Preface Calculus Topic Background/Motivation Learning Objectives

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2 Prfac On of th goals of th EXCEL program is to nhanc undrstanding of mathmatics (primarily calculus) for Scinc, Tchnology, Enginring and Mathmatics (STEM) studnts. Mathmatics is considrd th cornrston of th succss of any studnt pursuing a dgr in a STEM disciplin. On way of achiving this goal at UCF is by introducing studnts to th Applications of Calculus courss. This book is th ttbook matrial for th Applications of Calculus II cours that you will b taking in th spring smstr of your frshman studis at UCF. It is comprisd of a sris of chaptrs writtn by EXCEL faculty from various disciplins of th scincs, nginring and mathmatics. Ths chaptrs and th prsntations that th individual EXCEL faculty mmbrs will mak in th Applications of Calculus II class will show you how matrial that you will b larning in your Calculus II class is actually usd in th disciplin of th faculty prsntr. Ths prsntations ar corrlatd with th sctions of your calculus book and will b prsntd just shortly aftr you covr thos sctions in your Calculus II class. Each of th chaptrs of this book is organizd in th sam way. Th chaptrs bgin with a sction titld Calculus Topic which includs at last on calculus problm takn dirctly from your calculus tt along with a dtaild solution. Th purpos of including a problm from your calculus tt is to ) rinforc th tchniqus usd to solv problms from that particular sction and ) to plain how a particular mathmatical concpt is rlatd to a particular physical (ral world) problm. Th nt sction in ach chaptr is a Background/Motivation sction. This sction provids you with th basic undrstanding of th fild of study that will follow and why th mathmatics (in particular, a Calculus II topic) is ncssary in that fild of study. Following th Background/Motivation sction you will find a sris of Larning Objctivs. Each Larning Objctiv is followd by tt, figurs and tabls which lad th radr through th procss of undrstanding that objctiv. Each of th Larning Objctivs shown in a chaptr will b a focus of th faculty prsntation. Th faculty prsntrs ar outstanding ducators and rsarchrs in thir rspctiv filds. Each faculty mmbr authord his/hr chaptr as wll as th prsntation that will b givn in class. W hop that you, th radr and studnt, find this book to b hlpful in your journy through this cours, that th cours offrs a usful nhancmnt to your undrstanding of Calculus and that th wkly prsntations broadn your horizons and pk your intrsts in topics of mathmatics, nginring and scincs that you may hav nvr othrwis studid on your own.

3 Acknowldgmnts First and formost, w would lik to acknowldg th support of th National Scinc Foundation grant DUE , who has ntrustd us with th opportunity to ducat th STEM youth of tomorrow. W also want to acknowldg th gnrous support from Provost Hicky, Vic Provost Schll, th UCF Offic of Rsarch and Commrcialization (in particular, th Vic Prsidnt of Rsarch Dr. Soilau), th Collg of Enginring and Computr Scinc Dan s Offic (in particular, Dan Gallaghr, Associat Dan Nayfh and Ms. Falls), th Collg of Scincs Dan s Offic (in particular, Dan Panousis, Associat Dan Johnson and Ms. Kirkpatrick), th Offic of Oprational Ecllnc and Assssmnts (in particular, Drs. Pt Armacost, Krist and Lancy), th Institutional Rsarch Offic (in particular, Ms. Ramsy), UCF Admissions Offic (in particular, Dr. Chavis and Ms. Costllo), th First Yar Advising Offic (in particular, Ms. Prist and Ms. Datta), th UCF Orintation Offic (in particular, Mr. Richi and Mr. Hicks), th UCF Divrsity Offic (in particular, Dr. King), th UCF Housing Offic (in particular, Mr. Paulick, Mr. Novak and Ms. Rutkowski) and Associat Provost Poisl, amongst many othrs. This book would not hav bn possibl without th hard work and ddication of th EXCEL faculty: Drs. Gorgiopoulos and Young (EXCEL Dirctors), who hav bn involvd with all aspcts of this projct from planning, writing, coordinating, schduling to th mor thanklss jobs of handling th financs of th projct. Drs. Gigr, Hagn, Islas and Winningham (EXCEL Coordinators), all of which spnt a trmndous amount of tim in planning and dircting this projct as wll as thir continuing projcts. Th EXCEL faculty who hav contributd thir knowldg and prtis in ordr to mak this book a rality ar Drs. Kocak (Chaptr ), Ou (Chaptr ), Rollins (Chaptrs 3 and 5), da Vitoria Lobo (Chaptr 4), Efthimiou (Chaptr 6) and Turgut (Chaptr 7). Th assssmnt and valuation matrials associatd with this class wr dvlopd by th Faculty Cntr for Taching and Larning undr th guidanc of Drs. Morisson Shtlar and Crous, and this is a contribution for which w ar vry thankful. Spcial thanks go to th EXCEL graduat studnts: Ptr Bacopoulos, Erin Holland, Chris Sntll, Todd Smith and Tomasz Wlodarczyk, who rviwd and critiqud vry chaptr in this book. Ths studnts gav thir tim and nrgy to rviw all th chaptrs in this book and providd cllnt advic to th EXCEL faculty in changing th chaptr writ up to nhanc its undrstandability by th EXCEL studnts. Finally, th load of intgrating th information from svn diffrnt EXCEL faculty into a cohsiv volum of work fll upon Tomasz Wlodarczyk and Ptr Bacopoulos, who did an outstanding job undr th guidanc of th EXCEL coordinator, Dr. Hagn. W ar vry gratful for thir ffort.

4 Tabl of Contnts Sigmoid functions and thir usag in artificial nural ntworks... Laplac Transforms and initial valu problms... 3 Mathmatical Modling of th Population Growth of a Singl Spcis I Continuous Intractiv Graphics Curvs using Paramtric Equations Mathmatical Modling of th Population Growth of a Singl Spcis II Discrt Th Gomtric Sris and Som Applications Multipath Effcts in Wirlss Ntworks... 7

5 Chaptr Sigmoid functions and thir usag in artificial nural ntworks by Dr. Taskin Kocak Wks: /8 and /5 of Spring 007

6 Sigmoid functions and thir usag in artificial nural ntworks Calculus Topic: Hyprbolic Functions Sction 7.6 #9: Prov idntity tanh tanh = + cosh sinh cosh sinh tanh tanh = + = + = = = = In this sction w will discuss sigmoid functions, which blong to th family of hyprbolic functions. W will also prsnt thir usag in artificial nural ntworks. Background/Motivation A sigmoid function is a mathmatical function that producs a sigmoid curv, i.. a curv having an S shap. An ampl sigmoid function, which is th spcial cas of th logistic function, is givn blow and it is plottd in Figur. A logistic function modls th growth of som st. Th initial stag of growth is approimatly ponntial; thn growth slows and at maturity, growth stops. t t + = ) ( sig ()

7 Sigmoid functions and thir usag in artificial nural ntworks Figur : Sigmoid function. In gnral, a sigmoid function is ral valud and diffrntiabl, having a non ngativ or non positiv first drivativ, on local minimum, and on local maimum. Th logistic sigmoid function is rlatd to th hyprbolic tangnt as follows = tanh + () Sigmoid functions ar oftn usd in artificial nural ntworks to introduc nonlinarity in th modl. A nural ntwork lmnt computs a linar combination of its input signals, and applis a sigmoid function to th rsult. A rason for its popularity in nural ntworks is bcaus th sigmoid function satisfis a proprty btwn th drivativ and itslf such that it is computationally asy to prform. Drivativs of th sigmoid function ar usually mployd in larning algorithms. d dt sig( t) = sig( t)( sig( t)) (3)

8 Sigmoid functions and thir usag in artificial nural ntworks 4 Larning objctiv #: Dtrmin th rlationships btwn th biological and artificial nural ntworks Modrn digital computrs prform bttr than humans whn it coms to numric computation and rlatd symbol manipulation. Howvr, humans can bat th world s fastst computr in solving compl prcptual problms such as rcognizing a man in a crowd from a mr glimps of his fac at a high spd. Th rmarkabl diffrnc in thir prformanc is du to th biological nural systm architctur, which is compltly diffrnt from th computr architctur. A nuron (or nrv cll) is a spcial biological cll that procsss information (s Figur ). It is composd of a cll body, or soma, and two typs of out raching tr lik branchs: th aon and th dndrits. Th cll body has a nuclus that contains information. A nuron rcivs signals (impulss) from othr nurons through its dndrits (rcivrs) and transmits signals gnratd by its cll body along th aon (transmittr). Figur : A sktch of a biological nuron. Inspird by biological nural ntworks, Artificial Nural Ntworks (ANNs) ar massiv paralll computing systms consisting of an trmly larg numbr of simpl procssors with many intrconnctions. McCulloch and Pitts [] proposd a binary thrshold unit as a computational modl for an artificial nuron (s Figur 3).

