Electromagnetic Waves ECE 341
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ELECTROMAGNETIC WAVES Uniform plane waves in lossless media 2 Solutions timedomain Ert aEE cosat Br aE E cosat a a Br Ert aE E cosat 3x cosl9X y cosl9y z cosez Solutions frequencydomain Ert Re E e1quott Era aEE atquot The electric and magnetic vectors perpendicular and both are perpendicular to the direction of propagation H aB x 1E or use Faraday s law 77 E 77H x aB or use Ampere s law 739 2 39T v f 2 general 13 3 s E x H lossless only 1 J All kinds of applets on transmission lines waves etc L t Mm httn39lmnAnv Example TEM wave propagation E An Efield is given by E a2 5Ocos109t 5 Vm M Me Find i direction of travel ii velocity iii wavelength iv wave or intrinsic impedance v H 341 lecture notes Energy conduction current displacement current From the pom form othm s aw J OE two pomts can be apprecrated 1 We conducnon current and the e ectnc new vary ahke cnanges m the e ectnc new produce We vanatrons m the conducuon curren 2 I versus as snown be ow From Jou e s aw i E J dv Gwen that E and J are m pnase and paraHeL the powerdrssrpated pervomme m conductwe matenax rs Swmpwthe product onne magmtudes ofE an 7EJ dv g9 341 mm nozes Displacement current density is the time rate of change ofthe electric ux density c vector The ux density can be broken into two pieces the rst being sDE whi associated with the separation of bound change DsEssaEsaEs15DE DsaEP JD a s E z a a a low 1 u k r y u me Hello iui u g r Below The power per E JD I I JD EDat 341 ieame notes lfthe material under consideration have both displacement and conduction currents presence there will be both energy storage and average power dissipation The net current will be between 0 and 90 outofphase with the electric eld This situation can also occur in dielectrics at high frequencies in which the time 39 39 quot 39 L L quot L no longer be neglected with respect to the time rate of change ofthe eld Displacement current at highfrequencies I J aD t E n dPdv k Lr b Wm m v X The result in either case whether the material has both conduction and displacement current or whether the frequency is suf ciently high so that the displacement current is not in phase with the electric eld the result is that there is an average power lost 39om the eld in the material Another loss mechanism to be treated later is radiation loss which occurs quot L 39 39 39slossan 39 39 39 39equency and geometry are such to lead to ef cient radiation Radiation appears in the treatment of antennas AI 4 conductivity and an equivalent conductivity can be assigned to each 347 iecme notes Waves in ossy material 1 Taking Ampere and Faraday with no charge or current sources Note no current sources DOES NOT mean there are no currents n conductive materials 0 i O currents exist in response to an electric field in conductive material WHJ8EUESE at at 0H V E X at Combining these equations a 0 0E VVEVVoEV2E VH oE X X ag X ital sat 0E 02E VZE gunquot was O In the frequency domain Vz jmm asz VZE wzyg 41 VZE wzyg ji 0 we we V2E w2y EO where 5 s 1 ji is the complex permittitivty me In the frequency domain solutions are E aE Ee 39r 039 where 39y 7ay lwzys a7 ijw lsy 1 i a ma Terminology 7 complex propagation vector 9 complex propagation constant a attenuation constant equal to zero for ossess materials 8 phase constant 341 lecture notes Details of solution The solution must satisfy the wave equation VZE wzy E O E aE Ee 39r E a Eel e a j ayr E a j x cos X y cosay z cosaz E aE Eel e 02 02 02 39 i iaj x0056X ycos y zcos z W ayz E aEEem e X j wzlu aE Eewi eizz j x cosa ycossr zcosazj 0 Each derivative pulls down ia j cos Q from the exponential The derivatives hear are second derivatives iaj cos92 iaj cos92 iozj cos92 Ew2y EO lt gt x lt gt y lt gt 2 PM j6 2cos29X cos29y cos292 E wzy E O The sum of the squares of the directional cosines must equal zero PM j6 2 coszeX cos29y cos292 E wzy E O 2 7Ew2y EO ya2u The result is the propagation constant is complex a j The propagation vector also indicates direction of travel it 5 a a j a Propagation along a7 E aE Ee 39r E aE Eeiaw39r cosat 6 ay r W Propagation along a7 E aE Ee39739r E aE Eeawq cosat 6 ay r f 341 lecture notes Plane waves in lossy materials E aE Ee 39r E aE Eewi eizz j x c056X ycos y zcos z E aE Eeialayre ayrej In the time domain E aEEe0 ayr coswt i ayr i ax cos y y cos l 2 20562 EaEEe7 cosat i 6 xcos9X ycos9y zcos92 Hf EM waves in lossy material are exponentially decaying sinusoids 1 Consider an electromagnetic wave pointing in aX direction traveling in a lossy material in the az direction assume 4 O E aXEequotz 0 72 wzms J a 7ij2 ejg am In the time domain E aXEe39 zcosat z The intrinsic impedance is complex E and H are NOT in phase for lossy materials H a x E or use Faraday s law 77 E Fl x aB or use Ampere s law 1 8 v fl general V Iossless only 341 lecture notes Wave impedance in a lossy material What is the relation between E and H in the diagram shown 1 EM wave propagation in lossy material Ex t 0 A E Ee39azcoscntBz r 0 V V V Hy A H He39azcosmtpzen OVVV N 341 lecture notes Example Wave propagation in Iossy material E An electric field is given as E ax100e397z Vm is traveling through material with c01Qm w1 s4 The frequency is 245 GHz Find 0c and B and the dBm attenuation in the material 341 lecture notes Power flow in a lossy material if E points along ax traveling in az direction Take phase of Ezo 0 i0 S E x H aXEe39 zcosat 82 x ayHe39 zcosat 6 2 977 azEZ He39z z cos977 cos2at 282 97 EM wave propagation in lossy material t0 1AEHe Z Izlcoswn cos2mtZBzen timeaverage Poynting vector 1 N N 1 zzz j6 z 1399 zzz j z S EReE x H EReaXEe e x ayH e quote e 1 19 2az 1 2zzz azReEEHe quote azEEH cos977 e S zav 341 lecture notes Special Cases 1 High loss good conductors 039 gt we MaxE 111 2 ME 1391 we we 392 N mm J owae 071a 71j m Low loss good dielectric 039 ltlt we EM wave propagation in a good conductor EM wave propagation in a good dieloctric low loss yij1jiEJW E139j2sj Eh gt Haws am quot V 341 lecture notes Good dielectrics low loss and good conductors high loss 1 Note that in the expression for complex permittivity the imaginary term within the parentheses gives the ratio between conduction current and displacement current JdQgE Jdjw5ngE at at Jc i5 L 11 Jd jws law me good dielectricsquot are defined by ljdl 1 good conductorsquot are defined by JC gtgtJd Example Good conductor Good dielectric Find the propagation constant for f 1OOHz c 10Qm i151 s10 Is the material a good conductor or a good dielectric Neither Example Copper At what frequency does Cu cease to be a good conductor c58107Qm p1 s1 341 lecture notes Skin depth and its effects in circuits 1 The distance a wave must travel into a good conductor for its amplitude to be reduced by a factor of e391 is called skin depth 5 Recall that for a good conductor conduction current is much greater than displacement current 6 gtgt ms Jc cE Jd dDdt s dEdt 039 72 wz s J ms rmz raw 2 7 a J J 2 1 2 0c is the attenuation constant and is the reciprocal of skin depth Consider an