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. ε < |y| < xn, where K cancels on whole spheres. 2. xn < |y| < R for some large R, which is a bounded region on which K is well-behaved. 3. |y| > R, on which the hemisphere cancellation of K is useful. The details of the argument are left to the homework. 3 MIT OpenCourseWare http://ocw.mit.edu 18.156 Differential An...
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Linear Spaces we have seen (12.1-12.3 of Apostol) that n-tuple space has the following properties : V, Addition: 1. (Commutativity) A + B = B + A. 2. (Associativity) A + (B+c) = (A+B) + C. 3. (Existence of zero) There is an element -0 such 'that A + -0 = A for all A. 4. (Existence of negatives) Given A, there is...
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will be concerned only with n-tuple space and with certain of its subsets called "linear subspaces" : Vn -Definition. Let W be a non-empty subset of Vn ; suppose W is closed under vector addition and scalar multiplication. Then W is called a linear subspace of V n (or sometimes simply a subspace of Vn .) To say W ...
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subset of Vn . consisting of all vectors X of the form X = cA is a subspace of 'n It is called the subspace spanned by A. In the case n = 2 or 3 , it can be pictured as consisting of all vectors lying on a line through the origin. Example 3. Let A and B be given non-zero vectors that are not 1 parallel. The su...
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cl+dl)A1 f * * * + (ck+dk)Akr ax = (ac ) A + * * * + (ack)Akf 1 1 so both X + Y and ax belong to W by definition. Thus W is a subspace of Vn. - Giving a spanning set for W is one standard way of specifying W. Different spanning sets can of course give the same subspace. Fcr example, it is intuitively clear that, f...
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so are X + Y and cX. It is easy to see that W is the linear span of (1,0,0) I and (O,l,O). Example 6. The subset of V3 consisting of all vectors of the form X = (3a+2b,a-b,a+7b) is a subspace of V3. It consists of all vectors of the form X = a(3,1,1) + b(2,-1,7), so it is the linear span of 3 1 , and (2,-1,7). ...
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? n n n 2. In each of the following, let W denote the set of I I all vectors (x,y,z) in Vj satisfying the condition given. (Here we use (x,y,z) instead of (xl,x2,x3) for the general element of V3.) Determine whether W is a subspace of V j . If it is, draw a picture of it or describe it geometrically, and find ...
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c A + ... + ck+ 1 1 for some scalars ci. If S spans the vector X, we say that S spans X uniquely if the equations X = ciAi i=l and imply that ci = di for all i. It is easy to check the following: Theorem 1, Let S = < A ~ ,.. . be a set of vectors of Vn; let X be a vector in L(S). Then S spans X uniquely if a...
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A] ,...,%J of vectors of V is said to n be linearly independent (or simply, independent) if it spans the zero vector I uniquely. The vectors themselves are also said to be independent in this situation. If a set is not independent, it is said to be dependent. Banple 8. If a subset T of a set S is dependent, the...
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set of vectors is called an orthonormal set. The coordinate by the vectors Ai vectors E l form such a set. n B.mple A set ccnsisting of a single vector A is independent if A # Q. A set consisting of two non-zero vectors ARB is independent if and I only if the vectors are not parallel. More generally, one ha...
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of the vectors Ai . as follows: We do so, using a "double-indexing" notation for the coefficents, j Multiplying the equation by x and summing over j, and collecting terms, we j have the equation In order for <x .B to equal 2 , it will suffice if we can choose the x j so that coefficient of each vector Ai in t...
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less than n, then the solution space con- tains some vector other t2ian 0. I'ronf.. We are concer~tcd here only v i t h proving the existence of some solutioli otlicr tJta11 0, not with nctt~nlly fitidirtg such a solution it1 practice, nor wit11 finditig all possildd solutiot~s. (We 11-ill study the practical prob-...
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the coefiicielits of st is nonzero, attd 1t.e rncly s~tj~poscfor cortvetlier~ce that the equations have beerr arranged so that this happetls ill the first ec~uation, with the result that 0 1 1 + 0. We rnultiply the first crlt~ation 1)y the scalar azl/afl and then suhtract i t from the second, eli~nitiat~itlg tghe X...
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io~lo t l ~ c rthan the zcro vcctor). $Ire prove this as follows: Sr~ppose(d2, . . . , d.,) is a solutiolr of t l ~ o stna1.ller system, different from , . . . ,. We su1)stitutc itito tllc first equation and solve for XI,thereby obtailiirlg the follo~ving vector, w1ricI1 yo11 may verify is a s o l ~ ~ t ~ i o nof t...
