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2k  k  k  k  2k  k  2 2 1 4 ik0x3i0t  ik0x3i0t  2ik0x  k0e k0e 1 2 1 4 1 4 1 2 1 1 2 1 1 ik0xi0t  2 2 2 e 0 0 0 0 0 0 0 1 ik0xi0t  1 2ik0x4i0t   e  2 x 2 e 1 e 2 ik0xi0t  dx   2k0e 2ik0x4i0t  1 2 ik0xi0t k0e  dx   k0e 3ik0x3i0t ...
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for this particle to be flying left (in the opposite direction of x axis)? Here we need to find the probability that particle has negative momentum. Indeed from plane wave expansion we find that our particle can have momentum of k0 since it has a plane wave component eik0 xi0k . The coefficient in front of this...
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   AˆBˆ  BˆAˆ  0 Geometrical interpretation: Recall from your linear algebra recitation and the problem set 1. When two matrices commute they share eigenvectors. Example: Free space Hamiltonian and momentum: Hˆ     2 2     2m x 2 , i      x   x  2 2  2  2m x  i  x  x ...
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i  e x ikx  k  eikx And from the previous lectures we remember that the plane waves are also eigenfunctions for the free space Hamiltonian:   eikx   eikx 2 2 2m x 2 2k 2 2m Fundamental theorem of algebra: If two operators A and B commute one can construct a basis of the state space with eigenfunct...
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measurement error) must obey the following relationship:  x  p   2  xˆ, pˆ 2 x  p  This relationship is called Heisenberg’s Uncertainty Principle. In fact it also holds for the uncertainties of energy and time: E  t   2 Symmetries, conserved quantities and constants of motion – how do we ident...
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fest Theorem: d dt d dt Aˆ  1 Aˆ, Hˆ     i Aˆ t Aˆ x  x    x     1 i Aˆ, Hˆ     x      x Aˆ t    x Consequently in order for a physical quantity to be a constant of motion the corresponding observable has to obey the following relationships: 6  A...
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��       0 Reminder: d dt Hˆ  0  E  const  uE x    x, t   i uE   x e t E  E Schrodinger’s equation:   2    2m    r     i    V   r, t   t r, t   t  . Let’s substitute it into the Schrodinger’s equation above: , t  r       t  i...
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   i  t t    t  r 2  2m  V r  V r  t    r 1    t Note that the left side of the equation only depends on position r and the right side of the equation only depends on time t. This can only be true when both sides of the equation are constant E – for energy. Then the equation above split...
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corresponding eigenfunctions, the solution for this system is a linear combination of all the possible solutions corresponding to different energies:       r, t  CEuE E i E t   r e  , where CE are the coefficients that can be determined from the initial and boundary conditions. This is a very ...
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��  0       x  pˆ, Hˆ   pˆ t   0 d dt pˆ  0 Therefore the momentum is a conserved quantity and its eigenvalues can be used to label the states. Then the unique labels for the eigenfunctions above would be: uE,k  x  e , k  ikx uE,k    eikx x 2mE 2  In fact we represent all the e...
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       1 The eigeenfunctions of the paritty operator all are eitheer odd or evven. f (−x)= f (x) even f f (−x)−= f (x) odd� f Does Hammiltonian foor Simple Haarmonic Osc illator commmute with thee parity operrator? ˆ H x 2 2   2  m2 x 2  HˆH x     2m x 1 22  22     2  x 11 22 ...
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��     x   2 2   2  2m x 2       2m  2  x   mm2 x     1 2 2 2 22  x      2mm 2 2 x 2   m2 x  x  0  1 2    1 2 m2 2   x  x    This meaans that one can always ffind a set of eigenfunctioons commonn to Hˆ and ˆ . In fact, ll...
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6.241 Dynamic Systems and Control Lecture 9: Transfer Functions Readings: DDV, Chapters 10, 11, 12 Emilio Frazzoli Aeronautics and Astronautics Massachusetts Institute of Technology March 2, 2011 E. Frazzoli (MIT) Lecture 9: Transfer Functions Mar 2, 2011 1 / 13 Asymptotic Stability (Preview) We have seen tha...
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1Bu[i] + Du[t] � i=0 or y (t) = C exp(At)x(0) + C � t 0 exp(A(t − τ ))Bu(τ ) dτ + Du(t). However, the convolution integral (CT) and the sum in the DT equation are hard to interpret, and do not offer much insight. In order to gain a better understanding, we will study the response to elementary inputs of a for...
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oli (MIT) Lecture 9: Transfer Functions Mar 2, 2011 4 / 13 Output response to elementary inputs (1/2) Recall that, y (t) = CeAt x(0) + C � t 0 e A(t−τ )Bu(τ ) dτ + Du(t). Plug in u(t) = u0est : y (t) = CeAt x(0) + C � t 0 e A(t−τ )Bu0e sτ dτ + Du0e st �� t � = CeAt x(0) + C e(sI −A)τ dτ e At Bu0 + Du0e st...
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transforms an input e → st G (s) is also known as the transfer function: it into the output G (s)e . st E. Frazzoli (MIT) Lecture 9: Transfer Functions Mar 2, 2011 6 / 13 Laplace Transform The (one-sided) Laplace transform F : C defined as � +∞ F (s) = f (t)e−st dt, → C of a sequence f : R ≥0 → R is for al...
