 2 Corollary 1....
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A big first theorem for us to prove is Schauder’s. Recall the idea of Holder-regularit ¨ y. For 0 < α ≤ 1, we have the semi-norms [f ]Cα := supx,y | f (y) f (x) − |x − y|α | , and corresponding norms (cid:12)f (cid:12)Cα := [f ]Cα + (cid:12)f (cid:12)C0 , where these are defined on some space of functions. Then Theorem 1...
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If u ∈ C 3 ( nRcpt ), then (cid:12)∂2u(cid:12)L2 = (cid:12)Δu(cid:12)L2 . 3 Proof. This follows from two applications of integration by parts: (cid:7) (cid:6) (cid:12)∂2u(cid:12)2 L2 = (∂i∂ju)(∂i∂ju) i,j Rn (cid:7) (cid:6) = − i,j (cid:7) (cid:6) = = i,j (cid:6) Rn (cid:7) ( Rn i 2 = (cid:12)Δu(cid:12)L2 (∂2 i ∂ju)(∂j...
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cients of Δ − L are aij − δij, which are bounded in absolute value by (cid:3), so continuing the chain of inequalities, we have (cid:12)∂2u(cid:12)2 ≤ (cid:12)Lu(cid:12)2 L2 L2 + 2(cid:3)n (cid:6) |Lu||∂2 u| + (cid:3) n2(cid:12)∂2u(cid:12)2 2 L2 . Applying the Cauchy-Schwarz inequality to the middle term, this yields (...
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| 2 |∂2(ηu)|2 |∂2(ηu ) |2 B (cid:6) 1 ≤ 4 | L(ηu) 2 | B1 (cid:6) (cid:6) (cid:6) ≤ 4 η2 Lu | |2 + C(n) |∇u||∇η| + C(n) |∂2η||u| B1 The first term is zero, and for the other two, an application of Cauchy-Schwarz separates the u and η terms in into different integrals so that we get in total (cid:12)∂2u(cid:12)2 2 L2(B ) ≤...
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the class is Schauder’s inequality. 2.2 Solving Δu = f (a story from physics) On our way to Korn’s inequality, we want to understand how we can write explicit estimates on the second partial derivatives of u in terms of Δu. Here, the story begins with some classical physics theory. We will make the mathematics a bit mo...
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(cid:5) Br Fz(x)dz. By symmetry, V (x) is always parallel to x. It suffices to check that (cid:6) (cid:6) V · nor = F0 · nor|Br|. ∂B(R) ∂B(R) The RHS is easily seen to be 4π|Br| (by the calculation we stated at the end of the lemma), while the LHS is (cid:11) (cid:6) (cid:8)(cid:6) (cid:9) (cid:6) (cid:10)(cid:6) (cid:6)...
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MIT OpenCourseWare http://ocw.mit.edu 18.156 Differential Analysis II: Partial Differential Equations and Fourier Analysis Spring 2016 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
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MIT OpenCourseWare http://ocw.mit.edu Resource: Calculus Revisited Herbert Gross The following may not correspond to a particular course on MIT OpenCourseWare, but has been provided by the author as an individual learning resource. For information about citing these materials or our Terms of Use, visit: http://oc...
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Massachusetts Institute of Technology 6.270 Autonomous LEGO Robot Competition IAP 2005: Attack of the Drones Workshop 5 — Servos and Advanced Sensors Monday, January 10, and Tuesday, January 11, 2005 1 Items to Bring • Handy Board with Expansion Board 2 Reading Chapter 5 and Appendix 5 of the course notes 3 S...
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Sensors • Phototransistor. This sensor alone is unreliable. Although useful for detecting the starting light, it should be calibrated for each different lighting environment to which the robot is subjected. The phototransistor is very sensitive to light, and should be shielded carefully. We suggest using the black he...
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for break­ beam purposes (remember that IR is susceptible to red light and color). 1 Updated January 10, 2005 Massachusetts Institute of Technology 6.270 Autonomous LEGO Robot Competition IAP 2005: Attack of the Drones 5 Activity This activity will help your robot orient correctly and dependably. Wire up a pho...
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be? Now try doing some readings without the LED, and also with varying light conditions. Try shining a flashlight on the table or shading the table with your hand. What difference does it make? This is a fairly simple activity, but these are all considerations you need to make when attaching sensors to your robot, an...