electric wave pointing in the xdirection and traveling in the zdirection E This implies that the magnetic eld must lag the electric eld J w4450 E FI 039 Wave impedance 341 lecture notes Example Skin effect if The distance in a good conductorthat is required to reduce the amplitude of the field vectors by a factor of e391 is called the skin depth 5 Assuming copper EZ EZ Eoe39 z EoeE Eoe391 forz 5 aquot 1 2 1 f 1 00661 5 0 g 071a 27139 58f Jf copper nonmagnetic p 110 4n10397 Hm o 58107 Sm frequency Hz 60 Hz 1 KHz 1 MHz 1 GHz skin depths 85 mm l 21 mm 661 pm 21m 341 lecture notes Skin effect in circuits 2 Consider the important special case in which the skin depth is much smaller than the cross sectional dimensions of the conductor In this case it is convenient to suppose the current density is uniformly distributed over a skin depth Ru a Al The AC resistance of conductors is a function of frequency since the skin depth is a function offrequency 5 i circular conductors 3 lt gt lt cirwlar rem me mndudur ii square conductors iii stranded Wire 3417 lecture roles Example Skin effect if Avoltage vst 10 cos 21cft V is applied to a 100 Q load and a solid copper conductor as shown The radius of the conductors is 2 mm and their length is 5 cm a What power is dissipated in the copper conductor at 60 Hz l lt 5 cm I 100 9 v5 conductor load b What is the resistance of the conductors at 4 GHz What power is dissipated in the copper conductors at 4 GHz A voltage vst 10 cos 21rft V is applied to a similar conductorload this time the conductors consist of a 10 pm copper coating with the interior of the conductors being a good insulator like Teflon a What is the resistance of the Cucoated conductors at 60 Hz What power is dissipated by the Cucoated conductors at 60 Hz lt 5 cm gt 100 9 v5 conductor load b What is the resistance of the Cucoated conductors at 4 GHz What power is dissipated by the Cucoated conductors at 4 GHz 341 lecture notes Poy nting s theorem 2 80 John Henry Poynting E VXHE JEj Using this and the vector identity V EXH H VXEE VXH 347 lecture HOIeS Poynting s theorem 1 Looking at the integral form of Poynting s theorem it is readily seen that the theorem is an expression of conservation of energy IF Poynting s vector is power flux density in Wm2 Conservation of energy is the justification for the interpretation of S ExH as power flux density Example units What are the units of a ED b HB c S Example Calculations of power flow An electric vector pointing in the ydirection travels in the zdirection At z0 the electric vector has an amplitude of 2 Wm f10 MHz with material properties m1 s4 c10395Qm Calculate the terms in the integral form of Poynting s theorem for the cube Oltxlt2m0ltylt2m0ltzlt2m 341 lecture notes 20 341 lecture notes 21 Example Power density 1 Calculate Poynting s vector andthe timeaveraged Poynting s vector for the plane wave below traveling in air Z E ax 2eJ2y Vm 341 lecture notes 22 Example wave propagation in Iossy material E An EM wave travels along az in nonmagnetic material with a complex permittivity m 1010 rs of 5 so 20 j15 258061369 It is known that E points in the ax w 20 629az A12 i a What is the material s conductivity b Can the material be classified as a good conductor or as a good dielectric direction and that S ii What is the material s complex propagation constant give in rectangular form iii What is the material s intrinsic impedance give in polar form 341 lecture notes 23 Example cont An EM wave travels along az in nonmagnetic material with a complex permittivity m 1010 rs of 5 so 20 j15 25809136 It is known 20 629 a2 V12 iv Give the electric and magnetic vectors in the frequency domain that E points in the aX direction and that S assume the phase of E is 0 at z0 v Give the timedomain electric and magnetic vectors vi Find the Poynting s vector S and the timeaveraged Poynting vector vii a What is the wave s wavelength b What is the wave s phase velocity 341 lecture notes 24 EM waves incident on boundaries normal incidence 1 How are EM waves reflected and transmitted when a plane dielectric boundary is encountered Consider the situation below X medium 1 71181411 medium 2 11232 c10 E 620 incident wave gt H E1 IEr reflected wave 4 H transmitted wave gt Hi Define a transmission coefficient T and reflection coefficient P Let Iii aXEie39j giz Boundary conditions E an 20 Emquot 20 Hmquot 20 Hmquot 20 341 lecture notes 25 Example transmission and reflection E Find the reflection and transmission coefficients for a wave traveling in a printed circuit board made from FR4 most common PCB material fiberglass epoxy base into air FR4 s44 i151 air s1 1 341 lecture notes Polarixat39 n Io T Putahzattph referstu the path traced bythe eteetheveetpr m a travehng h m what path tstraeeu by the eteethe vemur m the gtltry ptah as UmE advances7 th geherah a Wave travehng m the ax meetch eah have h chumpu eht a u a yr heh The Me a h h epmpp t resu ht vemurtraces ah Epse4llip calpu a zatmnim the gtltry ptah as tttrave s the shape ufWhmh depends uh E E t a d ch E axExcusmr z 4 aVEvcusmr Z a Linear putahzattph ts a speetat ease e E 7 at y x YLX Wh h Ex EV a d ts a d ch dtffer by saw the resmt ts circular putahzatmh t ftrhanded Dr V mtatmh at E the thump ts m the meetch at prupagatmn Examples 7 What tsthe putahzattph fur E axZ cusut r 52 my 7 aVW u cusmr 527 am LHEF aVZ cusntr 527 45 7 RHCF39 aV cusntr 51 my 7 LR t 18 43 What tsthe putahzattph fur E axZ cusut r 52 my What tsthe putahzattph fur E aXS cusnt 752 my 341 mm hates 27 EM waves incident on boundaries oblique incidence 1 Now consider an EM wave traveling in region 1 which encounters a second region at an angle other than normal In this case we must consider two types of polarization 1 Perpendicular polarization E I to plane of incidence POI 2 Parallel polarization E II to POI E H to POI Definitions POI plane formed by propagation vector B and the vector normal to the boundary between regions 1 and 2 Polarization in general any nonrandom orientation of an electric and magnetic field In particular polarization more usually describes the path the electric vector takes in planes of constant phase as the wave travels Note We re considering only linear polarization where the electric vector points in one direction It reaches its maximum positive grows smaller becomes zero then negative becomes maximum negative grows smaller becomes zero etc That is the vector remains on a straight line thus linear polarization Any TEM wave can be expressed in terms of linearly polarized waves 341 lecture notes Refection and transmission at dielectric boundaries E L POI x 111311111 39 quotnZ EZ PZ 0391 5392 15 give expressions for the electric and magnetic vectors 28 if Then to find reflection and transmission coefficients require tangential H and E to be continuous at the boundary 2 O 341 lecture notes 29 Refection and transmission at dielectric boundaries E L POI 341 lecture notes 30 Refection and transmission at dielectric boundaries E H POI if x 7111511111 396 39 7121921112 341 lecture notes 31 341 lecture notes 32 Example