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it to you to show that this statement holds.(Be sure you ccnsider the case where one or more or all of the coefficents are zero.) a -E2ample 13. We have already noted that the vectors El,...,E span all It. follows, for example, that any three vectors in V2 are dependent, n of Vn. that is, one of them equals a lin...
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depend ecce of A , ..A for A i # 0, we can solve this equation contradicting the fact that Ai+l does not belong to L(Alt....A,). r while if c ~ + ~ 1 Cc~ntinuing the process just described, we can find larger and larger independent sets of vectors in W. The process stops only when the set we obtain spans W. Does...
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0 alone, then the dimension of W: is zero. Example 14. The space Vn has a "naturaln basis consisting of the vectors E1,...,E . It follows that Vn has dimension n. (Surprise!) There n are many other bases for Vn., For instance, the vectors form a basis for V,, as you can check. I I 1 i ~xercises 1. Consi...
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ose All.. .,A,, is a basis for V; let B1p.afBk be arbitrary vectors of W. (a) Slim? there exists a linear transformation T : V + W such that T(Ai) = Bi fc>r all i. (b) Show this linear transformation is unique. 7. L e t W be a subspace of Vn: let Al,...,% be a basis for W. Let X, Y be vectors of W. Then X = 2 xjAi ...
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1 + " CmBm Bnt+l Show that Bmcl is different from 2 and that L(B1,. ..,B ,B L(B~,...,B ,A orthogonal to each of B ..,B, . m m+l + ) - Then show that the ci mmy be so chosen that Bm+l is m m+l) = Steep 2. Show that if W is a subspace of Vn of positive dimension. then W has a basis consisting of vectors that are m...
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of another row, say rcm m. (3) Multiply row i of A by a non-zero scalar. These operations are called the elcmentary & operations. Their usefulness cclmes from the following fact: Theorem 6. Suppose B is the matrix obtained by applying a sequence of elementary row operations to A,successively. Then the row spaces ...
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first column of your matrix. I (I) If this column consists entirely of zeros, nothing needs to ba I done. Restrict your attention now to the matrix obtained by deleting the first column, and begin again. % (11) If this column has a non-zero entry, exchange rows if necessary to bring it to the top row. Then add ...
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appear at the "inside cornersw of the stairsteps are often called the pivots in the echelon form. Yclu can check readily that the non-zero rows of the matrix B are independent. (We shall prove this fact later.) It follows that the non-zero rows of the matrix B form a basis for the row space of B, and hence a basis ...
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The final step is to multiply each non-zero row by an appropriate non-zero scalar, chosen so as to make the pivot entry into 1. This we can do, because the pivots are non-zero. At the end of this process, the matrix is in what is called reduced echelon form. The reduced echelon form of the matrix C above is the m...
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Proof. Let D and D 1 be two reduced echelon matrices, w5ose rows span the same subspace W of Vn. We show that D = D'. Let R be the non-zero rows of D ; and suppose that the pivots (first non-zero entries) in these rows occur in columns jl,...,j k t respectively. ( a ) =ow that the pivots of D 1 wrur in the colw...
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,, ' The equation ( * ) is often called a parametric equation for the line, and t is called the parameter in this equation. As t ranges over all real numbers, the corresponding point X ranges over all points of the line L. When t = 0, then X = P; when t = 1, then X = P + A; when t = $, then X = P + %A; and so on. ...
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0 . We can solve these equations for P in terms of Q and B: P = Q + t 2 B - tlA = Q + (t2-tl&JB. Now, given any point X = P + tA of the line L(P;A), WE can write X = P + tA = Q + (t2-tlc)B+ tcB. Thus X belongs to the line L (Q;B). Thus every point of L (P;A) belongs to L (Q:B) . The symmetry of the argument show...
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i s p a r a l l e l t o L. By Theorem 8 , any o t h e r l i n e c o n t a i n i n g Q and p a r a l l e l t o L i s e q u a l t o t h i s one. 0 Theorem - 11. Given - two d i s t i n c t p o i n t s P - and Q , t h e r e -i s e x a c t l y --one l i n e c o n t a i n i n g -them. P r o o f . L e t A = Q - P ; t...
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i s a p o i n t of Vn B a r e independent v e c t o r s of V n r w e def i n e t h e p l a n e through P determined 2 A -and B t o be t h e s e t of a l l p o i n t s X of and i f . A and t h e form where s and t run through a l l r e a l numbers. We denote t h i s p l a n e by M ( P ; A , B ) . The e q u a t ...
