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# Introduction to Statistical Orbit Determination 1 ASEN 5070

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This 127 page Class Notes was uploaded by Laila Windler on Friday October 30, 2015. The Class Notes belongs to ASEN 5070 at University of Colorado at Boulder taught by George Born in Fall. Since its upload, it has received 63 views. For similar materials see /class/232164/asen-5070-university-of-colorado-at-boulder in Aerospace Engineering at University of Colorado at Boulder.

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Date Created: 10/30/15

Notes on the Term Project State Transition Matrix ASEN 507 0 October 17 2003 George H Born Introduction The state vector for the term project is given by r r He 1 M F51 J2 r X or X where 3 J2 y rs2 C0 13 C0 3 1 r 1 51 7 X 53 9x1 r52 r 2316 m m m list mm m m 3 2L lZ L gt gilt lZ EL git I is a vector of constant force model parameters and y is a vector of constant measurement model parameters in this case station coordinates Hence the submatrices in tt0 have the following values 67 av 07 art 0W Jill 0 This is because 1 05 1quott and are independent of y Also al6l Bro at and o 61 at 6 because I and y are independent of each other and 1 0 and 1 0 Finally 611 av1 6p because I and y are constants Consequently the only portion of tt0 which must be solved for by numerical integration is the upper left most 6X9 portion In order to work with square matrices the upper 9X9 portion of d3 t to At t to generally is numerically integrated An approximation for tt0 For values of At t to less than a few minutes some insight into the elements of mo may be obtained as follows The equations of motion are given by Her I r t E 1 0 J rt3 I 2 Ignoring the term of order J2 and assuming that t to is small yields f0 E u r0 3 Dre 1 0 Integrating this yields 1quott 1quot0 At i0 and A12 rt 1 0 Atr0 711 We may now differentiate these solutions to obtain the elements of tt0 ie 6rt1m are Bro 2 Bro but g 0 and D 0 when compared to I Bro 6 0 Hence 6xt 6xt 6xt 6x0 ayquot 62 1 0 0 artD Liei 6yt 6yt 0 1 0 Bro 6x0 Bye 620 0 0 1 62t 62t 62t axo aye 620 Also tar I At and arc will be small for small At 61 Next 6120 At 0 and 6101 61 0 Bro are 3 The matrices g and g are given by 513 5 ar 6rt 6rt 6rt 611 6rt 6rt 6rt al le aJz 6CD a al le aJz 6CD They may be evaluated using the expressions given for rt and rt For example 6rt aJ 2 aJZ39 2 The Xcomponent of this is given by 2 2 uexo 3 R 20 x 1 J 5 r03 A 2 r02 r02 we y R2 5i 6J2 r03 2 r 2 r02 0 Evaluating this for the term project initial conditions yields 6 xl23gtlt10 3 2 6 s 2 At At 20 seconds 6x 2 6J2 2 6J2 246 gtlt10 3 km 246 m am 6 6 Also at At 20 seconds the 9 component of At is given by Z mm 6 6 246gtlt10393 E 246 3 S S The corresponding numbers on the ASEN 5070 web page for these partial derivatives at At 20 seconds are 249 and 248 respectively Hence these partial derivatives work well for small values of At Approximations such as these would be used in an extended Kalman lter processing data closely spaced in time In summary for small At d3 mo can be approximated by 61 1 MIL 01M ltmgt 0 1 01M 0 0 1 01m 0 0 0 11M 61quot where Z E and a can be evaluated as previously illustrated STATISTICAL ORBIT DETERMINATION ASEN 5070 LECTURE 1 1 92206 Cnlnla ln cum In Asnnnynzmcs Research 12 Unmale m Cnlnlznn hen pmennaI mmmn cunIamsIessevaI and secIuvaI havm ms Is consmm such than Wu can besImwn am an energ 41L mmng exlsxs known as the Jambi Inwgml u 23 24 where K Is he Jacobl Constant and 1 Is In speed or the sareIIne with xesped m the mung System I y 1 AnoIIxex 01m of II 1ng is gn en by Band and AIImzm I996 Much Is suntabIe fm II nonmtaung formulation 2 7W h 7 4N K 2315 r where V Is he ECI speed 15 and h me also expxessed m the EC conrdmme sysIemTIxc1eIMion betwern EC I VeIocny VI and KT veinle W15 Vv xr Cnlnla ln cum In Asnnnynzmcs Research 12 Unmale m Cnlnlznn Alternate Derivation for I using Taylor Series Expansg X3 xt Define xt Xt X t then 60 0X0 Assume that we can write the solution for Xt based on initial conditions Xt Expand the true solution about X I FXtand retain 15 order terms aX TKO xtxz0i FXz xt0xro emrrm X0 xtolt1gtttoxo Colorado Center for Astrodynamics Research The University of Co orado Alternate Derivation for I using Taylor Series Expans For example assume that x then y FWD 8Xt aXz x0 x 8Xt0 8X0 y 1 and my my my my my Note that 63 aye 63 aye 60 the last row eg 9 m m wwe 1010 axo ayo 0 0 0 0 1 6 6x0 6y0 6 Colorado Center for Astr The University of Co orad odynamics Research a J State Transition Matrix Differential Equation for a W General State Vector Let f represent a jxl vector of force model parameters and y a vector of measurement model parameters and y are constants X XFX I Eio r 7 0 57Lx3 0 WAVsz 1 63sz 0 577ka 0 A aFXt EVZFLK EVAVLXK EZELJ BZVLW 0 IE kxl kx jk Colorado Center for Astrodynamics Research The University of Colorado State Transition Matrix Differential Equation for a General State Vector Hence At may be written as 0ng 1ng A 0747 0le 0 01X rlm 0 0 0 0 0 0 0 0 I 0 0 A21 A22 A23 0 0 0 0 0 0 0 0 0 6jc gtlt6jk Colorado Center for Astrodynamics Research The University of Colorado 4 State Transition Matrix Differential Equation for a g 7 General State Vector Then i ACID yields 0 1 0 0 11 u 13 0 1 A21 A22 A23 0 21 22 23 0 0 0 0 0 0 0 I 0 0 0 0 0 0 0 0 I 6jkgtlt6jk M3 M3 M1 01 Ammmlxs swam 1X Amwmzsh olxk 0 0 0 0 0 0 0 0 Colorado Center for Astrodynamics Research The University of Colorado State Transition Matrix Differential Equation for a General State Vector Hence we need only to integrate the 6x6j matrix of differential equations d 21 22 23 1 A21 11 A22 21 A21 12 A22 22 A21 13 A22 23 Mm q1togtto le6 Gig The remaining elements of 1 simply are the elements of an identity matrix Colorado Center for Astrodynamics Research The University of Colorado sguares with apriori Information Ifan apriori value is available for fk call it fk and an associated symmetric weighting matrix Wk the weighted least squares estimate of Ack can be obtained Colorado Center for Astrodynamics Research The University of Colorado wquares with apriori Information Given ykas xk xk 771 Where 77k is the error in Yk and its accuracy is reflected in the weighting matrix Wk and ylnxl 3 1x1 Choose fk to minimize the performance index 1 1 J xk EsTws EnkwknkT Colorado Center for Astrodynamics Research The University of Colorado sguares with apriori Information Writing Jock explicitly in terms of xk Jarky kafwy kafk xkT1vr a W 4324 M 0 Results in See Eq 874 Bxk a39axk y kaT wH fk xkka 0 xk T T T T T y wHxk H wH xk wkxk wk0 Colorado Center for Astrodynamics Research The University of Colorado wquares with apriori Information Solving for xk yields 56k x HTwH wk yTwH Y wk fr yT14Hffi42kHTwHwk 1 fck HTwH wk1HTwy wkfk 4325 Colorado Center for Astrodynamics Research The University of Colorado sguares with apriori Information Note that HTwHwk 71 is symmetric also J HTwH w 8x k which will be positive de nite if a solution exists Colorado Center for Astrodynamics Research The University of Colorado Computational Algorithm for the Batch Processor in 92k HTWH HTWy W93 nxl Look atHTWH W1 0 0 H1 H5 H Hf 0 0 I 0 0 W H1 HITW1H1HZTW2HZ 2szle Likewise quot1 l AWMEAWM 11 and 1 71 ge rgww 11 l EHWmsz 11 HtTWth Colorado Center for Astrodynamics Research The University of Colorado Juliq Computational Algorithm for E the Batch Processor Hence the computational algorithm involves forming the indicated summations performing the Matrix inversion and solving for Xk We will see later that there are more computationally efficient ways to solve for Xk without inverting the normal matrix Colorado Cenler for Ash39odynamics Research The University of Colorado LEO Orbit Determination Example I The batch or least squares processor processes all observation data at once and then determines the best estimate of the slate deviation thereby estimating X I The best estimate ofx is chosen such that it minimizes the sum ofthe squares ofthe calculated observation errors I Estimating xk begins With the following state propagation and observationstate relationships xm LItrkrk 139H ks 1 Hm rlt1gtrr I The H matrix relates the state deviation vector at an epoch time to the observation deviation vector at another time CW3 Copyright 2066 16 LEO Orbit Determination Example The normal equations are formed m N R is the observation error covariance matrix Eqn 13 may now be solved for fa The RMS of the observation residuals is given by m S 139 71139un l3 RIMS I U 7H Copyright zoos m m 139 71 71 r 71 717 HR 53 liva 31 in 10 13 13 in and 17B are the a priori state estimate and state error covariance respectively LEO Orbit Determination Example l 1 In umampzmaemuqmq l A primumxa in LA pn39al39llxwldmu m mamLinux m hwunfumuchmndnm mum mum mu cm a e r39 Fx39z 41 quotnuax uAc A tuml l Form H wtx wax In K 60 4 mivwm 394 f L I I Pu warm H1Ki xf m V II c 2 t p m A KM Copyrigit 2006 The Batch Least Squares Algorithm Updm it u 2 2 Repeat initialization using 0 e reference trajectory Set xquot xquot and use the original value of E Reprocessa observations s the initial The Batch Least Squares Algorithm Copyright 