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1 Multilevel Memories Joel Emer Computer Science and Artificial Intelligence Laboratory Massachusetts Institute of Technology Based on the material prepared by Krste Asanovic and Arvind CPU-Memory Bottleneck 6.823 L7- 2 Joel Emer CPU Memory Performance of high-speed computers is usually limited by memory bandwidt...
https://ocw.mit.edu/courses/6-823-computer-system-architecture-fall-2005/29244173f6b6a78eed99d3b2b50196b6_l07caches.pdf
electrode (VREF) Ta2O5 dielectric word access FET Image removed due to copyright restrictions. bit Explicit storage capacitor (FET gate, trench, stack) poly word line W bottom electrode access fet TiN/Ta2O5/W Capacitor October 3, 2005 Processor-DRAM Gap (latency) 6.823 L7- 6 Joel Emer 1000 e c n a ...
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in flight! October 3, 2005 DRAM Architecture 6.823 L7- 8 Joel Emer N s s e r d d A w o R r e d o c e D N+M M Col. 1 bit lines Col. 2M word lines Row 1 Row 2N Memory cell (one bit) Column Decoder & Sense Amplifiers Data D • Bits stored in 2-dimensional arrays on chip • Modern chips have around 4 logi...
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processor are often highly predictable: … loop: ADD r2, r1, r1 SUBI r3, r3, #1 BNEZ r3, loop … PC 96 100 104 108 112 What is the pattern of instruction memory addresses? October 3, 2005 Typical Memory Reference Patterns Address n loop iterations linear sequence Instruction fetches Stack access...
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storage, e.g., memory – Address usually computed from values in register – Generally implemented as a cache hierarchy • hardware decides what is kept in fast memory • but software may provide “hints”, e.g., don’t cache or prefetch October 3, 2005 A Typical Memory Hierarchy c.2003 6.823 L7- 16 Joel Emer Split ...
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HIT Not in cache a.k.a. MISS Return copy of data from cache October 3, 2005 Read block of data from Main Memory Wait … Return data to processor and update cache Q: Which line do we replace? Placement Policy 6.823 L7- 21 Joel Emer Block Number 0 1 2 3 4 5 6 7 8 9 1 1 1 1 1 1 1 1 1 1 0 1 2 3 4 5 6 7 8 ...
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g a T t k c o B l t e s f f O b October 3, 2005 HIT Data Word or Byte Replacement Policy 6.823 L7- 26 Joel Emer In an associative cache, which block from a set should be evicted when the set becomes full? • Random • Least Recently Used (LRU) • LRU cache state must be updated on every access • true imp...
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29 Joel Emer Average memory access time = Hit time + Miss rate x Miss penalty To improve performance: • reduce the miss rate (e.g., larger cache) • reduce the miss penalty (e.g., L2 cache) • reduce the hit time What is the simplest design strategy? October 3, 2005 Write Performance 6.823 L7- 30 Joel Emer T...
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Row Column Precharge Row’ [ Micron, 256Mb DDR2 SDRAM datasheet ] October 3, 2005 Data 400Mb/s Data Rate
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MIT OpenCourseWare http://ocw.mit.edu 2.161 Signal Processing: Continuous and Discrete Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. Massachusetts Institute of Technology Department of Mechanical Engineering 2.161 Signal Processing - Continuous and Di...
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set {fn} unless it is known a-priori that f (t) contains no spectral energy at or above a frequency of π/ΔT radians/s. • In order to uniquely represent a function f (t) by a set of samples, the sampling interval ΔT must be sufficiently small to capture more than two samples per cycle of the highest frequency componen...
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angular frequency a and a + 2πm/ΔT , for any integer m, will generate the same sample set. In the figure below, a sinusoid is undersampled and a lower frequency sinusoid, shown as a dashed line, also satisfies the sample set. f(t) o o t o Δ T o 0 o 0 o o 10–2 This phenomenon is known as aliasing. After ...
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3) (cid:4) (cid:6) (cid:3) (cid:3) (cid:4) (cid:6) The following figure shows the effect of folding in another way. In (a) a function f (t) with Fourier transform F (j Ω) has two disjoint spectral regions. The sampling interval ΔT is chosen so that the folding frequency π/ΔT falls between the two regions. The spectru...
