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Numerical Analysis

by: Manuela Boyer PhD

Numerical Analysis EEE 244

Manuela Boyer PhD

GPA 3.66


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This 13 page Class Notes was uploaded by Manuela Boyer PhD on Monday October 5, 2015. The Class Notes belongs to EEE 244 at California State University - Sacramento taught by Staff in Fall. Since its upload, it has received 21 views. For similar materials see /class/218828/eee-244-california-state-university-sacramento in Electrical Engineering at California State University - Sacramento.

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Date Created: 10/05/15
CHAPTER 23 231 x x X12 0261799388 0965925826 XH 0523598776 0866025404 x 0785398163 0707106781 X1 1047197551 05 X2 1308996939 0258819045 true isin7r4 7070710678 The results are summarized as firstorder second0rd er Forward 7079108963 7072601275 711877 72674 Backward 7060702442 7071974088 14154 71787 Centered 7069905703 7070699696 1138 0016 233 x x X12 18 6049647464 XH 19 6685894442 x 2 7389056099 X1 21 8166169913 X2 22 9025013499 Both the rst and second derivatives have the same value truth 82 7389056099 The results are summarized as firstorder 7401377351 0166750 7395215699 00 361 First derivative Second derivative 235 The true value is lx 15 19459171098612 D2T secondorder 7389031439 0000334 7389047882 0000111 02 0211824 D0 1791759261386294 0202733 D 30202733 7 0211824 0199702 237 At x 7 x1 Eq 239 is 2x e x e m x171 7x x171 7 11 2x 7 x171 7 x11 x1 7 x171x1 7 x11 f39xfx1 fx 2x 7 x 7 x fxx1 71 x11 7 x171 x1 7 x1 For equispaced points that are 11 distance apart this equation becomes xrhxxh h 7h f x 7 magi foe 2 h 11 7117211 fx117 211k 7fx71 0 fx11 fx117fx171 2h 2139 211 239 The rst forward difference formula of 0h2 from Fig 231 can be used to estimate the velocity for the rst point at t 0 f0 7 58 432 7 30 149 225 s The acceleration can be estimated with the second forward difference formula of 0h2 from Fig 231 7 78 458 7 532 20 km u 0 f 25 7000967 S2 For the interior points centered difference formulas of 0h2 from Fig 233 can be used to estimate the velocities and accelerations For example at the second point at t 25 f 25 7587 0 1167km 225 s H 58 7 232 0 km f 25 2 70009687 For the nal point backward difference formulas of 0h2 from Fig 232 can be used to estimate the velocities and accelerations The results for all values are summarized in the following table t y v a 0 0 140 00096 25 32 116 00096 50 58 092 00096 75 78 068 00096 100 92 044 00096 125 100 020 0 0096 231 1 Here is a VBA program uses Eq 239 to obtain rstderivative estimates for unequally spaced data Option Explicit Sub Te5tDerivUnequal0 Dim n As Integer i As Tnte Dim x100 As Double y10m Rangequota5quotSelect n ActiveCellRow SelectionEndxlDownSelect ctiveCellRow 7 n Rangequota5quotSelect For i 0 To n ActiveCellVa e ActiveCellOff5et0 lSelect yi ActiveCellValue ActiveCellOff5etl 71Select x i r As Double dy100 As Double Ne t For i 0 To n dyi DerivUnequalx y n xiH Next i Rangequotc5quotSelect For i ActiveCellValue dyi ActiveCellOff5etl 0Select Next i End Sub Function DerivUn Dim ii A5 Tntege If xx lt x0 Or xx gt xn Then DerivUnequal l5e equalx y n xx r quotout of rangequot If xx lt xl Then DerivUnequal Dnyxx x0 xl x2 y0 yl y2 Elself xx gt xn 7 1 Then DerivUnequal 7 7 Dnyxx xn 7 2 Else For ii 1 To n 7 2 If xx gt xii And xx lt xii 1 Then If xx 7 xii 7 l lt xii 7 xx Then 391f the unknown is closer to the lower end of the range 39xii will be chosen as the middle point DerivUnequal 7 Dnyxx xii 7 l Else 39Otherwise Xltii ii 1 yltii 7 1 yltii yltii 1 if the unknown is closer to the upper end iil will be chosen as the middle point DerivUnequal 7 nynxlxx xlu xlu 1xl11 2 Vin ylu 1 Vin 2 End runotron runmon uyuxlx XEI x1 x2 yEI y1 y2 DnyyE l2 xx1xZlx x1lx xz7 y1 l2 xx x2lx1xUlx1xz7 yZ l2 xx x1lexUlx2x1 unu runmon When the program ls run the result ls shown below Mama solutron fx aseix e IMQquot The results can be dlsplayed gmphlcally below where the computed values are represented as polan and