2005 LEO Orbit Determination Example Instantaneous observation data is taken from three Earth fixed tracking stations over an approximate 5 hour time span light time is ignored 1 p xhyhzhx y i 42x 2xxSi yy51cose 2xySi stiSin9 ZZZsi xxyyzZ esxyysxcos96xx5iyysxsin6xySI yxsisir19xysl yxslcos ZzSi p P Where x y and 2 represent the spacecraft Earth Centered Inertial ECI coordinates and are the tracking station Earth Centered Earth Fixed ECEF coordinates Wit Copyright 2005 LEO Orbit Determination Example 9 80L85 85 0 5L550000 1sz8st5st511000 11LL581 91005z Sma1 ga ueg LOLZZIEILZSJSVLEOO 0 9Z9ZL99LOL 51 0 m azueg v9z zz0 8m ZiL gm 1W W sautex sm quumu uonsmesqo 00v 009 002 00L 0 m umu uonEuesqo 00v 009 002 00L 0 WWW 0W0 E0w mum uonsmesqo 00v 009 002 00L 0 Aequmu menswequ 00v 009 002 00L 0 m umu uonEuesqo 00v 009 002 00L 0 qu umu uonEuesqo 00v 009 002 00L 0 lt0 mm MM 51m sxanpxseH emH eBuEH SIBanSQH sembs 1s29391 110mg aldumXH 11011911111119190 11110 031 m sxanpxseH eBuEH m InnIth r mm 9002 qupAdoo 2m g pm 2 1 sasszd m Imnunaq anus Ipl LOO39QHH6810LSSGVC39S 8009112 LQLIGIZ 839 80095L89586L S005 LOO39QZ6SIILLWOLELS39Z 6826692817 9 9f391 ISSSZSSLGQSOEZE39O39 989CTL86I LL39O 12920810180281 390 EOICV99 LCZZI I 9tL 08E981 WHY 9002 13919103 1 LOOQEQLQSQGOOEOEV39Z39 80098UZ68 CZOELO39G EIIW39QSV8OCCEOFIQVtZ39Z IO989LV1L969OIE9V399 EIO39QECI8SHLLLC1VEL398 LUO39WSZEEI909OV86939V39 SID3810365 LLC39C IGSSCI E EII 9139 010393L8FC6 S8EIGGOGV39Z OIO39QEC8ZOS62888Ct639I OIO39QSSESLZOEWFZSET39 LOO39QSt COLCOfSEOftQI LOO9605688LISLW5939Z39 LOO918L9L9LCH 0fQI39Z g sued L0 L81886t510200 88850598319990 LOO960S5808L8f0f quotC 90098016C16f609 8 39139 900399L909Lf69F0L69839I39 66EISZ96L8SHWU39O 800399817t9t8D50188i39C EfSSS E39LH9OEEE39 238017 6E 08 000390 EGO3 Et9609fZ SIIEOOO39O39 IZLOStOO88 6608039039 Et1911f908 390 QRW WW quot quot10 0 z ssud 600WL UCC 8 6 VSSZSISSLSEQS39OI39 LOO986HIOF80092 90095599188656C8Lf391 900996090250699298391 If 1719 CLH39O LOO91D091702L1719ELS39939 80992639K0i9t639 1 I A HO39O39 E659996E68fl 0390 S09EOEI6F9F 60F00 COIO9ZIO8EIO8I39O39 LVC8CZ L8 Z 17LC39O fLC8 ILCI 8 9 039O39 w o I ssud 5 umwuw ems aldumXH 11011911111119190 mm 031 g Q Given the following system y12x1x2 1 y24x12x2 2 and 2 1 1 0 4 39 y X a H 5 W a Y 1 x2 4 2 0 2 y2 10 a Could we obtain a solution for 2 using the least squares equation Why or why not Will minimum norm yield a solution 7 1 7 b Assume we are given a priori information X I and W I What is the least squares solution for X Colorado Center for Astrodynamics Research The University of Colorado g y12xlxz 1 y2 4361 2162 2 Colorado Center for Astrodynamics Research The University of Colorado p 4 Problem 19 Given the observationstate relation will Find the best estimate of x 139 1 2 3 and the observation sequence at 2 1 y1 2 and at t 2 y2 1 Colorado Center for Astrodynamics Research The University of Colorado 4 Problem 19 X1 Findiuiczaicg X X2 X3 3 yt1 X1 X2 X3 yfz 22 X 2x1 4X2 8X3 11 yt1 1 1 1 X1 2 X2 t 2 4 8 1 y 2 X3 We have 2 observations and 3 unknowns hence solution given by Eq 4313 ull Note the rank of H is 2 we use the minimum norm Colorado Center for Astrodynamics Research The University of Colorado 4 Problem 19 lg A 1 X HT HHT y Note that because X 2 X ie the state vector is not a function of time ltIgt I and H H 1 2 1 2 A 1 1 1 2 X 1 4 1 4 2 4 8 1 1 8 1 8 1 2 1 5 25 2 1 4 25 05357 1 A A 0 1 8 s y HX 0 x1 186 A A Remember that we derived X with X2 039964 the constraint that 5 0 X3 039821 ie y Hx0 Colorado Center for Astrodynamics Research The University of Colorado 4 Problem 19 What if we had apriori information eg 2 100 i1 W 010 0 001 Can we now have a least squares solution Colorado Center for Astrodynamics Research The University of Colorado 4 Problem 19 6 rank 3 but their sum must be rank 3 Note that the rank of AB is less than or equal to the rank of A plus the rank of B Hence in this example the rank of W and HTH need not be Colorado Center for Astrodynamics Research The University of Colorado L at Given x1 axl bx2 x2 0x1 6x2 with xl x2 and a b c and e are constants X xl 5c 2 a Determine the A matrix b lfabce0 whatisqgttt0 Assume initial conditions c10x20xmx20 are given at to 0 Colorado Center for Astrodynamics Research The University of Colorado lfthe H matrix is not of full rank ie HTHy1 does not exist can we make HTWH 1exist by the proper choice of a weighting matrix W 2 1TorF 2 Justify your answer The rank of the product AB of two matrices is less than or equal to The rank of A and is less than or equal to the rank of B Hence the answer to 1 is false Colorado Center for Astrodynamics Research The University of Colorado in 16 Given range observations in the 2D flat earth problem ie 2 2 1 i0 xO th x5 yo yoti jgt y Assume all parameters except x0 and x0 are known We can solve for both x0 and ice from range measurements taken simultaneously from two well separated tracking stations T or F 2 Justify your answer in terms of the rank of H Colorado Center for Astrodynamics Research The University of Colorado H ax axn apn ap P12 x0 x01 x52 apn ax 6pm ax pxl pxz yo y39all kgtf ys 1 2 2 x x t x1 x x t x1t pxl C Ct x jc t xZ x jcnt xzt K Kt pxz Colorado Center for Astrodynamics Research The University of Color do l The differential equation Xa5cbx5c0 is choose all correct answers gt9 NT 2 d order and 2nd degree 2nd order and 1st degree linear nonlinear Colorado Center for Astrodynamics Research The University of Colo rado STATISTICAL ORBIT DETERMINATION Error Elligsoid BPlane ASEN 5070 LECTURE 35 120406 Colorado Center lor A rodynamics Hessardl The University or Colorado The Probability Ellipsoid O I quot20737715 I IIH Z lll IS an UI39UIDI39IOHTIHI sysnem OI nlgenvecmrs associated respectively with the Eigenvalues 1 AZ Aquot of an n x n symmeu39ic positive de nite matrix P and if UI11xu2yIU1Imltm then AI 0 0 0 2quot39 0 UrPU D1 A2 An 4162 0 0 n o U PU isadiagonalmauixoonminingtlleEigenvaJuesofP The normalized vectors IL ll i u we called principal axes ofP Urnvurr i run um muuum vcuun x mm quotman x mm vmmuwwuvmlmluc r ul pullulpm Cnpyngh ioo axes x are given by x UT x 4163 2 The Probability Ellipsoid o m 39 P P EEx39 4 Yx39 4 7V UTEx x TOT U UTPU 2 D A1 An 4164 0 Let Ax lepiesentdlees ma onmvecmde nedby P EAxAxT 4165 1m Cnpyngh 2006 3 2 51321 g 2 4166 2 o Thed pWs Ill2 and3mcalled 1elo 20 andka 11m rm n 161 39 L 1A1 391 5 17 i lAz 3 2 4168 1 A3 239 Q 0 z 1 1 j a fizz 4169 Wk Cnpyngh 2006 The Probability Ellipsoid Theaxesofthelo ellipsoidal egivmbysolvingEqGJGSHOIlland 4 395 39L 39 ip mmamh semimgiotazdsuseti amdizquot0 The axis of the 039 ellipsoid are given by a mquot b m c m3 4 1 2 3 030 7 093m 7am 7 093M 593 U 30M 050m ism cgo ov isgce 41611 565w 590 Cs Capynglqt 2mm The Probability Ellipsoid m o TheEuletangles 4t OandValeda nedby z zm cam mm Figure 4162 Euler angles de ned The Probability Ellipsoid 9 o andaleoomwned vm atau2 U 0 3 3 27r 41612 7U23 9 aces U33 0 S 9 3 7r 41613 1 atauz 0 g 1 3 2n 41614 U32 Wit Capyngh 2mm Ellipsoid 9 Exaxuple Gymnast nounally dbl39nhuled 2D Isndxn wags x where x a A IU P 1 p sketch u 1 And 30 pruhxlbll y outh The elgenvalues ue given wy he pol39ylhrc nlal and Ipi uI 0 a A A 39IT 2 2 21 hence Az OA 10 and M 52w x 0 714 The canesrondmg Elgenvk oxs am given by P I39 n 112 Thu ntlmcmud gemNIH m awn m 61 am m music mum um pmupu anduuguml um sysmu u okmmd by m ugmung um Ilk amnnum Irnnlthuwlan mm H 31m by um mu b am mummy L umal 14 mm m va in ms 4 mm o m ll wu C39 Hmw um m A munan xlumn ma m g m m mmmwwsn and mmnrms a m pwu w u A Viguu A 7 qmnm 2 7 131 BMW The Probability Ellipsoid whh numerical Va lues The Probability Ellipsoid Example OonxiduaaimplecasewhemdleEigmvecm 1 0 0 U 0 i 0 0 0 1 100 0 0 UTPU 0 10000 0 DHAZA3 0 0 900 I Plot this ermr ellipsoid using Matlab Semi 1111 5111101 841139 13 gnl m PlotEJJipsoid U Semi The Probability Ellipsoid Views of Error Ellipsoid view 00 azimuth 0 elevation 0 W3 view down the negative yaxis Cupynghtl viewazimuth elevation azimuth is a clockwise rotation about the positive zaxis 12 View 9000 mi View down the positive xaxis Capynght 2mm The Probability Ellipsoid View 090O WEquot View down the positive zaxis cam m The Probability Ellipsoid Views of Error Ellipsoid r n in an m View 375 0 W standard ma ab View Cnpyngh zoos The Probability Ellipsoid Body Plane BPlane L l39mhulnm y mm mm 5 y MWMWHW m m m o Impomnt parameters are m m B T B R and LTOF Linearized Time ofFlight Viking Launch Trajectory Viking Trajectory 200 SCALE MILLIONS OF KM Fig 13 Wing 1 hallocerm39lc hame r 39Mh i i l d l m 4 OD solutions before L 12h S 99 IAUNcII DISPERSION nurse ALL SOLUTIONS AFTER L 12h m MARS IMPACT LAUNCH AIMPOINT mus 62076 I707 urc ARRIVAL was f yua nrge W1 5 a nor Lmad Mar 1 QAJ Eaara Timwry y Mrd Cour j 200 400 600 53 r A 1 M 4 Cigars a of 3903 km 1625 urc i Viking 1 Departure Control lt DEFAKIIJIE as Unce ainty in time of 3 mo closest approach m 2521212212 MANEUVER EXECUrIbN 2 min ORBIT DETERMINATION I2 mini YARGET Fox MARS ORBIT INSERTION ICA 5 9 15251 FINAL MARS APPROAO39J CONTROL ACCURACY R MT A0lt 5 Aklt 700 km 30 Error ellipse for 1St mid course maneuver 20 1O Mars Impact Radius 6000 R km 5500 500 O T km 1 Solution using original AMD data 2 Solution using ccrrecled