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) (cid:13) (cid:17) (cid:13) (cid:16) (cid:8) (cid:19) (cid:21) (cid:9) (cid:6) (cid:11) (cid:12) (cid:6) (cid:8) (cid:17) (cid:18) (cid:15) (cid:8) (cid:10) (cid:17) (cid:19) (cid:16) (cid:20) (cid:6) (cid:11) (cid:10) (cid:21) (cid:13) (cid:17) (cid:13) (cid:16) (cid:8) (cid:19) (cid:21) (cid:9) (cid:7) (cid:3) ...
https://ocw.mit.edu/courses/2-161-signal-processing-continuous-and-discrete-fall-2008/2934569026a399e9062597924c730a78_lecture_10.pdf
) (cid:6) (cid:4) (cid:7) 1. Selecting a sampling interval ΔT sufficiently small to capture all spectral components, or 2. Processing the continuous-time function f (t) to “eliminate” all components at or above the Nyquist rate. The second method involves the use of a continuous-time processor before sampling f (t)....
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3) (cid:4) (cid:2) (cid:5) (cid:15) (cid:15) (cid:15) (cid:7) (cid:4) (cid:8) (cid:2) (cid:5) (cid:10) (cid:4) (cid:8) (cid:2) (cid:5) (cid:10) (cid:11) (cid:17) (cid:20) (cid:26) (cid:6) (cid:9) (cid:4) (cid:6) (cid:27) (cid:3) (cid:15) (cid:16) (cid:28) (cid:9) (cid:13) (cid:9) (cid:13) In practice it i...
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cid:16) (cid:15) (cid:4) (cid:6) (cid:7) (cid:3) (cid:8) (cid:4) (cid:6) (cid:2) (cid:5) (cid:3) (cid:8) (cid:4) (cid:6) (cid:2) (cid:4) (cid:6) (cid:7) (cid:10) (cid:3) (cid:8) (cid:4) (cid:6) (cid:7) (cid:3) (cid:8) (cid:4) (cid:6) (cid:7) (cid:9) (cid:9) (cid:4) (cid:8) (cid:2) (cid:5) (cid:9) (cid:8) (ci...
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30) (cid:21) (cid:30) (cid:8) (cid:15) (cid:18) (cid:21) (cid:6) (cid:12) (cid:18) (cid:12) (cid:8) (cid:19) (cid:20) (cid:17) (cid:16) (cid:17) (cid:15) (cid:18) (cid:21) (cid:6) (cid:12) (cid:7) (cid:11) (cid:3) (cid:8) (cid:10) (cid:4) (cid:6) (cid:4) (cid:7) (cid:9) (cid:9) (cid:4) (cid:2) (cid:5) (cid:9) (cid:...
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y(t) = F −1 {F (cid:2)(j Ω)H(j Ω)} = F −1 {F (j Ω)} = f (t). The filter’s impulse response h(t) is h(t) = F −1 {H(j Ω)} = sin (πt/ΔT ) , πt/ΔT (cid:18) (cid:4) (cid:2) (cid:5) (cid:19) (cid:23) # (cid:4) (cid:6) (cid:23) $ (cid:4) (cid:6) (cid:23) " (cid:4) (cid:6) (cid:23) (cid:4) (cid:6) (cid:12) (cid:4) (c...
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nΔT − σ) dσ sin (π(t − nΔT )/ΔT ) , π(t − nΔT )/ΔT 10–5 ! (cid:15) (cid:20) (cid:16) (cid:20) (cid:16) (cid:12) (cid:15) (cid:15) (cid:15) (cid:15) (cid:11) (cid:21) (cid:10) (cid:6) (cid:8) (cid:8) (cid:15) (cid:25) (cid:12) (cid:11) (cid:16) (cid:17) (cid:20) (cid:3) (cid:9) (cid:9) (cid:17) (cid:9) (cid:17) (cid...
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0.3 2 The Discrete Fourier Transform (DFT) We saw in Lecture 8 that the Fourier transform of the sampled waveform f ∗(t) can be written as a scaled periodic extension of F (j Ω) F (cid:2)(j Ω) = 1 ΔT � ∞ � � F j Ω − n=−∞ �� 2nπ T We now look at a different formulation of F ∗(j Ω). The Fourier transform of...
