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157), and if we apply the inverse Fourier transform to the result- ing equation, we obtain 0 0 U N ^ 2 w(x , z , t ) ik ( (k) s i n ( k z + k U t + k x )dk N/U 2 ; 8 U ; 0 s Z { 1 : 2 N (7.158) + ik ( (k) e x p ( k z ) sin(+k U t + k x )dk ^ 2 0 N/U s ; ; U Z 9 } In terms of the fxed reference frame, the vertical velo...
https://ocw.mit.edu/courses/2-062j-wave-propagation-spring-2017/2870676df2a0100adddd62ce6d852589_MIT2_062J_S17_Chap7.pdf
Next, we consider an example of a localized topography, the \witch 0 ; of Agnesi", for which ( (x) (fxed reference fram e) is given by the equation A 0 ( (x) , (7.160) 1 + ( x/b) 2 and its Fourier transform is given by the equation ^ ( (k) b exp( A kb ): (7.161) 0 ;j j For this particular example, the vertical veloci...
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1 (x, z ) A b exp( b k ) sin ( k z + kx )dk 0 8 ; j j ; U 0 s N/U 2 N 2 Z { 1 : 2 2 N + exp( b k k z ) sin(+kx )dk N/U s ; j j ; ; U 9 Z } 41 (7.164) By inspecting equation (7.164), if the buoyancy frequency N is zero, the stream function 1 has a simple expression given by the equation 1 (x, z ) , (7.165) 1 + [ x/(b +...
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the stream with horizontal constant speed U passing over a localized topography z ( (x). We assume the buoyancy frequency N constant along the atmosphere. For a reference frame fxed on the ground, we have the governing equation (7.137) for the fow vertical velocity w . For this governing equation we h a ve the boundary...
https://ocw.mit.edu/courses/2-062j-wave-propagation-spring-2017/2870676df2a0100adddd62ce6d852589_MIT2_062J_S17_Chap7.pdf
42 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 s i x a z -10 0 10 20 x axis Figure 8: Stream lines for the case of zero value for the buoyancy frequency N . A 1 0 and b . 4 43 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 s i x a z -10 0 10 20 ...
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Fourier transform pair given by equations (7.141) and (7.142). For w^ (k , z , t ) (Fourier transform of w(x , z , t )), we consider the time dependence 0 w^(k , z , t ) w^ (k , z ) ex p ( i! t): (7.167) 0 ; 0 The governing equation for w^ (k , z ) is given by equation (7.146) and it has also to satisfy the boundary co...
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) , (7.170) ( ! ; ; or we can obtain a dispersion relation, which follows kN ! : (7.171) 2 2 p k + m This dispersion relation will be necessary to obtain the group velocity o f t h e w ave distur- bances generated by the topography, which will b e necessary to discuss how to deform ...
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(U t + x ))dk (7.173) ; 2 sinh(mh) ;1 Z This inverse Fourier has closed form solution, which is basically the sum of the residue of part of the poles of the integrand in equation (7.173). To obtain these p o l e s , w e use 1 2 z sinh z z 1 + (7.174) 2 2 l l=1 Y Therefore, we can write sinh(m(h z )) ; 1 m (h;z ) 2 2 ...
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1, there is no real N l� U h 2 poles, which implies no waves associated with the localized topography. In this case we have just a local evanescent wavefeld close to the localized topography. To evaluate the integral in equation (7.173), we consider a closed contour, which is the original integration contour along the ...