AMD data 3 Target 3s lgmc on Error Ellipse AMD Angular Momentum Desaturation 21 EELir Ellipse in the Bplane E Assume we have done an OD solution at some point t in the interplanetary phase of the mission and we wish to project our estimation error covariance onto the Bplane BBody 1 Compute the time of Bplane penetration based on conic motion this is known as the linearized time of lter LTOF 2 Map P to this time using PLOTF 15C I ILOTF 2 ts CDT ILOTF a ti I39 I QFT 3 Generally P will be expressed in ECI J2000 coordinates Rotate the position portion of LOTF into the STR Frame PSTR yggmmy gj 0 pSTUSaR pSRaSaR A A A PSTR 0T2 pTRUTUR R SXT 2 UR Colorado Center for Astrodynamics Research The University of Colorado Lari 4 Use the 2X2 portion of Pm corresponding to T332 irrEllipse in the Bplane 5 Compute the Eigenvalues and normalized Eigenvectors 6 Compute the semimajor axis and semiminor axis ofthe ellipse X y and the rotation angle Plot the 12 or 30 error ellipse on the Bplane a I Am bNAmm 213 9 atan2UmUlz For a 2D error ellipse the probability of being inside the No error ellipse N123 is given by N pNa1 ei 2 Colorado Center for Astrodynamics Research The University of Colorado Lamar Ellipse in the Bplane For a bivariate normal distribution N 17 N0 1 3935 SMAA 2 8647 A 3 9889 k39 T 4 9997 7 Bplane The B plane parameters are chosen because the variation of the B and S vectors are nearly linear With respect to midcourse orbit parameters Colorado Center for Astrodynamics Research The University of Colorado Example of Kalman Filter Application ASEN 5070 October 18 2005 George H Born Problem 430 of text Given Martial illziiil Yltrgt iillf lliiiiHiiiil Hence 01W 1 EFL and Note that both the state propagation and observationstate equations are linear Assume the following a priori information is given togt l l ltrogt1 z t l 0 Rt10 2 for all values ofz39 Find 01 00 P01 and Pt0 using the sequential Kalman lter Verify the value of Aqto and Pt0 by using the batch processor algorithm Step 1 7 Do a time update at t1 Note that there is no observation at to If this were the case we would rst do a measurement update Xt1 t1t0t0 3 iii 3 Pt1q3t t tt0 Step 2 7 Measurement update at t1 Kt113t1197t11 5117 t1R 1 t it 2H 2 2 1 L 1 2 21 13 0 1 We may now map 01 to to to obtain 00 16 Xtoq3totlftl 1 1 1 2 01 Note that t0t1 qb l 1110 Pt0 is given by Pt0 t0t1Pl Tt0tl 1 1 14 X1 1 0 0 1 M 1 1 The batch processor may be used to compute 00 m HTIOR 1HIOE171H7t0R 1Yt11i 1it0 where mg 4 m 2H m 2 am fllil PO 0 1100 101 0 0 6 1 0 1 19 A4 1 1 39 A A Hence the values of Aqto and Pt0 from the batch processor agree with the results of the Kaman or sequential lter L IE STATISTICAL ORBIT DETERMINATION Statistical Interpretation of Least Sguares ASEN 5070 LECTURE 15 100206 Colorado Center for Astrodynamics Research The University of Colorado Wnterpretation of Least Square Show that 1 is an unbiased estimator of x ie E x where y Hx 8 Assume that lSt two moments of the probability density function for 3 are known ie T E8Ee0 RE8 88 8 E88T From the least squares solution we have that 2 HTH 1HTy Ec HTH 1HTEHx 3 But Ex x Le x is the true value Heme Ec HTH391HTHx and I is an unbiased estimate ofx Colorado Center for Astrodynamics Research The University of Colorado mgllnterpretation of Least Square Determine the mean and variancecovariance matrix of the estimation error 5 for the weighted least squares estimate x 5 fc x Because 56 is unbiased E6E 6 0 V th variancecovariance matrix usually simply referred to as the covariance matrix P PE6 66 6T E657 From the normal equation HTwHc HTwy 1 Colorado Center for Astrodynamics Research The University of Colorado t Wnterpretation of Least Square Cont Substitute y Hx 8 into Eq 1 HTWHE HTwHxe HTWHX xHTW8 5 HTwH 1HTwe P E 5N HTwH 1HTweeTwH HTwH1 If w R 1 Then P HTR lH 1 Colorado Center for Astrodynamics Research The University of Colorado mgllnterpretation of Least Square If apriori information f and 13 are given where fxn xsTruevalue f E Apriori value Assume r 5 Error in f En 0 Then E1mT P W 1 We have shown that the normal equation in this case is HTR IH P4 HTR Iy 134 Recall that the estimation error is given by 62 x Colorado Center for Astrodynamics Research The University of Colorado Wnterpretation of Least Square It can be shown that E c x and E6 0 hence P a E 55f and HTR 1H13 15 HTR 1 Hx e13 1 1H1 HTR 1HI3 1 xHTR le 1341quot Colorado Center for Astrodynamics Research The University of Colorado mgllnterpretation of Least Square Define Y E fIZF1271IP77171 Then 5 JE x 1HTR 18 13391n P E 55f yE HTR le 13417 eTR lH nTF1y Assume that s and 17 are uncorrelated then EM EW and P yHTR391RR391H 13 11313 1y W39ly y Colorado Center for Astrodynamics Research The University of Colorado 5 ti al Interpretation of Least Square Hencet e OBSerVathn error COVarIanCe matrIX In glven T 71 71 1 P 11 H R H P The quantity HTR39lH 134 is called the information matrix A ie A HTR IH 131 For example if then 2 x 0 pxya y ma a y 2 3 I P pxnyGy 0y X x y39 2 ma a 0a a Colorado Center for Astrodynamics Research The University of Colorado mgllnterpretation of Least Square Recallthat lull luxy pxyaxay GenerallyF will be a diagonal matrix If we do not wish to estimate a parameter say a we may leave it out of the state deviation vector or set its apriori variance to a very small number Le a D 0 sat10quot This will result in Q l 9 Colorado Center for Astrodynamics Research The University of Colorado Wnterpretation of Least Square Demonstrate this for a 2D case Let x X assumeP a 8 Where 8 ltlt1 gtgt1 Note The assumption that 5 gtgt1 is not necessary for this demonstration f and X O so o and a sD0 HTH 0 ilHTyy 0 X 0 0 s 5 Colorado Center for Astrodynamics Research The University of Colorado 9 mgllnterpretation of Least Square let b d l ll b c 6 then 7 a1 b 1d3 X b c e bUt a ac d de Thus 71 a b d b V 07 S 8 Colorado Center for Astrodynamics Research The University of Colorado Wnterpretation of Least Square d A 1 b X bZ a X d bY A s A A with IZDquots a 96 my bd A A d bY 7 x A A D b o bds 7 AOl a HencedDo A d b5 Note that 95 7 and is affected bya Colorado Center for Astrodynamics Research The University of Colorado igaltlnterpretation of Least Square 50 The state vector is given be x where x1 and x2 are constants An k x2 a priorvalue Xx x1 is given with its relative accuracy described by the matrix x2 W Three observations of the state are given by y1 x2 x2 s y2 2x1 2x 52 ys x12 xz3 53 The relative accuracy of the three observations is given by the weighting matrix W 1 Set up a step by step algorithm describing how you would solve for the weighted least squares estimate X including use of the a priori information Define the H H and 1 matrices ie what is each element of the matrices Colorado Center for Astrodynamics Research The University of Colorado 7 39 tical Interpretation of Least Square Answer the following questions 2 Is the observation state relationship linear or nonlinear 3 Was it necessary to use both a state and observation deviation vector 4 Do the 131 and H matrices differ for this problem Why or why not 5 Was it necessary to generated a computed observation for this problem Why or why not Colorado Center for Astrodynamics Research The University of Colorado 39calnterpretation of Least Square ll 20 Given the system 539 3X 4X 0 X with the state vector defined by X and the deviation vector defined by 6X 6X 6X Where 6 indicates a small deviation from a reference value a Write the linearized equations in state space form 6X A6X b How would you determine the state transition matrix for this system What Additional information is needed to generate the state transition matrix Colorado Center for Astrodynamics Research The University of Colorado STATISTICAL ORBIT DETERMINATION Potter and Choles AI orlthms ASEN 5070 LECTURE 22 102306 Oulmndn 0am cu Asumlynnmlcs Resenlch mg Unmalslly m Oulmndn E IP 2am rm when is he maximum eigemalue of P and m value In ba e 10 nrirhme c m p signi caul digns nu matrix Hoe mm Map Hence numencal dif cumes shou Id um be encoumered mm CO CHquot w W project Infor 1043 UP 1W Oulmndn 0am cu Asumlynnmlcs Resenlch mg Unmalslly m Oulmndn Mratrix condition number A m is me minimum eigenr s mericzl dimcuuies invexsion and me precision of x may be encountered as In er for 39 Ion number for term mauo 1575 with n mamx IS 57 Tm SQLHRE ROOT MEASUREMENT UPDATE 39 ms ALGORI39H Lng has idem he tmuriance meawremem update eqlmlloll Eq 147 m mu be exprcx sed in squnm mm form as ronnws 11717an HTH 17quotH77 1574 Nov 1m 1 H H39T and maketlussubsmmmn m Eq 576 to obmm 1 7 Inquot H T 577 ru WH TIITHW 11711 n n39T Usmg the following de ninons f WTFITu4fT 1711 Colorado center lor Aslrod ynzmlcs naezrcn The unlvasny of Colorado 571 Tm sound Ron r ME Sl llEllquot39l39 UIDME Auzmunnh moo mcnsuxcnmn updmc cquanou Eq 4710 1 follows Usmg these Mons um covm L can be expressed m square mm lonn s 1 THT UT11T11T411139UIT H 701 Now let P H39er uml make nu mlmilulion in Eq m en lo obluin 1 WW7 u n39Tiu39n39Tn HU39WTH R HH39H39T 577 Lumg me mmmmg de nllimh THTHI3TIRquot 57x The unlvasny of Colorado Colorado center lor Aslrod ynzmlcs naezrcn 5 7 l Tm Sou Rr ROOT MEASUREMEYI39 UPDATE Q icom139nrs Eq 57 7 can he