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subspace of Vn span&: by A and B. J u s t as f o r l i n e s , a p l a n e has many d i f f e r e n t p a r a m e t r i c r e p r e s e n t a t i o n s . More p r e c i s e l y , one h a s t h e following theorem: Theorem 12. - The p l a n e s M ( P ; A , B ) - and M(Q:C,D) - are have a p o i n t - i n common and ...
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of C and D. Symmetry shows that C and D lie in the linear span of A and B as well. Thus these linear spans are the same. Conversely, suppose t h a t t h e p l a n e s i n t e r s e c t i n a p o i n t R and t h a t L(A,B) = L(C,D). Then P + s l A + t B = 1 R = Q + s 2 C + t 2 D for some scalars si ard ti. We can...
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s a y t h e p l a n e s M(P;A,B) and M(Q;C,D) a r e . p a r a l l e l i f L(A,B) = L(C,D). Corollary 13. TKO distinct parallel planes cannot intersect. corollary - 14. Given - a plane M - - and a point Q, there -is exactly one - plane containinq Q -- that is parallel - to M . Proof. Suppose M = M ( P ; A , B )...
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P and B = R - P; then A and B are independent. The plane M(P; A,B) cGntains P and P + A = Q and P + B = R * Now suppose M(S;C,D) is 'another plane containing P, Q, and R. Then 'I . Subtracting, we see that the vectors for some scalars s Q - P = A and R - P = B belong to the linear span ~f f2and D. By and ti i...
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L be the line in Vj through the points P = (1.0.2) and Q = (1,13). Let L1 be the line through 2 parallel to the vector A = ( 3 - 1 ) Find parametric equations for the line that intersects both L I and Lf and is orthogonal to both of them. Parametric equations for k-planes Vn. 432 Following t h e p a t t e r n f...
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t h e k - p l a n e M (P;Al,. . .,Ak' i f and o n l y i f X - P i s i n t h e l i n e a r s p a n o f A1, ...,Ak. - Note that if P = 2, then t h i s k-plane is just the k- dimensional 1 linear subspace of Vn s p ~ e dby A1, ...,%. J u s t a s w i t h t h e c a s e o f l i n e s ( 1 - p l a n e s ) and p l a n e s...
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t h e r e -i s e x a c t l y -one k - p l a n e -i n V, Q , p a r a l l e l - t o M.. Vn 5 p o i n t c o n t a i n i n g Q . ..- 4 -- Lemma 19. Given points POt..,Pk in Vn, they are contained i n a plane of dimension l e s s than k i f and only i f the vectors I P1 - Po,..., Pk- Po are dependent. Theor...
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both the point P and the point Q = (-1,0,2)? 4. Given the 2-plane MI. in V4 containing the points P = (1,-I, 2,-1) and Q = (O.l,l,O)wd R = (lrl,0,3).Find parametric equations for a 3-plane in Vq that containsthe point S = (l,l,l.) and is parallel to MI. Is it unique? Can you find such a 3-plane thatcontains both S ...
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Verilog L5: Simple Sequential Circuits and Verilog L5: Simple Sequential Circuits and Acknowledgements: Materials in this lecture are courtesy of the following sources and are used with permission. Nathan Ickes Rex Min L5: 6.111 Spring 2006 Introductory Digital Systems Laboratory 1 Key Points from L4 (Sequential Bl...
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logic,cd: contamination or minimum delay through logic network L5: 6.111 Spring 2006 Introductory Digital Systems Laboratory 3 System Timing (I): Minimum Period System Timing (I): Minimum Period CLout D Q Clk In D Q Combinational Logic CLK IN FF1 CLout Clk Th Tsu Tcq Tlogic Tcq,cd Tl,cd Th Tsu Tcq Tcq,cd Tsu2 T > Tc...
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Spring 2006 Introductory Digital Systems Laboratory 6 Importance of the Sensitivity List Importance of the Sensitivity List (cid:132) The use of posedge and negedge makes an always block sequential (edge-triggered) (cid:132) Unlike a combinational always block, the sensitivity list does determine behavior for synthes...
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ilog supports two types of assignments within always blocks, with subtly different behaviors. (cid:132) Blocking assignment: evaluation and assignment are immediate always @ (a or b or c) begin x = a | b; y = a ^ b ^ c; z = b & ~c; 1. Evaluate a | b, assign result to x 2. Evaluate a^b^c, assign result to y 3. Evaluate...