Reflection and transmission 1 A uniform plane wave in air impinges on a dielectric material at 6 45 The transmitted wave propagates with 6 30 The frequency is 300 MHz and the electric vector is perpendicularto the plane of incidence i Find 82 in terms of so assume p1p2po ii Find the reflection coefficient F Find the transmission coefficient T iv Give expressions forthe incident E and Flfields the reflected E and Fl fields and the transmitted E and Flfields v 341 lecture notes 33 EM waves normally incident on good conductors 1 Consider EM waves encountering a conducting material x medium 1 conducting material E I incident wave gt Hi 1539 re ected wave lt H l39 341 lecture notes 34 Example Reflection and transmission from a metal 1 A 1M Hz plane wave in air is incident on a copper sheet The amplitude of the incident Efield is 100 Vm Find F T 5 and the E and H fields at a distance of 5 into the copper 341 lecture notes 35 Standing Waves 1 Consider normal incidence In steadystate the incident and reflected waves interact to produce in region 1 a traveling wave and a standing wave x medium 1 medium 2 gt 4 j E1 Ei Er choose reference phase Ei Eie 2 e39J39gz 4 4K E1 Eie 2 e39J39gz FeWFEie 2 e z 4 j 4E 4E E1 Eie 2 e39J39gz FEie 2 e z FEie 2 e39J39gz FEie 2 e39J39gz 341 lecture notes 36 Standing Wave Ratio SWR if SWRE1F Emin 1r SWR1 rSWR1 hlrl The SWR ratio can readily be measured so too F For no reflection SWR1 As more of the incident wave is reflected the SWR ratio grows 341 lecture notes 37 Capacitor if capacitor The ratio of the charge stored on the plates Q and V is the capacitance of the structure In this model no current flows I 0 from the DC voltage source Since I O and O there is no reason that the parameter Cdc should be suspected of having any particular relation to the capacitance from lumped element circuit theory defined by tCac Are Cdc and Cac really the same If so are there limits to the validity For instance does the capacitor operate at high frequencies as it does at low frequencies What are low frequencies What are high frequencies If there are changes at high frequencies what are they 341 lecture notes 38 Inductor if inductor If the plates are shorted and a DC current I is input into as shown the magnetostatic field is The ratio of the magnetic flux mm to the current I is the inductance of the one turn structure de Vm In this model no voltage exists V 0 across the PEG plates due to the DC current Since V O and 0 there is no reason that the parameter de should be suspected of having any particular relation to the inductance from lumped element circuit theory defined by d t i vtL ac dt Are de and Lac really the same If so are there limits to the validity For instance does the inductor operate at high frequencies as it does at low frequencies What are low frequencies What are high frequencies If there are changes at high frequencies what are they 341 lecture notes 39 Maxwell s equations Gauss law VD pV Conservation of magnetic flux V B O Ampere s law V x H J Faraday s law V x E at Constituent relations J cE D SE B pH In EM Fields 340 and in the beginning of EM Waves 341 wave propagation effects were assumed to be negligible For example when discussing the magnetic field due to a changing current it was assumed the field appears identically everywhere at the same instant This cannot be so EM waves travel fast but their speed is finite Forthe past couple of weeks in EM waves we ve been discussing electromagnetic waves When does the transition occur When must one begin accounting for wave phenomena What are the effects These important questions are answered by quasistatics which is the study of the nether region between statics and waves QuasiStatics if Maxwell s equations can be expanded in a power series in 03 about 03 O The first two terms the static solution and the 1s order term are often referred to as the quasistatic solution There are several motivations for developing the quasistatic technique 1 finding the solution to Maxwell s equations can be a daunting task and using the techniques that have been developed for the quasi static field we can use solutions from statics to approximate the solution for the dynamic situations 2 what is meant by low frequencyquot can be defined 3 the emerging effects seen when leaving the low frequencyquot regime can be found 341 lecture notes 40 Power series expansion of Maxwell s equations 1 2 2 m n n no BBwE 10 Zw0Bn ZwquotBn m390 M HFO 2 0a HFO quot0 n M HFO quot0 0B Faradayslaw V x E E V x EjaB Looking at the two sides of the equation for Faraday s law consider them in the form of power series in 03 Since each power of m is independent once can write a series of equations one for each power of m V x jgquot En jaiaquot i0 i0 Since each power of m is independent equations for each power must hold This gives rise sets of equations in the zerothorder firstorder etc In the frequencydomain V x min 1mquot am for n 0 V x E 0 0B In the timedomain V x Equot at for n O V x E0 O zerothorder firstorder secondorder coefficients of m0 coefficients of m39 coefficients of m2 VXE00 VXE10B0 VXEZ0B VxHo 10 at at V39Dopo VXH1J1 VXH2J20FD1 VBo 0 at VJ0O VID1p1 V392p2 v31 o VB2 0 0p 0p 1 2quot boundary conditions an xE1 E2m 0 an xH2 H2m Km an D2 m psm an B2quot O 341 lecture notes if parallel plate model with distributed source along y at z Capacitor quasistatic solution 41 341 lecture notes 42 O horder fields 1 WED anz any ax anx 0E0z a any anx a 0 By 02 02 0x y ax By E vovgo50g0 EOXWOE02 p0V ax ay 02 0H 0H WHO Wozwax0 m0quot39ma OY0H XazJ ay 02 02 0x Y ax ay 0H v B0 v yoHo o0x ay v 039z 0 341 lecture notes 43 1s order fields 1 VXE1 399E1z aE1y ax 0E1x aE12 a aE1y aE1gtlt a2 0H0 6y 02 02 0X y 0X 6 0E V D1V 8 E180jpw Z 6H 6H VXH1 6H12 W ax anx 6H12 a W anx az J1 80 6E0 6y 62 62 6x y 6x 6y 6t anx aHW 6sz 0 62 VB V H 1 01 luoax 6y 341 lecture notes 44 2quotdorder fields 1 6H 6H WHZ 5H222v ax W2x5H2zjay 2v5H2x aZJ2805E 6y 62 62 6x 6x y at 6H v B2 V 0H2 O 6HX Ty L 0 Z VXE aE22 0E2Y a aE2gtlt 0E22 a aEZY 0sz a 0H1 2ayazxazaxyaxayz at 0E V0D2VogoE2gojp2l z 341 lecture notes 45 capacitor quasistatic solution and 2quotdorder corrections 1 341 lecture notes 46 Transmission lines 1 Transmission lines are a special case of solutions to Maxwell s equations when the properties of the physical system do not change in one direction typically taken to be the zdirection x I l y Maxwell in the frequency domain current and chargefree V x E V x Fl VDO V o x 5 Solutions are solutions to the Helmholtz wave equation V2 m282 0 Plane waves are solutions Exyza l xyaeij398z Flxyza llxyaeij398z E 02 W2 mzyg Egg 0 where V V2 3 TEM solutions exist in transmission lines coaxial lines stripline twinlead for which the az component of E and H are zero The microstrip transmission line carries waves that are TEM to a good approximation 341 lecture notes 47 TEM waves 1 The fields are of the form Exyzw xywei z Hxyzw FIxyweiW The key is that the Helmholtz equation splits into two pieces 2 02 2 2 2 El 2 2 02 V augBFl 0 whereV V VlEl 2 2 El O and a 8 8 O VEH H