expressed ui H39H39T 1TI JUIT39TWT 579 fr mnmx can he found such lhm UTIintT39T 57 Im rlren Eq IS 9 can be expressed as nuquot WHTWT 57 II Hence 1139 111 51m ooloraao came 10 As1lor1ynamlcs Research The umversryor coloraao 571 T111 591 mu Ruo39r Musumzmzxr Urmrrc 6 Aunmnuus The square root measurement update ulgurulun can he expressed us follows fuJHT n 1R FT 1 1 39Y My 7 HTL where 1 I in ll3 The primary dri fereuces in the various algorithms for compuuug the measurement updale in square root form he in Ihe manner in which he um 39i 15 corupuled The method rst used In pmcice Is that given by Potter 8mm 10 oorouaao Caner 1o Asuoaynamucs Research The umversryor coloraao i uare Root Filter Iutl39udllrtiuu In general square root lial ale more luuueneully slable tllau llle couveulloual Kalluml fl lEl39 X019 lllal llle toudllluu number for llle square mol of a covanauce uluu 1s the square root oflhe coudlllou uumbel of llle cul ul39lauce ulall lx Hence the square l39DDl flller 39111 be less affected by uumerltal problems le fll slsqualeluul 39 39 l quot quot J le mlnufm39llle l J llmlledeaseuf l ll l l nu u lugtllllf1mlslrtllll 1 an the Lunar Excursmu Module LEM39I for the Apollo Program Colorado Cenlerlor A rodynamls Hessardl The Umversly at Colorado m rSquare Root Filter Puller AJgarirhm Derivnlian um update 5 from H a K and drop the llldsces 3 DPLP De ne lhe square ml ofP WW P From Eq 1 and 2 13 WW zp in Where Wlt1gtW Colorado Cenlerlor A rodynamls Hessardl The Umversly at Colorado tigers uare Root Filter 06 Next A 1 m quot 39 upl mp llPllllenlllofW l r Y r l p l JUH milligram i zlwlr rlllllwrlm l W119 0quot 15 the V39 l39l llCF ofthe 059111111011 El39l39DI V Colorado Comer lor Aslrodynamics Research The University ol Colorado Le Y a aEHW39H7H 6 6 where a IS a scalar De ne F WFHquot 47 hen kl F aauquot s and K 1 9 Colorado Comer lor Aslrodynamics Research The University ol Colorado S uare Root Filter The measumnem updme forP 15 P WW 171051 2 711117131 h771177 I I r ui n 7 mziai rw 10 Polter absexved that 1fa mamx 2 could be found such hm 12217u quot 11 then P TW W 11 Colorado Center lor Aslrodynamics Research The U 39 ado nlversily 0 Color B r Square Root Filter To nd 3 mtmdure the scalar y what If 1 yaIEFTH1 ya lq 4pmquot Solving for y I aFFTI lya f i 14 Colorado Center lor Aslrodynamics Research The Universin 0 Colorado De ne 1151 where 9 Isa scalar Then 571mm a r y1aiF 0 a i o 15 a nmal 3011mm Colorado Centerlor Amrodynamics Research The University at Colorado ng Square Root Filter W Tlllls a y I M10 16 Usnlg llle 5011111011 01 a quadrant qulaIan alley some algebra 17 whale the A slgll 15 1110551110 prevent llle posslblllly that y 11 when an 1 Colorado Centerlor Amrodynamics Research The University at Colorado ht Sguare Root Filter Recall om Eq ll l 11a W WA A 1393 H llraFF l and K W 19 Hence VVKIEI 30 luck 15 the measurement update for W Colorado Center lor Aslrodynamics Research TheUniversily ol Colorado Llamas S uare Root Filter The mm Cumpmaliounl Algnri un Gwen PE xk yk 1 1 where a WKW Compute 6 WFWMKF 1er Colorado Center lor Aslrodynamics Research The University ol Colorado Qr n mare Root Filter 77 Integate the reference orbquot and d A21 forward 0 k1 s Timeupdme 1K and 39 m k1 9 Remm to StEp 1 mm k kel Colorado Center for Astroaynamus Rsearm The University or Colorado E n Qauare Root Filter y H 3 HS comams 39 39 r 1 u er N um 39 39 y we would r e 1 proceed m m tune npdma of step 8 me 11m uullke P Wmmmus nquot dmmu elements u 2 W13 um innmen m The 2111an algumlm reduces Wm an upper 01 low tunngl1annalnxsee Bxermau 1977 Colorado Center for Astroaynamus Rsearm The University or Colorado WERE Instability of Kalman Filter Q Potter Algorithm Summary of P2 Results 1 28 1 3 1 2 248 Exact to order 8 Conventional Kalman l2 13s 1 1 1 13 24 12 1 4 Joseph Batch 12s 13s 128 13 13s 28 13s 24s Colorado Center for Astrodynamics Research The University of Colorado Square Root Solution Method The solution to the least squares estimation problem including a priori information I7 and f is represented in the normal equation form as which can be expressed as Computational problems may be encountered in forming and inverting the matrix M H TWH 17quot CW3 Copyright 2005 HTWHF 1quotH7W3F 1X 511 may 512 Square Root Solution Methods I An orthogonal transformation approach can be used to write Eq 512 in the form R 7 513 R is an nxn upper triangular matrix can be obtained by backward substitution without inverting R Using the orthogonal transformation approach accuracy can be achieved with a singleprecision computation that is equal to the accuracy obtained by invelting M using doubleprecision computations 21 Copynght 2005 Square Root Solution Methods Cholesky Decomposition Before examining the orthogonal transformation approach we will describe the solution to the normal equations using Cholesky Decomposition We begin with the normal equation M N 512 M is a symmetric positive definite matrin and let R be an upper triangular matrix computed such that RTR M 521 22 Copyright 2005 Square Root Solution Methodsp Cholesky Decomposition 0 Thus RTR39 522 o De ne Rf 523 0 Then Eq 522 can be Written as A RTN 524 0 The components of Z can be determined rising a forward recursion relation 0 Then Eq 523 can be solved using a backward recursion to obtain the elements of o The elements of the error covariance matrix P HT WH 13quot 4 Mquot can be obtained from the condition Tr mm r P Mquot 0712quot RquotR39T SST 525 23 Copyright 2006 Square Root Solution Methods The Cholesky Algorithm 0 The Cholesky algoritluu for R is derived by equating the elements of R7 R to the elements of M o The resulting algorithm is lZ H I ZI 12 17 H 526 1i Alvequrk 397rl 11 39 k1 o The elements of z are obtained from an expansion of Eq 524 x1 3 S39n r quotu 317 11 24 Copyright 2006 Square Root Solution Methods The Cholesky Algorithm i may now be solved for 117 Nil 1 528 h A 39 1 i Z lf Fm The elements of S are obtained from an expansion of SR 2 I Examples of the application of this algorithm will be giVen later 25 Copyright 2005 Square Root Solution Methods The Square Root Free Cholesky Algorithm Note that the Cholesqu Algorithm requires square root operations which are computationally inef cient A square root free Cholesky algorithm is given in Section 522 of the text and may be derived from the factorization M UDUT 5210 Where U is unity upper triangular and D is diagonal 26 Copyright 2005 Solution of the Linearized Equations of Motion ASEN 507 0 42202 G H Born Introduction The equations of motion for a satellite are given by X FX r where l X XYZXYZ 17 X is the state vector containing siX position and velocity elements and an m vector represents all constant parameters such as gravity and drag coefficients that are to be solved for Hence Xis a vector of dimension nm6 Equation 1 can be linearized by expanding about a reference state vector denoted by X 9 ie 6230 6X0 Xe Xt Xt Xt h o t 2 The indicates that the quantity is evaluated on the reference state By ignoring higher order terms h o t and de ning xEXt Xta 3 we can write Eq 2 as aim xt6XI xt 4 De ne AU E 62km 6X0 then x0 Atxt 5 Equation 5 is a linear system of first order differential equations withA being an n x n time varying matrix evaluated on the known reference state X Note that 0 sothat 0 03 6X0 Because Eq 4 is linear1 and ofthe form Atxt the solution can be written as 6x0 xt x 6x0 0 It is also true that 6X0 xt x 6 6X0 0 This follows from the fact that the reference state does not vary in this operation i e 6x0 6Xt Xt 6X0 6x0 6X0 X0 6X0 39 It is easily shown that Eq 6 satisfies Eq 5 Differentiating Eq 6 yields x7 is a constant vector x0 22 x0 7 Equating Eq 7 and Eq 4 and using Eq 6 yields x0 650 x0 a tx0 Atxt 8 Substituting Eq 6 and 7 into Eq 5 yields 6Xt 6X0 6X0 At 6X0 9 1 A differential equation of any order order of highestordered derivative is said to be linear when it is of d 2 x dtz the first degree in the dependent variable and its derivatives and their products e g x 73t2isa I dx second order equation of first degree hence linear while x 31 4 is second degree and nonlinear t De ne the state transition matrix to be tt0 10 The differential equation for the state transition matrix is given by Eq 9 axe 6X 0 d3tt0 At tt0 11 with initial conditions r0rg1 Consequently Eq 6 may be written as xt t to x0 12 The State Transition Matrix Insight into the mm state transition matrix can be obtained as follows Let 6rt 6X mtg 7 ttg56 t5 mm 13 7 mm a K Note that 3 t to is an m x 71 matrix of constants partitioned into an mx6 matrix of zeros on the left and an mxm identity matrix on the right where m is the dimension of and X is of dimension n Because of this it is only necessary to solve the upper 6x71 portion of Eq 1 1 Equation 1 1 also can be written in terms of a second order differential equation This can be shown by differentiating Eq 13 arm 6X0 6X 2 0 BX 052 0 dgtt t0 arm 6X0 am 6X0 14 6X0 6X0 0 mm In the above equations 0 represents an mxn null matrix Notice from the rst of Eqs 14 that a 1 2 15 Hence we could solve this second order system of differential equations to obtain mg ie m 6X0 g I to 20 2 Bid axe 6170 are am 16 0 7X0 6va are are a m a 0 1 6X0 