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of F ∗(j Ω) in a single period, from Ω = 0 to 2π/ΔT , that is at frequencies Ωm = 2πm N ΔT for m = 0, 1, 2, . . . , N − 1 (cid:9) (cid:9) (cid:4) (cid:8) (cid:2) (cid:9) (cid:17) (cid:13) (cid:15) (cid:9) (cid:8) (cid:21) (cid:13) (cid:17) (cid:15) (cid:10) (cid:17) (cid:20) (cid:21) (cid:18) (cid:15) (cid:21...
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N m=0 Fm e j 2πmn/N for n = 0, 1, 2, . . . , N − 1 which is known as the inverse DFT (IDFT). These two equations together form the DFT pair. 10–7 (cid:7) (cid:5) (cid:20) % (cid:30) (cid:12) (cid:23) (cid:8) (cid:10) (cid:6) (cid:19) (cid:17) (cid:18) (cid:15) (cid:8) (cid:6) (cid:10) (cid:11) (cid:17) (cid:12)...
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Fourier transform case, we adopt the notations DFT {fn} ⇐⇒ {Fm} {Fm} = DFT {fn} {fn} = IDFT {Fm} to indicate DFT relationships. 10–8
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MIT OpenCourseWare http://ocw.mit.edu 6.080 / 6.089 Great Ideas in Theoretical Computer Science Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 6.080/6.089 GITCS 21 February 2008 Lecturer: Scott Aaronson Scribe: Emilie Kim Lecture 5 1 Administriv...
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care about that. 3 Oracles Oracles are a concept that Turing invented in his PhD thesis in 1939. Shortly afterwards, Turing went on to try his hand at breaking German naval codes during World War II. The first electronic computers were used primarily for this purpose of cracking German codes, which was extremely d...
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A : {0, 1}∗ → {0, 1} where the input is any string of any length and the output is a 0 or 1 answering the problem, then we can write M A for a Turing machine M with access to oracle A. Assume that M is a multi-tape Turing machine and one of its tapes is a special “oracle tape”. M can then ask a question, some stri...
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be solvable. If we solve the Diophantine problem, can we also then solve the halting problem? 5-2 � This was an unanswered question for 70 years until 1970, when it was shown that there was no algorithm to solve Diophantine equations because if there were, then there would also be an algorithm for solving the hal...
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comes K¨onig’s Lemma. Let’s say that you have a tree (a computer science tree, with its root in the sky and grows towards the ground) with two assumptions: 1) Every node has at most a finite number of children, including 0. 2) It is possible to find a path in this tree going down in any finite length you want. Claim: ...
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number of possible choices. Further, we’ve assumed that you can tile any finite region of the plane, which means the tree contains arbitrarily long finite paths. Therefore, K¨onig’s Lemma tells us that the tree has to have an infinite path, which means that we can tile the infinite plane. Therefore, the tiling problem i...
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Duper Halting Problem, the Super Duper Duper Halting Problem... Is there any problem that is in an “intermediate” state between computable and the halting problem? In other words, is there a problem that is 1) not computable, 2) reducible to the halting problem, and 3) not equivalent to the halting problem? This wa...
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a more powerful system, and then there will be statements within that system that can’t be proved, and so on. 5-5 ComputableHalting Problem(Diophantine, Tiling,...)Super Halting ProblemComputableHalting Problem(Diophantine, Tiling,...)Super Halting Problem? Second Incompleteness Theorem: No consistent, computable sys...
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sentence is provable, and is therefore a provable falsehood! That can’t happen if we’re working within a sound system of logic. So the sentence has to be true, but that means that it isn’t provable. G¨odel showed that as long as the system of logic is powerful enough, you can define provability within the system. Fo...
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is consistent. G¨odel shows that no sound logical system can prove its own consistency. The key claim is that Con(S) G(S). In other words, if S could prove its own consistency, ⇒ then it actually could prove the “unprovable” G¨odel sentence, “This sentence is not provable”. Why? Well, suppose G(S) were false. Then ...
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Ocean Acoustic Theory • Acoustic Wave Equation • Integral Transforms • Helmholtz Equation • Source in Unbounded and Bounded Media • Reflection and Transmission • The Ideal Waveguide – Image Method – Wavenumber Integral – Normal Modes • Pekeris Waveguide 13.853 COMPUTATIONAL OCEAN ACOUSTICS 1 Lecture 3 U(t...