9 Sigmoid functions and thir usag in artificial nural ntworks 5 j This mathmatical nuron computs a wightd sum of its N input signals, =,, K, N, and gnrats an output of if this sum is abov a crtain thrshold T. Othrwis, an output of 0 rsults. Mathmatically, N y = Θ w j j= j T, (4) whr Θ(.) is a unit stp function (to b plaind in th lctur) at 0, and w j is th synaps wight associatd with j th input. Thr is a crud analogy hr to a biological nuron: intrconnctions modl aons and dndrits, connction wights rprsnt synapss, and th thrshold function approimats th activity in a soma. Figur 3: McCulloch Pitts modl of a nuron. Th McCulloch Pitts nuron has bn gnralizd in many ways. An obvious on is to us activation functions othr than th thrshold function, such as picwis linar, sigmoid, or Gaussian mainly bcaus drivativs ar takn in th most common larning algorithms, which rquir that th thrsholds ar smooth functions. Th sigmoid function is by far th most frquntly usd in ANNs. Qustion. What ar th similaritis btwn th biological and artificial nural ntworks? Larning objctiv #: Usag of Artificial Nural Ntworks Th OR ampl In this sction, w will utiliz th McCulloch Pitt modl to train a nural ntwork to larn th logic OR function. Th OR function w will us is a two input binary OR function givn in Tabl.

10 Sigmoid functions and thir usag in artificial nural ntworks 6 I I Output Tabl : OR function. First, w will us on nuron with two inputs as illustratd in Figur 4. Not that th inputs ar givn qual wights by assigning th wights (w s) to. Th thrshold, T, is st to 0 in this ampl. Figur 4: On nuron modl for th OR function. W calculat th output as follows: ) Comput th total wightd inputs X = I i w i (5) i= X = I w +I w = I +I = I +I (6) ) Calculat th output using th logistic sigmoid activation function O = + X (7) Now, lt s try it for th inputs givn in Tabl. For I = 0 and I = 0; X = 0, O = = = (8) + + For I = 0 and I =, and I = and I = 0; X =,

11 Sigmoid functions and thir usag in artificial nural ntworks 7 For I = and I = ; X =, O = = (9) O = = (0) For all cass th rsults ar corrct assuming that 0.5 and blow ar considrd as 0 and abov as. Qustion. If th wights wr 0.5 rathr, will th ntwork still function lik OR? Qustion 3. In groups of two studnts, discuss whthr th sam ntwork can b usd to larn th AND function? (Hint: You may chang th thrshold (=0.5) if ncssary.) I I Output Tabl : AND function. Larning objctiv #3: Th classic XOR problm If w us th sam on nuron modl to larn th XOR (clusiv or) function, th modl will fail. Th XOR function is givn in Tabl. Th first thr cass will produc corrct rsults; howvr, th last cas should also b considrd as, which is not corrct. I I Output Tabl 3: XOR function. Th solution is to add a middl (hiddn in ANN trminology) layr btwn th inputs and th output nuron as shown in Figur 5.

12 Sigmoid functions and thir usag in artificial nural ntworks 8 Figur 5: ANN with th hiddn nurons for th XOR problm (not that thr will b a thrshold function at th final output (O)). Choos th wights w=w=w=w=. Us a diffrnt sigmoid function, which is givn with a crtain thrshold for ach nuron: sig sig sig ( ) = + ( ) = + ( ) = + H H O ( 0.5) (.5) ( 0.) () Confirm by calculating th nuron outputs for ach possibl input combinations that this nural ntwork is indd functioning lik an XOR. (Hint: Th output qual or blow 0.5 is considrd 0, othrwis ) Rfrnc [] W. S. McCulloch and W. Pitts, A logical calculus of th idas immannt in nurons activity, Bull. Math. Biophys., vol. 5, pp. 5 33, 943.

13 Chaptr Laplac Transforms and initial valu problms by Dr. Miaojung Ou Wks: / and /9 of Spring 007

14 Laplac Transforms and initial valu problms Calculus topic: Intgration by parts Sction 8. #4: Evaluat th intgral 3 d. Solution: Rpat intgration by parts (IBP) thr tims to gt 3 d IBP = 3 ( 3 ) d = 3 3 d ( ) IBP = 3 3 d = d ( ) IBP = d = c. In this sction, w discuss how to comput th on sidd Laplac transform for functions dfind on [0, ) and of ponntial ordr. Intgration by parts tchniqu will b applid to study th rlation btwn th Laplac transform of a function and that of its drivativ. This rlation maks th Laplac transform mthod a powrful tool in solving initial valu problms of ordinary diffrntial quations. Background/Motivation Initial valu problms aris whn th information for currnt tim is usd to prdict th valu of a crtain physical quantity in th futur. For ampl, if w know quantitativly how a spac shuttl is going to b acclratd by its powr systm at any tim (law of chang) and if w know whr th shuttl is launchd (initial location) and how fast it is launchd (initial vlocity), th prdiction of its location in spac aftr launch is by natur an initial valu problm. If th law of chang is dscribd by a rlation btwn th drivativs of th dsird physical quantity and itslf, thn th law is calld a diffrntial quation and th initial valu problm (IVP) associatd with it is an IVP for diffrntial quations (vs. IVP for othr typs of quations such as diffrntial algbraic quations). For ampl, th hight h of a ball thrown into th air with initial upward vlocity of m/s on a 3 mtr high platform satisfis th following initial valu problm: { h (IVP) (t) = 9.8m/s, h(0) = 3m, h (0) = m/s. Th Laplac transform can b applid to solv a class of initial valu problms for diffrntial quations. As w will s in th following sctions, Laplac transform

15 Laplac Transforms and initial valu problms 3 provids a powrful tool to rwrit th diffrntial quation togthr with th initial conditions as on algbraic quation through intgration by parts. Th solution (prdiction) is thn obtaind by solving th algbraic quation followd by invrs Laplac transform. Larning Objctiv #: Calculation of Laplac transform of lmntary functions Dfinition. Th on sidd Laplac transform of a function f dfind on [0, ) is afunctionofs such that L{f}(s) := Th domain of L{f} is whr th intgral ists. 0 st f(t)dt. Eampl. Calculat th Laplac transform of f(t) =. L{}(s) := 0 st dt N := lim st dt N 0 st t=n = lim N s t=0 ( sn = lim N s ) = s s. So th Laplac transform of th constant function f(t) =is s for s>0. Eampl. Calculat th Laplac transform of f(t) = t. L{t}(s) := 0 := lim N IBP = lim N = lim N st tdt N t st dt 0 [ t st t=n s t=0 ( N sn s = for s>0. s N 0 st s dt ) sn s + s Not that L Hospital Rul was applid to valuat th first trm in th limit. ]

16 Laplac Transforms and initial valu problms 4 Qustion. Us similar idas to calculat th Laplac transform of f(t) =t. Do you idntify som pattrn? Us th pattrn to mak your bst guss on L{t n } for n = 3.What about for n = 0? Eampl 3. Calculat th Laplac transform of f(t) = sin t. W valuat N 0 N 0 st sin tdt L{sin t}(s) := 0 := lim N st sin tdt N 0 st sin tdt. st sin tdt by using intgration by parts as follows: IBP = st t=n sin t N st cos t s dt t=0 0 s ( ) sn sin N = 0 + N st cos tdt s s 0 ( IBP = sn sin N + st t=n ) cos t N st sin t s s s + dt t=0 0 s = sn sin N + ( ) sn cos N s N st sin tdt. s s s Combining th N 0 st sin tdt trms and divid, w obtain sn sin N N + ( ) sn cos N st s s s sin tdt = 0 + s = s sn sin N ( sn cos N ). s + squz thorm N = lim st sin tdt = for s>0. N 0 s + Qustion. Calculat L{sin 3t} in a similar way. Do you idntify any pattrn? 0

17 Laplac Transforms and initial valu problms 5 Eampl 4. Calculat th Laplac transform of f(t) = t. L{ t }(s) := 0 := lim st t dt N st t dt N 0 N ( s)t = lim N = lim N = lim N = s 0 ( s)t N s for s 0 ( ( s)n s ) s for s<0, i.. for s>. Qustion 3. Calculat L{ t } and L{ t }in a similar way. pattrn? Do you idntify any Larning Objctiv #: Calculation of Laplac transform of th drivativ of a function Dfinition. Afunctionf(t) is said to b of ponntial ordr a if it grows at most as fast as th function at for larg t. For ampl, f(t) =t is of ordr bcaus lim t t t = 0 by L Hospital Rul. Eampl 5. Calculat L{f }(s) for a continuous function f = f(t) ofponntial ordr a. L{f }(s) := lim N IBP = lim N N st f (t)dt 0 [ st f(t) N N0 ( s st f(t)dt )] ( = lim sn f(n) f(0) + s N = sl{f}(s) f(0) for s>a. 0 N 0 ) st f(t)dt Qustion 4. Us th abov formula to calculat L{f }(s) =?(Hint: f is th drivativ of f.) With this drivativ formula, w can asily calculat L{cos t} from L{sin t}, which is alrady calculatd in Eampl 3, bcaus cos is th drivativ of sin t. Eampl 6. L{cos t}(s) =sl{sin t}(s) sin(0) = s ( ) = s s + s +.