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6 Introductory Digital Systems Laboratory 10 Use Use Nonblocking for Sequential Logic Nonblocking for Sequential Logic always @ (posedge clk) begin always @ (posedge clk) begin q1 <= in; q2 <= q1; out <= q2; end q1 = in; q2 = q1; out = q2; end “At each rising clock edge, q1, q2, and out simultaneously receive the o...
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c; end endmodule Nonblocking Behavior a b c x y Deferred module nonblocking(a,b,c,x,y); (Given) Initial Condition a changes; always block triggered x <= a & b; y <= x | c; Assignment completion 1 1 0 1 1 0 1 0 1 1 0 1 0 1 1 x<=0 0 1 0 1 1 x<=0, y<=1 0 1 0 0 1 input a,b,c; output x,y; reg x,y; always @ (a or b or c) ...
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Introductory Digital Systems Laboratory 14 Verilog The Ripple Counter in Verilog The Ripple Counter in Single D Register with Asynchronous Clear: module dreg_async_reset (clk, clear, d, q, qbar); input d, clk, clear; output q, qbar; reg q; always @ (posedge clk or negedge clear) begin if (!clear) q <= 1'b0; else q <...
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will retained by a D Register (cid:132) Next value of counter (N) computed by combinational logic C3 C2 C1 N3 N2 N1 0 1 0 0 0 1 0 0 1 1 1 0 1 1 1 0 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 0 1 1 0 0 1 1 0 0 0 0 1 1 1 1 0 N1 := C1 N2 := C1 C2 + C1 C2 := C1 xor C2 N3 := C1 C2 C3 + C1 C3 + C2 C3 := C1 C2 C3 + (C1 + C2 ) C3 := (C1 ...
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1111 (conditioned by T); used for cascading counters Synchronous CLR and LOAD If CLRb = 0 then Q <= 0 Else if LOADb=0 then Q <= D Else if P * T = 1 then Q <= Q + 1 Else Q <= Q 7 10 P T 163 2 6 5 4 3 CLK RCO D C B A QD QC QB QA 15 11 12 13 14 9 1 LOAD CLR 74163 Synchronous 4-Bit Upcounter L5: 6.111 Spring 2006 Introduc...
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Introductory Digital Systems Laboratory 19 Simulation Simulation Notice the glitch on RCO! L5: 6.111 Spring 2006 Introductory Digital Systems Laboratory 20 Output Transitions Output Transitions (cid:132) Any time multiple bits change, the counter output needs time to settle. (cid:132) Even though all flip-flops sh...
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fine up to 8’b11101111: VDD 1 1 1 1 QA QB QC QD T ‘163 RCO DA DB DC DD CL LD P VDD 1 P 0 1 1 1 QA QB QC QD T ‘163 RCO DA DB DC DD CL LD 0 0 0 0 QA QB QC QD T ‘163 RCO DA DB DC DD 0 P CL LD CLK Problem at 8’b11110000: one of the RCOs is now stuck high for 16 cycles! VDD 0 0 0 0 QA QB QC QD T ‘163 RCO DA DB DC DD CL LD P...
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24 Summary Summary (cid:132) Use blocking assignments for combinational always blocks (cid:132) Use non-blocking assignments for sequential always blocks (cid:132) Synchronous design methodology usually used in digital circuits (cid:134)Single global clocks to all sequential elements (cid:134)Sequential elements al...
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OpenCourseWare 1. Overview and some basics 9 August 2006 Spoken language conveys not only words, but a wide range of other information about timing, intonation, prominence, phrasing, voice quality, rhythm etc. that is often collectively called spoken prosody. These aspects of an utterance are sometimes called supr...
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. 1992, Lehiste 1987). On the one hand, this string can be produced with a prominence on broke and a phrase boundary just after that word, corresponding to the orthographic representation It BROKE, out in WASHington. <broke1> On the other hand, it can be produced with a prominence on out and a phrase boundary just a...
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the intonation of an utterance, i.e. changes in the pitch that are caused by changes in the frequency of vibration of the vocal folds, often called f0. An f0 that is high in a speaker’s range on a salient syllable can mark a pitch accent, but so can a lower (but still high-for-this-speaker...
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. At this writing, the nature of variation for prosodic categories (and thus for ToBI labels) is not fully understood, and this results in some reasonable disagreement in prosodic parses in some renditions of some utterances, particularly in spontaneous speech. ToBI labelling is a new endeavor compared to, say, phon...