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Harvard-MIT Division of Health Sciences and Technology HST.951J: Medical Decision Support, Fall 2005 Instructors: Professor Lucila Ohno-Machado and Professor Staal Vinterbo From Propositions To Fuzzy Logic and Rules Staal A. Vinterbo Harvard­MIT Division of Health Science and Technology Decision Systems Group, BWH...
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�, ∨, (, )}. Definition An expression in PL is any string consisting of elements from the sets V and S, i.e., any string of variables and symbols. An expression is either a well formed formula (wff) or it is not. The following wff fomation rules allow us to define wff: Definition � A variable alone is a wff � If α...
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pretation: Truth Value of Expressions Definition We define a setting s as a function s : V → {0, 1} assigning to each variable either the value 0 or the value 1, denoting true or false respectively. Definition An interpretation is a function that takes as input a wff and returns 0 or 1 depending on the setting used...
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) Fuzzy Stuff HST 951/MIT 6.873 11 / 56 Staal A. Vinterbo (HST/DSG/HMS) Fuzzy Stuff HST 951/MIT 6.873 12 / 56 Propositional Logic Semantics Propositional Logic Semantics Propositional Logic Semantics Semantics of Operators: Infix notation Propositional Logic Semantics Computing the Interpretation I Usua...
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∧ is called conjunction (“and”) (a ∧ b) = ∼ (∼ a∨ ∼ b) def (a ∧ b) is often called the “conjunction of a and b”. → is called implication (“implies”) � def (a → b) = (∼ a ∨ b) Is (∼ (∼ a∨ ∼ b)) = ∼ (∼ 1∨ ∼ 0) = ∼ (0 ∨ 1) =∼ 1 = 0 � Left side is the antecedent, right side is the consequent. We also let (b ← a) =...
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of variables a and b and the resulting value for (a b). → � Is (a → b) valid? Satisfiable? Valid: No. Satisfiable: Yes. Note:Tables can become Large. Staal A. Vinterbo (HST/DSG/HMS) Fuzzy Stuff HST 951/MIT 6.873 17 / 56 Staal A. Vinterbo (HST/DSG/HMS) Fuzzy Stuff HST 951/MIT 6.873 18 / 56 Propositional Logic ...
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r ))) → q) valid? (( p 1 4 ∧ 1 2 ( p ↔ ( q ∧ r ))) → q ) 1 6 1 5 1 8 1 7 1 9 0 1 0 3 Answer: Yes. The settings underlined pose a contradiction. Note: If we during the process shown are allowed alternatives, we need to show a contradiction in all the possible alternative settings in order to declare our...
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/b] is ((c ∧ d ) → ((c ∧ d ) ∨ c)). Modus Ponens (also called the rule of detachment) is sometimes written as α α β β→ If α and α the truth­functional meaning of →. → β are theorems, then by MP so is β. This simply reflects Staal A. Vinterbo (HST/DSG/HMS) Fuzzy Stuff HST 951/MIT 6.873 23 / 56 Staal A. Vinter...
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Propositional Consequence: Example By showing how we would do without the rule: (1) (2) (3) (4) (5) (6) (7) (8) ↔ (β ∧ γ))given αgiven (α ((α → (β ∧ γ)) ∧ ((β ∧ γ) → α))US (a→ ((((α → (β ∧ γ)) ∧ ((β ∧ γ) → α))) → ((α → (β ∧ γ)))US ((a∧b)→ (α (β ∧ γ) ((β ∧ γ) → β → (β ∧ γ))MP (3)+(4) β)US ((a∧b)→a) MP (1)+...
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. � Furthermore, f is the characteristic function of the subset S of U such that S consists exactly of the elements x in U such that f (x ) = 1. � Formally, S = f −1(1) = {x ∈ U f (x ) = 1}. We will denote the | characteristic function for the set S ⊆ U as χS . Staal A. Vinterbo (HST/DSG/HMS) Fuzzy Stuff HST 95...
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over Sets Semantics: Truth Sets Propositions over Sets Semantics: Truth Sets for “Syntactic Sugar” The semantics of p over U is based on truth sets. We define truth sets of wff of PL(U) according to the following rules: For p ∈ F , and wff α and β | � T (p) = {x ∈ U p(x ) = 1}, � T (∼ α) = U − T (α), and � T (α ∨...
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number” over the natural numbers modeled by the characteristic functions even and prime with the usual definitions. Let α = even ∧ prime. Then we have that T (α) = T (even) ∩ T (prime) = {2}, and I(α, x ) = 1 if and only if x = 2. Staal A. Vinterbo (HST/DSG/HMS) Fuzzy Stuff HST 951/MIT 6.873 35 / 56 Staal A. Vi...
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is a proposition HeightTall over the set of all people. We now formulate the if­then rule as propositions over sets: (HeightTall ∧ HairDark) → LookHandsome The application becomes: (HeightTall ∧ HairDark) (HeightTall ∧ HairDark) → LookHandsome (LookHandsome) In other words we set I(β, x ) = Effect: We infer th...
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Fuzzy Stuff HST 951/MIT 6.873 42 / 56 Fuzzy Sets Fuzzy Sets Fuzzy Sets Crisp Set Operators Definitions Fuzzy Sets Fuzzy Set Operations Example Let A and B be two subsets of some set U. We define union, intersection, difference, and complementation using in terms of χA and χB as follows: Example Definition χA...
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interbo (HST/DSG/HMS) Fuzzy Stuff HST 951/MIT 6.873 45 / 56 Staal A. Vinterbo (HST/DSG/HMS) Fuzzy Stuff HST 951/MIT 6.873 46 / 56 Fuzzy Relations Fuzzy Relations Fuzzy Relations Fuzzy Composition Definition Let X , Y and Z be three sets and let R and R� be two fuzzy relations from X to Y and Y to Z , respecti...
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47 / 56 Staal A. Vinterbo (HST/DSG/HMS) Fuzzy Stuff HST 951/MIT 6.873 48 / 56 4 A Restricted Fuzzy Logic A Restricted Fuzzy Logic Fuzzy Logic Defining the Fuzzy Logic Language Fuzzy Logic Semantics Recall: For PL(U), the interpretation I(α, x ) is given by I(α, x ) = χT (α)(x ). Definition (Fuzzy Proposit...
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(α β)(x ) = ↔ min(max(1 − µT (α)(x ), µT (β)(x )), max(1 − µT (β)(x ), µT (α)(x ))). Definition (Fuzzy Interpretation) If we let WFF (U) be the set of wffs of FPL(U) we define the interpretation I(α, x ) of a wff α with respect to an element x in U to be � I(α, x ) = µT (α)(x ). Staal A. Vinterbo (HST/DSG/HMS) Fuz...
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characteristic functions of subsets of U. � that a truth set for a given wff is the set for which the interpretation is a characteristic function. � that a propositional rule essentially is the application of modus ponens to an implication called the rule. Summary Fuzzy Sets and Logic We have learned � that fuz...
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Lecture 10: Supercurrent Equation Outline 1. Macroscopic Quantum Model 2. Supercurrent Equation and the London Equations 3. Fluxoid Quantization 4. The Normal State 5. Quantized Vortices October 13, 2005 Massachusetts Institute of Technology 6.763 2005 Lecture 10 Macroscopic Quantum Model 1. The wavefunctio...
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closed contour within a superconductor: The line integral of each of the parts: Therefore, flux integer n = - n’ Massachusetts Institute of Technology 6.763 2005 Lecture 10 Fluxoid Quantization The flux quantum is defined as And the Fluxoid Quantization condition becomes Fluxiod Experiments testing fluxi...
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Foundations of Applied Superconductivity. Reading, MA: Addison-Wesley, 1991. ISBN: 0201183234. Massachusetts Institute of Technology 6.763 2005 Lecture 10 Induced Currents To have flux quantization, currents must be induced in the cylinder to add to or oppose the applied magnetic field. Induced flux= L i Image rem...
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conductors Image removed for copyright reasons. Please see: Figure 6.1, page 259, from Orlando, T., and K. Delin. Foundations of Applied Superconductivity. Reading, MA: Addison-Wesley, 1991. ISBN: 0201183234. Massachusetts Institute of Technology 6.763 2005 Lecture 10 The Vortex State nV is the areal density of ...
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current density is then the electrons become normal. This ξ In the absence of any current flux, the superelectrons have zero net velocity but have a speed of the fermi velocity, v . Hence the kinetic energy with currents is F Massachusetts Institute of Technology 6.763 2005 Lecture 10 9 Coherence Length x The...
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MIT OpenCourseWare http://ocw.mit.edu (cid:10) 6.642 Continuum Electromechanics Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. (cid:13) 6.642, Continuum Electromechanics, Fall 2004 Prof. Markus Zahn Lecture 7: Pressure–Velocity Relations for Inviscid,...
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α (cid:3) , v x ( ) Δ = (cid:3) α v x 6.642, Continuum Electromechanics Lecture 7 Prof. Markus Zahn Page 1 of 5 Θ ( ' 0 ) β (cid:108) = Θ (cid:3) ( , v 0 x ) = (cid:3) v β x (cid:108) ( Θ x ) = (cid:108) Θ α sinh kx (cid:108) − Θ β ( sinh k x − Δ ) sinh k ...
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Θ dx (cid:3) p = j ( ρ ω - k U z ) (cid:108) Θ ⎡ ⎢ ⎢ ⎣ (cid:108) Θ α (cid:108) Θ β ⎤ ⎥ ⎥ ⎦ = 1 k ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ − coth k Δ − 1 sinh k Δ 1 sinh k coth k ⎤ ⎡ ⎥ Δ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ Δ (cid:3) v (cid:3) v α x β x ⎤ ⎥ ⎥ ⎥ ⎦ ⎡ (cid:3) p ⎢ ⎢ (cid:3) p ⎣ α β ⎤ ⎥ ⎥ ⎦ = ( ρ ω − j k U z ) k ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ − coth k Δ − 1 sinh k Δ 1 inh k ...
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Courtesy of MIT Press. Used with permission. II. Gravity – Capillary Dynamics 6.642, Continuum Electromechanics Lecture 7 Prof. Markus Zahn Page 3 of 5 Equilibrium: gx P + −ρ a 0 x 0> 0P = ; v 0= , 0ξ = −ρ b gx P + 0 x 0 < Perturba...
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⎦ (cid:3) v x1 (cid:3) v= x4 ≡ 0 (rigid boundaries) Interface: v = x ∂ξ t ∂ + v y ∂ξ y ∂ + v z ∂ξ z ∂ ⇒ (cid:3) v x2 (cid:3) v = x3 (cid:3) = ωj ξ Force Balance ' P 3 ( ) ξ ' - P 2 ( ) ξ = - ⎛ γ ⎜ ⎜ ⎝ 2 ∂ ξ 2 y ∂ + 2 ∂ ξ 2 z ∂ ⎞ ⎟ ⎟ ⎠ P 3 ( ) ξ = P 30 ( 0 + ξ ) ' + P 3 ( ) ξ = P 30 ( 0 ) + dP 30 dx x 0 = ξ + ( ' ...
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� (cid:3) coth kav ⎤ ⎦ x2 = - 2 ρ ω a k (cid:3) coth ka ξ 6.642, Continuum Electromechanics Lecture 7 Prof. Markus Zahn Page 4 of 5 2 ω k ⎡ ⎢ ⎢ ⎣ ( ρ a coth ka + ρ b coth kb + g ) ( ρ − ρ b a ) 2 - k γ (cid:3) ξ = 0 ⎤ ⎥ ⎥ ⎦ Dispersion Relation:...
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( ρ − ρ b a ) coth ka ≈ 1 ka coth kb ≈ 1 kb 2 ω 2 k ρ a a ⎛ ⎜ ⎝ + ρ b b ⎞ ⎟ ⎠ = g ( ρ − ρ a b ) 2 ω 2 k = 2 v P = ( g ρ b ρ − ρ a ρ + a a b b ) Non-dispersive gravity wave 6.642, Continuum Electromechanics Lecture 7 Prof. Markus Zahn Page 5 of 5
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Lecture 4 PN Junction and MOS Electrostatics(I) Semiconductor Electrostatics in Thermal Equilibrium Outline • Non­uniformly doped semiconductor in thermal equilibrium • Relationships between potential, φ(x) and equilibrium carrier concentrations, po(x), no(x) –Boltzmann relations & “60 mV Rule” • Quasi­neutral situa...
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0 What is n o(x) that satisfies this condition? no, Nd partially uncompensated � donor charge Nd(x) + no(x) net electron charge - Let us examine the electrostatics implications of n o(x) ≠ Nd(x) x 6.012 Spring 2009 Lecture 4 5 Space charge density ρρρρ(x) = q Nd(x) − no(x) [ ] no, Nd ρ partially unc...
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µµµµn • Dn dφφφφ 1 = dx no dno dx • Using Einstein relation: q dφφφφ • kT dx = d(ln no) dx Integrate: q kT (φφφφ− φφφφref )= ln no − ln no,ref = ln no no,ref Then:   q(φφφφ− φφφφref )   no = no,ref exp   kT   6.012 Spring 2009 Lecture 4 9 Any reference is good In 6.012, φref = 0 ...
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p o = −(25m )• ln 10) ni ( • log po ni Or p o φφφφ≈ −(60 m )• log ni EXAMPLE 2: 18 −3 no = 10 cm ⇒ po = 10 cm ⇒ φφφφ= −(60m ) × −8 = 480mV 2 −3 6.012 Spring 2009 Lecture 4 12 Relationship between φφφφ, no and po : po, equilibrium hole concentration (cm−3) p-type intrinsic n-type 1019 1018 1016 101...
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7 cm −−−−3 ) −−−− φφφφ(no ==== 1015 cm −−−−3 ) ==== 120 mV Example 4: Compute potential difference in thermal equilibrium between region where po = 1020 cm­3 and po = 1016 cm . ­3 φφφφ( po ==== 10 20 cm −−−−3) ==== φφφφmax ==== −−−−550mV φφφφ( po ==== 10 16 cm −−−−3 ) ==== −−−−60 ×××× 6 ==== −−−−360mV φφφφ( po ==...
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.012 Spring 2009 Lecture 4 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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Maximum Likelihood Estimation Parameter Estimation Fitting Probability Distributions Maximum Likelihood MIT 18.443 Dr. Kempthorne Spring 2015 MIT 18.443 Parameter EstimationFitting Probability DistributionsMaximum Likelihood eliho od 1 Maximum Likelihood Estimation Framework/Definitions Outline 1 Maximum Likel...
https://ocw.mit.edu/courses/18-443-statistics-for-applications-spring-2015/2a08e2ffa8c87f3187d3969cc24c168b_MIT18_443S15_LEC4.pdf
X1, X2, . . . , Xn are observations of a time series {Xt , t = 1, 2, . . .} Joint density of X = (X1, X2, . . . , Xn) is given by: f (x1, . . . , xn | θ) = f (x1 | θ) × f (x2 | θ, x1) × f (x3 | θ, x1, x2) × · · · = =⇒ lik(θ) = ×f (xn | θ, x1, x2, . . . , xn−1) n n f (xi | θ, {xj ; j < i}) i=1 n n f (xi | θ...
https://ocw.mit.edu/courses/18-443-statistics-for-applications-spring-2015/2a08e2ffa8c87f3187d3969cc24c168b_MIT18_443S15_LEC4.pdf
i=1 MIT 18.443 Parameter EstimationFitting Probability DistributionsMaximum Likelihood elihood 5 Maximum Likelihood Estimation Framework/Definitions Specifying the MLE Example 8.5.A: Poisson Distribution X1, . . . , Xn i.i.d. Poisson(λ) f (x | λ) = λx ex! (cid:80) n £(λ) = i=1 MLE λˆMLE −λ [xi ln(λ) − λ − ln(x!...
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(µ, σ2): ˆ σ2 θMLE = (ˆµMLE , ˆMLE ) £(θˆMLE ) maximizes £(θ) = £(µ, σ) ˆ = 0 and θMLE solves: ∂£(µ, σ2) ∂µ ∂£(µ, σ2) ∂σ2 = 0 MIT 18.443 Parameter EstimationFitting Probability DistributionsMaximum Likelihood elihood 7 Maximum Likelihood Estimation Framework/Definitions MLEs of Normal Distribution Param...
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18.443 −n = + n ln(λˆ) + ln(xi ) n n i=1 n n + n ln(α) − n ln(X ) + ln(xi ) (cid:80) n i=1 ln(xi ) i=1 =⇒ 0 = Γ/ (α) Γ(α) Parameter EstimationFitting Probability DistributionsMaximum Likelihood elihood 10 Maximum Likelihood Estimation Framework/Definitions MLEs of Gamma Distribution Parameters Note: ...
https://ocw.mit.edu/courses/18-443-statistics-for-applications-spring-2015/2a08e2ffa8c87f3187d3969cc24c168b_MIT18_443S15_LEC4.pdf