18 Laplac Transforms and initial valu problms 6 f(t) =L {F (s)}(t) F (s) :=L{f(t)}(s) t n, n =,,... sin bt cos bt s, s>0 n!, s>0 sn+ b s + b, s>0 s s + b, s>0 at s a, s>a t n at, n =,,... at sin bt at cos bt n!, s>a (s a) n+ b (s a) + b, s>a s a (s a) + b, s>0 f (t) sf (s) f(0) f (t) s F (s) sf(0) f (0) Tabl : Brif Tabl of Laplac Transforms. Not: L dnots th invrs Laplac transform. Now, w summariz all th Laplac transforms w hav calculatd so far, togthr with a fw usful ons which can b calculatd in a similar fashion, in th following tabl: Larning Objctiv #3: Us Laplac transform to solv Initial Valu Problms of diffrntial quations In this sction, w will us th Laplac Transform Tabl to solv initial valu problms for diffrntial quations (IVPDiffEq). As can b sn in th following ampls, th stratgy is to transform th (IVPDiffEq) in variabl t to an algbraic quation in variabl s, followd by solving th algbraic quation in s and invrs Laplac transform back to t. Eampl 7. Us Laplac transform to solv th initial valu problm y y +5y = 8 t, y(0) =, y (0) = 0.

19 Laplac Transforms and initial valu problms 7 Lt Y (s) :=L{y}. Us th Tabl to obtain L{y } = s Y (s) sy(0) y (0) = s Y (s) s. L{y } = sy (s) y(0) = sy (s), L{ 8 t } = 8 s +. Applying Laplac transform to both sids of th DiffEq lads to th following algbraic quation for Y (s): Us partial fractions s Y s (sy ) + 5Y = 8 s +, Y (s) = s s 0 (s +)(s s +5). s s 0 (s +)(s s +5) = A s + + Bs + C s s +5 to gt Y (s) = s + + s 5 s s +5. Sinc y(t) =L {Y (s)}, w nd to us th Tabl to invrs transform ach th right hand sid of th abov quation. { } L s + { } s 5 L s s +5 { } = L = t, s + { } (s ) 3 = L (s ) + { (s ) = L (s ) + =??? } 3 L { } (s ) + Qustion 5. Us th Tabl to find th solution to th abov ampl. How do you chck if your answr is corrct? Th last ampl is from an application problm from control thory. W considr a srvomchanism that modls an auto pilot. Such a systm applis a torqu to th string shaft so that an airplan will follow a prscribd dirction (angl) g(t). As can b pctd, th tru dirction of th airplan y(t) is not always to b th sam as th dsird g(t) at all tim t. Whn this happns, th srvomchanism will masur th dviation btwn y(t) and g(t) and fd back to th string shaft a torqu which is proportional to th dviation (t) := y(t) g(t) but with opposit sign. In this way, th airplan can stay in th dsird cours. On th othr hand, according to Nwton s scond law of motion, th total torqu of a rotating objct is rlatd to th

20 Laplac Transforms and initial valu problms 8 momnt of Inrtia (dnotd by I) of th objct and th angular acclration y (t). Hnc th srvomchanism following th following rul in corrcting th dirction of an airplan Iy (t) = k(t), Hr k is th positiv proportional constant of th corrction systm. Eampl 8. Dtrmin th rror (t) for th auto pilot if th string shaft is initially at rst (y (0) = 0) in th zro dirction (y(0) = 0) and th dsird dirction is givn by g(t) = a (i.. a fid dirction=a straight lin). Th initial valu problm is Iy (t) = k(t), y(0) = 0, y (0) = 0. () Lt Y (s) = L{y(t)} and E(s) = L{(t)}. Sinc (t) = y(t) g(t) = y(t) a, w hav E(s) =Y (s) L{a} = Y (s) a s. Laplac transform both sids of th diffrntial quation () to gt L{Iy (t)} = I[s Y (s) sy(0) y (0)] = Is Y (s) = ke(s). Rplac Y (s) withe(s)+ a s,wobtain Thrfor, E(s) = Problms ( Is E(s)+ a ) = ke(s). s Isa. Invrs Laplac transform E to conclud that Is + k { } ( ) (t) = al s k = a cos s + k I t. I. Us Laplac transform to solv th initial valu problm y y +5y = 8 t, y(0) =, y (0) = 3.. Considr th autopilot string problm givn in th last sction. Calculat th rror (t) in th systm if th string shaft is initially at rst in th π/ dirction (i.. y(0) = π/, y (0) = 0) and th dsird dirction is g(t) =a, whr a is a constant.

21 Chaptr 3 3 Mathmatical Modling of th Population Growth of a Singl Spcis I Continuous by Dr. David Rollins Wks: /5 and / of Spring 007

22 Population Growth of a Singl Spcis I Continous Modls 3 Calculus Topic: Intgration of Rational Functions Sction 8.4 # 0: Evaluat th intgral (t +4)(t ) dt Solution: Th dnominator of th intgrand is alrady factord with th factors bing distinct, so (t +4)(t ) = A t +4 + B t. Whn th right hand sid is rcombind with a common dnominator w hav (t +4)(t ) = A t +4 + B t A(t ) + B(t +4) =. (t +4)(t ) This mans that =A(t ) + B(t +4) which must b tru for all valus of t. Sowhnt =,whav=b( + 4) = 5B and thn B =. Similarly whn t = 4, w hav = A( 4 ) = 5A which mans 5 A =. Th intgral is now 5 ( 5 t +4 + ) 5 dt = t 5 dt t dt t (Why?). Ths intgrals ar thn valuatd using th natural log antidrivativ function. Th rsult is (t +4)(t ) dt = 5 (ln t ln t +4 )+C = 5 ln t t +4 +C whr of cours it is vry important that w includ th intgration constant C. Background/Motivation In biology, mathmatical modls hav bn usd for many yars to dscrib th population of a particular spcis, whthr it b human, bactria or an ndangrd spcis. Whil such modls may not always giv th act quantitativ bhavior, it is important that thy giv th corrct qualitativ bhavior. Th important quantity is th population of th spcis as a function of tim. W shall assum that th population of a givn spcis can b rprsntd by a continuous, diffrntiabl function of tim. This will b dnotd by (t). This is not quit ralistic sinc can only tak intgr valus but is a good approimation if th population is larg. On of th simplst ways to modl a population is with a diffrntial quation (DE). A diffrntial quation is an quation whr th unknown is a function that must b found. A diffrntial quation contains not only trms involving th unknown

23 Population Growth of a Singl Spcis I Continous Modls 3 3 function but also it s drivativs. An ampl of a DE with unknown (t) that also contains th first drivativ is (t) =4(t)+t Th subjct of diffrntial quations is a rich on but w will focus on ampls that can b solvd by intgration. Larning Objctiv #: S how a population can b modld by an initial valu problm and s how to solv it by using th partial fraction intgration mthod. Simpl Growth Modl. Th simplst modl will only concrn itslf with th ffcts of births and daths. Lt (t) b th spcis population at any tim t. Th rat of chang of (t) with rspct to tim is givn by d dt = birth daths Assum that th birth trm is proportional to which is quit rasonabl as w pct th numbr of births to incras with. Similarly assum th dath trm is also proportional to. Thn our diffrntial quation for (t) is d dt = b μ =(b μ) whr b is th pr capita birth rat and μ is th pr capita dath rat. Both ths constants hav th units of tim. It is natural to dfin a nt growth rat constant r = b μ and thn th diffrntial quation for (t) to b solvd is d dt = r. Bfor w solv this lt us notic som qualitativ faturs of this quation. Rmmbr that th population satisfis (t) 0with(t) = 0 only whn th spcis bcoms tinct. Suppos th pr capita birth rat is largr than th dath rat so that r>0. Thn w hav d dt 0, which mans that is an incrasing function of t. This maks sns sinc thr ar mor births than daths and so th population will grow. Qustion. Suppos r<0. What dos this man about births and daths? In this cas is an incrasing or dcrasing function of tim? Eplain your answr.