mm or am am am 1tat0arw 1tatg6ft 1tatg a 3tatg With initial conditions 1toto Ilsa 03xn73 1roro mug 01m 11 01W 1 We could solve Eq 17 a 3 x 71 system of second order differential equations instead of the 6 x 71 rst order system given by Eq 1 1 Recall that the partial derivatives are evaluated on the reference state and that the solution of the m x 71 system represented by 3 tt0 0 is trivial ie 3tat0 01 11mm In solving Eq 17 we could write it as a system of n x 71 rst order equations ie 1rr0 2rr0 am am am zttg 6r 1htgMOMUUH a MUG 18 Mme 0 It is easily shown that Eq 18 is identical to Eq 1 l i e W 01m 11 01W M g t tg 2 am am am 07 3 0 am arm a 3 739 MUG quotquot MUG 01m 01m 0le Or d3tt0 At t t STATISTICAL ORBIT DETERMINATION Concept Exam 2 ASEN 5070 LECTURE 20 101806 Colorado Center for Aslrodynamics Research The Universin of Colorado ASEN 5070 g CONCEst EXAM 2 10 17 06 T1115 or False 1 If x and 3 are not independem Variables and E o Ihen E y 0 u Gwenme Join lensxty function fy Where 4 assume values between 0 and x the marginal density funenon of x is given by gm Ix r dd n3 3 In problem 2 he mean Value of X is given by an fyd o Colorado Center for Aslrodynamics Research The Universin of Colorado ASEN 5070 i 7 CONCEPTS EXAM 2 10 1706 4 Given ftxr and gr where 10lt 39lt 10 C ltr the probability that 0lt lO given that fr lt t39 quot0 is 190 I f it39ldt39rlr WHEN4 ylt0 m lgrrl 390 5 Given observation data and a priori formation for the state vectori with variance covariance P if i 0 the n prmr39r itrfonttatiort will not in uence the rninirtmrn variance estimate of xi Colorado Center for Astrodynamics Research The University of Colorado ASEN 5070 CONCEPTS EXAM 2 i 101706 6 If i is a biased estimator of it ie E i x b the estimation error covariance matrix is given by PErarxtbrtt xt bl i If the state is a rector of constants fl H 8 i re diagonal elements of the estimation error covariance matrix are the variances of the errors in the estimates of elements of the state vector Colorado Center for Astrodynamics Research The University of Colorado ASEN 5070 m CONCEPTS Exm 2 101706 9 If a linear relationship exists between two state variables the information matrix will be singular The ruininruru variance estimator chooses the value of x that minimizes the observation error covariance matrix Q l l The weighting matrix for the rniniruurn variance estimator is the observation error covariance matrix Colorado Center for Astrodynamics Research The University of Colorado 7 g 6 STATISTICAL ORBIT DETERMINATION Probability and statistics review ASEN 5070 LECTURE 12 13 and 14 92506 Colorado Center for Astrodynamics Research The University of Colorado Given 561 axl bjc2 562 0x1 6x2 with xl x2 and a b c and e are constants 561 562 a Determine the A matrix b If a 1 c e 0 what isqgttt0 Assume initial conditions c10x20xmx20 are given at to 0 Colorado Center for Astrodynamics Research The University of Colorado lfthe H matrix is not of full rank ie HTHy1 does not exist can we make HTWH 1exist by the proper choice of a weighting matrix W 2 1TorF 2 Justify your answer The rank of the product AB of two matrices is less than or equal to The rank of A and is less than or equal to the rank of B Hence the answer to 1 is false Colorado Center for Astrodynamics Research The University of Colorado 16 Given range observations in the 2D flat earth problem ie 2 2 1 i0 xO th x5 yo yoti jgt y Assume all parameters except x0 and x0 are known We can solve for both x0 and ice from range measurements taken simultaneously from two well separated tracking stations T or F 2 Justify your answer in terms of the rank of H Colorado Center for Astrodynamics Research The University of Colorado H ax axn apn ap P12 x0 x01 x52 apn ax 6pm ax pxl pxz x x t x1 x x t xZ yo y39all kgtf ys 1 2 2 J xn at x1t pxl C Ct xnjc t xzt K Kt pxz Colorado Center for Astrodynamics Research The University of Color do l The differential equation Xa5cbx5c0 is choose all correct answers gt9 NT 2 d order and 2nd degree 2nd order and 1st degree linear nonlinear Colorado Center for Astrodynamics Research The University of Colo rado The Ulll Colo aqo my rubu A rumlum mllcumc m n c 4 mule u vl39 m man h39 mum at stplc mine on c m nmcn um I r5 nch A u h r Um Vnr lepcImNu Lemh m momm mum um nwr 3cm numcncul nlucs 0K rculmnunx 01 the min unable Example A mm h lllpprd u um he domain I 39 w mm Appcdn in let denou Ilk numbcr ul umo he I ml numhch Example 2 A pmnl on vhu Mum m Ihc mm u down on mnIum Lu X and 339 More m Inmuur uml longmm The mum or me mmm vommlc m r 5 he pan 1 numst 2 Y27 V x l Calla lol As1lodynamics Hesealch lado vels39lyol Colo s on paws m mo rectangle 4 mun m hunznnm mm wgum B mm m wmcm mm mgmn AH pmuu m m hm rognm 11 B an palms m mm am pm B pm PB lAB mm lolado Coma lo Asuodynamics Research The Univels39lyol Colorado I and Statistics Review Axioms of Probability pA39 quot A an Ain mmm mheadsappeamonthe ipofcoin p AB is the numerical probability that both events A and B oecun o TheAxiomsomebahiJity 1 pA Z 0 2 pS1 S is the certain event a pA B pA pB pAB 693 Capynght 2mm and Statistics Review 9 Axioms of Probability Conditional Pmbability 14 Two events we independentif 14 pAandP 123 ie pAB pApB Capynght 2mm Probabilit and Statistics Review 9 Probability Density amp Distribution Functions Probabilityovu39uleima val xxdx isde nedin p S X S as 45 A4i1 a Fa iecm nmusnndomva nbleaxiomsland2become 1 f I Z 0 A42 2 L fxdz 1 A43 o Thethildax39mmbecomes 3 pa g X g c dr AA4 0 whichforaltbltc pasxscgtquotfzdwjbquot fwdw paSXSbpbSX L39 A45 n Copyright zoos Probabilit and Statistics Review Probability Density amp Distribution Functions Out of interest in the event X S a we introduce F x the disrributionfuncr tion of the continuous random variable X and de ne it y pX g ft it A46 It follows that F7oo 0andFoo 1 From elementary calculus at points of continuity of F 1Fc M f I AM y 394 Copyright zoos and Statistics Review Probability Density amp Distiibution Functions r WI Example pc s X 3 24112 I T L 73 Density function of a Distribution function of n rnanunm randnm vnrinhla pnn mimm mnrlnm vnvinhlc n Ling lav quotLm b pangbfzdzFb7Fa A43 V k 39 Emmaxiomslandzwe nd cdnpynglnizoo o g Fz g 1 Ai4i9n and Statistics Review Expected Values memudvaluemdlemeanoinsw nenEXandisde nedby EX W mfzdz A51 mekthmomentofXabmndieotiginis EXquotk ntfndz A6l a Wealsomayspeakof iekmmommofXabwl iemcanhInlhiscasewe de ne HICEEM why inc Aimisz Ab2 Notematu ulsodenotedasazvf scalledthevntiameofx W3 2 EX 7A1 ziAi zm A53 and Statistics Review A v mevalianoeofbwillbemud ngetdlandmofa oapmm 200 Expected Values 0 02 M quot u A 39 L vuiuv x M 11 X z I M i a b L 11Note m Expected Values o Amw u vixwa rumu mueaamuixmnnwope m Em bX a bEX A64 whenaandbmoonsummAlso Also uz MAUJWW L 2zz 2zhlfzdm Ag ZAf tAf A2 r A A66 miln 2 Mom w m cammot a a a 1 x MM and Statistics Review a The Gaussian 0r Normal Density Function n n M exp 7 92 foo lt I lt on A87 The Gaussian density function is depicted graphically as f 1 xl o 11 1145 oapmm zoos A1a p1 g a g x 3 A1 0 mm 68268 A 0 A1 7 217 g X 3 A1 20 95449 p1 7 3039 g X 3 A1 36 39730 i oapmm zoos and Statistics Review A7 MOMENT GENERATING FUNC lONS Consider the panicuInt once of EnX grfi rlf A7l for which my where H is a dummy variable Since W 11Lr A12 substituting Eq A7Z into Eq A7 it results in WA an Etc nv1T2 A73 WI Capynghtl Thus Ere may be said to generate the moments An A V A of the random variable X It is called the moment geuz rvlring mmion of X and is wcitten AI 9 Note that Ah A74 hen Accepting the feet the the moment generating function for the function 12 X is e39 en by Monte e quot frdr A751 1et 71X X e Alethen Ankmat Hinge A76J A77l Capynghtl Probabilit and Stati From Eqs A73nnd A16 Miximlgl 87ml lt and for example lmAl e all H20 A2 7 Ag recall that WE cnmmm stics Review 9 92A 6 0l7l12 ll091 Millin 7 Alt 1A1 was H Il va ableamaybewnnenas Fz y 2 MI y E f x y dz dgy 2 oapmm 2065 Probabilit and Statistics Review 9 Two Random Variables In analogywi iasinglemndom vatiabls 1edenaityfunc on mttwomdom pSXSzdwySYSydyfIydIdy A96 o Inmmmaryfmtwomndomvariables joint distribution function of X Y joint density function of X Y joint probability element of X Y Probabilit and Statistics Review Marginal Distributions given the ueuaviu uf two pX S zuo condition on Y F I 00 when m Fzooj 7fnudvdu 7fuvdv du gudu oc Wk oapmn me We often want to determine the probability behavior of one random Variable Ai101 A102 Probabilit and Statistics Review Marginal Distributions and g0 f1ydy u Similarly My Hz mm isthemal gimldenxityfunc onofl capyng n me A103 A104 Probabili Independence of Random Variables Xandx 39 39 ii ML 11 gzhy A111 where g m and h 3 are the marginal density functions ofX and y 33 Cnpynght zoos and Statistics Review 9 and Statistics Review Conditional Probability mmaingymmmeaimpieevemAmdammwede m4 by 9 f m 11 91 gigs h h1 7 MAB pBA FM I A12i1 n1 n 1 39 39 39 X mld V withdenn yfunc om fx y i g Win 3 by govy an hwz Am IfXandeindepmdentwehave fix 1 9Ihy y1y To 7M9 91 A125 Probabilit and Statistics Review 9 Expected Values of Bivariate Functions TheexpeaedvulueE X Y ofanarbiu39aryfunction X Yoftwo con nuousmndomva ablesXaninsgiwnby wa Y1 wyfzydzdy Am