Fund EM & Trans Lines
Fund EM & Trans Lines ECE 3300
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ECE 3300 Maxwell s Equations in Free Space and Plane Waves Maxwell s Equations General Time Domain and Phasor Forms To convert from time domain to frequencyphasor domain assume that EH are functions of elm Then ddt jm V05pv VXE a B jaau at VE0 35 VXHJCa O39E608EJCECE t 39 l u 86 08 Jaw 8 18 Here Jc conduction current density Amz 2c is complex dielectric constant Wave Equations in ChargeFree Currentfree Medium If the material is lossless 60 then there is no charge and no current This is called a perfect dielectric An insulator is a good or nearperfect dielectric VOE0 Vszij Vo 0 inlzjang Your book derives the general wave equation in the time domain So we will derive the wave equation in the frequency phasor domain instead Derivation of WuveEquution Take Vx VgtltVgtltE jaiuVgtlt Substitute Vx WE jaaujwe th Vector property VxVxE VVoE V2E Substitute VxVxE V0 V2E V2E Combine wztiecf VP 2 0 Vi 72E 2 0 2 2 t r a 80 a 16 2 2 2 Lossless k a 8 6 2 2 V E k E 0 Similarly 2 V H 72 H 0 These wave equations can be used to describe any type of wave planar spherical etc No approximations or limitations have been made yet Plane Waves are a special type of wave 0 Planar wavefront of infinite extent 0 TEM EH are perpendicular to eachother and to the direction of propagation 0 Plane waves are like transmission line waves which are also TEM Look for the similarities Here are the transmission line wave equations How do we solve problems with the wave equation 1 Write wave equation for instance in rectangular coordinates 2 Separate the vector components into Xyz Equate vector components which means to do a separate equation for each vector component 3 For plane wave front at wavefront moving in the z direction ddx and ddy 0 Substitute this into 2 above 4 Verify that Ez or Hz 2 0 PLANE WAVE HAS NO FIELD COMPONENTS in direction of propagation It is a TEM wave like the ones we have studied in transmission lines General solution for a TEM wave such as Plane wave This applies to vector fields AND to each component separately E ET 7m r m Z ZE Z E0 6 E0 ze Now for a plane wave traveling in the z direction Ez0 but EX and Ey may not be zero Solve for H from Maxwell s equations see text and we obtain 7 1 HyO Ex0 Ex0 aw 77 Where 1 is Intrinsic Impedance of medium 112377 for air nzreal for lossless materials E and H are perpendicular and in phase ncomplex for lossy materials E and H are perpendicular an out of phase Duality to Transmission Lines E 9 V recall also V integral E d1 H 9 I recall also I 2 closed integral H d1 1 9 20 Other Similarities 1 Waves can be broken into and 7 traveling waves 2 Propagate in z direction as esz kz including effect of phase change and attentuation Waves will re ect depending on the material discontinuity they hit a Re ection coefficient transmission coefficient b Standing Waves 6 Smith Chart solution 3 Differences 1 V and I are not vector quantities E and H are This means we can have polarization ECE 3300 Displacement current Displacement Current Time varying electric field produces magnetic field a7D Vgtlt a To relate this to current a5 7 i 7 Leftls LVgtltHdS StokesTheorem E d2 C Amperel sLaw 3 d2 1 C WhichGives J 87D 0 d 0 d2 Displacement 7 Current 3 at C g 0 d 2 Conduction 7 Current Displacement 7 Current zmloszd Total 7 current 0E 87D 0E Bel 039 E at a at For e7 3 7 a7 W So time dependence 7 a r r Total7current 039 jaEE 875 E ng E Complex7 8 0ja jwel jwe e j039aeo 8 808 jUweo 805 5 Why all this Conduction Current carries charges Displacement current represents displacement rotationally of molecules but no actual motion Parallel Plate capacitor example Vst 2 V0 cosot Time varying voltage Conduction current in wire Ic C dVdt CVo osinot Displacement current in wire 2 0 because DE0 in metal wire Electric field between the plates EchdyVocosotd Displacement Current between the plates 85 7 a A 8V0 A 1d SEas Abe d cosaxj 0yds Voasinat CV0asinwt 1 Conduction Current is owing down the wire and