24 Population Growth of a Singl Spcis I Continous Modls 3 4 Solution:. Divid both sids by : d dt = r. Intgrat both sids with rspct to tim t. Not that oftn, diffrntial quations ar solvd by intgrating. d dt dt = Th lft hand intgral can b simplifid by changing th variabl of intgration to by using th fact that d = ddt: dt d = rdt Th right hand intgration is vry simpl (rmmbr that r is a constant!). Th lft hand intgration is a vry simpl ampl of partial fractions with th intgrand alrady a partial fraction. W gt rdt ln (t) = rt + C whr C is th intgration constant. Not that th absolut valu symbols ar not rquird sinc (t) > 0. W would lik to solv for (t), so w tak th ponntial of both sids and gt (t) = rt+c = C rt = C 0 rt. Not that C is a constant which w will dfin as C 0,thatwdon tknow yt. To find C 0 w will hav giv on mor pic of information which is calld th initial condition. Suppos th population is givn at som particular tim (prhaps by counting) thn C 0 can b found. Suppos (0) = 000 is th starting population. Thn w hav (0) = 000 = C 0 0 = C 0 and th solution is (t) = 000 rt. A diffrntial quation plus initial condition is calld an initial valu problm. 3. Intrprtation. Suppos r>0 (births most important) thn (t) isagrowing ponntial function with lim (t) =. t This mans a population plosion with no way to stop th growth. But w know that othr factors will com into play. W hav not includd th fact that thr ar limitd rsourcs of food and land into our mathmatical modl.

25 Population Growth of a Singl Spcis I Continous Modls 3 5

26 Population Growth of a Singl Spcis I Continous Modls 3 6 Not that th quilibrium solutions ar = 0and =. procdur Following th abov. Divid both sids by ( ): d ( ) dt = r. Intgrat both sids with rspct to tim t. Not that oftn, diffrntial quations ar solvd by intgrating. d ( ) dt dt = As bfor, th lft hand intgral can b simplifid by changing th variabl of intgration to : ( ) d = rdt Th lft hand intgration is don by using partial fractions: rdt ( ) = ( ) = A + B This implis =A( ) + B and givs A =andb =. So Combining th logarithms and invrting w hav ( ) dt =ln ln = rt + C. ln = rt + C = C rt = C 0 rt. Lt s suppos th initial population is (0) =, thn th constant C 0 can b found = C 0 0 C 0 =. and thn /( ) = rt. Sinc w don t hav (t) givn plicitly w algbraically solv to gt (t) = rt rt = rt which is th population of our spcis for any tim t.

27 Population Growth of a Singl Spcis I Continous Modls 3 7

28 Population Growth of a Singl Spcis I Continous Modls 3 8. Imagin a spcis that is huntd or fishd with a yarly quota spcifid. In this cas, th diffrntial quation modl is modifid as follows d dt = r( /K) H whr H is a constant and is calld th harvsting rat. Writ down th initial valu problm in th cas whn th nt growth rat is, th carrying capacity is 00, th harvsting rat is and th initial population at tim zro is 50. (a) What ar th quilibrium solutions? (b) Find (t) by intgration. (c) Us algbraic manipulation to find an plicit formula for (t) including th valuation of th intgration constant. (d) Find th larg tim limiting bhavior and intrprt th rsult from a biological prspctiv. How did th harvsting trm affct th rsults compard with qustion?

29 Chaptr 4 4 Intractiv Graphics Curvs using Paramtric Equations by Dr. Nils da Vitoria Lobo Wks: /9 and /6 of Spring 007

30 Intractiv Graphics Curvs using Paramtric Equations 4 Calculus Topic: Calculus with Paramtric Curvs Sction. # 3: Find an quation of th tangnt to th curv at th point. = t 4 +, y = t 3 + t, t = Solution: W gt d dt =4t3 and dy dt =3t +. Hnc, th slop of th tangnt any tangnt is givn by dy d = dy/dt d/dt = 3t + ; 4t 3 and at t =, th slop simplifis to. Now, whn t =, w gt = and y =. So, using th point slop form of an quation for a lin, namly, y y 0 = m( 0 ), w gt y ( ) = ( )( ) or y =. Application: Intractiv Graphics Curvs In this sction, w discuss how intractiv graphics curvs can b providd to dsignrs and othr usrs by mploying paramtric curvs (also s th Laboratory Projct at th nd of Sction.). Intractiv graphics curvs ar oftn ndd that ar smooth, and can b convnintly and spdily gnratd and modifid. On such systm is calld Bézir Curvs, namd aftr Pirr Bézir (90 999) a mathmatician in th Frnch automotiv industry. A pic of smooth curv is spcifiabl by giving two points P 0 ( 0,y 0 )andp 3 ( 3,y 3 ), and two guidpoints (lik string handls) P (,y ) and P (,y ). Th curv must pass through P 0 and P 3. Th sgmnt joining P 0 and P dfins th dirction of th tangnt at P 0, whil th lngth of this sgmnt dfins for how long th curv starting at P 0 should stay clos to th tangnt. Similarly, for th sgmnt joining P to P 3. Figur shows an ampl of such a curv. Th usr can chang th position of P and P to chang th curv. Figur shows svral additional ampls of Bézir Curvs. Background/Motivation Intractiv graphics curvs ar usd anywhr artists nd to dsign curvs that ar thn rproducibl automatically. On such ampl is th dsign of fonts and othr symbols for printrs. Othr ampls includ th dsign of consumr goods such as automobils, cll phons, furnitur, and th lik. In gnral, complicatd curvs will b producd by a chain of svral simplr curvs, ach a Bézir Curv. This allows th usr to chang on portion of th complicatd curv with littl or no ffct on othr portions of th ovrall curv. Onc th chain of points is spcifid for th ovrall curv, th usr controls th shap of a portion btwn two conscutiv points by placing two guidpoints for this portion. Th formula to gnrat th curv is vry spdy to cut on a computr, and so th usr can mak changs to th dsign and s th ffcts of ths changs instantly.

31 Intractiv Graphics Curvs using Paramtric Equations 4 3

32 Intractiv Graphics Curvs using Paramtric Equations 4 4

33 Intractiv Graphics Curvs using Paramtric Equations 4 5

34 Intractiv Graphics Curvs using Paramtric Equations 4 6 Qustion. Draw a symbol similar to th archs in th McDonald s M. Thn, plac guidpoints that would hav bn ndd to obtain this symbol. Larning Objctiv # 3: To manually follow th plotting procdur for a givn spcifid curv For th thr curvs spcifid nt, giv th (, y) coordinatsforfourintrmdiat points pr curv. Th first curv is spcifid by: P 0 (3,6); P (3.3,6.5); P (.8,3.0); P 3 (,). Th scond is spcifid by: P 3 (,); P 4 (.5,.5); P 5 (5.8,5.0); P 6 (6,6). Th third is: P 6 (6,6); P 7 (5.0,5.8); P 8 (5.5,.); P 9 (5,). Not that your four t choics pr curv will b 0., 0.4, 0.6 and 0.8. On graph papr, plot th (6) points, and turn this in along with a list of all th (, y) coordinats. Th manual procdur you ar following is on that is automatd by Bézir Curv Systms, such as thos usd in industry. Thr ar svral intractiv dmos for drawing simpl Bézir Curvs on th Intrnt. Us sarch trms Bézir curv dmo to amin som of ths. If you lav th word dmo out, you will also gt additional sits whr you can rad mor about ths intrsting curvs.