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Lecture 9: Transfer Functions Mar 2, 2011 8 / 13 Matrix geometric series Recall the formula for the sum of a geometric series: k−1 � i m = i=0 1 − mk 1 − m . For a matrix: k−1 � M i = I + M + M 2 + . . . M k−1 . i=0 k−1 � M i (I − M) = (I + M + M 2 + . . . M k−1)(I − M) = I − M k . i=0 i.e., k−1 � M...
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a complex number. The function G : z describes how the system transforms an input z → k into the output G (z)z . k G (z) is also known as the (pulse, or discrete) transfer function: it E. Frazzoli (MIT) Lecture 9: Transfer Functions Mar 2, 2011 10 / 13 Z-Transform The (one-sided) z-transform F : C → C of a seq...
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⎥ ⎢ ⎦ ⎣ 0 1 � C = b0 b1 � . . . bn−1 D = d G (s) = C (sI − A)−1B + D G (s) = bn−1s n−1 + . . . + b0 s n + an−1s n−1 + . . . + a0 + d E. Frazzoli (MIT) Lecture 9: Transfer Functions Mar 2, 2011 12 / 13 Models of discrete-time systems DT DT System DT x[k + 1] = Ax[k] + Bu[k] y [k] = Cx[k] + Du[k] ⎡ ...
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Control Spring 2011 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .
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2.092/2.093 — Finite Element Analysis of Solids & Fluids I Fall ‘09 Lecture 4 - The Principle of Virtual Work Prof. K. J. Bathe MIT OpenCourseWare Su = Surface on which displacements are prescribed Sf = Surface on which loads are applied Su ∪ Sf = S ; Sf ∩ Su = ∅ Given the system geometry (V, Su, Sf ), loads (...
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τij nj = fi Sf on Sf Compatibility: ui = u S i u on Su and all displacements must be continuous. • • Stress-strain laws This is known as the differential formulation. Example Reading assignment: Section 3.3.4 • Equilibrium EA EA d2u + f B = 0 dx2 � du � � � dx x=L = R 2 (a) (b) Lecture 4 The Prin...
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dx = dδu dx du dx 0 External virtual work }| { f Bδudx z Z L + 0 Virtual work due to boundary forces z }| { (cid:12) (cid:12) Rδu (cid:12)L dx are the virtual strains, du where dδu on Su, since we do not know the external forces on Su. To solve EA d2u where d2u dx2 exists ( du where only u is continuous. dx are the rea...
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that for any compatible virtual displacement field imposed on the body in its state of equilibrium, the total internal virtual work is 4 Lecture 4 The Principle of Virtual Work 2.092/2.093, Fall ‘09 equal to the total external virtual work. Note that this variational formulation is equivalent to the differential for...
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Metals and Insulators • Covalent bonds, weak U seen by e-, with EF being in mid-band area: free e-, metallic • Covalent or slightly ionic bonds, weak U to medium U, with EF near band edge – EF in or near kT of band edge: semimetal – EF in gap: semiconductor • More ionic bonds, large U, EF in very large gap, insulator...
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Note that in both photon stimulated promotion as well as thermal promotion, an equal number of holes and electrons are produced, i.e. n=p ©1999 E.A. Fitzgerald 5 Thermal Promotion of Carriers • We have already developed how electrons are promoted in energy with T: Fermi-Dirac distribution Just need to fold this i...
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2m ⎞ 2 1 e ⎟⎟ 2 ⎜⎜ 2 2π ⎝ h ⎠ 3 2 E k b F T E − kb e g T NC EF −Eg n = N C e kbT 7 ©1999 E.A. Fitzgerald Density of Thermally Promoted of Carriers • A similar derivation can be done for holes, except the density of states for holes is used • Even though we know that n=p, we will derive a separate expre...
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the Fermi level sits about in the center of the band gap 3 kbT ⎛ ⎞ 2 (me p or n = ni = 2 ⎟ ⎜ 2 ⎝ 2πh ⎠ * )3 * mv 4 −E g e 2kbT ©1999 E.A. Fitzgerald 9 Law of Mass Action for Carrier Promotion 3 kbT ⎛ ⎞ 2 = np = 4 ni ⎟ ⎜ 2 ⎝ 2πh ⎠ (me * )3 * mh 2 E − kb e g T ; 2 = NC NV e ni k b E − g ...
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− Eg This can be a measurement for Eg For Si, Eg=1.1eV, and let μe and μh be approximately equal at 1000cm2/V-sec (very good Si!) σ~1010cm-3*1.602x10-19*1000cm2/V-sec=1.6x10-6 S/m, or a resistivity ρ of about 106 ohm-m max •One important note: No matter how pure Si is, the material will always be a poor insulato...
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r e t p a h 4 C 4 Matrix estimation Over the past decade or so, matrices have entered the picture of high-dimensional statistics for several reasons. Perhaps the simplest explanation is that they are the most natural extension of vectors. While this is true, and we will see exam- ples where the extension from vectors t...