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les. This decision is associated if we have waves downstream or upstream of the localized topography, and to carry it out, we need to compute the group velocity, which is given by 2 2 m N ! m C g , (7.177) 2 2 3/2 (m + k ) k k ; ! j j where ! /k is the phase velocity of the wave following the topography, which has valu...
https://ocw.mit.edu/courses/2-062j-wave-propagation-spring-2017/2870676df2a0100adddd62ce6d852589_MIT2_062J_S17_Chap7.pdf
. Therefore, the only contribution comes from the poles inside the closed contour illustrated in the fgure 10. Now, the expression for the vertical velocity w(x , z , t ) can be written as follows: 0 Case (U t + x ) > 0. We assume that the frst L poles are real, and the poles for 0 • l > L are pure imaginary. We have t...
https://ocw.mit.edu/courses/2-062j-wave-propagation-spring-2017/2870676df2a0100adddd62ce6d852589_MIT2_062J_S17_Chap7.pdf
U ( (k) e ; 2 2 ; 2 sinh(m(k , N /U )h) j k= (N/U ) ;(j� /h ) p j=1 X L � � i sinh(m(k , N /U )(h z )) ^ (ik(U t +x )) 0 + Res ikU ( (k) ; e 2 2 ; 2 sinh(m(k , N /U )h) j k=; (N/U ) ;(j� /h ) p j=1 X 1 � � sinh(m(k , N /U )(h z )) (ik(U t +x )) 0 ^ + iRes ikU ( (k) e ; 2 2 ; j sinh(m(k , N /U )h) k=i (j� /h ) ;(N/U ...
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of l is one, the critical speed for a given value of the buoyancy frequency N is U . For current values U > , there is no wave � � N h N h disturbance downstream of the localized topography. To illustrate the fow for this problem, we change from the moving reference frame to the fxed reference frame (x + U t x), and we...
https://ocw.mit.edu/courses/2-062j-wave-propagation-spring-2017/2870676df2a0100adddd62ce6d852589_MIT2_062J_S17_Chap7.pdf
Case x > 0. We assume that the frst L poles are real, and the poles for l > L are • pure imaginary. We have that 49 L i sinh(m(k , N /U )(h z )) ^ 1 (x, z ) Res ( (k) exp(ikx) ; 2 2 ; 2 sinh(m(k , N /U )h) j k= (N/U ) ;(j� /h) p j=1 X L � � i sinh(m(k , N /U )(h z )) ^ + Res ( (k) exp(ikx) ; 2 2 ; 2 sinh(m(k , N /U ...
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184) The equations (7.183) and (7.184) can b e written in a simple way in terms of the quantities e , 1 and a defned in the appendix A. j j j Case x > 0, where the frst L poles are assumed real numbers, and the other poles • are in the upper part of the complex k plane. L 1 1 (x, z ) / e sin(a x) / 1 exp( a x), (7.185...
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1 (x, z ) / 1 exp(a x), (7.187) j j j ; j=L+1 X We chose for ( (x) the same topography we considered in the previous section. The stream lines for this fow are illustrated in fgure 11, 12 and 13. For stream speeds that approach the critical values from below, the group velocity l� N h C for the l-th lee wave approache...
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/U ) (j /h ) (A.188) j 2 2 ± ± ; p For N/U /h we have, j ia i (j /h ) (N /U ) (A.189) j 2 2 ± ± ; p ...
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a fnite atmosphere with U /N 2 and A 1/2. We expect to see a superposition of four lee waves for 0 this value of N /U . 53 30 25 20 s i x a z 15 10 5 -10 -5 0 5 10 15 20 25 x axis 30 35 40 45 50 Figure 13: Stream function...
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Next, we give the expression for the residues in equations (7.183) and (7.184). Case x 0. We consider in this case the real p o l e s for which j 1, : : : , L . We • 2 frst give t h e residue at the real poles. 55 Res ( (k) exp(ikx)dk ( ( a )e , (A.190) ; k=±a j j j ^ ^ sinh(m(k , N /U )(h z )) � � sinh(m(k , N /U )h) ...
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cos(j ) (j /h ) (N /U ) 2 2 ; p Case x 0. We consider for this case the poles in the lower part of the complex • k plane. The residue at these poles follows: Res ( (k) exp(ikx)dk ( ( ia )( i1 ) ; k=;ia j j j ^ ^ sinh(m(k , N /U )(h z )) � � sinh(m(k , N /U )h) j ; ; (A.194) ...