35 Chaptr 5 5 Mathmatical Modling of th Population Growth of a Singl Spcis II Discrt by Dr. David Rollins Wks: 3/5 and 3/9 of Spring 007

36 Population Growth of a Singl Spcis II Discrt Modls 5 Calculus Topic: Convrgnc and Divrgnc of Squncs Sction. #9: Dtrmin whthr th squnc convrgs and divrgs and if it convrgs, dtrmin th limit. a n = n 3 n+ Solution: Th first fw trms in th squnc ar as follows a 0 = 3,a = 3 = 9,a = 3 3 = 4 7,a 4 = = 8 8,... By looking at th dcimal approimations of ths numbrs (0.333, 0., 0.48, 0.098,...) w might pct that th squnc will convrg to zro in th limit of larg n. This is mad prcis by valuating th limit lim a n ) n n = lim = lim n n 3n+ =0 n 3( 3 sinc /3 < and th limit of r n as n is zro if r <. Background/Motivation Earlir w saw that mathmatical modls can b usd to dscrib th population of a particular spcis, whthr it b human, bactria or an ndangrd spcis. W usd a continuous diffrntiabl function to rprsnt th population of a spcis at any tim. W drivd a diffrntial quation that whn solvd by intgration gav th population at any tim. It is also possibl to rprsnt th population as a discrt variabl n which givs th population at th n th tim intrval. So w can idntify th subscript as rfrring to th tim. Gnrally, w tak n =0,,,..., corrsponding to tim 0, on tim unit latr, two tim units latr and so on. This can b intrprtd as th tim unit btwn succssiv gnrations of th population, so n givs th population of th n th gnration. Th starting population is 0. Th population is mathmatically a squnc, with th convrgnc or divrgnc proprty of this squnc dtrmining th bhavior of th population for larg tim (n ). Whil diffrntial quations ar usd in th continuous cas, diffrnc quations ar usd in th discrt cas. Th solution to a diffrnc quation is a squnc. Diffrnc quations offr th possibility of mor complicatd population bhavior than th diffrntial quations w considrd. For ampl, chaotic bhavior is possibl which was not possibl for th simpl diffrntial quations w studid in th continuous cas. Chaotic bhavior appars to b random and would man wild fluctuations in th spcis population.

37 Population Growth of a Singl Spcis II Discrt Modls 5 3 Larning Objctiv #: S how a population can b modld by a diffrnc quation problm and s how to find th squnc and dtrmin th bhavior. Simpl Growth Modl. Th simplst modl will only concrn itslf with th ffcts of births and daths. Lt n b th population of th n th gnration. Assum that th birth trm is proportional to n which is quit rasonabl as w pct mor births whn n is larg. Similarly assum th dath trm is also proportional to n. Thn our diffrnc quation for n is n+ = n +(b μ) n =(+r) n whr b is th pr capita birth rat, μ is th pr capita dath rat and r th nt growth rat constant. This quation says that th population at th nt gnration is sum of th currnt population ( n trm) plus nw growth (or loss) (r n trm) du to th nt growth rat. Suppos th initial population 0 =andr =so n+ = n. Thn th squnc is givn by =4, =8, 3 =6, 4 =3,... W can clarly s by amining th first fw trms of th squnc that n = n+. W can vrify that this in fact solvs th diffrnc quation by plugging it in: n+ = (n+)+ = n+ = n+ = n Now w ask whthr this squnc convrgs or divrgs. lim n = lim n+ = n n which mans th sris divrgs. From a population growth point of viw, this mans thr is a population growth plosion which is not that ralistic as it dos not tak into account th limitd rsourcs that ar availabl. Lt s now solv th diffrnc quation for any choic of 0 and r. =(+r) 0, =(+r) =(+r) 0, 3 =(+r) 3 0,... W prdict that in gnral n =(+r) n 0 which will convrg or divrg dpnding on th valu of r. Furthrmor not that if + r<0 i.. r< that th squnc altrnats btwn ngativ and positiv valus for th population, which is physically irrlvant. So this modl is not usful if r<. Whn +r <, that is whn <r<0 th squnc convrgs to zro sinc lim ( + n r)n =0 in this cas. This mans population bcoms tinct which fits th fact that th nt growth rat is ngativ. If r>0, thn th squnc divrgs to infinity sinc lim ( + n r)n =

38 Population Growth of a Singl Spcis II Discrt Modls 5 4

39 Population Growth of a Singl Spcis II Discrt Modls 5 5

40 Population Growth of a Singl Spcis II Discrt Modls 5 6 Eampl 3. Suppos 0 =,r =.5 andk = 0. Thn th diffrnc quation to calculat th valus of n is: n+ = n +.5 n ( (n /0) ). For ampl = ( ( 0 /0)) = +5( (/0)) = 6. Similar calculations show that approimatly =, 3 =6, 4 =, 5 = 6 and so on. This squnc has lmnts that oscillat back and forth btwn 6 and so th squnc dos not convrg. This has th biological intrprtation of a population that is priodic in tim and rpats itslf vry two gnrations. Othr priods ar also possibl for this diffrnc quation. Problms. For th modl abov (a) Tak 0 =0andr =0.5, find th formula for th population n. (b) Dos th squnc of population valus convrg or divrg. (c) What can you conclud from a biological point of viw about th long tim bhavior of th population?. A population is dfind by th diffrnc quation n+ = r n n + A whr r and A ar constants. (a) Givn 0 =, r = 0 and A = find th nt four lmnts in th squnc. (b) What do you think happns to th population as th numbr of gnrations incrass? (c) Givn 0 =, r = and A = 0 find th nt four lmnts in th squnc. (d) What do you think happns to th population as th numbr of gnrations incrass? () Mak an argumnt similar to that in Eampl to s what th possibl act limits should b for this squnc for any valu of r and A.

41 Chaptr 6 6 Th Gomtric Sris and Som Applications by Dr. Costas Efthimiou Wks: 3/6 and 4/ of Spring 007

42 Th Gomtric Sris and Som Applications 6 Calculus Topic: Th Gomtric Sris and its Convrgnc Background matrial Sction. as introduction to sris. Pay spcial attntion to Eampl 5, pag 75. Th matrial prsntd in th following discussion is basd on sction.6 of your book. Th ratio and root tsts applid to th gomtric sris. sction. it is said that th gomtric sris In ampl 5 of S = n () n=0 convrgs if <. In fact, its limit is n = n=0. By stting = y, from th discussion in th sam ampl w s that th sris ( ) n y n also convrgs for y < tothsum n=0 S = ( ) n y n = n=0 +y. () Notic that th two rsults ar idntical although thy may sm diffrnt: whn is positiv, y is ngativ and vic vrsa. For ampl, stting =/ in (), w find ( ) n =. n=0 Stting y = / in (), w find th act sam quation. Whn = /, quation () givs ( ) n = 3. n=0 This rsult is also rcovrd from quation () for y =/. Now, w would lik to r amin th convrgnc of th gomtric sris () from anothr prspctiv: that of sction.6 of your book. Th gnral trm of th sris is a n = n.thn a n+ = n+ an =. n

43 Th Gomtric Sris and Som Applications 6 3 W notic that according to th ratio tst, th sris S is absolutly convrgnt whn < and divrgnt for >. If = ±, th ratio tst cannot draw a conclusion. Similarly, w notic that n an = n n =. Thrfor th root tst also arrivs at th sam conclusion: th sris S is absolutly convrgnt whn < and divrgnt for >. If = ±, th root tst cannot draw a conclusion ithr. Two spcial sris. sris As notd abov, th root and ratio tsts cannot say if th two +++, +, convrg. Howvr, if w look at th partial sums, w can asily say what th bhavior is. For th first sris, th partial sums ar s =,s =+=,s 3 = ++ = 3,... Thrfor s n = n and lim s n =. For th scond sris, th partial sums ar s =, n s = =0,s 3 = +=,s 4 = + = 0. W notic that s n = {, if n=odd, 0, if n=vn. Howvr if a squnc {a n } convrgs, thn lim a n+ = lim a n. n n (S rcis 5, pag 747 of your book.) If th sris was convrgnt, th abov quality (applid for th squnc of partial sums) would imply = 0. Thrfor th sris is not convrgnt. Larning Objctivs In th following sction that includs applications of th gomtric sris, you should larn th following:. How sris appar in physical problms. Th problms w will prsnt ar rlatd to motion of objcts. Morovr, for simplicity, w hav trid to focus on th gomtric sris only. Howvr, many diffrnt kinds of sris appar in virtually all topics in physics.. Addition of an infinit numbr of quantitis dos not ncssarily rsult in an infinit rsult: On of th most common misconcptions among th studnts who study physics and hav not mastrd calculus is that whn an infinit summation is involvd, th sum will always b infinit. Howvr, sris tach us that this is not th cas. Adding an infinit numbr of quantitis can giv in som cass a finit numbr.

44 Th Gomtric Sris and Som Applications Mmorization of th basic formulæ and thorms: In scinc thr is an normous amount of information. It is crtainly impossibl for somon to mmoriz all rsults known to scintists. On th othr hand, it is of paramount importanc for somon to rcogniz th most common situations and know basic thorms and rsults. As you work on this projct, mmoriz th rlatd thorms for sris. Also, mmoriz th basic rsults and formulæ of th gomtric sris; it will b a lif savior for th futur. As you rad th nt two sctions, com back to ths larning objctivs and nsur that you hav ralizd thm. In class,. I will prsnt svral diffrnt situations, and I will ask you to plain in which of thm a sris will appar.. I will giv you svral sris and ask you to apply th root and ratio tst to tst convrgnc. 3. I will giv you som gomtric sris and ask you to find th sum. Znon s Parados Znon of Ela (fl. c. 460 BC) was an ancint Grk philosophr who invstigatd som problms daling with tim and spac. Thir thortical invstigation basd on common sns smd to contradict vry day rality. Znon s parados rmaind unrsolvd in his tim. Unfortunatly for him, h was lacking many notions that calculus introducd much latr. In particular, h was lacking th notion of limits and th mthods to sum an infinit numbr of quantitis. Such notions wr first introducd by Archimds (Archimds mthod of haustion) and bcam popular and formal by Nwton and Libnitz. I will introduc to you, two of Znon s parados. Howvr, I will provid no solutions! I will lt you think about thm on your own (and rturn a writtn rsolution of on of th two during th scond mting). Thir rsolution will lad you to a dpr undrstanding of limits and sris. In sction I will dscrib anothr problm and its solution. You may us this problm as th guiding light for Znon s parados. Spdy Achillas and Turtl. It is said that th grat hro of th Trojan war Achillas could run with th amazing spd of 60 kilomtrs pr hour. On day h arrangd a 0 kilomtr rac against a fast turtl which could run at th spd of 30 kilomtrs pr hour. To mak th rac fair, th turtl was givn a had start of 30 kilomtrs. Th turtl was sur that sh was going to win th rac. Hr rasoning was as follows. Achillas starts bhind m. By th tim Achillas rachs point T 0, I will W actually now know that Archimds had dvlopd and advancd calculus to a grat lvl 900 yars ahad of Nwton and Libnitz but his book was lost. A copy of this work was rcntly discovrd. Is this a rally fast spd? Estimat th spd of a 00 mtr sprintr to compar with.

45 Th Gomtric Sris and Som Applications 6 5

46 Th Gomtric Sris and Som Applications 6 6

47 Th Gomtric Sris and Som Applications 6 7

48 Th Gomtric Sris and Som Applications 6 8 Th sum insid th parnthsis is a gomtric sris that convrgs to a finit numbr. Thrfor th total tim is / = 60 sconds. Qustion 3. If th ball aftr ach collision with th floor maintains a fraction f (f <) of th nrgy it had bfor th collision, how much tim dos it tak until it stops? Commnt Finally, a commnt that may b of som us whn you solv th problms: Notic that in th Achillas Turtl rac, th spd of Achillas is twic as much as that of th Turtl. Similarly, in th dart targt problm, w hav slctd midpoints. Although ths rlations ar not significant for th concptual aspct of th problm, thy simplify your calculations.

49 Chaptr 7 7 Multipath Effcts in Wirlss Ntworks by Dr. Damla Turgut and Chris Sntll Wks: 4/9 and 4/6 of Spring 007

50 Multipath Effcts in Wirlss Ntworks 7 Calculus Topic: Taylor and Maclaurin Sris Sction #57: Find th sum of th following sris. n n ( ) π +. n+ n= 0 (n+ )! 4 Solution: Lt f ( ) = sin f ( 0) = 0 f f f f ' '' ( ) = cos f ' (0) = '' ( ) = sin f (0) = 0 ''' ( (4) ( ''' ) = cos f (0) = (4) ) = sin f (0) = 0 Th Taylor pansion of a function f about th point a is For ( ) sin ( ) ' '' ( k) f ( a)( a) f ( a)( a) f ( a)( a) f ( ) = f ( a) L+!! k! n= 0 ( n ) ( )( ) f a a = n! f =, th pansion about a = 0 is sin ( ) n n= 0 n n ( ) ( n + ) = + + L =! 3! 5! 7!! k Now tak = π, so that 4 π sin =. Now w hav th formula 4 n= 0 n n ( ) π + n+ ( n + ) = 4! Background/Motivation In this assignmnt, w ar going to discuss Taylor sris as thy apply to multipath in 80. wirlss ntworks. Multipath propagation occurs whn radio frquncy (RF) signals go through diffrnt paths from a sourc to a dstination. A part of th signal travls dirctly to th dstination whil anothr part can bounc off an obstruction such as mtal, coatd glass, or ston, just as light rflcts from a shiny surfac bfor raching th dstination. As a rsult, part of th signal arrivs at th dstination via a longr path, incurring dlay. This phnomnon is calld multipath distortion. Th ffcts of multipath

51 Multipath Effcts in Wirlss Ntworks 7 3 distortion can rang from harmlss to th cancllation of th signal, dpnding upon th paths takn by th signal. If th signal rcivd via th longr path has sufficint dlay it can intrfr with th signal from th dirct path rsulting in cancllation. On th othr hand, th multipath ffct can also boost th rcivd signal whn both paths arriv in phas such as whn multipl transmitting antnnas ar usd. Multipath intrfrnc can b vry compl, for ampl, in an urban nvironmnt thr ar plnty of opportunitis for rflctions from building structurs and automobils. In th hom, th walls, appliancs, and furnitur introduc numrous opportunitis for rflctions. In Figur, th ffct of multipath in an offic with a stl dsk can b radily obsrvd. This figur is cratd by simulating th ffct of rflctions from th RF gnratd by a transmittr within an offic. This is a simpl simulation in that only th four walls and stl dsk rprsnt opportunitis for rflction, whil in an actual offic th situation would b far mor compl. Th variations in color rprsnt th fluctuations in signal strngth rsulting from multipath cancllation as rflctions intrfr with th nrgy rcivd dirctly from th transmittr. All of th dark rd aras rprsnt aras whr, if an 80. rcivr wr locatd, th signal strngth would b poor whil light blu aras rprsnt aras whr signal strngth would b high. Th dark rd rctangls rprsnt th stl dsk within th room, which rprsnts an ara of low intnsity bcaus RF nrgy dos not pntrat mtal objcts. As this shows, a wirlss rcivr can princ larg variations in rcivd signal strngth vn as it is movd about a small offic. Figur : A dpiction of multipath in a standard offic with a dsk looking down on th offic. Th chang in color rprsnts variation in fild intnsitis that occur vn in a static nvironmnt du to multipath ffcts within th room. Rd indicats wak intnsitis whr cancllation has occurrd and blu rprsnt strong intnsitis.

52 Multipath Effcts in Wirlss Ntworks 7 4 Larning Objctiv #: Multipath Intrfrnc Th following figur dpicts th gomtry dscribing a multipath rsulting from a simpl rflction incidnt with th ground whr T rprsnts th transmittr and R is th rcivr. If w assum that th ground is a simpl flat surfac, th point of rflction point can b found by finding a spot whr th angl of incidnc and angl of rflction ar qual. In fact, th vry sam physics govrning rflctions occur in optics. Obsrv that th distanc travld by dirct lin of sight, d, is shortr than th total distanc travld by th rflctd path, d. It is this diffrnc in distanc, w will find, that crats intrfrnc and subsquntly multipath ffcts. W will us this figur to dvlop a mathmatical prssion for multipath, and w will b rfrring to this figur in th rmaining portion of this lsson. T d R h t h r incidnc angl ψ d ψ rflction angl d Figur : Multipath gomtry for a simpl rflction from th ground. Multipath can b plaind by studying th cosin function and obsrving th ffct whn two cosin functions with diffrnt phass intract as shown in Figur 3. From trigonomtry, th cosin function, of cours, is rprsntd, mathmatically, as ( ) cos( π θ ) r t = A ft+, () whr f is th frquncy (Hz or /sc), A is th amplitud (volts), and θ is th phas angl (radians). Figur 4 shows two cosin functions that ar 80 dgrs out of phas. In this instanc, th two cosin wavs, if thy hav th sam amplitud, will compltly cancl ach othr out. Th combind prssion for th addition of two cosin functions is () cos( π ) cos( π θ) r t = A ft + A ft+, ()

53 Multipath Effcts in Wirlss Ntworks 7 5 whr θ (radians) rprsnts th diffrnc in phas btwn th two cosin wavs and thr is a diffrnt amplitud associatd with ach wav. For two cosin wavs that ar 80 dgrs or π radians out of phas, w gt, () = cos( π ) + cos( π + π) = Acos( π ft) Acos( π ft) = ( A A ) cos( π ft) r t A ft A ft (3) and, of cours, if th amplitud ar qual, thn this rsults in a solution of zro Amplitud (Volts) Tim (sconds) Figur 3: Dpiction of two cosin functions with on having a phas shift lss than 80 dgrs rlativ to th othr Amplitud (Volts) Tim (sconds) Figur 4: Dpiction of two cosin functions that ar 80 dgrs out of phas.

54 Multipath Effcts in Wirlss Ntworks 7 6 In th cas of wirlss ntworking, w ar oftn concrnd with knowing th avrag powr of a rcivd signal. Th avrag powr (Watts) can b found with th following quation P avg T = limt T T () r t dt (4) Th trm r() t priod (sconds) of th wavform, rprsnts th instantanous powr of th signal. Th valu T is th r( t ). By intgrating th instantanous powr ovr th priod T and dividing by th priod, w obtain th avrag powr. Aftr calculating th intgral, th limit is takn as T approachs infinity, which rprsnts th avrag powr ovr all tim. This limit allows us to comput th avrag powr for a signal that is, prhaps, infinit in duration. Th following prssion for r t dscribs th addition of two cosin functions having diffrnt amplituds and diffrnt phas. Solving for, ( ) () cos( π ) cos( π θ) r t = A ft + A ft+ (5), w obtain th following P avg T Pavg ( A, A, θ) = lim Acos( π ft) + Acos( π ft+ θ) dt T T T T Pavg ( A, A, θ) = lim A cos ( π ft) + A cos ( π ft + θ) + A A cos( π ft) cos( π ft + θ) dt T T T A A Pavg A A AA (,, θ) = + + cos( θ) Of cours, all of th stps hav not bn shown hr, but this is a simpl intgral to solv involving a fw trigonomtric idntitis and a fw substitutions. Aftr solving th intgral, th priod T drops out prvnting th nd for solving th limit. This maks sns from th prspctiv of a cosin function whr th avrag powr within th intrval T will b qual to th avrag powr ovr all tim. W lav th solution of th intgral as an rcis for th motivatd studnt. Qustion. Graph th avrag powr as a function of θ if A = A =. Qustion. What happns if th two amplituds ar not qual, i.., A =, A =? Graph th avrag powr as a function of θ.

55 Multipath Effcts in Wirlss Ntworks 7 7 Larning Objctiv #: Simpl Multipath Eampl In th cas of multipath, th radiatd nrgy from th wirlss dvic, th accss point in th ampl, can b rprsntd as a cosin wav with a frquncy of.4 GHz or Hz. Looking again at Figur, w can compar th dirct path distanc to th rciving nod vrsus th distanc of th indirct path whr nrgy is rflctd from th ground. W s that th nrgy rflcting from th ground must travl a longr distanc. Th longr distanc implis that th tim of arrival is slightly dlayd for th rflctd path vrsus th dirct path sinc th RF nrgy must travl at th spd of light. It turns out that this dlay can b writtn as a phas as follows. ( ) = cos π ( + τd ) () = cos[ π + () = cos( π + θ) r t A f t r t A ft π fτ r t A ft d ] (6) whr, θ = π fτd. Thrfor, th phas of th rcivd signal is a function of th tim dlay and th frquncy of th signal. Of cours, th tim dlay can b computd givn th distanc and th fact that RF nrgy must travl at th spd of light as, d τ d =, (7) c whr d is th distanc in mtrs and c is th spd of light, which is qual to mtrs/sc. In our cas, w ar rally intrstd in th tim dlay diffrnc btwn th dirct and rflctd nrgy as a function of th diffrnc in distanc. Thrfor, w can writ a nw prssion of phas as follows with d rplacd by d. d θ ( d d ) = π f, (8) which shows that if w know th diffrnc in distanc covrd btwn th dirctd and rflctd nrgy, w can comput th phas diffrnc btwn th two signals. Thrfor, th intrfrnc rsulting from multipath rflction off of th ground can b writtn as, c ( d d ) r() t = Acos( π ft) + Acos π ft+ π f c (9) whr th first trm on th right hand sid of th prssion is th nrgy rcivd dirctly and th scond trm is th nrgy rcivd indirctly.

56 Multipath Effcts in Wirlss Ntworks 7 8 Th path diffrnc can b computd by mploying Pythagoran s thorm, as shown in Figur 5. h h t r T d R h t h r h + h t r d d h r Figur 5: Using Pythagoran's thorm to comput th path diffrnc btwn th dirct and rflctd path. d This rsults in th following prssion, ( ) ( ) Δ d = d d = d + h + h d + h h t r t r (0) whr ht is th hight of th transmittr (mtrs) and hr is th hight of th rcivr (mtrs). W would lik to simplify this compl prssion. Using th Taylor s sris, w can rprsnt this function as an infinit sris and, thn, by only using th first two trms of th infinit sris, w can form an approimation to th function which will mak th final formulation for our multipath ampl much simplr. First, w put th prssion into a form that will b a littl asir to work with for th Taylor s sris approimation. ht + hr ht hr Δ d = d + d + d d () W can simplify th prssion undr th radical, for th purposs of finding th Taylor s sris approimation, as Δ = + +, () d d d

57 Multipath Effcts in Wirlss Ntworks 7 9 ht + hr = whr d and ht hr = d for ach part of th prssion in (). Th gnral Taylor s sris pansion is ( ) ( ) ( ) ( ) f a f a f ( ) = f ( a) + a + a +L (3)!! Sinc w ar trying to approimat (), w can substitut a = 0 in th abov quation, that is, w ar finding an pansion around th point = 0. ( 0) f ( 0) f f ( ) = f ( 0) L (4)!! Although th Taylor s sris rprsnt th function as an infinit sris, th infinit sris allows for arbitrary prcision and w ar only intrstd in a suitabl approimation, thrfor, w will tak only th first two trms of th Taylor s sris approimation. Equation (4) is also rfrrd to as th Maclaurin sris. In gnral, w can find an approimation to th prssion + as, ( ) 0 (5) Thrfor, w can now rwrit ach radical in () ht + hr ht hr Δd d + d + d d (6) and, simplifying this prssion, w driv th following,

58 Multipath Effcts in Wirlss Ntworks 7 0 ( h + h ) ( h h ) Δd d + d d ( h + h ) ( h h ) t r t r t r t r = d ( ) ht + hr + hh t r ht + hr hh t r = d 4hh t r = d hh t r = d d (7) Thrfor, th approimation is W can, thn, combin this with (8) to obtain hh r t Δd (8) d 4π fhh t r and our prssion for th rcivd wavform bcoms, θ = (9) cd 4π fhh t r r() t = Acos( π ft) + Acos π ft+ cd (0) Of cours, w know th gnral prssion for rcivd powr lvl as a function of phas from th prvious larning objctiv, for which w can now substitut in th actual phas for th multipath gomtry A A 4π fhh t r P( A, A, d) = + + AA cos cd () It turns out that th amplitud rcivd via th rflctd path can b writtn in trms of th amplitud rcivd via th dirct path. In fact, th rflctd nrgy princs a 80 dgrs phas shift as it is rflctd from th ground in our problm. Thrfor, w can chang th prssion for rcivd powr assuming th rflctd amplitud is th ngativ of th dirct amplitud. A A 4π fhh t r P( A, d) = + A cos cd ()

59 Multipath Effcts in Wirlss Ntworks 7 4π fhh t r P( A, d) = A A cos cd (3) 4π fhh t r P( A, d) = A cos cd (4) Thn, using a trigonomtric idntity, w can transform quation (4) to π fhthr P( A, d) = A sin cd (5) Rcivd Avrag signal strngth Powr (powr) (Watts) Distanc (mtrs) Figur 6: Rcivd signal strngth as a function of distanc from th transmittr. In Figur 6, a plot is mad whr th amplitud is Volt, A =, th frquncy is GHz, and th transmittr and rcivr ar both at mtr abov th ground. As sn in th figur, th signal strngth oscillats at nar distancs bfor hitting a pak, at about 3 mtrs, whr a transition from oscillatory bhavior to an ponntial dcay occurs. Th point at which th signal first transitions from th oscillatory bhavior is rfrrd to as th first Frsnl zon. W can find th transition point, or first Frsnl zon, by solving for th first drivativ of quation (6) and stting this qual to zro as follows, ( ) dp A, 8π fhthra π fhthr π fhthr = sin cos 0 dd cd = cd cd (6)