The result is that the form of the waves in transverse directions are solutions are static solutions E and Fl are solutions to Laplace Example 1 coaxial lines 1 inner radius a outer radius in r r Y 341 lecture notes 48 Example 1 coaxial lines continued 341 lecture notes 49 Voltage amp Current Waves Telegrapher s equations 1 zAz vzAz From Faraday 341 lecture notes 50 Voltage amp Current Waves Telegrapher s equations 1 zAz vzAz From charge conservation 341 lecture notes Coaxial lines Source for table is E Kuester U of Colorado 2000 Type Graphic fm ax Notes Phone jacks 100 KHZ Phone jacks and plugs are also known as TS TipSleeve for two conductor connections or TRS TipRingSleeve for threeconductor connections They are widely used with musical instruments and audio equipment These are really only suitable for audio frequencies RCA jacks 10 MHZ A round presson connector commonly used for consumergrade audio and composite video connections The jacks are sometimes colorcoded red audioright black or white audioLeft and yellow composite video Generally not a constant characteristic impedance connector F video Type N 250 MHz to 1 GHz The F series connectors are primarily utilized in television cable and antenna applications Normally these are used at 75 ohm characteristic impedance 38 32 coupling thread is standard but pushon designs are also available 2 GHz or higher Bayonet NeilConcelman or British Navy Connector The BNC is used in video and RF applications to 2 GHz Above 4 GHz the slots may radiate signals so the connector is usable but not necessarily mechanically stable up to about 10 GHz Both 50 ohm and 75 ohm versions are available 12 GHz or higher The Ntype connector was designed by Paul Neill of Bell Laboratories and offers high performance RF performance with a constant impedance The connector has a threaded coupling interface to ensure that it mates correctly to provide the optimum performance Both 50 and 75 ohm versions are available the 50 ohm version being by far the most widely used T The connector able to withstand relatively high powers when compared to the BNC or TNC connectors The standard versions are specified for operation up to 11 GHz although precision versions are available for operation to 1BGHz Type N uses an internal gasket to seal out the environment and is hand tightened There is an air gap between center and outer conductor A 75 ohm version with a reduced center pin is available and is used by the cableTV industry 12 GHz or higher The SMA Subminiature A connector is intended for use on semirigid cables and in components which are connected infrequently It takes the cable dielectric directly to the interface without air gaps A few hundred interconnect cycles are possible if performed carefully Care should be taken to join connectors straighton 18 GHZ The APC7 Amphenol Precision Connector 7 mm offers the lowest reflection coefficient and most repeatable measurement of all 18 GHz connectors These connectors are designed to perform repeatably for thousands of interconnect cycles as long as the mating surfaces are kept clean 341 lecture notes 52 24mm 50 GHz This design eliminates the fragility ofthe SMA and 292mm connectors and by increasing the outer wall thickness and strengthening the female higher ngers The inside of the outer conductor is 24 mm in diameter and the e 9 outside is 47 mm Because they are not mechanically compatible with SMA 35mm and 292mm precision adapters are required in order to mate to those types The 24mm product is offered in three quality grades general purpose instrument and metrology General purpose grade is intended for nomy use on compone ts cables and microstrip where limited connections and low repeatability is acceptable The higher grades are appropriate for their respective applications Microstrip Widely used in PCBs Resource ht tn39lmics 39 quot 39 html Stripline Widely used in PCBs Stripline provides some self shielding Resource hl tn39lnhiwan cs nrisu nndak t d quot 39 39 39 39 htm Twin lead Twin lead characteristic impedance is commonly 300 Q Resource hi tn39lmAAAAtechnick 39 quot39 39 39 39 dnane nhn mim n dputi inductance wire 2 All kinds of applets on transmission lines waves etc hi tn39lvwwv 39 htm 341 lecture notes 53 Voltage amp Current Waves Telegrapher s equations 1 zAz ja ja V 02 02 029 0i 022 1ME a ja V 02V 2 Jw 462 w 162 462 162 022w V0 VVe e Ve e Ve Ve azi 0V g meg ma a I 2 0 236 I o I V e39mz V el z 02 Z0 Z0 These are both wave equations with 8 anL E Although technically NOT a general result for transmission lines this formula can typically be used since most transmission lines if practical ARE lowloss The lowloss case gives this result 341 lecture notes 54 Consider a voltage wave traveling in the 2 direction and determine the relation between voltage and current This relation defines the characteristic impedance htt obiwancsndsunod2kedu Nekhan mes to arms std htm 341 lecture notes 55 Lossy Transmission Lines 1 Real transmission lines are lossy First there is loss due to the current which is modeled as a series resistancemeter For coaxial lines this resistance can be found using techniques already developed j i 1 1 0395 P P inner outer The other loss is associated with the voltage between conductors and is due to dielectric loss This loss is modeled as a conductancemeter z m BAZ zAz VZ eAz eAz VzAz 4 721127 modefor my line Physical basis for C Review of mechanism for dielectric current loss due to radiation loss due to lattice coupling 341 lecture notes 56 Transmission ine equations 1 07 0i J L M V 02 1w 02 Q J 029 0i 022 72 jw az J ja jw V Defining impedancemeter and admittancemeter zyzjw ygjme The following form is obtained 2 0VZV 022 239quot Mum i z 7 y c J Lowloss 72 ltlt me C ltlt me 341 lecture notes 57 Reflected Waves 1 ii L l m gt Vr Ir Z21 131 l ZCZ39BZ z0 r l r L T L V z0 V z0 The boundary conditionsquot are KVL and KCL r Zc2 Zc1 T 22oz Zc2 Zc1 Zc2 Zc1 341 lecture notes 58 Lumpedelement loads 1 The reflection coefficient would be unchanged if instead of a second medium with characteristic impedance Zcz there were a lumped element with this same impedance ZL Zcz Matched load ZL Zc1 F 0 no reflections Short circuit ZL O F 1 reflected voltage cancels incident voltage V1 0 Open circuit ZL F 1 total voltage is twice incident voltage Or why you don t want to be at the end of the power line in an area with frequent lightning storms F for a passive load is a complex number with a magnitude less than or equal to one 341 lecture notes 59 Transient waves on lossy lines 1 When loss must be taken into account high frequencies see a different environment due to the frequency dependence of the loss mechanism skin effect and dielectric loss This complicates the analysis for the transient solution The most important things here are its qualitative characteristics and the physical reasons for these characteristics Dispersion and attenuation E In general or and vp are functions of m in a lossy transmission line It is true that in the lowloss approximation vp is not a function of m but strictly