X Y X Ym EX Y39quot z ymf2ydcdyEAlm A132 PM 2 quot written A die bu moment ofX Y about the origin Probabilit and Statistics Review Expected Values of Bivariate Functions The bu momemabout lemeanisobminedbysel ng a3 X Y X Am Y g Amyquot This results in E X in Y AM 1 Aiullly Am39 fz ydz dy E mquot A133 y 394 cg ht zoos pm 22 Aan 1 Am E XL the mean ofX 01 E Y its mean of 1 11 E iiX 7 E ltXgtitY 7E YD the Covariance ofX and y 20 7 m am the variance on 0 0 1 0 Hon i 0 2X the variance ofX z Comma as an emnpie he cumpulzlliun of 11EX HEIXY 01 I 7 we 7 Amiftr yidrdi J39i Amy Aoir 10Ulf ltyidf 11 A11 7 21tn1 10 Am 11 Alaoi Al34 WE Cupyngqizooa and Statistics Review Htample Problem J Given kx2y 05x5105y51 f xgty 0 elxewhere Find a k b The marginal density functions of X and y c The probability that x 3 d p x y 5 egt perm f Whetherx and yare independent g p0ltyltltxltdx h poltyltxgtgt Colorado Center for Asrodynamics Research The University of Colorado Example Problem 9 a kx yq39ydx 1 kxyy2 kx1 kx1dx 1 1 kx72x0 439 k Colorado Center for Astrodynamics Research The University of Colorado LsExample Problem b 1 gx x2ydv xyy2 4 gx x1 1 0 ltgt 21lt gt 2 x2 1 k y x2y dx 2xy 3 32 0 Colorado Center for Astrodynamics Research The University of Colorado Example Problem 0 pxsFoofgxdx 6 Afltx1gtdx 2 x2 x 3 2 0 21 1 5 3 8 2 12 6 Colorado Center for Astrodynamics Research The University of Colorado Example Problem d p xys Wax f f x2ydydx A fxyyz7xdx z 2 6 3f idx3 30 2 4 3 4 40 24 also Vy I7 xys f f x2ydxajz and p xy21 pxys as x ranges from 0 andy ranges over 0 gt x Colorado Center for Astrodynamics Research The University of Colorado Example Problem g e pxy0 f fxy x2y xl 2y a gxhy Hence they are not independent f 1 y g P0ltyltltxltdx g dy wi Colorado Center for Astrodynamics Research The University of Colorado Example Problem h p0ltxltxgt p0ltyltltxltl 170 2m x i x 2 dxd fy3 y yi gxdx x1dx 2 Colorado Center for Astrodynamics Research The University of Colorado Probabilit and Statistics Review The VarianceCovariance Matrix The symmetric matrix EY E or 430012 X 7 WOW 400 Y EltYgtX W Y E YW 200 11 P m 020 Al4i1 oapmm zoos Probabilit and Statistics Review The VarianceCovariance Matrix The oomlation wef cient is de ned as Eux E on Y mm m EIX EX212EYEY2 2 M11 00060 A142 The Variancecovariance matrix for an ndimensional random vector X can be written as Vi P120102 quot1117010quot 91217152 397 quot39P2ni7217n P z I A143 2 Pln lan Pznrfz n 39 quot 0n Wk whemp isameasmofdledegi eeof nearconeh onbetweenX39anXm Cnpyngi39it 2m Probabilit and Statistics Review 9 Properties of the Correlation Coef cient 15pysi o If p tl there is a linear relationship between X and Y ie X a l bY where a and b are constants Wk Cnpynght zoos Probabilit and Statistics Review 9 Properties of Covariance and Correlation From the de nition of M11 M11 EX EXY EYl IflargevaluesofXalefoundpahedgmu aleMIhlargevaluesonJndif mallvaluesofXarepairedwithmallvaluesonuu andhmcepx willbe positive If large values of X are paired with small values of Y then it and hence p will be negative Ifsomelargeandsmallvalnesomedepaired m I 0 mm Cnpynght zoos 20 Probabilit and Statistics Review Properties of Covariance and Correlation Copyright zoos Determine the VarianceCovariance matrix for the example problem fxyx2y Osxsl Osysl 0 elsewhere We have shovm that the marginal density functions are given by 2 g x EEC 1 my 1 2 yZ Colorado Comer ror As1rodynamics Research ample Problem continued The Univers39ry at Colorado 21 mple Problem continued g The elements of the variancecovariance matrix are computed below 1 1 Ex Am fxgxdx xx17 x 0 5 0 10 Ex 6 1 2 1 Ey 01fyhydy fy2y yy 0 0 1Ey 02 x M20 ix Aw2gxdx x1dx 13 E Colorado Center for Astrodynamics Research The University of Colorado mole Problem continued Cont 02 y M02 iym2hydy iyZ 2ydy M20 4 M02 Gzy Mn Jib 2 y 2m fxydxdy 211 5 11 7 k i 2 dxd 3lllx glly 18lltxygt y i 243 Colorado Center for Astrodynamics Research The University of Colorado M11 22 Example Problem continued The variancecovariance matrix P is given by 01 x Mm H11Xgty 02 y 162 55243 00802 4226 55 4 0226 0148 243 A7 The correlation coef cient for random variables x and y is given by Co 1100060 226l0802 148 00247 The conventional OD expression for the variancecovariance matrix has variances on the diagonal covariance39s in the upper triangle and correlation coef cients in thelowertriangle 0802 1226 0247 0148 Colorado Comorror As1rodynamics Hessardr The University or Colorado p Hence P and Statistics Review Bivariate Normal Distribution The bivan ate normal demityfunctian is given by 2 foo lt at lt oo 2 717 14 fwd 27razn11p2 exp 21 p 01 oo ltylt 00 I 1tu Am y Mir 2 7 A171 wary no Nonethatifp 0 fx y mybefamwinm Ma y 30 W Hence p 0 is a suf cient condition for statistical independence of bivariate normal variables 4 Copyright zoos Probabilit and Statistics Review Marginal Density Function 39 L Ynnd 39 39 mtmalmargimldenxityfunc ons By carrying out the integial 9a f1ydy A18l Itcanbeshown ux yz 1 exp 1 2 A182 27raz 2 111 Sim nlesultsexistfotdiemarginaldmsityfunc onofl ca ht 2006 pm 47 Probability and Statistics Review a Conditional Density Function 39 LL LA r andme normalie 1 y on payvzr Mall 2 xexp i y lipz A184 quotmmL r l EelI Au fly 2 Anal 0 alldesz am mm 7 p2 cg ht zoos pm 48 and Statistics Review a Conditional Density Function realizations ofX 39 y hm nhmh hm muma EMx 3 This would be the estimate of y given that z r3 and Statistics Review FA 4141 Run Am quot14 Tina at A estimate of Y we have several choices including 1 The mean the quotcenter of the probability massquot distribution 2 39n g u I Q of the density function 3 TL r a v lie 01 J4 ueu tum unu me piuuauinit left and halftn his rich I J x cnmm 2m and Statistics Review The Multivariate Normal Distribution o 39 39 39 39 cgquot a The multivariate nnrmai demity mclinn is given by l in 239117W8 12XATV 1XA 00 lt M lt oo A191 is the p X p variancecovariance matrix of the vector X Vl is the detemlinant of V A is a p X 1 vector of mean values of X i Cnpynght zoos and Statistics Review 9 The Multivariate Normal Distribution 0 The matrix Vis de ned as VEX7AX7AT A192 interns of ncm elationcoe icimt L A193 a m piza11azz iuv11011 91p0110zy pxuwm Mm can ma mlr nl mww nww pv01 171vP2F IQJUIPJ 02 Cnpynght zoos and Statistics Review a Conditional Distribution for Multivariate onna Varia e mmg g X is manually distributed with mean X is partitioned into two subvectois such that s i VnVag a9 2 Xi i X 3 and V 1 7 2 then the conditionalp distribution of the the vector x2 x2 is the multivariate rimmal distribution with mean and Statistics Review 9 Theorem 4 The covariance matrix of the conditional distribution am given X2 x does not depend on xg Proof The proof of this is obvious from an examination of Eq Al95 that is vwxz XQ V11 7 V12 if Vz From Theorem 3 we also have EXiXa x2 Al H3291le 7 A2 A196 Ifwe weie attempting to estimate X1 its mean value given by Eq A 196 would be a likely value to choose Also because the covariance of the conditional density function is independent of xo we could generate the covariance without actually knowing the values of X2 This would allow 6 a X in in i 1 an X2 V Cupynght z39nu and Statistics Review 9 Central Limit Theorem Ifa large number of sets of samples of Slze n are taken of a random variable with any given density function the mean of these samples will be normally distributed as w a Cupynght 2m Example Central Limit Theorem and Statistics Review 0 Perform alarge number of experiments Where for each experiment We de a large number of samples n gt 30 ofx from the density mction in Figure a If We compute the mean f x a Density mction of x 28 and Statistics Review 77x with b Histogram of sample means assuming many samples for each experiment b Number of Experiments If we take the limit as the number ofbins and experiments approach 00 f0 f 39 39 39 c Dens1t functlon of Copynght 2006 y 57 P bb39l39t dSt t39t39 R 39 Central IImIt theorem I 52 Elementary Stallstlcs Sim i In V quot n 3 r a falt5 a a we Mo 97L jaswd 2 134 n f a g m L g n z 1 rd FIGURE 225 Distribuuuns of Y 390 large numbers of Samples DI Size 71 selected 51 Elid0l1 From nonnarmal populallons Central limits hennarn Insure lhal distribunon 039 Xlends toward n El dlstrlbu Ian 5 n increas 5 Pa papulallons39 Om whl IQ re drawn are shown In a Distribunons m 7 are shown to b samples of sze 0 Samples OI SIZE 77 i 4 and 1 Samples of size N 25 After Lapin 1982 From Dan39s 7 aquot 75 rtlr ri39ca Dava w yr d A 260 033 5 0 ueyl l9a 6 Copyright 2006 58 29 Lu W STATISTICAL ORBIT DETERMINATION Kalman Filtering ASEN 5070 LECTURE 21 102006 Colorado Center ior Astrodynamics Research The University at Colorado m lonsiderations for the Kalma Fina 4 7 The Sequential Estiirmtion Algal39itiun 205 Time and measurement update Of the estimation error covariance or r mark matrix I n39n I39 n my wrbrirul 1 10 H f 4 I D k 4 Ug U U i i I Iw IiE i i Figure 472 Illustration of the behavior of the trace of the state estimation error covariance matrix Colorado Center ior Astrodynamics Research The University at Colorado Magical Considerations for the Kalman lllter In order to prevent these problems several alteninte algorithms have