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for the needs of particular sites.) The four labelling tiers each appear in their own window: (1) the Tone tier, for transcribing tonal events (2) the Orthographic tier, for transcribing words (3) the Break-Index tier, for transcribing boundaries between words (4) the Miscellaneous tier, for recording addition...
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a constriction in the vocal tract), and zero (or nearly zero) when there is no speech signal. The rate of vibration of the vocal folds is what we hear as the pitch, and this is represented in the second window as a semi-continuous blue line superimposed on a different representation of the speech signal, the spectr...
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perienced ToBI labellers often use the spectrogram to help find the location of successive sounds, syllables and words. This in turn helps them to keep track of where changes in the pitch track occur across the words and syllables of the utterance. Learning how to use this information may take some time, so at the b...
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end of each word for the subjective strength of its association with the next word, on a scale from 0 (for weakest perceived boundary/strongest perceived conjoining, as in doncha for don’t you) to 4 (for the most disjoint boundary, i.e. at the end of the highest-level intonationally marked phrase). These categories ...
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correspond to which words in the utterance. The Miscellaneous tier is the bottom white box in this display. It is essentially a 'comment' tier that can be used to mark events such as breaths, coughs, laughter, long silences and other non- speech events. These are traditionally marked with angle brackets (e.g. <cough...
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of prosody which are predictable from other parts of the transcription or from auxiliary tools, such as dictionaries, that can be used to determine the location of lexical stress within words. The categorical aspects of prosody which we try to capture completely (according to the first principle) are of two types. ...
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aspect of prosody which we leave out (in accordance with the third principle) because it should be fairly predictable is the marking of the lexically stressed and unstressed syllables within each word. By this level of stress we mean the word-internal alternation between more stressed and less stressed syllables, wh...
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MIT OpenCourseWare http://ocw.mit.edu 6.013/ESD.013J Electromagnetics and Applications, Fall 2005 Please use the following citation format: Markus Zahn, 6.013/ESD.013J Electromagnetics and Applications, Fall 2005. (Massachusetts Institute of Technology: MIT OpenCourseWare). http://ocw.mit.edu (accessed MM DD, YYYY...
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��� 0 k ¯ × k ¯ × E ¯ˆ = k ¯ k ¯ · E ¯ˆ − E¯ˆ(k ¯ · k¯) = ωµ k ¯ × H ¯ˆ = −ω2�µE ¯ˆ � � 2 |k¯ | = kx 2 = ω2�µ A ¯ × (B ¯ × C¯) = B¯(A ¯ C¯) − C¯(A ¯ B¯) · 2 + ky · 2 + kz S ¯ˆ = ˆ¯ S = 1 ¯ˆ H¯ˆ ∗, H ¯ˆ = E × 2 � 1 ωµ (¯ E¯ˆ) k × ˆ E ¯ × 1 2 � 1 k × Eˆ∗ = ¯ ¯ ωµ 1 2ωµ � ¯ E ¯ · E¯∗) − ¯ Eˆ · k¯) k( ...
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j(ωt−kxr x+kzrz)¯iy � H¯ r = Re � � Eˆr (cos(θr)¯ix + sin(θr)¯iz)ej(ωt−kxr x+kzrz) η Boundary conditions require that kxr = k sin(θr), kzr = k cos(θr) 2 From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission. Eˆy(x, z = 0) = 0 = Eˆyi(x, z = 0) + Eˆyr(x, z = 0...
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ix − sin(θ) e−jkz z + e +jkz z ¯iz ej(ωt−kxx) � � � � � � � = 2Ei [cos(θ) sin(kzz) sin(ωt − kxx)¯ix − sin(θ) cos(kzz) cos(ωt − kxx)¯iz] H ¯ = Re � ˆ � Ei e−jkz z + e +jkz z ej(ωt−kxx)¯iy η � � = 2Ei cos(kzz) cos(ωt − kxx)¯iy η Kx(x, z = 0) = Hy(x, z = 0) = 2Ei cos(ωt − kxx) η σs(x, z = 0) = −�Ez(x, z = 0) =...
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= − Eˆt cos(θt)e−jkxtx 1 η2 kxi = kxr = kxt ⇒ k1 sin(θi) = k1 sin(θr) = k2 sin(θt) θi = θr k1 k2 sin(θt) = sin(θi) = ωc2 � ωc1 � sin(θi) = c2 c1 sin(θi) (Snell’s Law) Index of refraction: Reflection Coefficent: R = Transmission Coefficent: T = Eˆt Eˆi B. Brewster’s Angle of No Reflection �rµr √ √ η2 √ ...