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� = λj ujvj⊤ , j=1 X , λ1, . . . , λr} r diagonal matrix with positive diagonal entries IR that are orthonormal and V is r} ∈ IRn that are also orthonormal. Moreover, { m × where D is a r U is a matrix with columns a matrix with columns it holds that u1, . . . , u { v1, . . . , vr} ∈ j uj , { AA⊤uj = λ2 and A⊤Avj = λ2 ...
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�Ax . x ∈S n−1 Norms and inner product Let A = aij} { in the following notation. and B = { be two real matrices. Their size will be implicit bij} Vector norms The simplest way to treat a matrix is to deal with it as if it were a vector. In particular, we can extend ℓq norms to matrices: |q = A | a q ij | | 1/q (cid:17)...
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. ∞ ∞ We are going to employ these norms to assess the proximity to our matrix of interest. While the interpretation of vector norms is clear by extension from the vector case, the meaning of “ kop is small” is not as transparent. The B following subsection provides some inequalities (without proofs) that allow a bette...
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However, we will see that in several interesting scenar- ios, some structure is shared across coordinates and this information can be leveraged to yield better prediction bounds. The model Throughout this section, we consider the following multivariate linear regres- sion model: Y = XΘ∗ + E , (4.1) ∈ T is the matrix of...
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franchise. As a result, one may assume that the matrix Θ∗ has each of its T columns that is row sparse and that they share the same sparsity pattern, i.e., Θ∗ is of the form: IRn ∈ ∈ × 0 0 0 0  • • • • 0 0 . . . . . . 0 0 • •  • • 0 0 . . . . . . 0 0 , Θ =          indicates a potentially nonzero entry.  ∗ ...
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matrix Θ∗ is low rank or approximately so. As a result, while the matrix may not be sparse at all, the fact that it is low rank still materializes the idea that some structure is shared across different tasks. In this more general setup, it is assumed that the columns of Θ∗ live in a lower dimensional space. Going back ...
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|0 is small. enables us to use thresholding methods for estimating Θ∗ when However, this also follows from Problem 4.1. The reduction to the vector case in the sGMM is just as straightforward. The interesting analysis begins when Θ∗ is low-rank, which is equivalent to sparsity in its unknown eigenbasis. Θ∗ | Consider t...
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can always choose − , it holds u N2 be a 1/4-net for N1 be a 1/4- It follows from Lemma 1.18 . Moreover, for any T − S 12d and 1 |N | ≤ 2 |N | ≤ S 1, v T 1 2T 1. 1 − − d ∈ S ∈ S u⊤Av ≤ ≤ ≤ It yields 1 max x⊤Av + max u⊤Av 4 u x ∈N1 d−1 y max max x⊤Ay + max max x⊤Av + max u⊤Av x 1 ∈N ∈S ∈S 1 4 x 1 max max x⊤Ay + max max ...
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≤ k (cid:0) (cid:1) (cid:0) (cid:1) or f 4σ log(12)(d t ≥ ∨ T ) + 2σ 2 log(1/δ) . p p The following theorem holds. Theorem 4.3. Consider the multivariate linear regression model (4.1) under the assumption ORT or, equivalently, the sub-Gaussian matrix model (4.2). Then, the singular value thresholding estimator Θsvt wit...
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τ } | S λj ujvj⊤ and note that and Sc λj| ⊂ { j : j : | j λj| ≤ k F kop ≤ τ . It . A ˆΘsvt k Θ∗ 2 F k − ∈ ˆ2 Θsvt k P ≤ ¯ Θ k − 2 F + 2 ¯ Θ k − Θ∗ 2 F k (4.4) Using Cauchy-Schwarz, we control the first term as follows ˆΘsvt k ¯Θ k 2 F ≤ − ˆ rank(Θsvt ¯Θ) k − ˆΘsvt ¯Θ 2 op ≤ k S 2 | |k − ˆ Θsvt ¯Θ 2 op k − 4.2. Multivar...
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� 2 F ≤ k 144 τ 2 + 2 λj | Sc | jX∈ S jX ∈ 2) and on Sc, Since on S, τ 2 = min(τ 2, | ˆΘsvt k λj| Θ∗ 2 F ≤ k − 432 3 min(τ 2, 2), it yields, λj| | | 2 λj| ≤ min(τ 2, j X rank(Θ∗) 2) λj | | 432 τ 2 ≤ j=1 X = 432 rank(Θ∗)τ 2 . In the next subsection, we extend our analysis to the case where X does not necessarily satisfy...
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:17) Proof. We begin as usual by noting that Y X ˆ Θrk 2 F + 2nτ 2 rank(Θrk) ˆ k − which is equivalent to k Y − ≤ k XΘ∗ 2 kF + 2nτ 2 rank(Θ∗) , XΘ∗ X ˆ r Θ k 2 F ≤ Next, by Young’s inequality, we have ˆ E, Θ h X rk X − − k 2 k Θ∗ i − 2nτ rank(Θ ) + 2nτ rank(Θ∗) . ˆ rk 2 2 2 E, X ˆΘrk h − XΘ∗ = 2 E, U h 2 + i i 1 2 k ˆ ...