speaking vp is a function of 03 even in the lowloss case it s a secondorder effect in the lowloss case Dispersion occurs when vp is a function of m If one looks sufficiently close vp is almost always dependent on 0 Aside from effects due to loss 8 and 3 vary with frequency or more fundamentally a and u vary with frequency For transmission lines or typically increases with m Rac increases as 03 increases From Fourier analysis the harmonic functions form a complete basis set and we can use superposition to construct any waveform That is we can expand any signal in terms of harmonic functions Looking at the problem in this light there are two things going one simultaneously 1 the timedelay suffered by components varies with their frequency and 2quot the highfrequency components of waveforms suffer greater attenuation as they propagate The net result is that waves any wave that is not a pure sinusoid changes shape as it propagates Sharp corners become rounded rise times become longer fall times lengthen 341 lecture notes 60 Transient Waves on lossless lines 1 V th no loss all harmonics experience the same propagation environment The result is that waves retain their shape as they propagate z0 zd To track the transient response in this case two reflection coefficients can be defined 1quots at the source end and FL at the load end There is also a time delay associated with propagationtd P 341 lecture notes 61 341 lecture notes Bounce diagrams E This process is often analyzed using bounce diagrams 20 1 5 62 zd FL td 2td std 4td 5td 6td 7td 2 2 1 s11 Fifi 3 3 1 11 341 lecture notes 63 Example 341 lecture notes 64 Transient Waves on lossless lines 1 How are voltage waves reflected when transmission lines are terminated in reactive elements z0 d For this case the capacitor the reflection will vary with time Initially when the positivelytraveling voltage wave encounters the capacitance it will be uncharged At that instant the capacitor will act as a short circuit so will require the sum of the incident wave and reflected wave at that instant to be zero Forttd Foap1 For long times the capacitor will be fully charged and so act as an open circuit For large t 11351 How does the reflection coefficient move between these two points Answer It moves between them as a firstorder circuit with a time constant 1 ZcC Analysis 341 lecture notes 65 What would the situation if the line were terminated with an inductance 341 lecture notes 66 Example 1 Find the reflected wave in the timedomain at zd 341 lecture notes 67 Timedomain reflectometry if TDR is a widelyused experimental probe to investigate material properties and transmission line discontinuities Although workers in the field have 39 39 39 many I 39 quot quot 39 39 I the basic idea is simple The standard experiment in TDR for transmission lines is to inject a pulse and observe the reflected wave For meaningful location data to be obtained with TDR time must be measured very accurately The available information includes 1 location of discontinuity and 2 the character of the discontinuity Example A10 volt pulse is introduced into a 75Q transmission line having vp 108 ms 113 us later a minus two 2 volt pulse is observed at the TDR unit What information is available about the discontinuity and its location 341 lecture notes 68 SteadyState Waves 1 z0 d The source introduces a positively traveling wave of amplitude and phase 70 The frequencydomain components sinusoidal steadystate are 7z flue 71z Voe39j szfse39jz d nz Voe39j z llezse39jz39gd n no 39Bz 3963926d 396 3926d Ve Vz Zvntzvoel 2FLFSe V0e zZ1quotL1quotSeJ n 0 n 0 n 0 1 rLrse 7 V z ew Vquotequot398z 0 arseW Similarly it can be shown that 739z quot1quotLe39jz39gdej39gz 341 lecture notes 69 The ratio between the positivelygoing voltage wave to the negativelygoing voltage wave as a function of z is f2 1 Lequot392quot dej2398Z fin fz0 fLe39jz39Bd What does this mean It means the ratio of the voltage going in the 2 direction to that going in the z direction is constant in magnitude but that their relative phases vary as 2 varies we re neglecting loss here This should not be surprising after all the two voltage waves are traveling in opposite directions 7z 7z 739z quotequotquot8z fLe39jz dej z lz lz l39z Zl Jrequotquot8z fLe39jz dej z 2 92 eJJ z fLejz dej z Z ej z fLej2 dej z Z ej232 fLe j23d z c 2dc2 2d Z VeJ39 z 1 ejz dej z 9W 39 rLeJ 6 emz 91 62 39 rLeJ 6 L ZC 1 f eW Zin Zz0 Zc 42 1 Re 341 lecture notes 70 What are some consequences forthis expression for 2quot It means for instance that a short circuit 1L 1 can look like a open circuit when e39jz39gdejz39gz 1 that is when 26 zd 227r1zd7r The input impedance for a transmission line of length N4 when terminated with a short circuit looks like an open circuit The input impedance for a transmission line of length N2 when terminated with a short circuit looks like a short circuit Similarly The input impedance for a transmission line of length N4 when terminated with an open circuit looks like a short circuit The input impedance for a transmission line of length N2 when terminated with an open circuit looks like an open circuit VSWR Just as with fields there are standing waves present anytime there are reflections The voltage standing wave ratio VSWR 7z Pei3952 739z quot1quotLe39jz39gdej39gz VSWR VW Vmin 1 lle VSRW 1 39 In Voltage minima spaced N2 apart become more pronounced and narrower as fL grows see page 52 in text 341 lecture notes 71 SteadyState Voltages and Currents if 500 cos211051 mV 1 z34 m Find 8 21 d Z Zquot fin 341 lecture notes 72 341 lecture notes 73 Transmission Line Examples 1 A1 volt drop occurs across a load of 2L 130 j80 Q which is connected to a 127 m length of lossless transmission line The line has Zc 53 Q and vp 250 mps the signal generator has a Thev nin resistance Rs 100 Q and operates at 250 MHz Calculate i the power delivered to the load ii the transmission line input reflection coefficient iii the transmission line input impedance iv the transmission line input voltage V v the opencircuit voltage of the generator v the power delivered by the generatorto the input of the transmission line 341 lecture notes 74 341 lecture notes 75 Example 1 i Calculate the input impedance of an opencircuited transmission line of M8 length ii Calculate the input impedance of a shortcircuited transmission line of N8 length iii Calculate the input impedance of a 3M8 line with ZC5O Q which is terminated by a load of 2L100j150 Q iv A 50 Q transmission line of length 0225 X has an input impedance of Zquot 75 j75 Q Calculate the load impedance 341 lecture notes 76 Example 1 A generator with Voc10 V and Rth50 Q is used to test various transmission lineload combinations The generator is connected to the input to the line and the input voltage is observed on an oscilloscope The oscilloscope patterns for several tests are shown below Fill in the table 10 v v0C Zc ZL 1 us what is the load 5 V 3 V l l 341 lecture notes Electromagnetic Foundations Gauss law V D p Conservation of magnetic flux V B O Ampere s law V X H J Faraday