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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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ariabl tic  v e  characteris liminate  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  < ...
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y  >  0.                                               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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125m, α = k ρcp = (cid:2) 35 (cid:3) kg 0 m3 .8 (cid:4) kJ kg K (cid:5) 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. prehe...
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when Θ = 0.1 @ x = 0 and y = 0 By Symmetry: Θ(x, t) = f (F0,x) Θ(y, t) = f (F0,y) Θ(x, y, 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...
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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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Δu(cid:3))(x) = (Δu)(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 ...
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constant is u(x). (cid:5) − u is constant, and Sr This proof is somehow 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. A...
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x, which is only found in the domain of integration, not the integrand. One sees that (cid:6) 1 ∂iu(x) = (cid:10)nor, ∂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 de...
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they are worth attention. One might wonder whether the Maximum Principle, Mean Value Property, or Regularity results that harmonic functions enjoy are shared by variable coefficient operators. We will be discussing in what ways this is the case. A big first theorem for us to prove is Schauder’s. Recall the idea of Holder-...
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related problem. Suppose u : B1 → R is C 2 with |Δu(x)| ≤ 10−6 and |u(x)| ≤ 1 on B1. Then can we say |∇u(0)| ≤10 6? Lecture 2: 6 February 2015 2.1 Almost Harmonic Functions Proposition 2.1. 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 integrat...
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∂2u(cid:12)L2 ≤ 2 (cid:12)Lu(cid:12)L2 . |∂2u|2 |Δu|2 |Lu + (Δ− L)u|2 (cid:6) (cid:6) |Lu|2 + 2 |Lu||(Δ − L)u| + |(Δ − L)u|2 Now, the coefficients 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(c...
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| + |u|(cid:12)L (B1) 2 for some C(n) depending only on n. 4 Proof. The idea is to apply Proposition 2.2 to a localized version ofu. Namely, take η a smooth bump function supported on B1 with η = 1 on B1. Then (cid:12)∂2u(cid:12)2 L2(B1) = = ≤ (cid:6) B1/2 (cid:6) B1/2 (cid:6) 2 | ∂ u | 2 |∂2(ηu)|2 |∂2(ηu ) |2 B (cid:...
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does hold in the Lp-norm, and this is the Calderon-Zygmund inequality from the 1950s. 3. The inequality does not hold int the C 1-norm. 4. The inequality does hold in the C α-norm. This is Korn’s inequality from the early 1900s. This will be proved in Lecture 5. Our goal over the next week will be to work especially to...
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3R with smooth boundary, and z ∈ Ω, then (cid:6) ∂Ω Fz(x) · nor = +4π. (If z ∈/ Ω, the same proof will show that this integral is just 0.) Proof. Set U = Ω \ Br(z) where r is small enough so that the entire ball lies inside Ω. Then the divergence theorem shows that (cid:6) (cid:6) (cid:6) (cid:6) 0 = divFz = Fz · nor =...
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, we have that for a domain Ω as in Newton’s Theorem, (cid:6) (cid:6) Δu = ∇u · nor. Ω ∂Ω We can take this instead to be our definition of the Laplacian in the sense of distributions. Doing this, note that from the Lemma, we had (cid:6) ∇Γ · nor = Ω (cid:12) 4π, 0, ∈ Ω 0 0 ∈/ Ω so that in the distributional sense, ΔΓ = ...
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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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the mount on the servo. 4 Advanced Sensors 4.1 Reflectance 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...
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IR LED/Phototransistor. Useful for both breakbeam and reflectance sensing—probably better 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 Ac...
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best readings. Does the angle at which the LED hits the table surface affect the reading? How close to the table does your sensor need to 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 do...
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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...
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s October 3, 2005 One Transistor Dynamic RAM 6.823 L7- 5 Joel Emer 1-T DRAM Cell TiN top 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 ...
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put (T) = Number in Flight (N) / Latency (L) CPU Misses in flight table Memory Example: --- Assume infinite bandwidth memory --- 100 cycles / memory reference --- 1 + 0.2 memory references / instruction ⇒ Table size = 1.2 * 100 = 120 entries 120 independent memory operations in flight! October 3, 2005 DRAM A...
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charges bit lines to known value, required before next row access Each step has a latency of around 20ns in modern DRAMs Various DRAM standards (DDR, RDRAM) have different ways of encoding the signals for transmission to the DRAM, but all share the same core architecture October 3, 2005 Multilevel Memory Strateg...
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latency: Register << SRAM << DRAM why? why? • bandwidth: on-chip >> off-chip On a data access: hit (data ∈ fast memory) ⇒ low latency access miss (data ∉ fast memory) ⇒ long latency access (DRAM) Fast mem. effective only if bandwidth requirement at B << A October 3, 2005 Management of Memory Hierarchy 6.823 L...
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533MHz, 32-bit bus, 2. • 1GB/s1GHz, 2x32-bit bus, 16GB/s • Up to 8GB DRAM, 400MHz, 128-bit bus, 6.4GB/s • North Bridge Chip • PCI-X Expansion, 133MHz, 64-bit bus, 1 GB/s October 3, 2005 18 Five-minute break to stretch your legs Inside a Cache Address Address Processor CACHE Data Data Main Memory copy o...
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7 Fully (2-way) Set Associative Associative Direct Mapped block 12 can be placed anywhere anywhere in set 0 (12 mod 4) only into block 4 (12 mod 8) October 3, 2005 Direct-Mapped Cache Tag Index Block Offset t V Tag k Data Block b t = HIT October 3, 2005 2k lines Data Word or Byte Direct ...
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• FIFO with exception for most recently used block This is a second-order effect. Why? October 3, 2005 Block Size and Spatial Locality 6.823 L7- 27 Joel Emer Block is unit of transfer between the cache and memory Tag Word0 Word1 Word2 Word3 Split CPU address block address 4 word block, b=2 offset b 32-b b...
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005 6.823 L7- 31 Joel Emer Write Policy • Cache hit: – write through: write both cache & memory • generally higher traffic but simplifies cache coherence – write back: write cache only (memory is written only when the entry is evicted) • a dirty bit per block can further reduce the traffic • Cache miss: – n...
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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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intervals ΔT , can not be unambiguously reconstructed from its sample 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 cap...
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nΔT + φ where m is an integer, giving the following important result: Given a sampling interval of ΔT , sinusoidal components with an 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 ...
https://ocw.mit.edu/courses/2-161-signal-processing-continuous-and-discrete-fall-2008/2934569026a399e9062597924c730a78_lecture_10.pdf
6) (cid:3) (cid:3) (cid:4) (cid:6) (cid:3) (cid:3) (cid:4) (cid:6) (cid:3) (cid: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 ch...
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, there is nothing that can be done to eliminate the effects of aliased frequency components. The only way to guarantee that the sample set unambiguously represents the generating function is to ensure that the sampling theorem criteria have been met, either by 10–3 (cid:6) (cid:5) (cid:6) (cid:7) (cid:7) (cid:6) (...
https://ocw.mit.edu/courses/2-161-signal-processing-continuous-and-discrete-fall-2008/2934569026a399e9062597924c730a78_lecture_10.pdf
(cid:10) (cid:6) (cid:8) (cid:8) (cid:15) (cid:25) (cid:12) (cid:11) (cid:16) (cid:17) (cid:20) (cid:14) (cid:15) (cid:3) (cid:8) (cid:6) (cid:5) (cid:9) (cid:11) (cid:9) (cid:12) (cid:13) (cid:9) (cid:9) (cid:14) (cid:2) (cid:10) (cid:4) (cid:8) (cid:2) (cid:29) (cid:3) (cid:4) (cid:2) (cid:5) (cid:15) (cid:15) ...