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k XΘ∗ k − Note that rank(N ) equality, we get ≤ E, U h 2 = i Φ⊤E, N/ h Φ⊤E k ≤ k ≤ rank(N ) k 2 op Φ⊤E N 2 kF i k 2 N 1 2 k op k 2 N F k k Φ⊤E rank(Θrk) + rank(Θ∗) . 2 op ˆ k k Next, note that Lemma 4.2 yields ≤ k (cid:2) Φ⊤E 2 op ≤ nτ 2 rank(Θrk) + rank(Θ∗) . nτ 2 so that (cid:3) k ˆ k E, U h 2 i ≤ Together with (4.5)...
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) ≤ min IRd×T k − k Θ ∈ rank(Θ) Y XΘ 2 F k ≤ can be solved efficiently. To that end, let Y = X(X⊤X)†X⊤Y denote the orthog- onal projection of Y onto the image space of X: this is a linear operator from IRd T . By the Pythagorean theorem, we get for any Θ T into IRn IRd T , × × × ¯ ∈ k ¯ Next consider the SVD of Y: − k Y ...
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k. − Once XΘrk has been found, one may obtain a corresponding Θrk by least 2 F over matrices of rank at most k ¯ 7→ k XΘ Y ˆ ˆ squares but this is not necessary for our results. Remark 4.5. While the rank penalized estimator can be computed efficiently, it is worth pointing out that a convex relaxation for the rank penal...
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contains information about the moments of order 2 of the random vector X. A natural candidate to estimate Σ is the empirical covariance matrix ˆΣ defined by ≻ ∈ (cid:2) (cid:3) n ˆΣ = 1 n XiX i⊤ . Using the tools of Chapter 1, we c i=1 X an prove the following result. Theorem 4.6. Let X1, . . . , Xn be n i.i.d. sub-Gaus...
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k Idkop ≤ 0, by a union bound, − x,y ˆ 2 max x⊤(Σ ∈N Id)y − I dkopk − Σ1/2 kop be a 1/4-net for Let Lemma 4.2 that N d − 1 such that |N | ≤ 12d. It follows from the proof of So that for any t ≥ ˆ IP Σ k (cid:0) It holds, Idkop > t − ≤ (cid:1) IP x⊤ ˆ(Σ x,y X ∈N (cid:0) Id)y > t/2 . (4.6) − (cid:1) ˆ x⊤(Σ Id)y = − n1 n ...
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− 2 ed Cauchy-Sc(cid:0) (cid:0) exp (cid:0) IE i (cid:1)(cid:1) Z 2 − − IE[Z 2 ] − 1/2 , (cid:16) (cid:0) where in the last inequality, we u(cid:1)s(cid:1)(cid:3) subG d(1), we have Z+, Z (cid:17) hwarz. Next(cid:1),(cid:1)(cid:3)s ince X subG(2), and it follows from Lemma 1.12 that (cid:0) (cid:2) (cid:2) ∼ Z 2 + − IE...
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In particular, the right hand sid − − ≤ ∧ (cid:17) h i ( ) . 144d exp n 2 2 t 32 t 32 t 32 ≥ 2d n 2 n (cid:16) es our proof. This conclud log(144) + log(1/δ) log(144) + log(1/δ) 2d n ∨ (cid:17) (cid:16) 2 n (4.7) (0, 1) if ∈ 1/2 (cid:17) Theorem 4.6 indicates that for fixed d, the empirical covariance matrix is a consis...
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⊤u) d = (cid:12) (cid:12) δ. . Σ k kop r (cid:16) d + log(1/δ) n d + log(1/δ) n ∨ (cid:17) with probability 1 − The above fact is useful in the Markowitz theory of portfolio section for IRd such that example [Mar52], where a portfolio of assets is a vector u |1 = 1 and the risk of a portfolio is given by the variance V...
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” half a million dimensions into only two? The answer is that the data is intrinsically low dimensional. In this case, a plausible assumption is that all the 1, 387 points live close to a two-dimensional linear subspace. To see how this assumption (in one dimension instead of two for simplicity) translates into the str...
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and v ∈ S d − 1. The vector v is called the spike. 4.4. Principal component analysis 95 Courtesy of Macmillan Publishers Ltd. Used with permission. Figure 4.1. Projection onto two dimensions of 1, 387 points from gene expression data. Source: Gene expression blog. v Figure 4.2. Points are close to a one dimensional sp...
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| − v 2 2 ≤ | 2 sin2 ∠(v˜, v) 8 ˜ Σ Σ 2 . θ2 k − kop ≤ min 1 ∈{± ε } (cid:0) (cid:1) 1, it holds under the spiked covariance model Proof. Note that for any u that d − ∈ S u⊤Σu = 1 + θ(v⊤u)2 = 1 + θ cos2(∠(u, v)) . Therefore, v⊤Σv − v˜⊤Σv˜ = θ[1 − cos2(∠(v˜, v))] = θ sin2(∠(v˜, v)) . Next, observe that v⊤Σv v˜⊤ − ˜ ≤ = ...
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˜of Σ and in the last one, we used the fact that the matrix v˜v˜⊤ vv⊤ has rank at most 2. − Next, we have that v˜v˜⊤ k − vv⊤ k 2 F = 2(1 − (v⊤v˜)2) = 2 sin2(∠(v˜, v)) . Therefore, we have proved that 2 ∠ θ sin ( (v˜, v)) ˜ 2 Σ ≤ k Σ kop sin(∠(v˜, v)) , − so that sin(∠(v˜, v)) 2 ˜ θ ≤ k Σ Σ − kop . To conclude the proof...