s law V X E Traveling waves Taking the curl of Faraday 2 E VgtltVgtltE VV E V E VX at atVXB In a chargefree p0 and currentfree JO region VV E V2E V2E VXB yVxH VZE 71VXH Substituting for VxH using Ampere s law J0 we obtain the wave equation 6 6E V2E 8 at at J 62E VZE 8 O This equation is called the wave equation A wave equation for H can be found as well Taking the curl of Ampere s law and then substituting Faraday s law Again in a charge free pv0 and currentfree JO region 62H VZH 0 Together these equations predict that an electromagnetic wave can travel through space These vector equations one in E and one in H would each give three scalar equations one for each component ofthe vectors for a total of six scalar equations Using Cartesian coordinates one scalar equation for EX would be 62EX 62EX 62EX 62EX 2 2 2 39 8 2 6X 6y 62 6t Waves traveling in the z direction would have the form 2 62 1 62 g g t i 2 g 7 v Two waves are possible one traveling in the az direction with a speed of v 1lyg the other traveling in the az direction with the same speed Plane Waves Assuming the electric vector to only have a component in the Xdirection EX which depends only on 2 and t 62EX 62E 622 ug at2X Solutions are functions where the 2nd derivative with respect to z are equal to their 2nd derivative with respect to t aside from a constant multiplying factor Sinusoids are an obvious choice this particular choice travels in the az direction EaX Ecosa2t zq Substituting this possible solution z aXE cos wt z 0 81 m2 aXE cos wt z 0 2 8y m2 3 a 8y Just as for sinusoidal steady state analysis in electrical circuits harmonic analysis is provides a convenient notation E aX E cos wt z 0 ReaXEequotquot equot z em ReE em That is for elds that vary harmonically with time each Fourier component can be described as Ercosa2tq Re Erequot l ReErejV ej E ReEejquot E lt gt E Just as in phasor analysis the derivative with respect to time corresponds to a joa multiplier in the frequency domain 6 6 jg jg EE EReEe ReJmEe 3E lt gt wa at In the frequency domain Maxwell s equations become V X E jm VXFI J U H O Dz Example 1 Suppose as above an electric vector pointing in the aX direction travels in the az direction How can the magnetic vector be found Answer Faraday s Law VXE jmyH N N up up HVXEL 6Oax ay az my my ay 62 62 6X 6X 6y Fl i39J39 Ee39mza iEe39jpza m 8 Ee39j za my y my y my y Notice that E and Fl are related by the eld analog of Ohm s law E J2 Fl 77 Fl where 77 J2 is the intrinsic or wave impedance in Q 8 8 The wave impedance for an electromagnetic wave in free space nearly the same for a wave traveling in air since in air 8 s goand y yo is 77 5 37m 80 The relative directions of E and Fl can be determined from the fact that E X H is in the direction oftravel In fact S E gtltH is known as Poynting s vector and is the instantaneous power ux density with units ofWm2 transmitted by an EM wave For periodic fields the timeaveraged power ux density 3 ReltE gtlt FI Example 2 An electric eld of amplitude 377 Vm points in the az direction and travels along I 45 in the Xy plane The wavelength is 700 nm and the medium is air i Give the frequency domain description ofE and H Take the phase of Eto be zero degrees ii What is the frequency of the waves Give their timedomain descriptions y H 4539 Solution i The wave number or phase constant 3 is found from the wavelength 27239 27239 9106 m391 1 700 nm E a 377 ej9106xc0545 ysin45 Vm The magnitude ofthe magnetic vector can be determined from the eld analog of Ohm s law and the direction ofthe vector can be determined from the fact that E x H is in the direction of travel as shown in the diagram above The magnetic vector travels with the electric vector after all we have a traveling EM wave of which the two vectors are part Fl aysin 45 axcos 45 3777Vggm ej9106xcos45 ysin45 e H 01aysln 45 axcos 45 e 9 10 ms mm Am v The speed of propagation can be determined from the physical properties ofthe medium in this case air Knowing the speed of propagation and the wavelength permitsthe frequency to be determined 1 1 v s forar F J 480 vp s 3108 ms 4885400 Fm 4723910397 Hm 3108 ms 9 42900 Hz 70010 m Reflection and Transmission How are EM waves re ected and transmitted when a plane dielectric boundary is encountered Considerthe situation below x medium 1 n1a1u1 medium 2 11232112 quot10 Ei 620 incident wave gt H gt1 15 f reflected wave 4 Hr H transmitted wave gt De ne a transmission coef cient T and re ection coef cient P Let E aXEequotquot z From EM theory the boundary conditions that must be met are that the tangential components ofthe electric eld and the magnetic eld assuming no surface currents are equal on either side of the boundary Boundary conditions EZ0 E2Z0 HZ0 H2Z0 The rst equation is obtained from the boundary conditions on the electric vectors E EaX Et aX 2 E FE T E The second is obtained from the boundary conditions on the magnetic vectors HHrayay 3 Egg 771 771 772 So the result is two equations in F and T 11 T and l l 771 771 772 Solving for F and T r T 2i 772771 772771 Using these coefficients and the intrinsic impedances ofthe materials one can find the incident re ected and transmitted electric and magnetic vectors in terms ofthe incident electric vector E aXEie z E axFEiej z Et axTEieJ z Fli ayEe39j39Blz Flr yr Elev z Flt ayT Eie39j r lZ 771 771 772 From the eld vectors the power densities can be determined 8 Reltl i gtlt Flf ReaXEiequot z gtlt ayEe39j39BlZ 771 8 lReaXEieM2 gtlt ayEeW azE i 2 771 2771 Similarly 2 2 2 2 Sr a2 F El and St a2 T E 2771 2772 Note if the materials are conductive the reflection and transmission coef cients as well as the intrinsic impedances are complex Standing Waves When re ection is present the incident and re ected waves will interfere to produce standing waves in region 1 E E E aXEequot398iz aX1 Eequot398iz ax11 equotquot8iZ aX1 Eej398iz e39j39giz E1 aX 1 1 equotquot5lZ aX2FEicos 1z where the 2quotd term in the standing wave In the time domain we have E1 Rel 1ej ax11 cosa2t 1z ax21 Ei cos 1z cos 1z The standing wave ratio SWR is commonly used to characterize re ections from a surface The SWR is defined as the ratio between the maximum amplitude electric eld in region 1 to the minimum electric field E 1F2F1F SWR1 SWR 3r E 1 r 1 r SWR1 min Where F is the magnitude of f Reflection and Transmission 62 0 X yI gtz medium 1 n181u1 medium 2 11252112 c10 E 624 incident wave gt H El Er re ected wave lt H H transmitted wave gt The conductivity in region 2 leads to complex values for the re ection and transmission coef cients as well as for the intrinsic impedance in region 2 This results in a phase shift between the incident wave and waves that are re ected and transmitted In addition the fact that the intrinsic impedance in region 2 is complex leads to their being phase shifts between the electric vector and the magnetic vector in region 2 E aXEie39J z E aXfEielmum E aXTEieK zz gar Fli ayEe AZ Flr ayFEiej iZ Flt ay TEielmz 114 771 771 772 E The incident power is unchanged from the previous case S a22 771 The reflected power density