been suggested A common alternative is to use the form of the equatiou given by Buoy and Joseph 1968 Pi I e Kka ml 7 KijIM IrkRaKE 4719 Note that this formulation will always yield o symmetric result for Pk al though it may lnse its positive de nite quality for a poorly observed system use a square root formulation to iiprl ate the comi39izmce matrix The square root algorhhms are dismissed in Chapter 5 ooioraoo center tor Astroaynarniee Research The University oi ooioraoo E Equivalence of Joseph and Conventional 39 of the Update of Need to show that I eKHTJIeKH KRK I Ky First derive the Joseph formulation YKy 751 subst y HX YKHxs 7H 2YKHX7YK subtract from both sides and rearrange gexexeKHexKe IeKHiexKe ooioraoo center tor Astroaynarniee Research The Unmerslly oi ooioraoo Equivalence of Joseph and Conventional Matteo of the Measurement Update of form P a Ec xc x7 PI KH13I KHTKRKT where 13Ef xf xf RE88T E f 9057 0 since EU is independent of s Colorado Center for Astrodynamics Research The University of Colorado Equivalence of Joseph and Conventional 1 nation of the Measurement Update of Next show that this is the same as the conventional Kalman I KH13I KHT KRKT IKH13 I KH13HTKT KRKT IKH13 13HTKT KH13HTKT KRK7 IKH13 13HTKT KH13HT RKT subst for K FHTH17HT R 1 IKH13 13HTKT 13HTH13HT R391H13HT RKT 1 KH13 13HTKT 13HTKT I IltH13 Colorado Center for Astrodynamics Research The University of Colorado Example Illustrating Numerical Instability W 39al Kalman Filter see 471 The following example problem from Bierman 1977 illustrates the numerical characteristics of several algorithms we have studied to date Consider the problem of estimating x1 and 62 from scalar measurements Z1 and Z2 21 x1 8x2 v1 22 x1 x2 V2 Where v1 and v2 are uncorrelated zero mean random variables with unit variance If the did not have the above traits we could perform a whitening transformation so that they do In matrix form 1 21 8 x1 v1 4720 22 1 1 x2 v2 Colorado Center for Astrodynamics Research The University of Colorado Example Illustrating Numerical Instability uential Kalman Filter x The a priori covariance associated with our knowledge ofX 1 is assumed to be x P0 02 72 0 1 Where a X and 0 lt s D 1 The quantity 8 is assumed to be small enough such that computer roundoff produces 152 1 4721 This estimation problem is well posed The observation Z1 provides an estimate of xi which when coupled with Z should accurately determine x However when the various data processing algorithms are applied several diverse results are obtained Colorado Center for Astrodynamics Research The University of Colorado Example Illustrating Numerical Instability uential Kalman Filter Let the gain and estimation error covariance associated with 21 be denoted as Cl and P respectively Similarly the measurement Z2 is associated with k2 and P2 respectively Note that this is a parameter constant estimation problem Hence ltIgtz tk1 1 e H H h 1 1 hz We will process the observation one at a time Hence Pk PIH Pk I kkhklk71 kk 4th hkPth 1 1 k 12 472161 Colorado Center for Astrodynamics Research The University of Colorado Example Illustrating Numerical Instability e uential Kalman Filter Note Since we will process the observations one at a time the matrix inversion in Eq 4721 a is a scalar inversion The exact solution is k Pohf I211th 1 1 71 21 21 0 s 1 5 81 1 1 2 2 k1 W 8 4722 whereas 1 The estimation error covariance associated with processing the first data point is P1 1k1h1P0 1 1 P I 1 s 02 1 1 252 a Colorado Center for Astrodynamics Research o The University of Colorad Example Illustrating Numerical Instability cg 1 uential Kalman Filter 1 2 0 10 02 4723 where a 1282 Processing the second observation k2 Rh thlhi 1 1 1 2 a 1 1 2 a 1 k 1 1 1 2 a a 0211H a a 021H1 1 81 28 X1882 where A1 28282282 1 Colorado Center for Astrodynamics Research The University of Colorado Example Illustrating Numerical Instability e uential Kalman Filter The exact solution for P2 is given by Pz I kzhz 1 1 252 1 5 2 4724 A 1 s 2 s The conventional Kalman filter yields using 1 22 1 1 k1 8 1 Z R 1 1 s a 1 s 0 a R 2 4725 0 0 Colorado Center for Astrodynamics Research The University of Colorado Example Illustrating Numerical Instability uential Kalman Filter Note that P1 is no longer positive definite The diagonals of a PD matrix must be positive However 3 1 does exist since a 0 Compute PZ using the Conventional Kalman k2 Rh hzfihzT 171 1 8 1 8 1 23 PzI kzhzPl 1 1 1 4726 1 28 1 1 NowP2 is not positive definite le 0 nor does it have positive terms on the diagonal In fact the conventional Kalman filter has failed for these observations Colorado Center for Astrodynamics Research The University of Colorado Example Illustrating Numerical Instability 5 uential Kalman Filter The Joseph formulation yields T Pz I kzhzPlI kzhz kzk P 12 13 Z 1 33 2 3 The Batch Processor yields 2 71 Pi yaw 1 H1 sym 1321 2 121 1 5 221 1 Colorado Center for Astr odynamics Research The University of Colorado Example Illustrating Numerical Instability cg 1 guential gKalman Filter szorthe 128 13e batch P 2 13e 24e To order s the exact solution for P2 is 1 15 18 2 P212s 12e 13e 13s 2mg Colorado Center for Astrodynamics Research The University of Colorado Example Illustrating Numerical Instability Tie uential Kalman Filter Summary of P2 Results Exact to orders Conventional Kalman 12s 13s 1 1 1 13s 24s 125 1 1 Joseph Batch 128 138 128 138 13s 28 138 24e Colorado Center for Astrodynamics Research The University of Colorado Notes on forming HTWH for the Batch Processor ASEN 5070 May 14 2002 George H Born Assume that the state vector is given by the general case y mcl where n33 m r and 1 are the 3 X 1 position and velocity vectors is an x1 array of constant parameters which do not appear in the observation state equations but do appear in the state propagation equation eg gravity and drag coefficients y is an m X l array of constant parameters which appear in the observation state equations but do not appear in the state propagation equations e g station coordinates and measurement biases The state transition matrix for this state vector is given by Bru am a 6 w w w m 2 6rn 61 y 6 6 I 2 0 W W W W El 07 6 6 6y 6y 6y 6 El 07 Simplifying Eq 2 by recognizing that some of these submatrices are either null or identity matrices yields 6r t la L quotM w m r 363 6i 3163 r w a L W 6 mad mm m Assume that range and range rate p observations are being taken The H matrix can then be written as 6W 6W 6W 6W arm arm a a 650 650 650 650 area arm a a m Equation 4 can be simpli ed by recognizing that p is independent of 1 and that all observations are independent of Hence 6 t 6 1 p0 all m 13 0 rm o 6150 6I50 0 6I50 Bray 1163 aria 1x3 M a 1m 2m The Ht matrix may now be formed from Eqs 3 and 5 ie HmHm mm 8pm arm 0pm arm 8pm arm 8pm BADFar 1x3 y Bra Big 1x3 Bra 0 1 a lxm 020 arm am arm am arm am arm 820 3 1x3 am 0 arm a M By 6 7 am am am arm 1 Bra Br Brit Br Bra Big Brit Big lxm x For simplicity designate the elements of H as E F G K 1x3 1x3 lad 1m 2m H A1163 31x3 Clad Dlxm I We may now form H TH We will ignore the weighting matrix because it only scales the elements of H TH and does not change anything said here Now ATET HTHBTFT ABCD CT GT EFGK DTKT ATA ETEM ATB ETF33 ATC 1376 ATD ETKW BTA FTE33 BTB FTF33 BTC 1176 BTD FTKW CTA GTE CTB GTFM CTC 676 C7D from DTA KTE DTB KTF DTC KTG DTD KTKW 8 mad mad ml m Had we included the weighting matrix as a block diagonal matrix with w 1 and w as the weights for p and prespectively ie w1 0 W 9 0 WZ 2x2 then the rst and second term in each submatrix of Eq 8 would be multiplied by W1 and w respectively eg the rst element would be ATwlA E7w2E and the second element would be ATwlB E7w2F Term Project Example Lets examine the matrix on the lower right of Eq 8 for the case of the Term Project DTDKTK z emf 6pm emf am 10 6y 6y 6y 6y In the case of the Term Project m 9 and y represents the 9 coordinates of the three tracking stations Hence 71 r 7 2 11 7 3 9161 where each vector yi 139 l 23 represents the three coordinates of one of the tracking stations Using the range measurement as an example we can write DI6prapr 6pm 6pm a 12 6r 67 67 67 and DTD of Eq 10 becomes r 6 6y 6y By 63 1x9 wr 0 WOT 6W 6ptT 6W WOT 6W 6 67 67 67 67 67 apafapa apafapa apafapa 13 6r 67 67 67 67 6 apafapa apafapa apafapa 6 6r 6r 67 67 6 However in the case of the term project at any given time no two stations can view the satellite simultaneously Hence all submatrices except those on the diagonal are null Thus Eq 13 becomes 3 63 03x3 03x3 DTD 0363 M3163 03x3 1 03x3 03x3 Q3163 where the symbols in Eq 14 are de ned by the corresponding terms in Eq 13 The same conclusion is drawn for KTK in Eq 10 Consequently when the information matrix N AZHFWIHH 15 11 is summed over all N observations the 9x9 lower right submatriX of Eq 8 will contain null 3x3 submatrices as shown in Eq 14 corresponding to all pairs of tracking stations which do not simultaneously track the satellite Question Does the covariance matrix contain null submatrices in the same locations ie are the estimation errors in station coordinates for stations that cannot simultaneously track a satellite uncorrelated 393 VJ i i Es A E lera u quot hi B MMmu