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) d + log(1/δ) n d + log(1/δ) n ∨ (cid:17) This result justifies the use of the empirical covariance matrix Σ as a re- placement for the true covariance matrix Σ when performing PCA in low di- n, mensions, that is when d the above result is uninformative. As before, we resort to sparsity to overcome this limitation. n. ...
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� Xn be IRd such that IE XX ⊤ = Σ and X Theorem 4.10. Let X1, . . . , n i.i.d. copies of a sub-Gaussian random subGd( vector X kop). Assume k further that Σ = θvv⊤ + Id satisfies the spiked covariance model for v such d/2. Then, the k-sparse largest eigenvector vˆ of the empirical that ≤ covariance matrix satisfies, ∈ |0...
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that i d matrix M , we defined the matrix M (S) to be the 1, . . . , d } ⊂ { Σ(S), vˆ(S)vˆ(S)⊤ vv = Σ(S) v(S)v(S)⊤ IRd, x(S) | × | IR| × − − − − ∈ ∈ S ˆ ⊤ i h S | v⊤Σv vˆ⊤Σvˆ − ≤ k ˆ Σ(S) Σ(S) kopk − vˆ(S)vˆ(S)⊤ − v(S)v(S)⊤ k1 . Following the same steps as in the proof of Theorem 4.8, we get now that min 1 ∈{± ε } εvˆ |...
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i − 2 (cid:17) where we used (4.7) in the second inequality. Using Lemma 2.7, we get that the right-hand side above is further bounded by exp n 2 ( − t 32 t 32 ∧ 2 (cid:17) ) + 2k log(144) + k log ed 2k (cid:0) (cid:1)i (cid:16) Choosing now t such that h C t ≥ r k log(ed/k) + log(1/δ) n ∨ k log(ed/k) + log(1/δ) n , fo...
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�M XΘ ∗ 2 kF . − σ2 rank(Θ∗) n T ) (d ∨ with probability .99. 2. Show that there exists a matrix n some estimator Θ and ˆ × n matrix P such that P M = XΘ for ˆ ˆ 1 n k X ˆΘ XΘ∗ 2 kF . − σ2 rank(Θ∗) n T ) (d ∨ with probability .99. 3. Comment on the above results in light of the results obtain in Section 4.2. Problem 4....
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MIT OpenCourseWare http://ocw.mit.edu 18.969 Topics in Geometry: Mirror Symmetry Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. MIRROR SYMMETRY: LECTURE 8 DENIS AUROUX Last time: 18.06 Linear Algebra. Today: 18.02 Multivariable Calculus. / 18.04 C...
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¨ complexified K¨ahler moduli space: if [B + iω] = � e∗ i B + iω. ahler cone, we obtain coordinates on the tˇiei, let ˇqi = exp(2πitˇi), tˇi = � Example. Returning to our example, ˇq = exp(2πi T 2 B + iω). Conjecture 1 (Mirror Symmetry). Let f : X → (D∗)S be a family of Calabi- Yau 3-folds with LCSL at 0. Then ...
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etc.). �p = � �m(p) ∂ Xˇ ∂ i ∂q , Xˇ , Ω) and the RHS to a (1, 1)­ ∂ = ∂tˇi = 1 � 2 DENIS AUROUX Remark. A more grown-up version of mirror symmetry would give you an equiv- T X) with its usual product structure and H ∗( ˇX, C) alence between H ∗(X, with the quantum twisted product structure as Frobenius alg...
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gives Xψ = Xξφ, so let z = (5ξ)−5 . Then z 0, i.e. ψ → ∞, gives a toric degeneration of Xψ to {x0x1x2x3x4 = 0}. This is maximally unipotent, as the monodromy on H 3 is given by → ∼ (4) ⎜ ⎜ ⎝ ⎛ 1 1 0 0 0 1 1 0 0 0 1 1 0 0 0 1 ⎞ ⎟ ⎟ ⎠ so it is LCSL. We want to compute the periods of the holomorphic volume form ψ. T...
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� should be given by the root of fψ which tends to 0 as ψ → ∞. We need to show that there is only one such value (giving us a simple degeneration rather than a branched covering). Explicitly, set x3 = (ψx0x1x2)1/4y: (6) i.e. (7) fψ = 0 ⇔ x0 5 + x1 5 + x 5 2 + (ψx0x1x2)5/4 y 5 + 1 − 5(ψx0x1x2)5/4 y y = y5 5 + ...
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Ωψ ∧ dfψ = 5ψdx0 ∧ dx1 ∧ dx2 ∧ dx3 ∂fψ at each point of Xψ. At a point where ∂x3 and = 0, ( x0, x1, x2) are local coordinates, (9) Ωψ = 5ψdx0 ∧ dx1 ∧ dx2 ∂fψ ∂x3 = 5ψdx0 ∧ dx1 ∧ dx2 4 − 5ψx0x1x2 5x3 Defining it in terms of other coordinates, we get the same formula on restrictions. We need to extend this to ...