is 3 lRel r gtlt Ft 1Re aXFEieJWW gtlt ayF Eiemzwg 2 2 771 S gt lRea FEejmzquot F X a F Eie39mzwq a FZE 2 y 771 2771 The transmitted power density is St lReEt x HI 1Re a TEe39mz39 X a EelWWW 2 2 X I y 772 2 2 St 1Re aXTEie39m z39m X a T EieJWW39w a2 T E cosq 2 y 772 2771 1 Standing Waves When re ection is present the incident and re ected waves will interfere to produce standing waves in region 1 E1 E E aXEie39j r lz aXFEiejWW ax11 equotquot81Z aX1 Ei ejWW e39j z E1 axlt1rgteW E ax11 equot3981Z ax21 EeJ 2 cos 1z where the 2quotd term in the standing wave In the time domain we have E1 Rel 1ejm ax11quotcosa2tb 12 ax21quotEi cost312 cos mt The standing wave ratio SWR is commonly used to characterize re ections from a surface The SWR is defined as the ratio between the maximum amplitude electric eld in region 1 to the minimum amplitude electric eld E 1F2F1F FSWR1 SWR 3 E 1 r 1 r SWR1 min Where F is the magnitude of f Example 3 i Find the transmission and re ection coefficients for the situation shown below if medium 1 is air and medium 2 is PCB material FR4 60 sr44 u1 ii Find the portion ofthe incident energy that is transported into the FR4 material iii Find the standing wave ratio in medium 1 X v medium 1 71161 medium 2 11252 cz0 610 E incident wave gt HI 5 re ected wave H H transmitted wave gt r Solution u u u 377 Q I Q 771 I go 772 1 82 1 4480 44 Knowing the intrinsic impedances F and T are readily calculated 377 Q 377 Q i44 1 444 quz z 772771 144 J44 T 2772 772771 1J44 T2E2 S 2 ii Q 0874 S E2 i ai 72 22771 This isjust the conservation of energy S S S 1 F s 206 1 F Notice that the standing wave ratio must be between 1 and 00 iii SWR Transmission Line Fundamentals Transmission lines are a special case of solutions to Maxwell s equations when the properties ofthe physical system do not change in one direction typically taken to be the zdirection Employing frequency domain analysis is convenient when working with transmission lines In frequency domain analysis one assumes that all elds vary sinusoidally in time Suppose a source axEscos oat produces in response an electric vector which travels in the aZ direction E aXEocosa2t z lfthe system is linear the complex source axEscos oat j sin oat would produce a complex response E aXEo cosa2t z j sina2t z Using Euler s formula ej z cos a j sin a this can be written as E a E ej quotpz a E e39mzej E ej Example 4 An example will illustrate the form that Maxwell s equations have in the frequency domain Using Faraday s law with E E ej t and B S ej 6 eJth VxE 2 Verl at Note that E and S have no time dependence and that ej t has no spatial dependence This allows the curl to pass through ej t and to pass through the time derivative 14 e V x E 369 ja2B e1 t 2 v x E jaB at In the frequencydomain version of Faraday shown above E and E are the frequency domain eld vectors They are not the time domain vectors The time domain vectors can be recovered by putting the time dependence back and taking the real part Taking the electric vector used in this example to demonstrate E Re E e Re aXEoe39mzem ReaXE ej quot z E ReaXE cosmt z j sina2t z aXEocosa2t z This result can be generalized to express Maxwell s equations in the frequency domain current and chargefree VXE ja VXFI V45 0 VB 0 Combining the equations as was done above in the timedomain one obtains the Helmholtz wave equation E V2 mzw O H Plane waves are solutions Exyzm EX y mej82 Xyza2 X y merjb z It is important to note that the Laplacian operator can always be split into two pieces one piece which depends only on a variable which varies in the direction of travel in this case 2 and the other piece which depends only on variables which vary in directions transverse to that of travel in this case X and y E 32 2 2 2 2 2 Vtayg ll O wherth VE Special types of solutions exist in transmission lines coaxial lines microstrip twinlead for which the az component of E and H are zero These solutions are transverse electromagnetic waves The equation above can be split into two pieces both ofwhich must be satis ed VEE0 and mzw 2g0 In the rst since Vf does not involve the variable along the direction oftravel E E E E v3 v3 3 f vf J 0 since VEEz foz 0 H Ht H2 Ht In the second since nonzero solutions for E and E are sought we must have wzyg 2 O fortraveling waves along a transmission line Transverse Electromagnetic TEM waves The elds are ofthe form EXyza2 EXya2eij398Z Flxyza2 IIXya2ej398Z Here the direction oftravel is az Waves having a zdependence of e39 32 travel in the az direction and waves with e 32 dependence travel in the aZ direction Since the transverse elds satisfy VfEt0 and VEFIFO the form of the waves in transverse directions are solutions are static solutions Example 5 inner radius a outer radius b ps i H 7 inner conductor Y i permittivity 8 outer conductor permeability it To nd the electric eld first note that by symmetry the eld must point in the aP direction using cylindrical coordinates and its magnitude can only depend on p This allows the integral form of Gauss law to be readily used Gaussian surface x p T r l I l r z f 7 7 innerconductor I I y 20 z L ps permittivity 2 Gauss law in integral form reads 8E ds Q gaussian surface inside Applying Gauss law for the Gaussian surface shown for a lt p lt b zDL 2 ZDL 27 J J39gEapp d dzap J J39psa d dz zzu 0 zzu 0 2angE 27raLps 2 E apa 8p To nd the magnetic eld rst note that by symmetry the field must point in the at direction still using cylindrical and its magnitude can only depend on p This allows the integral form of Ampere s law to be readily used Amperian path p I T l l x z 1 r e V x7 inner conductor Y permeability 11 Ampere s law in integral form reads cf H dl I Amperian path inside Applying Ampere s law for the Amperian path shown for a lt p lt b 27 I Hapd a l 2 H a J0 p p 2727 The transverse elds found using statics can are then used to nd the traveling wave In the frequency domain these elds are a N a 39 2 Et ap 3 E ap eim 8p 8p N i N i z Ht a 2 Ha eJ 2727 2727 V th transmission lines one is often not concerned with the details ofthe elds but rather is often more interested in the terminal current voltage equations In his work with transmission lines Oliver Heaviside developed the telegrapher equations which involve traveling voltage and current waves rather than traveling electric and magnetic fields Voltage amp Current Waves Telegrapher s equations I 1 dt Using Faraday s law E d B ds where CH and ds are related via path the right hand rule Using frequencydomain eld vectors JSE d ja ds path bounded bypah Carrying out the path integral Faraday s law shows that J ds 7zAz 2 1w 3 ds mimf J ds Where the quantity being a ratio of ux to current is clearly the inductance ofthe loop 7zAz 7z jal L By dividing the the interval A2 and taking the limit as AZ gt 0 one obtains quotm VZAz Vz A2 jal lim L jal Azgt0 AzaOAZ Where L is the inductance per unit length ofthe transmission line The result is the rst ofthe Telegrapher equations this equation if for a line without loss dQ d d From char e conservation lz l zAz 5 ds D ds 9 Jig06p