ya Elamwil Statistical Orbit Determination Software Packages and Current Research Brandon A Jones University of Colorado CCAR 393 j i ii in MiamiJ our lahu ymmm MW The Orbit Determination Tool Kit ODTK Brandon A Jones University of Colorado CCAR Introduction a Summary of ODTK Scenario Setup Process Data Processing Data Output Sample Results Brandon A Jones Boga momma m amilv mulls3 immui b ODTK Description 19 Provides OD and orbit analysis support Estimates satellite state Estimates environment parameters Pro le equipment characteristics Covariance analysis Integrated with Satellite Tool Kit STK Primary Tools Tracking Data Simulator Filter Capabilities Least Squares Estimator Sequential filter Filter Smoother GraphReport Generator Brandon A Jones scan Migrant vim illalwiw mini Myst ODTK Description B M junm sa Elamwib Residual editing Combines multiple observation sources to provide state estimate Includes vehicle attitude variations Advertises realistic covariance CCAR studies have shown this varies from satellite to satellite Brandon A Jones 4 73 ii 3quot Scenario Setup lahu yimmm MW Object Oriented Implementation Satellites Sensors GPS ReceiverAntenna pair FiltersSmoother Etc Object Browser and Properties window provide primary interface Brandon A Jones 5 Satellite Filter Properties osmium W i i7 i 7 Name l Vain Name Value Name Vania FuvczMndzl Pinuagamillonlinlx esaipli Giavily InlegiaLiDnMel Rio Ulbllslal Cartesian Degiewn 7U 39 SlapSIza EouvdFiame JZEJDD Tides Triqu l 5 min Epoch 13 Jun Z l 23 53 47 UEIEI Snlid39l iu2 TiuzAnuni deg XPnsnian 72933 34 km Time Ii 2 Ecce limit CI 04 r39F uxiliuri 7373 235 km riimSn Ephemeiis eneiaii ZPusiliuri Dues liuz Slair39lime lEltluri 20m 23 59 3 BED Welncily 5 2m M Mike4 MaxD ZD S an RED mm WElDDily 72 850quot km39sec l mm c l TimeSlep 1 min ZVelucily 3 51535 km xECAl GeneraIH iiue CiealeSTKFile lake Esiiriiaz ibil We i Variational 5 TK lename C STKJME phEmElldB S UilmElass i 2 dToS ii else PhysmalFmpevliex 7 Under 0 ScriplFileviame double click to edit a 522 kg PIECESXN Acceleratinn lake z Anide Use Based Elii libil lass DibiLlnL21lairily Swine File WW 7 FLwHia SUD Hi Filename C S T Kv6Fi3EHvDG2v l Emmi ingma 10000 in Sign im 1 i i i i i i i i i E iu2 Lannie DDni SlopTime i i S 1 Rdnlsigma I S iri sec W EEHVElU39MEFSI l Comm ldaLstgna U 4 m sec l X D m E Iqu CdULSlina 0 Z m sec l Y D m S l Rchniielaliun I 1 ThimBudies RCtaiielalmn U MSESWEMER PNJG Sun quotHE HRduLmiiela El Tiackiring lEIEIEI Maori qu Hld LEBlleiall Cl Meas was double click to edit Planets double click lo edii 7 RCdaLwiiela U RemaualEdiling Useln raise lEADDIlelallCM U NominalSi 3 Drag IRdnLcoiielali U Dynar i Ll 9 Based Eln Clibil ldolcuilelalinri I Enabled lalse WillUSEAll r lEdaLcmielali U Highs WEI AimDensil CIRA1572 CRduLcuiiala U mum in EsLimaleD lalsz ElduLwlielati 0 Mom 3 SpecMellri Max Ales Dd EEdoLcoiielal U Initial 0 min ED M6773 RdalldoLc ll 0 Thinningiimz 0 sec Area 0 7ZU35 ni 2 HdulEduLcnii 0 WWW 7n him i i mm n Brandon A Jones 6 i39 i Data Processmg W Two Primary Data in L Sources Simulation Data M External Data zmiiuisiaizmii mm i m u 7 minimum iimii ii 5 Several external data formats recognized RINEX More Data analysis automation through scripting u Monte Carlo Analysis Tim 5 mi italU lm ll 1 I Brandon A Jones 7 ammrm 3923 r Data Processmg Data simulation tool is capable of generating all data sources processed by ODTK Used for preliminary analysis and performance evaluation Assists in satellite and ground station design phase Helps determine operations requirements Brandon A Jones 8 Characterize filter amp smoother Aagi Intrack Position Uncertainty 095P Filter Current time process Smoother Postfit process Smoother Processing Direction rl Filter Processing Direction Prediction Error Growth l Filter Correction M lquot g at Tracking Data l ll Wquot l Data Gap Ml ml Smoother l i PostFit Solution u l H r lquotl l ll l i quot l l n 9 quotl w n l n l y ll lquot l rw quot mwquot in ii quotv kin rn rm llquot v J J u MW Ll V w it l m it v l39l Al y a milwht pvnwk li39xbjl i 1quot W 1 l 12 16 2b 24 28 3 2 56 4o 44 4s 52 ll J ll quot ill Pg 9 of 35 9 Data Output Smoother and Filter imp output as a STK ephemeris file Can output state and Z 39 F39 W covariance information Brandon A Jones 10 Data Output l r Easy import of ODTK output to STK Allows for analysis utilizing other STK tools Visual comparisons to another ephemeris J iiii Brandon A Jones 11 STK can be used to visualize OD Tool Kit process I slkcom Pg 12of35 Data Output E m StaticDynamic Product Builder Charts for visual output Reports for data output Multiple data formats MS Word PDF Text Reports allow for postprocessing of ODTK results Brandon A Jones If f RIC Errors RIC orbit differences radial RMS 041 m intrack RMS 054 m crosstrack RMS 024 m range RMS 072 m ABV mqqoq r r r 500 000 1500 Minutes past 18 Jun 2001 235947 Brandon A Jones 14 Measurement Resrduals Measurement Residual and Sigma E n E 9 m 13 E E E g 10 2 20 L 30 39 0 500 1000 1500 Minutes past 10 Jun 2001 235947 Brandon A Jones 15 Current Status v 40 Aagi OD Tool Kit will meet the accuracy requirements for most commercial and military satellite programs Force models complete through ocean tidal perturbations Tracking data models for most classical systems Achievable accuracy is limited by tracking data density and quality Force models limitations imply orbit error gt 10 cm OD Tool Kit has advantages over most custombuilt OD products Accounting fortimevarying biases Solving for corrections to maneuvers In the near future Force models will be extended to meet geodesy accuracy requirements Achievable orbit accuracy approaching 2 cm P9160f35 Summary ODTK provides most OD software required for data analysis Includes state estimate and covariance analysis capabilities Data export capabilities provide increased flexibility during data analysis process Questions Brandon A Jones 17 E i momma 15w ahux ymu sa immui b GlPSYOASIS GOA Brandon A Jones University of Colorado CCAR GOA Overview illalw yt mini Wyatt GPSlnferred Positioning SYstem and Orbit Analysis Simulation Software GlPSYOASIS Product ofJPLNASA SquareRoot Information Filter SRIF SRIF Smoother Advertises 12 cm level accuracy onorbit and terrestrial scenarios Brandon A Jones 19 10 a GOA Uses 393 5 E alw mmu ss Examui h Primarily processes GPS observations Aids in mission design process Provides capabilities to generate and process simulated observations Aids in operational OD Brandon A Jones 20 9 GOA vs STKODTK Advantages Pedigree le y mining lilaaaa ii Various modulesutilities have uses outside of GOA data processing GOA provides increased scenario customization Disadvantages Requires increased understanding of OD process Unix command line interface reduces user friendliness GUI is provided but reduces user control of OD processing Brandon A Jones 21 11 GOA Flowchart aim a i 9M iii 2 i Source GOA Tutorial Course Notes 5 22 Brandon A Jones foam 1 GOA InpUt Function inputs provided by FORTRAN namelist files Processes simulated and recorded GPS observations Recorded observation format RINEX Brandon A Jones 23 12 GOA Output ahux y numb3 Examui h Outputs filter state in FORTRAN binary file Includes utilities to convert output to text output in a variety of formats sp3 jpltext sp1 etc Outputs covariance in similar binary file Includes some graphical output capabm es CCAR studies utilize MATLAB to customize graphical output Brandon A Jones 24 9 P Q l dlun u aw er ymmm itaaaa il Expected OD Accuracy for High Altitude Highly lnclinated Satellites Using GPS Brandon Jones University of Colorado CCAR 13 Outline a Beaa mmer my 93M mugs3 Elamwi b Basic explanation of orbit determination OD and GPS Simulation development Summary of previous tests Results Future work Brandon A Jones 26 What IS OD ibbu ytmmm M35th Method used to determine the precise location of a satellite when given a collection of measurements Reference Trajectory 1 H KERN Trajectory IllK Estimated Trajectory Brandon A Jones 27 14 OD Measurements womb ahu y mulls3 Examui h Many different forms of measurements Range RangeRate from Earth based tracking systems Azimuth and elevation angles relative to Earth based tracking station Visual navigation measurements from star tracker Range Rangerate from satellite based tracking systems GNSS Brandon A Jones 28 9 GPS and oo 1 llaiw y mining itaaaa il Continuous measurement Coverage Range CA and Phase Rangerate DOD DOD High accuracy 12 cm D0 M Reduction in operation costs Earth based tracking not required Pedigree Brandon A Jones A 29 15 ahu y mulls3 Examui h GPS and oo 2 Satellite positions are known thus range measurements are used to triangulate the satellite position For realtime position estimation four satellites must be visible for position estimation Requires at least four equations for the four unknown values X Y and Z Time Brandon A Jones 30 GPS Visibility Tilllw y mining lilaaaa ii GPS satellites orbit at 20200 km altitude Primary signals broadcast in 278 deg cone Side lobes provide weakened signal M Limits satellite altitudes 39I for optimal visibility Acceptable for most LEO satellites Brandon A Jones 31 16

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