60 Multipath Effcts in Wirlss Ntworks 7 π fhh t r This prssion will b qual to zro whnvr sin cd or cos π fhh t r cd is qual to zro, or whn π fhth 0, π r,,... n π π fhthr = π =. W not th following lim = 0, cd d cd which implis that th minimum is at d =, and th smallst angl bfor raching 0 is π. As a rsult, w s that as d incrass, a maimum or minimum occurs at π fhthr π = n for n >. At n =, th last maimum occurs at th first Frsnl zon cd bfor th ponntial dcay bgins. Thrfor, th first Frznl zon bgins whn π fhthr π =, and th distanc (mtrs) of this transition is at, cd d frsnl 4 fhh t r = (7) c This calculats to b mtrs for our ampl sn in Figur 6. Thus far, w hav only shown th dcay of signal strngth rsulting from multipath intrfrnc and hav not considrd th natural rduction that occurs du to distanc. Th quation can b updatd to rflct this by adding a trm known as th fr spac loss as sn in th following quation and plottd in th following figur. Not that this trm mrly dcrass th signal strngth proportional to th squar of th distanc as can b sn in Figur 7. π fhthr c P( A, d) = A sin cd 4π fd (8)

61 Multipath Effcts in Wirlss Ntworks Rcivd Signal Avrag Strngth Powr (Watts) Distanc (mtrs) Figur 7: A plot showing signal strngth as a function of distanc whn th fr spac loss is includd. Qustion 3. In quation (5), how can th quation b furthr simplifid if th quantity π fhh t r is small? (Hint, how can th function sin ( ) b approimatd if ) Writ cd out th simplifid prssion. Qustion 4. What happns in th limit as th distanc d grows to infinity in quation (5)? Problms W hav a ral systm consisting of a singl accss point and a wirlss laptop that ar oprating outdoors in a rlativly cluttr fr nvironmnt (no objcts ar obstructing th lin of sight btwn th accss point and wirlss laptop). Th accss point has a Watt transmittr and is rsting on a platform 3 ft (.946 mtrs) abov th ground. Th wirlss laptop is hand carrid at a hight of 4 ft (.95 mtrs). Th amplitud for th Watt transmittr can b rprsntd as Volt or A = V. Th accss point is 9 transmitting at 450 MHz ( f =.45 0 Hz ). Using quation (6), plug in th constants from this problm to obtain a formulation of avrag powr (Watts) as a function of distanc (mtrs).. Using MS Ecl, comput and plot th rcivd avrag powr lvl for distancs btwn 5 and 75 mtrs at an incrmnt of 0. mtrs.

62 Multipath Effcts in Wirlss Ntworks Using quation (8), comput th distanc whr th first Frsnl zon bgins (curv transitions from th oscillatory bhavior to an ponntial dcay)? sin = for small to rwrit and simplify quation 4. Us th simplification ( ) (9). Not that th simplification is only valid for larg distancs. How dos th avrag powr lvl vary as a function of d in this quation (dscrib, no nd to plot)? Would this hav bn as asy to s if th Taylor s sris approimation had not bn usd to simplify quation ()? Why? Appndi: Using Microsoft Ecl MS Ecl can b usd to plot mathmatical functions and has th addd bnfit of bing lss costly than most of th mathmatical softwar tools such as MATLAB and Mathmatica. Du to its availability, it is important to undrstand th powr of MS Ecl in solving nginring problms. In this appndi, w will illustrat how to us MS Ecl for plotting a function. π In this rcis, w will plot th function () t = cos π ft+ whr f is th 4 9 frquncy in Hz and t is th tim in sconds. W will st f = 0 Hz for a frquncy of GHz. First, w want to stablish th domain of our problm or th coordinat. Th priod 9 t is T = f = 0 scor nsc. If w want to plot ± 4 priods of th function, of ( ) thn our tim should rang from sc to sc. If w want to plot a total of points, our incrmnt should b Δ t = =.6 0. Thrfor, w hav all of 50 th information ndd to crat our plot. Upon launching MS Ecl, w ar prsntd with th following.

63 Multipath Effcts in Wirlss Ntworks 7 5 Figur 8: MS Ecl with an mpty workbook at startup. W ar going to b crating two sris of data in th first two columns of th Ecl spradsht. W can start by labling th sris by typing Tim (sc) in cll A and typing (t) in cll B. Not that th column widths can b changd by slcting th lin btwn two columns such as column A and column B and right clicking and dragging th lin to th dsird location. Figur 9: Entring th column titls for th domain and rang.

64 Multipath Effcts in Wirlss Ntworks 7 6 Nt, ntr th starting valu for tim by typing 4E 9 in cll A. Also, th nt tim will b = basd upon th incrmnt valu. This can b ntrd as th nt tim by typing 3.84E 9 into cll A3. Figur 0: Entring th first two tim valus in column A. Using a shortcut within Ecl, th rmaining valus for tim can b automatically filld in. Slct both clls A and A3 by clicking on cll A and holding th lft mous button down whil dragging th slction ovr cll A3, and thn, rlasing th lft mous button so that both clls ar highlightd as follows Figur : Rsult of slcting clls A and A3. Not that thr is a filld in bo on th bottom right cornr of th slction bordr. Whn th mous hovrs ovr this bo, th cursor changs to a cross hair cursor. Click on th bo with th lft mous button and whil holding th lft mous button, drag th slction across 50 clls so that clls A through A5 ar highlightd with a hash bordr. Th following figur shows highlighting just clls A through A7.

65 Multipath Effcts in Wirlss Ntworks 7 7 Figur : Rsult of clicking th lowr right cornr of th slction in th prvious figur and dragging th mous to cll A7 whil clicking th lft mous button. Aftr all of th clls hav bn highlightd, rlas th lft mous button and th incrmntd valus ar automatically populatd. Th valu in cll A5 should b 4.00E 9. Figur 3: Rsults of automatically filling in th tim sris data.

66 Multipath Effcts in Wirlss Ntworks 7 8 π Now w ar rady to ntr th formula. Th formula will b () t = cos π ( 9 ) t+ 4, which can b spcifid in Ecl as, COS(*PI()*9*A + PI()/4), if th formula is placd in cll B and A rprsnts tim. To ntr th formula, typ =COS(*PI()*9*A + PI()/4) in cll B. Figur 4: Entring th formula into cll B. Onc this is don, a shortcut can b usd, onc again, to populat th rmaining valus. Do this by first slcting (clicking on) cll B and, thn, doubl click th lowr right cornr of th cll whr th filld in bo is locatd. At this point, all of th valus ar automatically populatd corrsponding to ntry of tim in th A column. Figur 5: Slcting cll B and obsrving th filld in bo at th lowr right of th slction bordr.

67 Multipath Effcts in Wirlss Ntworks 7 9 Figur 6: Rsult aftr doubl clicking th lowr right cornr of th slction in th prvious figur. To s what happnd, click on cll B3 and obsrv that th formula was automatically updatd to us cll A3 for tim instad of cll A. Figur 7: Obsrving th formula in cll B3. In ordr to plot th function, slct columns A and B by clicking on th A at th top of column A and dragging th slction to column B whil prssing th lft mous button.

68 Multipath Effcts in Wirlss Ntworks 7 0 Figur 8: Slcting columns A and B. Onc th columns ar slctd, slct th Chart Wizard using th button in th toolbar which dpicts a 3D bar graph. Figur 9: Th chart wizard button locatd in th toolbar. This brings up th wizard. W want to slct th XY (Scattr) plot sinc w ar plotting a graph containing both and y coordinats. From within th XY (Scattr) possibilitis, slct th chart sub typ which will provid a smooth curv without displaying th points in our data sris.

69 Multipath Effcts in Wirlss Ntworks 7 Figur 0: Slcting th XY scattr plot. Click Nt > to procd to th chart stup. Figur : Stp of th chart wizard, spcify data sris. Click Nt > to accpt th data rang, which w alrady slctd whn w highlightd columns A and B.

70 Multipath Effcts in Wirlss Ntworks 7 Figur : Stp 3 of th chart wizard, spcify titl and ais labls. Fill in th valus for th chart titl, and, y ais in stp 3 of th wizard and click Nt > to continu. Figur 3: Final stp of th chart wizard.