the true values as the curve 0 and T T ml NW mm ofapplyrngthatprogarntothrsproblern 2313 21 Create the following M function function yfnx ylsqrt2piexp XA22 Then implement the following MATLAB session gtgt x 2 gtgt yfnx gtgt Qquadfn ll 2 06827 gtgt Qquadfn 22 09545 Thus about 683 of the area under the curve falls between 71 and l and about 9545 falls between 72 and 2 b gtgt x 2l2 gtgt yfnx gtgt ddiffydiffx gtgt x l95ll95 gtgt d2diffddiffx gtgt x l9ll9 gtgt plotxd239o39 n u l 5 2 gnu Mew insevl 1m mm ew UnsusMAzW i Thus in ection points dZydx2 0 occur at 71 and 1 23J5 Program Integrate Use imsl Implicit None Integerirulel Realalblerrabs00errrel000l Realerrestresf External f Call QDAGfaberrabserrrelirulereserrest Print 39quot Computed quotF8439res Print 39quot Error estimate quotlPElO339errest End Program Function fx Implicit None Real x f Realpi Parameterpi3l415927 flsgrt2piexpx22 End Function Answers x l to l Computed 06827 Error estimate 4069E06 x 2 to 2 Computed 09545 Error estimate 7975E06 x 3 to 3 Computed 09973 Error estimate 5944E06 2317 MATLAB Script saved as pr0b2317m Numerical Integration of sintt function sintt Limits aO b2pi Using the quotquadquot and quotguadlquot function for numerical integration Plot of function t00l00l2pi yff2 t plottygt grid Integratlon format long a00l b2 Iquadquad39ff239ab Iquadlquadl39ff239ab function wff t wslnw t MATLAB execution gtgt prob23l7 Iquad 1 40815163720125 Iquadl 1 40815163168846 2319 8 lete Difference Approximation of slope 8 For fgtltegtltp2gtltgtlt 8 f39gtlt2egtltp2gtlt71 8 Centered amt dfdxf11f1rl 2dgtlt oltdxm 8 Fwd amt dfdx f124f1173f12dgtlt oltdxm 8 Bkwd amt dfdx3f174f171f1722dgtlt oltdxm x m fxegtlt 2gtltgtlt dfdx272exp2xil approx1matlon dx05 00101 for 1 engthdx X7 alues at 1edx and ezdx gtlt 1gtltdgtlt1 X2nl XZdgtlt l e5 at i7dx and 72dx fx7 fpiexp72xpi 7xpi f2piexp72x2pi 7x2pi fnlt 39 Xp72xni 7xni f2n39 exp72x2ni7x2n39 Finite Diff Approximations Cdfdxifpifni2dxi Fdfdxi7f2pi4fpi73fx2dxi Bdfdxi3fx74fnif2ni2dxi end dx00 plotdxFdfdx 397739dxBdfdx 397 39dxCdfdx 39739dx0dfdx2 39739 title39Forward Backward and Centered Finite Difference approximation 7 2nd Order Correct39 xlabel39Delta x39 ylabel39dfdx39 gtext39Centered39 gtext39Forward39 Forward Backward and Centered Finite Difference approximation 2nd Order Correct gtext39Backward39 X E quot6 O 005 DJ 015 02 025 03 035 DA 0A5 05 Deltax 2321 a t 7 t 7 Way A H1 m1 73 51055g dt 2h 4 s 2 t 72 t t 7 chqm x H1 xixi1 73 2623 51200522 dt 2 s b 7 xtH2 4xt117 3x03 7 8 473 7 363 0575E 2h 4 s a 7 7 x0 3 4xt127 5xt11 2x01 7 7 84 48 7 573 263 7 70 075 E 7 h2 7 22 7 I s2 C 7 3xt17 4xt1 xt2 7 363 7 451 34 7 MUSE 2h 4 s 7 2x0 7 5xt171 4xt27 xt3 7 2637 551 434718 7 70 2752 7 hz 7 22 7 39 Sz 2323 Use the same program as was developed in the solution of Prob 2311 Option Explicit Sub Te5tDerivUnegual0 Dim n As Integer i As Integer Dim x100 As Double y100 As Double dy100 As Double Rangequota5quotSelect n ActiveCellRow SelectionEndxlDownSelect n ActiveCellRow 7 n Rangequota5quotSelect For i 0 n xi ActiveCellValue ActiveCellOff5et0 lSelect Y39 ActiVeCellValue ActiveCellOff5etl 71Select Next i For i 0 n dyi DerivUnegualx y n xiH Next i Rangequotc5quotSelect For i o ActiveCellValue dyi ActiveCellOff5etl 0Select Nex 39 End Sub Function DerivUnegualx y n xx Dim ii As Integer If xx lt x0 Or xx gt xn Then DerivUnegual quotout of rangequot Else If xx lt xl Then DerivUnegual Dnyxx x0 xl x2 y0 yl y2 ElseIf xx gt xn 7 I Then DerivUnegual 7 x xx xn 7 2 xn 7 l xn yn 7 2 yn 7 l yn Else For ii I To n 7 2 If xx gt xii And xx lt xii I Then 39 lt xii 7 xx Then If xx 7 Xil 7 39If the unknown is closer to the lower e nd of the range ii will be chosen as the middle point 39x DerivUnegual 7 Dnyxx xii 7 l xii xii l yii 7 l yii yii 1 5e 39Otherwi5e if the unknown is closer to the upper end 39xiil will be chosen as the middle point DerivUnegual 7 xii xii 1 El ii 2 yltii yltii 1 yltii 2 End If End If End Functlon Function Dnyx XEI x1 x2 yEl yl y2 DnyyE 2xrxlrx2 xEIrxl xEIrXZ Y A l a l c l D E i F l G l H l l J l K 4 Push 2323 10 I x xt rm 3 4g n n 