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) c1 � 2 2c2 η1 sin2(θi) 2 c1 � µ1 �1µ1 �� �1 �2µ2 �� sin2(θi) 2 cos2(θt) = η1 � − η2 = η1 2 � 2 − η2 2 − µ2 �2 = µ1 �1 − µ2 �2 sin2(θi) = sin2(θB) = � µ 1 − � 2 1 µ 1 2 � �2 µ 1 − µ 1 2 θB is called the Brewster angle. There is no Brewster angle for TE polarization if µ1 = µ2. If c2 > c1, ...
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1 1 + �1 �2 θB + θt = π 2 + 1 ⇒ θC > θB 1 sin2(θB ) = 1 sin2(θC ) 8
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6 . 2 7 0 : A U T O N O M O U S R O B O T D E S I G N C O M P E T I T I O N • Assignment 2: General Comments • More on sensors Servos • • RF receiver • Robot control and state machines Threads Assignment 3 handed out • • LECTURE3: Advanced Techniques Delinquent Teams • Assignment 2 teams not finished: –...
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• p = 1800 W – Thirty 60-watt light bulbs • Lesson: ensure battery leads are well-insulated! i v R Phototransistors • It’s an art • Need to figure out an effective way of reading the color off the board or object – Factors: glossiness, ambient lighting – It’s not really color; it’s grayscale – Contest nigh...
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kΩ Pull-Up • Cut traces 5 4 2 (make sure you know where!) Distance Sensor • Range: 15-150 cm – 6-60 in Distance Sensor • You probably don’t need more than 3 • But if you’re really thatneedy, cut port 0 or 1 IN_16 IN_17 IN_18 IN_19 IN_20 IN_21 IN_22 IN_23 v = 5 V R = 47 kΩ to ADC VDD IN_0 GND ...
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match begins, and stop it when the match ends Voting Information • rf_vote_red • rf_vote_green • rf_vote_winner – You need some way of determining a winner when there is a tie – All automagically updated Position Information • rf_x0, rf_y0 • rf_x1, rf_y1 – Tell you the x and y coordinates of the two robots –...
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at a given point in time – Each state has predefined outputs – Transitions to other states depend on inputs • Why? – Effective way of thinking about your strategy – Define what to do for any combination of inputs Implementing a State Machine • Each action is a state – Moving forward – Turning • Actuators are o...
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on off drive wheel steering wheel steering and drive wheel phototransistor LED wall follow w all follo w Drive Mechanisms • Differential Drive • Synchro Drive (servos) • Rack-and-Pinion Drive (car) • Independent Drive (gearboxes; Assignment 2) differential steering synchro drive wheel steering wh...
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charge phototransistor LED Shaft Encoding • Works better on some ports: – Ports 7 and 8 have hardware counters (faster, more accurate) – Others use software counters – If you need more than 2, try using ports 2-6 • Both wheels may not turn at same speed • Use revolutions for feedback • Determine difference in...
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to see these things in advance • Test and debug incrementally Hints • Test sensors before mounting • Test small pieces of code before combining into larger procedures • Use the LCD screen • Remember mechanical reliability Error Detection • Your robot will mess up • How can it find out what’s wrong? • Timeouts ...
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true) { go forward wait until sensor pressed go backward wait until sensor pressed } } main() { while (true) { while (vote is tied) play tone 1 while (red is winning) play tone 2 while (green is winning) play tone 3 } } Example move() { while (true) { go forward wait until sensor pressed go backward wait unti...
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{ /* check for anything (like sensor inputs) */ if (you_really_need_to_leave_the_while_loop) break; } Your Winning Strategy • Sufficient sensors and AI to determine location of robot • Be able to react to potential problems that the robot might face • Be aware of your limitations – Amount of LEGO – Power and ...
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Heinrich Hencky (1885-1952) • Natural logarithmic strain measure; • Biography: ( !(t) = ln L(t) L0 )  High School (Humanistic Gymnasium), Speyer am Rhein, Germany  Technical University, Munich; Dipl. Eng. 1908  Technical University, Darmstadt; D. Eng. 1913 • Professor of Mechanical Engineering, M.I.T. 1930- 1933. ...
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Elasticity (normal stress differences)  “Second order fluids” (SOF)  Boger fluids $11 %$22 !" 2 !1( !") # • Fluid Memory (stress relaxation)  Relaxation time λ !0 s s e r t s r a e h S time G(t) = !12 (t) "0 ~ G0e#t $ 3 Natural Time Scale of Complex Fluids • Natural time scale • The Deborah number is a dimensio...