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dx2 where fψ has a unique pole at x3. The residue is precisely 5ψ (∂f /∂x3) , giving us (13) = � 5ψ T0 (∂f /∂x3) dx0dx1dx2 = � T0 Ωψ � DENIS AUROUX � T0 Ωψ = 1 2πi � (5ψ)−1(x5 0 + x5 dx0dx1dx2dx3 x0x1x2x3 dx0dx1dx2dx3 2 + x5 1 + x5 � 1 − (5ψ)−1 x0 3 + 1) − x0x1x2x3 5 + x2 5 + x1 x0x1x2x3 5 5 dx0...
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5 3 1 2 4 So (14) (15) � Ωψ = −(2πi)3 ∞ � n)! (5 5(5 ψ)5n (n!) T0 In terms of z = (5ψ)−5, the period is proportional to n=0 (16) Set an = (5n)! Then (n!)5 . φ0(z) = ∞ (5n)! n � z (n!)5 n=0 (17) (n + 1)4 an+1 = � d (5n + 5)! (n!)5(n + 1) � cnzn) = Setting Θ = z dz : Θ( (18) = 5(5n + 4)(5n +...
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Electric field lines in the space surrounding a charge distribution show: PRS02 1. Directions of the forces that exist in space at all times. 2. Only directions in which static charges would accelerate when at points on those lines 3. Only directions in which moving charges would accelerate when at points on ...
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⎥ 4 ⎦ ⎣ 3/ 2 ˆi ˆi E-Field of Two Equal Charges PRS02 G 1. E = ˆj 2k qs e ⎡ 2 d ⎤ ⎢ s + ⎥ 4 ⎦ ⎣ 2 3/ 2 There are a several ways to see this. For example, consider d→0. Then, G E → ke 2q ˆj 2s which is what we want (sitting above a point charge with charge 2 q) E-Field of Five Equal Charges PRS02 S...
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More rapidly than 1/r2 3) More slowly than 1/r2 4) Who knows? PRS02 E-Field of a Dipole (2) It falls off more rapidly We know this must be a case by thinking about what a dipole looks like from a large distance. To first order, it isn’t there (net charge is 0), so the E- Field must decrease faster. PRS02 A...
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3.012 Fund of Mat Sci: Bonding – Lecture 1 bis WAVE MECHANICS Photo courtesy of Malene Thyssen, www.mtfoto.dk/malene/ 3.012 Fundamentals of Materials Science: Bonding - Nicola Marzari (MIT, Fall 2005) Last Time 1. Players: particles (protons and neutrons in the nuclei, electrons) and electromagnetic fields (photons)...
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• Potential energy V 3.012 Fundamentals of Materials Science: Bonding - Nicola Marzari (MIT, Fall 2005) Polar Representation Diagram of the Argand plane removed for copyright reasons. See Mortimer, R. G. Physical Chemistry. 2nd ed. San Diego, CA: Elsevier, 2000, p. 1011, figure B.6. 3.012 Fundamentals of Materials Sc...
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h = 6.626 x 10-34 J s = 2π a.u.) See animation at http://www.kfunigraz.ac.at/imawww/vqm/movies.html Select “Samples from Visual Quantum Mechanics” > “Double-slit experiment” 3.012 Fundamentals of Materials Science: Bonding - Nicola Marzari (MIT, Fall 2005) Time-dependent Schrödinger’s equation (Newton’s 2nd law for qu...
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inger’s Equation (II) − ⎡ ⎢ ⎣ 2 h 2 m 2 +∇ r )( rV ⎤ r )( r ϕ = ⎥ ⎦ r rE )( ϕ 3.012 Fundamentals of Materials Science: Bonding - Nicola Marzari (MIT, Fall 2005)
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System Identification 6.435 SET 6 – Parametrized model structures – One-step predictor – Identifiability Munther A. Dahleh Lecture 6 6.435, System Identification 1 Prof. Munther A. Dahleh Models of LTI Systems • A complete model u = input y = output e = noise (with PDF). Lecture 6 6.435, System Identification 2 Prof...
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Examples …. ARMAX – ARMAX (Autoregressive moving average with exogenous input) • Description • Standard model • More general, includes ARX model structure. Lecture 6 6.435, System Identification 10 Prof. Munther A. Dahleh Examples …. ARMAX • One step predictor or • Pseudo-linear Regression past predictions where or s...
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A predictor model is a linear time-invariant stable filter that defines a predictor Def: A complete probabilistic model of a linear time-invariant system is a pair of a predictor model and the PDF associated with the prediction error (noise). In most situations, is not complete known. We may work with means & vari...
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. Munther A. Dahleh Model Structures Proposition: The parametrization (= set of parameters of A , B , C , D , F ) restricted to the set is a model structure has no zeros outside the unit disc Lecture 6 6.435, System Identification 28 Prof. Munther A. Dahleh Proof: Follows from the previous general derivative. Notice...
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OE: Suppose . Then & iff (B, F ) are coprime. (To do this cleanly, need to consider where & are the delay powers in both B & F ) Lecture 6 6.435, System Identification 34 Prof. Munther A. Dahleh Theorem: Identifiability is identifiable at iff 1) There are no common factors of 2) 3) ” ” Lecture 6 6.435, System Ident...