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) (cid:6) (cid:3) (cid:8) (cid:4) (cid:6) (cid:11) (cid:3) (cid:8) (cid:4) (cid:6) (cid:2) If it is assumed that the sampling theorem was obeyed during sampling, the repetitions in F (cid:2)(j Ω) will not overlap, and in fact f (t) will be entirely specified by a single period of F (cid:2)(j Ω). Therefore to recon...
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(cid:17) (cid:6) (cid:11) (cid:24) (cid:23) (cid:17) (cid:19) (cid:21) (cid:13) (cid:8) (cid:30) (cid:19) (cid:21) (cid:13) (cid:10) (cid:4) (cid:8) (cid:2) (cid:12) (cid:7) (cid:3) (cid:8) (cid:4) (cid:6) (cid:3) (cid:8) (cid:4) (cid:6) (cid:2) (cid:3) (cid:4) (cid:2) (cid:5) (cid:15) (cid:7) (cid:4) (cid:8) (ci...
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3) " (cid:4) (cid:6) (cid:23) (cid:4) (cid:6) (cid:12) (cid:4) (cid:6) " (cid:4) (cid:6) $ (cid:4) (cid:6) # (cid:4) (cid:6) (cid:2) and note that the impulse response h(t) = 0 at times t = ±nΔT for n = 1, 2, 3, . . . (the sampling times). The output of the reconstruction filter is the convolution of the input ...
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(cid:16) (cid:17) (cid:20) (cid:3) (cid:9) (cid:9) (cid:17) (cid:9) (cid:17) (cid:15) (cid:15) or in the case of a finite data record of length N f (t) = N −1 � n=0 fn sin (π(t − nΔT )/ΔT ) π(t − nΔT )/ΔT . This is known as the cardinal (or Whittaker) reconstruction function. It is a superposition of shifted sin...
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now look at a different formulation of F ∗(j Ω). The Fourier transform of the sampled function f (cid:2)(t) � ∞ F (cid:2)(j Ω) = f (cid:2)(t) e−j Ωtdt = � ∞ ∞ � f (t)δ(t − nΔT ) e−j Ωt dt −∞ � ∞ = −∞ n=−∞ f (nΔT ) e−j ΩnΔT n=−∞ 10–6 by reversing the order of integration and summation, and using the siftin...
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(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) (cid:25) (cid:15) (cid:7) (cid:15) (cid:15) (cid:4) (cid:8) (cid:2) (cid:5) (cid:7) (cid:3) (cid:8) (cid:4) (cid:6) (cid:3) (cid:8) (cid:6) (cid:3) (cid:8) (cid:4) (cid:6) ...
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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) (cid:12) (cid:2) (cid:4) (cid:10) • The DFT operations are a transform pair between two sequenc...
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m} {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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of integers. But who cares? That’s why Turing brought up the halting problem; we actually 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 comp...
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hard relative to which other problems. It can tell us things like “this problem can’t be solvable, because if it were, it would allow us to solve this other unsolvable problem”. Given: 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 wr...
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one by one in order, and then pause when we find the solution. However, if there is no solution, it would just go on forever. So our question to the oracle would be, “Does this program halt or not?”, and if so, the Diophantine equation would be solvable. If we solve the Diophantine problem, can we also then solve the...
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, if you could solve the halting problem, then can you decide whether a set of tiles will tile the plane or not? Can you tile a 100x100 grid? Can you tile a 1000x1000 grid? These questions can be answered by a Turing machine. But suppose every finite region can be filled, why does it follow that the whole infinite plan...
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subtrees must be infinite, and we can keep going on forever. The end result is an infinite path. So how does K¨onig’s Lemma apply to the tiling problem? Imagine that the tree is the tree of possible choices to make in the process of tiling the plane. Assuming you’re only ever placing a tile adjacent to existing tiles...
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same as the original halting problem. We can follow each machine step by step and ask the oracle, but the oracle won’t have the answer for the Super Halting Problem. If there were the Super Turing Machine that could solve the Super Halting Problem, then you could feed that Super Turing Machine itself as input and c...