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8.701 Introduction to Nuclear and Particle Physics Markus Klute - MIT 0. Introduction 0.5 Early History and People in Nuclear and Particle Physics 1 Early Developments in Nuclear & Particle Physics ~1820s: geologists and biologists have come to believe that the Earth is much older than 10s of thousands of year, p...
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1856-1950 Marie Curie 1867-1934 Pierre Curie 1859-1906 3 Early Developments in Nuclear & Particle Physics 1899: Paul Villard discovers a third component of radiation from uranium and calls them ɣ rays. 1901: The Curie’s measure the energy emitted by radioactive elements and discover that one gram of radium gives off...
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© Source unknown. All rights reserved. This content is excluded from our Creative Commons license. For more information, see https://ocw.mit.edu/fairuse. Eugene Marsden 1882-1936 Paul Dirac 1902-1984 This photo is in the public domain. Early Developments in Nuclear & Particle Physics 1931: Pauli and Fermi propose t...
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he 1906-2005 7 MIT OpenCourseWare https://ocw.mit.edu 8.701 Introduction to Nuclear and Particle Physics Fall 2020 For information about citing these materials or our Terms of Use, visit: https://ocw.mit.edu/terms.
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:28)(cid:16)%(cid:31) (cid:2)(cid:3)(cid:4)(cid:4)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9)(cid:10)(cid:11)(cid:12)(cid:8)(cid:6)(cid:13) : 2(cid:23)(cid:12)(cid:4)(cid:11)(cid:2)(cid:21)(cid:11)(cid:12)(cid:3)(cid:11)(cid:13)(cid:14)(cid:4)(cid:10)(cid:15)(cid:10)(cid:16)(cid:12)(cid:4)(cid:10)(cid:6)(cid:3)(cid:11)...
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cid:29)(cid:10)(cid:9)(cid:10)(cid:20)(cid:3)(cid:9)/ def greedy(items, maxCost, keyFunction): itemsCopy = sorted(items, key = keyFunction, result = [] totalValue, totalCost = 0.0, 0.0 reverse = True) for i in range(len(itemsCopy)): if (totalCost+itemsCopy[i].getCost()) <= maxCost: result.append(itemsCopy[i]) totalCost...
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:2)(cid:3)(cid:4)(cid:4)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9)(cid:10)(cid:11)(cid:12)(cid:8)(cid:6)(cid:13) &(cid:4) MIT OpenCourseWare https://ocw.mit.edu 6.0002 Introduction to Computational Thinking and Data Science Fall 2016 For information about citing these materials or our Terms of Use, visit: https://ocw....
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AN EXPOSITION OF BRETAGNOLLE AND MASSART’S PROOF OF THE KMT THEOREM FOR THE UNIFORM EMPIRICAL PROCESS R. M. Dudley February 23, 2005 Preface These lecture notes, part of a course given in Aarhus, August 1999, treat the classical empirical process defined in terms of empirical distribution functions. A proof, exp...
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. . . . . . . . 1.4 Proof of Lemma 1.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Proof of Lemma 1.2 . . . . . . . . . . . . . . . . . . . . . . . 1.6 1.7 Proof of Theorem 1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.8 Another way of defining the KMT construct...
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0 ≤ t ≤ 1, with EB(t) = 0 and EB(t)B(u) = t(1 − u) for 0 ≤ t ≤ u ≤ 1. Donsker (1952) proved (neglecting measurability problems) that αn(t) converges in law to a Brownian bridge B(t) with respect to the sup norm. Koml´ ady (1975) stated a sharp rate of convergence, namely that on some probability space there exist Xi...
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with specific constants, Theorem 1.1 below. Bretagnolle and Massart’s proof was rather compressed and some readers have had difficulty following it. Cs¨ o and Horv´ ath (1993), pp. 116-139, expanded the proof while making it more elementary and gave a proof of Lemma 1.4 for n ≥ n0 where n0 is at least 100. The purpose...
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Tusn´ady’s lemmas The main result of the present chapter is: Theorem 1.1. (Bretagnolle and Massart) The approximation (1.1) of the empirical process by the Brownian bridge holds with c = 12, K = 2 and λ = 1/6 for n ≥ 2. The rest of this chapter will give a proof of the theorem. In a preprint, Rio (1991, Theorem 5....
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and 0 < t < 1 let F −1(t) := inf{x : F (x) ≥ t}. Here is one of Tusn´ady’s lemmas (Lemma 4 of Bretagnolle and Massart (1989)). Lemma 1.2. Let Φ be the standard normal distribution function and Y a standard normal random variable. Let Φn be the distribution function of B(n, 1/2) and set Cn := Φ−1(Φ(Y )) − n/2. Then ...