becomes displacement current owing through the plates ECE 3300 Bnundary Cnndiu39nns Please read aboutthe denvatton of these boundary condltlons m the textbook Boundary seetton here Electric Field Bnnmlary Cnnditinns Region 1 Reg on 2 D2t Steps to salve hnundary cnnditinn prnhlems D in one ofthe two regions as shown above 2 Solve for the tangential eornponents like this D 51 0 En 522 52 Cause D 6E 3 Solve for the norrnal The norrnal eorn Din r Dzn 95 components like this ponents depend on the surface charge density p5 Cmz Om OR etEn e eaEae 9 Special Cases Perfect Dielectrics conductivity 0 L th m m V vi l are perfect dielectncs have no eonduetrvrtvthen p Perfect Cnnductnrs conductivity is in nite metals The eleetne eldlnslde the metal 0 so surface TANGENTLAL t 0 inside the metal and on its E 0 on surfaee of metal Magnetic Field Boundary Conditions Use the same gure as above but replace electric elds or ux density with magnetic elds H or ux density B Steps to solve boundary condition problems Typically you are given or have previously calculated the magnetic eld H or ux density B in one of the two regions 1 Break the magnetic ux density vector B into tangential and normal components as shown above 2 Solve for the tangential components like this The tangential magnetic elds depend on the surface current density most books call this Js some call it K This is the current density Mm2 owing ON THE SURFACE H2tHlt JsAm2K Ezy 3 Solve for the normal components like this Bln BZH 0R H1H1n H2 H2n Special Cases Perfect Dielectrics conductivity 0 Surface current density can only exist on a conductive surface so if both materials are perfect dielectrics have no conductivity then J s 0 Perfect Conductors conductivity is infmite metals The magnetic eld inside the metal 0 so Hn 0 inside the metal and on its surface NORMAL H 0 on surface of metal How to Apply This Concept There are three main ways we apply the concept of boundary conditions 1 As described above boundary conditions provide an understanding of how elds behave on the surface of metal The tangential E eld and normal H elds are always zero on a metal surface Given the elds in one region nd the elds in the other This calculation can be used to derive Snell s law for optics we will discuss this in the ber optic section and accounts for why light and other elds bend when they move from one material to another N V Given elds in one region how do we expect elds to behave in another region Please read the book chapter linked to this website don t worry it is short mostly pictures Observe that the electric eld is always weaker in the stronger L V dielectric larger 8 For instance when elds pass from air low a to skinfat low a to muscle high a the elds will be strong in skinfat and low in muscle Hyperthermia is a method of heating the body to treat cancer We have previously discussed that standing waves can be set up in the fat layer and overheat it Even without these standing waves we can expect the fat layer to have higher elds than the muscle layer because of the boundary conditions Turning up the power enough to get the elds we want in the muscle layer could overheat the fat layer The water bolus helps to eliminate this problem by reducing the standing waves and by removing the heat from the body through conduction Another interesting application of boundary conditions is in the use of petri dishes in a TEM cell for doing biological electromagnetic experiments A TEM cell is a large resonating chamber that produces a very uniform electric eld that is well known either through calculation or measurement If petri dishes are placed inside the TEM cell it could be tempting to assume that we know the eld in the petri dish because we just measured it in the TEM cell But the uid in the petri dish has reduced the eld inside of it and we no longer have either uniformity or the magnitude we expected The magnetic eld also produces interesting results The electric elds tend to circulate around the magnetic eld in the direction de ned by the ngers of the right hand with