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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 6.873/HST.951 Medical Decision Support Spring 2004 Evaluation Lucila Ohno-Machado Outline Calibration and Discrimination • AUCs • H-L stat...
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ectomy angiojet max pre stenosis max post stenosis no reflow Cases Women Age > 74yrs Acute MI Primary Shock Study Population Development Set 1/97-2/99 Validation Set 3/99-12/99 2,804 1,460 909 (32.4%) 433 (29.7%) p=.066 595 (21.2%) 308 (22.5%) p=.340 250 156 (8.9%) (5.6%) 144 95 (9.9%) (6...
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C f o r e b m u N 2000 1500 1000 500 0 Number of Cases 53.6% Mortality Risk 21.5% 12.4% 2.2% 0 to 2 3 to 4 5 to 6 7 to 8 9 to 10 >10 Risk Score Category 60% 50% 40% 30% 20% 10% 0% Evaluation Indices General indices • Brier score (a.k.a. mean squared error) 2 Σ(e - o ) i i n e = estim...
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“nl” “D” Sens = TP/TP+FN 40/50 = .8 Spec = TN/TN+FP 45/50 = .9 PPV = TP/TP+FP 40/45 = .89 NPV = TN/TN+FN 45/55 = .81 Accuracy = TN +TP 85/100 = .85 nl D “nl” 45 10 “D” 5 40 “nl” “D” Sensitivity = 50/50 = 1 Specificity = 40/50 = 0.8 threshold “nl” “D” nl D 40 0 10 50 50 50 40 60 nl TN diseas...
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50 70 30 “nl” “D” 1 y t i v i t i s n e S ROC curve 0 1 - Specificity 1 1 y t i v i t i s n e S ROC curve s d l o h s e r h T l l A 0 1 - Specificity 1 45 degree line: no discrimination 1 y t i v i t i s n e S 0 1 - Specificity 1 45 degree line: no discrimination 1 y t i v i t ...
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sick): 0.3 0.2 0.5 0.1 0.7 0.8 0.2 0.5 0.7 0.9 Estimates per class • Healthy (real outcome is 0) • Sick (real outcome is1) 0.3 0.2 0.5 0.1 0.7 0.8 0.2 0.5 0.7 0.9 All possible pairs 0-1 • Healthy 0.3 0.2 0.5 0.1 0.7 < • Sick 0.8 0.2 0.5 0.7 0.9 concordant discordant concordant concordant ...
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0.835 ROC = 0.50 0.00 0.20 0.40 0.60 0.80 1.00 1 - Specificity Calibration Indices Discrimination and Calibration • Discrimination measures how much the system can discriminate between cases with gold standard ‘1’ and gold standard ‘0’ • Calibration measures how close the estimates are to a “real” prob...
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1.3 sum of group = 3.1 0 0 1 0 0 1 0 1 1 1 sum = 1 sum = 1 sum = 3 Regression line Linear Regression and 450 line 1 p u o r g r e p s e u l a v d e v r e s b o f o g v A 0 Avg of estimates per group 1 Goodness-of-fit Sort systems’ estimates, group, sum, chi-square Estimated Observed ...
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3 sum = 0.8 0.5 0.5 sum = 1.0 0.7 0.7 0.8 0.9 sum = 3.1 Observed 0 0 1 0 sum = 1 0 1 sum = 1 0 1 1 1 sum = 3 Decomposition of Error The “ideal” model generates data D. A “learned” model is learned from D. Once learned, model M is fixed. After learning, I and M are conditionally independent given D. ...
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2 ⎤ ) ⎥⎦ ⎡ 1 + ⎢⎣ 2 ∑ B P ( 2 1 ) − ∑ B P ( 2 )2 ⎤ ⎥⎦ = Decomposition of Error = 1 2 + 1 2 − ∑ ) B P A P ( ( ) + ∑ P ( AB ) −∑ P ( AB ) + 1 ∑ 2 ( A P 2) − 1 ∑ 2 ( A P (B P 2) − 1 ∑ 2 (B P 2) = 2) + 1 ∑ 2 1 [∑ A P ( 2 2 ) − ∑ AB P ( ) + ∑ B P ( 2 ) ]= − = 1[∑ ) B P A P ( ( ...
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MIT OpenCourseWare http://ocw.mit.edu 3.23 Electrical, Optical, and Magnetic Properties of Materials Fall 2007 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 3.23 Fall 2007 – Lecture 9 BAND STRUCTURE 3.23 Electronic, Optical and Magnetic Properties of Materia...
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⎟ ⎟ ⎟⎟ ⎜⎜ ⎝⎝ CCq G+ 2 ⎠ ⎠ ⎟ ⎟ ⎟ 2 ⎟ ( q + 2G) ⎟ ⎠ V 2 − G V V− G 2 h 2m C C ⎛ − 2 ⎞ q G ⎜ ⎟ ⎜ q G ⎟ − ⎟ = E ⎜ C ⎟ ⎜ ⎟ C ⎜ q G ⎟ + ⎟⎟ ⎜⎜ ⎝ CCq G+ 2 ⎠ ⎝ ⎠ q 3.23 Electronic, Optical and Magnetic Properties of Materials - Nicola Marzari (MIT, Fall 2007) 4 Free electron dispersions, 1-d 3.23 Elec...