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that 0 ≤ j ≤ n and n + j is even, we have √ P (βn ≥ (n + j)/2) ≥ P ( nY /2 ≥ n(1 − 1 − j/n)), P (βn ≥ (n + j)/2) ≤ P ( nY /2 ≥ (j − 2)/2). (1.5) (1.6) √ (cid:4) Remarks. The restriction that n + j be even is not stated in the formulation of the lemma by Bretagnolle and Massart (1989), but n + j is always even ...
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lemma for n ≥ 8. We have An = exp((12n)−1 − θn/(360n3 )) where 0 < θn < 1, see Whittaker and Watson (1927), p. 252 or Nanjundiah (1959). Then by Taylor’s theorem with remainder, (cid:5) An = 1 + 1 12n + 1 288n2 + 1 6(12n)3 φne (cid:6) 1/12n exp(−θn/(360n 3 )) where 0 < φn < 1. Next, (cid:7) (cid:5) βn...
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2 288n (cid:6) . = 1 + 1 24n − 1 30n2 1 + 1 12n + 1 288n 2 Thus lim inf n→∞ βn ≥ 1 and βn → 1 as n → ∞. To prove βn ≥ βn+1 for n ≥ 8 it will suffice to show that 1 + 1 24(n + 1) + e1/108 6 · 144n2 ≤ 1 + 1 24n − 1 30n2 (cid:7) 1 1 + + 96 (cid:8) 1 288 · 82 3 or e1/108 6 · 144n2 + 1 30n2 (...
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(n k ) will be approximated via Stirling’s formula with correction terms as in Lemma 1.5. To that end, let := 0 for n + i odd. The factorials in (n := i/n. Define pni k )/2n and xi CS(u, v, w, x, n) := 1 + u/(12n) . (1 + v/[6n(1 − x)])(1 + w/[6n(1 + x)]) By Lemma 1.5, we can write for 0 ≤ i < n and n + i even pn...
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n) ≤ CS(βn, βn, βn, 0, n) ≤ CS(1, 1, 1, 0, n) = 1 + (cid:5) (cid:6) (cid:7) 1 12n 1 1 + + 3n (cid:8)−1 . 1 2 36n It will be shown next that log(1 + y) − 2 log(1 + 2y) ≤ −3y + 7y2/2 for y ≥ 0. Both sides vanish for y = 0. Differentiating and clearing fractions, we get a clearly true inequality. Setting y := 1/(...
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. Let j2 ≥ 2n, in other words xj ≥ 2/n. Recall that for t > 0 we have P (Y > t) ≤ (t 2π)−1 exp(−t2/2), e.g. Dudley (1993), Lemma 12.1.6(a). Then (1.10) follows easily when j = n and n ≥ 5. To prove it for j = n − 2 it is enough to show n(2 − log 2) − 4 2n + log(n + 1) + 4 + log[2 2π( n − 2)] ≥ 0, n ≥ 5. √ √ √ √ Th...
https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf
1.13) (1.14) Clearly fn > 0. To see that fn(x) is decreasing in x for 2/n ≤ x ≤ 1 − 4/n, note that (cid:13) 2(1 − x)f (cid:4) √ √ n/fn = 1 − 4n[ 1 − x − 1 + x], √ so fn is decreasing where 1 − x − (1 − x) > 1/(4n). We have √y − y ≥ y for y ≤ 1/4, so √ y − y > 1/(4n) for 1/(4n) < y ≤ 1/4. Let y := 1 − x. Also ...
https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf
√ J(x) := 4(1 − 1 − x)2 − g(x). 5 (1.15) (1.16) (1.17) Then J is increasing for 0 < x < 1, since its first and second derivatives are both 0 at 0, while its third derivative is easily checked to be positive on (0, 1). In light of (1.9), to prove (1.16) it suffices to show that (cid:5) 1 + (cid:6) 1 12n nJ (x...
https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf
) (0) = 7/2, so that the right side of (1.19) is the Taylor series of J around 0 through fourth order. One then shows straightforwardly that J (5) (x) > 0 for 0 ≤ x < 1. It follows since nx2 ≥ 2 and n ≥ 8 that nJ (x)/2 ≥ x/2 + 7/24n. Let K(x) := := (K(x) − 1)/x2 . We will next see that κ(·) is decreasing exp(x/2)/...
https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf