BME6360 Sp11 III P33-46

BME6360 Sp11 III P33-46 - Ill 33 Basic Vector Operations...

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Unformatted text preview: Ill. 33 Basic Vector Operations This is a brief review of some basic linear algebra l:] . inner product:<g,y>= 3Ty= 31ng X xi yi = 1 absolute value or length or Euclidean norm lzl = ll LII = §T L = (2(xi)2)1/2 How to make a unit vector (divide by length) u = g / M X I | | | Projection of 3: onto 5 3 14 ET A = ET g ° IX! Basis for vector space u1,u2, ,gn are abasis ifany gcan be written 3: a] u] + azuz -+ +an un Orthonormal Basis if if gj :0 for i¢j and ui _i=l III. 34 Some Examples of Linear Operations on Data Samples Taken from an Action Potential The examples below are intended to help you think about linear transformations in general and in particular, how they may be used to find patterns which are interesting. An essential idea is that, whenever you take the dot product of a weight vector with a data vector, you are “looking for data which looks like the weight vector”. Another way to think about it is that taking the dot product is a way of asking “how much does the data look like the weight vector”. Of course a linear transformation usually takes several dot products at the same time. -1 ix II Nv—awOOUN—‘l Standard Representation -- Identity Basis A tedious but correct way to describe the vector 5; is: gz—iu1—2u2+1u3+5u4+8u5+3 u6—1u7—2u8 where u 1 = oooooooi—A I: l\) II OOOOOOi—‘O Is: 00 ll #0000000 and the ui’s are the columns of the Identity Matrix I>—-4 ll III. 35 Haar Representation Another representation is related to the Haar Transform has alternations in sign, like a sinusoid, but uses the values of only 0, l, or ~l. (It could be considered the “poor man’s Fourier transform” -- you don’t need to do any multiplying -- just adding or subtracting -- and can find out frequency like information without a fast computer). Although the Haar transform is useful —— and has been used in the action potential sorting research literature —- the reason to introduce it here is to have you consider that there are lots of different linear transforms, each one of which emphasizes different qualities of the raw data. 1 l l l l l I l —l l l —l Q1: 1 -\/§ hz— -1 —\/§ h3— Q fill—1 l —l O l —l O l —l 0 0 l 0 0 —l O O O l 0 0 —1 D4: 1 +\/Z hs- 0 +45 he: 0 +N/5 1 O 0 -1 0 0 _1 0 0 O O 0 O 0 O O 0 Q7: 1 +‘\/5 big: 0 +Nl§ —l O 0 l O -1 The Haar representation has alternating signs. A quality of taking a dot product is that it provides a quantitative answer to the question: “how much does the signal vector look like ....?” The Haar functions “look like” action potentials with different widths, as you can see here. a= l—% 2% Fu— 19* W a: l—Ilj— / action potentials I) N‘—l til— i III. 36 Note the the hi’s are orthonormal and that 11TH = the identity matrix. J "10000000 thhrthh2~h1Ths 01000000 1gb] 00100000 Let E: h1h2 -L18 thenflTflz =00010000 00001000 ' - 00000100 hgm @118 00000010 000000014 Action Potential Representation Our example action potential can be represented by the following 23:39:11—1.8h2—45h3-l-7h4+0.7h5—2.8h6+3.5h7+0.7hg or, 23H = which is a shorthand notation for 3 represented on the H basis You can interpret this result as saying that our AP has an average value of 3.9, and looks a lot (value = 7) like the “square looking” action potential h4. Here is how you convert from one basis to another: Let £1: 5 represented on “normal” basis gH = 3 represented on Haar basis then 5 =fl; and 2g =fl"lx =HTX H I I H H H1. 