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1930, five years before Turing invented Turing machines, G¨odel showed that this was impossible. G¨odel’s theorems were a direct inspiration to Turing. G¨odel’s Incompleteness Theorem says two things about the limits of any system of logic. First Incompleteness Theorem: Given any system of logic that is consistent (c...
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He started out with the paradox of the liar: “This sentence is not true.” It can’t be either true or false! So if we’re trying to find an unprovable statement, this seems like a promising place to start. The trouble is that, if we try to express this sentence in purely mathematical language, we run into severe proble...
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we can define sentences that talk about their own provability. The end result is that “This sentence is not provable” gets “compiled” into a sentence purely about integers. What about the Second Incompleteness Theorem? Given any reasonable logical system S, let G(S) = “This sentence is not provable in S” Con(S) = “...
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provable in S. The system will never be able to prove its own consistency. 5-6
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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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Propositional Logic Syntax Components Formation Rules The PL language consists of � an infinite set of variables V = {a, b, . . .}, and � a set of symbols S = {∼, ∨, (, )}. 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 exp...
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6 Staal A. Vinterbo (HST/DSG/HMS) Fuzzy Stuff HST 951/MIT 6.873 8 / 56 Propositional Logic Semantics Propositional Logic Semantics Propositional Logic Semantics Setting: variable value assigments Propositional Logic Semantics Interpretation: Truth Value of Expressions Definition We define a setting s as a ...
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∼ {0, 1} → {0, 1} ∨ : {0, 1} × {0, 1} → {0, 1} ∼ (0) = 1 ∼ (1) = 0 ∨ 0 1 0 0 1 1 1 1 Table: Truth table for disjunction ∨ Staal A. Vinterbo (HST/DSG/HMS) 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 Propositio...
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1/MIT 6.873 14 / 56 Propositional Logic Semantics Propositional Logic Semantics Propositional Logic Semantics Example: Computing the Interpretation I Example I(∼ (∼ a∨ ∼ b)) = ∼ I(∼ a∨ ∼ b) = ∼ (∼ I(a)∨ ∼ I(b)) = ∼ (∼ s(a)∨ ∼ s(b)) If we let s(a) = 1 and s(b) = 0, then Propositional Logic Syntactic “Sugar...
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only if Is (α) = 1 for every setting s. A wff α is satisifiable if there exists a setting s such that Is (α) = 1, and unsatisfiable if no such setting s exists. Example The wff (α∨ ∼ α) is valid, while (α∧ ∼ α) is unsatisifiable. Propositional Logic Testing for validity: Truth Table Method The truth table for (a → ...
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( q ∧ r ))) → q ) 1 6 1 5 1 8 1 7 1 9 0 1 0 3 Staal A. Vinterbo (HST/DSG/HMS) Fuzzy Stuff HST 951/MIT 6.873 19 / 56 Staal A. Vinterbo (HST/DSG/HMS) Fuzzy Stuff HST 951/MIT 6.873 20 / 56 Propositional Logic Validity and Satisfiability Propositional Logic The PL Logic System Propositional Logic Examp...
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1 / 56 Staal A. Vinterbo (HST/DSG/HMS) Fuzzy Stuff HST 951/MIT 6.873 22 / 56 Propositional Logic The PL Logic System Propositional Logic The PL Logic System Propositional Logic The PL Logic System: Uniform Substitution Propositional Logic The PL Logic System: Modus Ponens � The result of uniformly replacin...
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’ ⇒’. As in: US: MP: � α ⇒� α[β1/a1, β2/a2, . . . , βn/an] � α, (α → β) ⇒� β Clear: we can manipulate wff by using the rules defining operators and semantics. Definition The wff β is a propositional consequence of wff α if and only if α ↔ β ∧ γ for some wff γ. We formulate this as a derived transformation rule:...
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not study, Alf has a good time. If Alf does not get good grades, Alf does not have a good time. What can we say about Alf? Staal A. Vinterbo (HST/DSG/HMS) Fuzzy Stuff HST 951/MIT 6.873 27 / 56 Staal A. Vinterbo (HST/DSG/HMS) Fuzzy Stuff HST 951/MIT 6.873 28 / 56 Propositional Logic Propositional Consequenc...