0150 5 i z 07 0425 Wquot 4 l 4 1 43 a 34 033 2 g 3 1 mus 0 jg lo a3 05w i 12 73 n 425 o 5 10 15 20 i u s n 275 j 15 34 IE h 2325 W a 1 39 be used to compute the derivative as in ith tt timeFquatiori239cari 8 n 164 273138 273135 277578 nn 3 mags 175x178 sslxsss 871x875 fl i Therefore the ow is equal to 295 cmZs 2327 The first forward difference formula of 003 from Fig 231 car he used to estimate the velocity for the first point at t 10 d 71754 248 73 352 110 701195 d 210 For the interior points ceritered difference formulas of 003 from Fig 233 car he used to estimate the derivatives For example at the second point at t 20 700885 20175352 dz 0 21 For the fmal point backvmrd differerice formulas of 0h2 from Fig 232 can be used to e ie H JUI all t c dcldt log c og dcldt 10 352 01195 0546543 092263 20 248 00885 0394452 105306 30 175 00625 0243038 120412 40 123 0044 0089905 135655 50 087 0031 006048 1 50864 60 061 0021 021467 467778 A log10g plot can be developed y 09946x 14527 R2 09988 2 The resulting best t equation can be used to compute k 10 1 45269 0035262 and n 0994579 The Rail Gun 7 Basic Design and Principles nf Operatinn lau39l Gun Dtxigi Eunduchng bar me armature slides alung m mm Eunmmng mus placed in a mug magin eld The moderating fume un me armature l5 pruduced by emem newmmugh me mus enameamemre Refenu Figure 1 heluwfnra magam emis cunceptml design Rm 4 i Nmzmle Ear 7 T 93 T Ran 4 I y v apphed elemnc palemia dmevence between A 1 me vans V Z quot Um Veda Dnedmns R vesisiance m we Avmamie am m r Iengm mmmaime may between mus my mrmzssonhemmamvelBalHkg a a magnetic m density m quot ebony at me mmame Ear Unis Avmemm um Dem K Loemmem m lnchnn Demeen Annalure and Hans Figum 1 7 Cnncqmlal Design nmie Rail Gun Ammome Eltcbndymnl39rx Cumiier meimee aning nn zdlglz eimge emiex qwivhinvh2Armalm Slhg e cnavge cammquunm a ruequotmama The Lorentz Force Law governs the electrodynamics here the force on the charge carrier N the electric eld intensity Vm the velocity of the charge carrier ms Um lt rm 11 H the magnetic ux density T This Lorentz Force can be broken down into an electric force E and a magnetic force Em where Ee qE andEm quE The initial electric force on q comes via the electric eld intensity produced by the voltage source E Vemf and therefore Ee After the bar begins moving along the rails with velocity u an effective magnetic counter force Fc often called a counter emf develops Fc q X B tuy The net ofthese two forces Enet is therefore E et Ee Ec tu Eq 1 n Consider now the forces on the Armature itself The magnetic force accelerating the Armature Em is a function of the current ow I through it and is give by Ampere s Law as Em If X E 11y X B2 HE A frictional force directly proportional to the weight of the bar opposes the magnetic force Er ng5 where we will consider K to be a constant coef cient of friction We can also express this force as a a a e d A d Fa Fm Fr ma md IEX Equat1ng these two we have IlBng md lt1 Eq 2 The current I at any time during the acceleration of the armature can be given by Ohm s Law as V F F 1 gm where V E l and E quote39 Therefore we can solve for I quot 39 1 effective effective effective Substituting this value of I into Eq 2 and solving for Fuel we get Fnet quduj Kqum 12B dt 12B qug Kqum 12B dt 12B 1 With a little algebraic rearrangement of this expression we end up with a rst order du lszu V B emf inhomogeneous D E with constant coef cients Kg Eq 3 dt mR mR This equation governs the velocity of the bar during acceleration from standstill As long as all the coef cients in the equation are constants the analytic solution can be easily found as D v B lB2 ut le39s whereD emf K andS E 4 S M g M q Substituting this value of Fuel into Eq 1 we get tu Some questions to ponder here 1 What physical assumptions and or simpli cations have been made in the analysis here 2 Why is it important to keep these in mind when trying to build a practical device 3 How can you nd the solution to Eq 3 if the coef cients are NOT constants


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