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( )q 2 h 2m V G V 2G V−3G VV −2G V−G ( ) q G + 2 2 h 2m V G − VV −3G V−4G ⎞ ⎟ ⎟ ⎟ CC −22GG ⎞⎞ ⎛ q ⎛ ⎟⎟ ⎜ ⎟ ⎟⎜ Cq G ⎟ ⎟⎜ C ⎟ ⎟⎜ ⎟ ⎟⎜ C q G+ ⎟ ⎟⎜ ⎟ ⎟⎝ Cq G 2 ⎠ ⎟ 2 ⎟ (q + 2G) ⎟ ⎠ V−2G V −G + q 2 h 2m − CC −22G ⎛ q G ⎞⎞ ⎛ ⎜ ⎟ ⎜ Cq G ⎟ = E ⎜ ⎟ Cq ⎜ ⎟ ⎜ C + q G ⎟ ⎜ ⎟ ⎝ C + ⎠ q G 2...
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of Materials - Nicola Marzari (MIT, Fall 2007) Fermi energy 2 -0.939 Copper A 1 Z 3 Q_ S 1 D 1 Z 2 Z 1 Z 3 Z 4 Z 1 1' 1 3 2' 5 2 3 1 A 1 2' 3 A 3 Q+ Q_ Q+ Q_ 3 A 3 Q+ 1 A 1 D 2 D D 5 2' D 1 S 3 S S 1 -0.539 G 12 G 25' G 1 S 4 S 2 K2 K4 K3 K1 K1 1 X W L G K Figure by MIT OpenCourseWare. 9 ρ r r ( ) = ∑r n k, fn k...
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Adjustable Voltage Power Supply +15V 0.1uf 270Ω 1N758 10v 10K V+ V- . Vo 1uf 6.091 IAP 2008 Lecture 4 Appendix p1 555 Block Diagram ready Threshold Control Voltage Trigger VCC 8 6 5 2 5k 5k 5k + Comp A _ + Comp B _ R S Flip Flop Q Inhibit/ Reset 7 3 Discharge Output +15 1 Gnd 4 Reset R C 1k Figure by MIT OpenCourseWar...
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Key Concepts for this section 1: Lorentz force law, Field, Maxwell’s equation 2: Ion Transport, Nernst-Planck equation 3: (Quasi)electrostatics, potential function, 4: Laplace’s equation, Uniqueness 5: Debye layer, electroneutrality Goals of Part II: (1) Understand when and why electromagnetic (E and B) interaction ...
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( ) x 2 RT zF ln + 1 e − κ x tanh − 1 e − κ x tanh ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤Φ⎛ zF ⎞ 0 ⎟ ⎜ ⎥ 4 RT ⎠ ⎝ ⎥ Φ⎛ zF ⎞ ⎥ 0 ⎟ ⎜ ⎥ 4 RT ⎠ ⎝ ⎦ κ , ⎛ = ⎜ ⎝ 2 2 2 z F c 0 ε RT 1/ 2 ⎞ ⎟ ⎠ Debye-Huckel approximation Φ = Φ ( ) x e κ− x 0 When zF Φ << 0 RT 8 6 4 2 ( ) c x c 0 Φ zF RT 0 = 2 c- (counterion) c+ (co-ion) 1.15 1.1 1.05 1 0.95 ( ) c x...
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⎝ c 2 c 1 ⎞ ⎟ ⎠ RT zF ln ⎛ ⎜ ⎝ c 2 c 1 ⎞ ⎟ ⎠ Nernst Equilibrium potential Diffusion of charged particles -> generate electric field -> stops diffusion of ions
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6.02 Fall 2012 Lecture #14 • Spectral content via the DTFT 6.02 Fall 2012 Lecture 14 Slide #1   Demo: “Deconvolving” Output of Channel with Echo x[n] Channel, h1[.] y[n] z[n] Receiver filter, h2[.] Suppose channel is LTI with h 1[n]=δ[n]+0.8δ[n-1] H1(Ω) = ?? = ∑h1[m]e − jΩm m So: = 1+ 0.8e–jΩ = 1 ...
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DTFT) for Spectral Representation of General x[n] If we can write h[n] = ∫ H (Ω)e 1 2π<2π> then we can write x[n] = ∫ X(Ω)e 1 2π<2π> jΩn dΩ where H (Ω) = ∑h[m]e − jΩm Any contiguous interval of length 2 m jΩn dΩ where X(Ω) = ∑ x[m]e − jΩm m This Fourier representation expresses x[n] as a weighted comb...
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�)X(Ω) Compare with y[n]=(h*x)[n] Again, convolution in time has mapped to multiplication in frequency 6.02 Fall 2012 Lecture 14 Slide #7 Magnitude and Angle Y (Ω) = H (Ω)X(Ω) | Y (Ω) |= |H (Ω) |. | X(Ω) | and < Y (Ω) = < H (Ω)+ < X(Ω) 6.02 Fall 2012 Lecture 14 Slide #8 Core of the Story ...
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