the thumb pointed in the direction of the magnetic eld Circulation can occur in multiple regions of the model if they are somewhat separated electrically See the model of the cow in the attached chapter These electric elds are once again reduced by high dielectric materials We also solve problems called boundary value problems in electromagnetic where we assume we know the shape of the incident eld a plane or spherical wave for instance and we solve for the electric andor magnetic eld in an easy tode ne object such as a layered cylinder sphere etc using boundary conditions You can learn more about this in an Advanced Electromagnetics Course ECE 3300 Snells Laws and Fiber Optics Snell s Law B m sqrt us for lossless material ei er sin Gil sin 91 51 Bi sqrt tell Mei Using Index of Refraction n c Vp sq us Hose sqn mar Assuming nonmagnetic material 112110 then sin Gil sin 91 nil ni Critical Angle When Bi gt Bl then for angles greater than the critical angle there is total re ection no transmitted power Sin 91 critical sqrt 131 H191 Brewster Angle At the Brewster angle the TM parallel polarized field is all transmitted no re ection The TE perpendicular polarized field still has a re ection For lossless nonmagnetic material 0 Brewster How Does all of this apply to Fiber Optics cladding Wave 1 Core W Light with angles less than or equal to acceptance angle can propagate down the fiber Fiber Core Lowloss material glass nf Cladding Lower index of refraction than core nC lt nf Fiber Optic Propagation Requires TOTAL INTERNAL REFLECTION The angle shown as 90 wave 1 hits the corecladding interface must be greater than or equal to the Critical Angle so 90 2 Gcritical 9 sin 90 2 nC nf 7106 Feeding Fiber Opical Cable We need to feed this at the right angle to create the total internal re ection required for propagation nO sin 9a nf sin 9b 7105 Combining 71056 gives the NUMERICAL APERTURE 22 nf nc n 0 This is also sometimes called the acceptance 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Cylindrical Spherical Coordinates Coordinates Coordinates Coordinate variables x y z r lt15 Z R 9 Vector representation A Ax 339Ay iAZ EA A iAz AR 9A9 A 2 2 2 2 MagmtudeofA IA A A AZj fA2A A ARA9A A Position vector 0P1 9061 3 izl Er 221 RRl for PXlylZl f0r1quotrl gtlZl for PRl91 1 Basevectorsproperties y Ii21 ird43221 r w e1 iyy22io r 22fo R9 R ilelzi fx i Rx0 frx2ir x2r oxdli ixizfl ixf XR9 DotproductAB AxBxAyByAsz ArBrA B AZBZ ARBRA939A B i i 2 r 3 2 ii 6 a Cross productA x B Ax Ay AZ A A Az AR A9 A4 Bx By 15Z 3 12 BZ BR 39 34 Differential length dl fidx y dy 2 dz f dr 8 d 2 dz RdR 5R d9 R sin6 Differential surface areas dsJr i2 dy dz ds f39r dd dz dsR RZ sin 0 16 dd dsy dx dz ds dr dz ng Rsin0dR dd dsz 2 dx dy dsz 2r dr dq dsq 43R dR d0 Differential volume dv dx dy dz r dr d dz R2 sin6 dR d0 dd 1 Table 342 Coordinate transformation relations Transformaiion Coordinate Variables 1 Unit Vecfrors Vector Cornponcms J Cartesian to icos sin Ar Axcos Aysin f cylindrical 45 tan 1yx J 4mm ycos Ad Ax sin A cos gt z z i 2 AZ Az Cylindrical to x r cos 5 i 2 recs 1 2 sind Ax Ar cos A45 sinq Cartesian yrsin yzrsin cos Ay Arsin A cos Z z i 2 AZ AZ Cartesian to R 7x2 y2z2 R isinecosq AR Ax sin6c05 spherical Isim93in gt 2cos6 Ay sinGsin AZ 0039 6tan 1 x2y2z cosecou A9Axcosecos cosesin 2sin9 Aycos sin Azsin9 tan yx isin 39cos A Ax sin Aycos Spherical to x R sine cos I Rsine cos gt Ax A R sichosqb Cartesian 9 cos6 cosqb ti sin A9 0039 c0345 A sinci yRsinesin Rsin9sin AyARsin0sin c036sin cos Agcosesin A cos zRcosB 2 cos9 sin9 AZARc036 Agsin6 Cylindrical to R 7 r2 z2 Ii 2 rsin6 20056 AR Ar sine Az cosB spherical 6 tan rz 3 i39cose isine A9 2 Ar c030 AZ sine 2 43 r13 Aqs 14 Sphericalto r Rsin6 rz sin6 cose ArARsin6A9cos6 cylindrical 4b 15 13 z 13 A4 A zRcos 22R0056 sin0 AZARcos0 Agsin6 ECE ma Plane WaveinLossy Media Re w onvsirg as milh Chan n2 R Remam Rama Ewan 3 m 01 a2 02 03 n 17mm mm chamom ic impedanees m sqmm 125 57 n a 596 9 Jam 2 77829 n 4225142om 2 2 2922 2 22 yr Ampfm sm 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