37 Discrete Sine/Cosine Transforms The same strategy can be used with the discrete sine and cosine transforms, which are the way Fourier transforms were computed before the era of computers. cos 0° cos 0° 0 O l I 00' || HHP—dt—db—‘h—db—‘b-d l 00' cos 135° cos 180° 19 II .in [\3 ll I—_————fi v 0' ll v a» mi L». L». mommmowm / cos 535° cosO° cos90 coslSO cos270 COS360 : cos450 cosS40 005630 :J I N H a, u 'l a Omom0mom ll cos 0° cos 135° IO U) H J? u = 0 (not used) cos 0 cos 180 cos 360 94 = j L III. 38 sin 0 sin 180 sin 360 sin 0 [$111 135] -3 sin 270 —-5 . .35 S3 = + 2 : 0 L61 Q: [9021929394. §1§2§3] Then LC = Q 31 or 3:39 QO—8.7_C_1 +3.5 Q2—0.3_C_3+1.1§_4+1.7 _S_1—1,0§2—.3§3 3.9 represented on COS/SIN basis —8.7 3.5 —O.3 0r Kc = 1.1 L (— “how much does AP look like sine wave fil” _ _ T Ill. 39 Fourier Transform Representation Here we’ve used some informal notation to help with the representation of the complex vectors used in the Fourier Transform. Otherwise, the methodology above is the same as above —- but the basis vectors are complex. em“ I] -> eio‘“ l —> 610: 1 __) 'Ivectors pointing in the 0° direction $00 1 _> in complex plane 1’10 = 6} —~l§ = +V§ = 6,10 l —> 610° 1 —> 6,103 1 —-—) Lew 1_ ——> ' 610° ' cos(Oc)+ j sin(0°) ‘ l 1 _ ewe cos(453) + j sin<455> ~5 + W a em cos(90°) + j sin(90°) j 2 C - ms“ ' ' C '5— T W : 1+JS] = 6 +Vrg: cos(l35 )+Jsm(l35) +Jg: V J ~ +x/g R i {2‘ also: cos(180°)+ j sin(180°) —1 <E 652256 c0592?) +j sin<225c) —«/§ — b5 ¢ 6137“ cos(270c) +j sin(2709) “j \ LCM J cos(315°)+ j sin(315°)J _-fi HWJ r eJOC 7 cos(05)+ j sinar) “ 1 ' [—91 619‘“ cos(90°) +j sin(90°) j T CW? cos(l800)+j 51110803) —1 s 7 ' a W 2703 +' ' 270° —‘ i = = e. c CO“ 3 W 3 mg = N5 61360 cos(360 )+] sm(360 ) 1 ‘9 €450: cos(4500)+j sin(450°) T ej540° cos(540°)+j sin(540°) ‘1 (— Let/>30" _ cos(630°) + j sin(630° )d —j_ _ H l’ 630 ' — l 1 WkO ejiasc —1 £5 = 313* Wm 61270 1 I ‘ Wu 61-4050 —1 V— \CGmpiex Wk; 1 J,21tkm __ _ :: + : ~ = _ 8 3’23- ejsw \/§ W4 1 8 conjugate) Wk wk4 ka wfge 61675” —1 ‘16: E; WkS 61810 1 WM: 19-15: L_.1 E7: 141* W _ J ‘— k7 111. 40 Representing the action potential with Fourier Coefficients aw = E31 3.9 —6.2+j1.2 §=3.9wo+(—6.2+j1.2)fl1 2.5—j0.7 +...+(—6.2—j1.2)m7 —0.2 -j 0.2 éw — 1.1 ~ ‘0-2 +1 0-2 (Since X is real the imaginary parts must 2.5 +J 0.7 cancel) ~62 —} 1.2 3.9 1 0° AP 6.3 416 170° 2.5 X —16° 0.3 )1 —130a 3W ’ 1.1 a: 03 0.3 1 130° A . h cosme Wit 2.6 160 a: a 1700 phase 6.3 l —170 Shift the AP looks a lot (amplitude 6.3) like a cosine with a 170° phase shift t = 0 Equivalence of Fourier and Cos/Sin Bases fl0=§0 fl1=Q1+131 fl2=§2+132 _W.3=Q3+JS3 w=§4 E5=Q3-153 916:22—152 fl7=.C_1—151 Ill. 41 Transformations are Rotations All orthonormal transformations to new orthonormal bases are rotations. This is easy to see in the 2- dimensional case. It also applies to Fourier transform, Haar transform, etc, but it is harder to imagine. the basis you are used to transformation rule cosG —sin6 cos 6 —sin6 r : _ 1‘ = R: . ‘1 sme ’2 +COSG Sine C056 5 can be represented by £1= 3 21 + 5 22 KR = 5.1 r] +2.8 :2 =r1):1+ r2) :2 Ill. 42 Distance Measures in Vector Spaces E2 3 (X1 — X2)2 + (Yr — y2)‘ for n dimensional n 1/2 (115 = E (Xi-Yi)2 = 11:qu = Ilwlz y d}; = Euclidean distance = dCBD = 2 [Xi*Yi| = Ila—XHCBD = Ila—EH] : — EH A (absolute value) Max (Error) Distance dA = .max lxi—yi' 1:1, n amb) = lll*¥llM b =|lé*¥ltm III. 43 Egui—Distance Lines Euclidean circles centered about gaumsiq >13 : KID means Maximum Error Distance squares about means Two Class