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.873 30 / 56 Propositions over Sets Propositions over Sets Propositions over Sets Propositions: Defined Propositions over Sets Syntax Now, a proposition over a set is a proposition that describes a property of the elements of that set. Such propositions are modeled by characteristic functions. Example Let N ...
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� T (β). Analogous to the PL case: � T (α ∧ β) = T (α) ∩ T (β), � T (α → β) = (U − T (α)) ∪ T (β), and � T (α ↔ β) = ((U − T (α)) ∪ T (β)) ∩ ((U − T (β)) ∪ T (α)). Example For the natural numbers and the proposition even the truth set is T (even) = {2, 4, 6, . . .}. Staal A. Vinterbo (HST/DSG/HMS) Fuzzy Stuff ...
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MS) Fuzzy Stuff HST 951/MIT 6.873 35 / 56 Staal A. Vinterbo (HST/DSG/HMS) Fuzzy Stuff HST 951/MIT 6.873 36 / 56 Propositions over Sets Propositional Rules Propositions over Sets PL(U) ⊇ PL Propositional Rules The implication view We state that PL is “contained in” PL(U). Indeed, PL is contained in PL({0...
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Tall ∧ HairDark) (HeightTall ∧ HairDark) → LookHandsome (LookHandsome) In other words we set I(β, x ) = Effect: We infer the unknown proposition LookHandsome. Definition Given a rule (α the computation of I(β, x ) as I(α, x ). → β). The application of this rule to a data point x is � I(α, x ) = 1, 1 0 otherwi...
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51/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∩B (x ) = min(χA(x...
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(χR (x , y ), χR� (y , z))} Staal A. Vinterbo (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 relati...
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7, .8)} 1, .5, .7} = . = max{. 7 Staal A. Vinterbo (HST/DSG/HMS) Fuzzy Stuff HST 951/MIT 6.873 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: Fo...
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actic Sugar” Fuzzy Logic Examples Example Analogous to the PL(U) case we can show that � µT (α∧β)(x ) = min(µT (α)(x ), µT (β)(x )), � µT (α→β) = max(1 − µT (α)(x ), µT (β)(x )), and � µT (α β)(x ) = ↔ min(max(1 − µT (α)(x ), µT (β)(x )), max(1 − µT (β)(x ), µT (α)(x ))). Definition (Fuzzy Interpretation) If w...
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a setting such that its interpretation is 1. Staal A. Vinterbo (HST/DSG/HMS) Fuzzy Stuff HST 951/MIT 6.873 53 / 56 Staal A. Vinterbo (HST/DSG/HMS) Fuzzy Stuff HST 951/MIT 6.873 54 / 56 Summary Summary Summary Propositions over sets We have learned � about the propositional language PL(U), over propositio...
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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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One can show that this is true as long as or Massachusetts Institute of Technology 6.763 2005 Lecture 10 3 Flux Quantization 3. Take the line integral of the supercurrent equation around a closed contour within a superconductor: The line integral of each of the parts: Therefore, flux integer n = - n’ Mass...
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is quantized in the bulk limit. Massachusetts Institute of Technology 6.763 2005 Lecture 10 5 Flux Quantization Experiments Flux trapped in hollow cylinder Deaver and Fairbank, 1961, measure and show that q*=2e; Cooper Pairs. Image removed for copyright reasons. Please see: Figure 5.3, page 249, from Orland...
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. The MQM wavefunction for the superconductor is the spatial average of this phase coherent wavefunction and is preserved with an applied field. The coherence persists over the macroscopic scale of the superconductor. In the normal state, the applied field causes dissipation; this energy loss causes the phase of t...
https://ocw.mit.edu/courses/6-763-applied-superconductivity-fall-2005/29bb21252c2fd3293111508de473d2ac_lecture10.pdf