Two Feature Decisions f / 2 1 / locus . / equidi / 1 / / / / / o 2 (l / f2 0 1 locus of <—-—-——— equidistant / points max CI‘I‘OI‘ ‘2 III. 44 Classification Problem: Which class does AP x belong to? distances from unit 1: c1131 = \/(6—3)3+(7—4)2 = \lfii dCBD1= l6-3l+l7-4l=6 dMl = max ([6 — 3|, |7 — 4|) = 3 (best according to Max Error Distance) distances from unit 2: DE2 = \l(6 — 7)2 + (7 — 11)2 = \Ifi (best accoreding t0 Euclidean Distance) dCBD2 = |6 — 7| + {7 — 111 = 5 (best according to City Block Distance) dM2 = max(|6—7|,|7—11|)=4 Separation Matrices a wav of represeting the pairwise distances between points distances between mean values of 2 features for each of4 classes III. 45 Gaussian Probability Density Function What happens when their is unequal variance? This can occur simply because the units are different —— for instance if one measure is in cm and another is in mm, it is likely that the scatter is unequal in the different directions. What happens when there is non—zero covariance? This happens when two variables tend to change together. The formula for Gaussian distributed variables is given below, as well as a graphical description of what happens in two—dimensions. PG) €XP[—%(§ — mlTlé—l (é — ml] / “iso-probability” curves width amplitude width fl. i a - cluster diagram each point is from one AP waveform iamplitude This leads to the problem highlighted below. a is clearly closer to the mean when measured with a ruler, while b is clearly closer in a probabilistic sense. What is needed is an algorithm which will “measure” the distance in a way which agrees with our intuitive understanding. This algorithm can be derived if it is assumed that the variables are Gaussian distributed. 8 Which is closer to In? a or b? in. Which is more likely to happen? a or b? b One way to do this is to use the following measure of distance: dMaln = (g — mfl 191 (1 — m) This is the Mahalanobis distance (of course, Mahalanobis did this firstl). It does exactly what we want. H1. 46 A second way is to transform the coordinate system using a linear transformation so that the distribution is no longer cigar shaped, but instead looks spherical. All contours are perfect circles <-—————- about mean (equal variance; covariance = 0) Here’s the procedure: assume that the matrix inverse K'1 exists . , 7 _ , T s » assume that you can take its “square root” K‘”‘ such that K 1= (K 1/2) K 1/“ then 1 1 _ i T _ , Pfé) = €XP[—;(§—E)T(K V') K UTE-31.)] or 1 l A A T A A 13(5) = CXP['E(§‘Q) (rmfl where >1 = Ema; Q = Eng In words, this means that if you do the transformations of the vector and the mean into the A coordinate system, then you can mesure the distances directly (“with a ruler”). The picture below shows that there are two things which happen: first, there is a rotation of the axes such that the principal axes are now along the x and the y directions; second, the axes have been either stretched or compressed so that one unit equals one standard deviation. Thus, this is a statistical representation. The distribution is said to have been “whitened” because the variation in any direction is equal to and independent of the variation in any other direction. Thus, the noise or variation is like white light -- its magnitude is independent of the measurement (analogous to wavelength). +/— o +/— 1 “whitened” distribution ...
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BME6360 Sp11 III P33-46 - Ill 33 Basic Vector Operations...

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