SUJJCS Using frequencydomain field vectors l5 ds lzlzAz39 l5ds397 m V l D ds Where the quantity WET is clearly the capacitance of the line section C iz lzAz ja7 o 2 lzAz iz ja7 o By dividing by A2 and taking the limit as AZ gt 0 one arrives at the second Telegrapher equation again forthe case of the lossless transmission line mm W jm7 m E Azgt0 AZ Azgt0 ja7 3 dz Lossless Transmission Lines circuit model governing equations I 8Az N z lzAz ij B dZ Vz 8A2 VzAz dl lt gt V 8 A2 dz la Taking a derivative with respect to z ofthe rst Telegrapher equation and substituting the second equation one obtains wv dl amp v B B e v dz2 W dz 3 dz2 W W r d QKCVo dz Similarly d4 dv d4 EWGE 3 Pzquot equot l d4 micio dz2 Solutions are traveling waves 7 06ij V39el z where 7 We and 7 V39eM l Pet l39emz where lquot Fem and l39 lei where a BC is the phase constant Consider a voltage wave traveling in the 2 direction and determine the relation between voltage and current The ratio between the voltage and current waves is the transmission line s characteristic impedance 20 using 7 7ejpz in the Telegrapher eq iii V 4a 3 z j Ve39i z ja T 3 T J Nejpz 39Jaiv ee freqW 39JWB a em2 2 Z 7 W 6 Lossy Transmission Lines Transmission lines serve to guide electromagnetic waves As the waves travel along the transmission lines there are three main sources of loss 1 Joule heating or copper losses 2 dielectric losses and 3 radiation Joule Heating as the EM wave travels along the conductors of the transmission line the electromagnetic wave enters the conductors typically a metal where it decays exponentially Even though the wave decays once it enters the TL conductors the EM wave does enter them to a degree Inside the conductor the EM field accelerates free electrons which then collide with metal atoms to make them vibrate The result is that the EM wave heats the metal This process is called Joule heating and is usually the dominant loss mechanism in transmission lines Dielectric Losses a dielectric is formed by fixed positive nuclei to which are bound negatively charged electrons In an ideal dielectric an applied electric field results in charge separation as shown below Notice that for an ideal dielectric the force and VVmax llmax l 2 3 A n normalized time motion are in the same direction for A of a period or equivalently the voltage and current have the same sign for A of a period and then are in opposite directions for the next A period The result is that work is done on the dielectric energy is stored for A of a period which is then given back in the next A period This is the physical basis of capacitance For a nonideal dielectric the charges have nonzero mass and their motion lags that shown The result is that the current is no longer 90 out of phase with the voltage which results in energy loss from the EM wave Radiation Losses anytime charge is accelerated an electromagnetic wave is emitted This loss mechanism tends to become important at higher frequencies or whenever the physical structure becomes a significant portion ofa wavelength especially when the current path is separated from its return path The transmission line model which includes loss is circuit model governing equations z 2A2 zAz N m 32 my dZ V52 eAz QAz VzAz 39 dl ltAz gt E 9 wegtv As with the lossless lines traveling waves result d2V dl d27 MN2 dzz 32 wag 2 dzz 32 jw C jwe v 2 v y v 2N N d 2 2V 0 dz Similarly for the current wave dzl 2 E 39 7 39 39 0 where i 52 jaw is the series impedance 1 C jme is the parallel admittance and 7 42 er is the propogation constant The propagation constant 7 is complex 7 a j where at is the attenuation constant and i is the phase constant Traveling wave solutions 7 7e3972 Ni39eiz Nquote39aze39mz Nl39e zzewz l lquote397z l39e7Z lquotequot12equot39 z l39e lzemz In the time domain vt Vequot12cosa2t z gm V39e lzcosa2t z 0quot it lequot12cosa2t z cpl I39e IZCOSat z pf Low Loss Transmission Lines When transmission lines are lossy nearly always they are lowloss 52 ltlt m3 and C lt we The general case is seldom and will not be treated here Forthe more common low loss case the expression forthe propagation constant 7 can be simpli ed 7 Jig l32 jaw jme jm13e 1j 1ji mi we 7 s jamBe 1ji 1ji forRltlta2 andqltltme 2m 2amp28 This can be further simpli ed 932 32 g s we 1 7 J 4amp22138 J2w sze The secondorderterm can be neglected 77 El nE 1jijij E jm 2w we 2 2 7aj janl e Most often copper loss dominates the attenuation constant In this case 7 a j s g junBe lowloss where a is determined by copper loss What is the nature of this propagation constant First the phase constant is the same as that found forthe lossless case This gives the important result that the phase velocity ofthe wave is the same as for the lossless case Second or is usually small which might lead one to question whether it is really necessary or whether one might assume that or 0 Since the attenuation factor is the product of x2 the answerto this question depends on 1 how long the transmission line is and 2 what magnitude changes are considered important in a particular application Example 6 Consider a transmission line with B SyHm e 002 uFm 32 001 Qm C 0 and f1 GHz i How long would the transmission line need to be for a voltage wave to suffer 5 attenuation v What is the speed of propagation iii What is the wavelength iv v What is the characteristic impedance Solutions Using these values the propagation constant can be determined 77 0 15 5 g junBe 0000316 j19869 m391 i Let the length ofline d e39 39000316d 095 2 d 1623 m In this case even though at is quite small ifd 1623 m there will be 5 attenuation at even if small cannot be ignored for long lines iv Strictly speaking the characteristic equation for a lossy line is complex 2 3 158j2510396 o 15849110396 o s 158 o 14 C we This example makes it clear that even though 26 is actually complex it is so nearly real that it hardly matters in nearly all practical cases Therefore to determine the characteristic impedance one would use zenEmma e Example 7 Consider the following RG58 coaxial cable operating at 100 MHz The speci cations provided by the manufacturer are iii v ii iii e 30 pFft vp 2108 ms ZC 50 Q at 45 dB100ft Convert eto pFm From e and 20 determine B could also use e and vp or vp and 20 Convert cc to nepersm then nd 52 830 E ii 1 984pFm ft 12m 00254m 26 g 3 13 zfe 50 of 984103912 Fm 0246 uHm a 0045 E 1 quoteper ii 1 0017 quoteper 0017 m391 ft 8686 dB 12 m 00254 m m forG0a E 3 R2aJ 2aZc17Qm 2 13 e 7 Check 25 mam e 98410 Fm Check v i I 1 20108ms P 136 24610397Hm 9841039 Fm 39 Time Domain Reflectometry Just as traveling electromagnetic waves are both re ected and transmitted at discontinuities so are traveling voltage and current waves With EM waves the boundary conditions were continuity of the tangential components of the electric and magnetic vectors V th voltage and current waves Kirchoffs Voltage Law KVL and Kirchoffs Current Law KCL furnish the necessary boundary conditions quoti39 39i l m gt Vquot Ir Zc139B1 Zcz z region 1 z0 region 2 f i r i Vi 20 i z0 777 3 1f T 111 3 3 i1fquotir c1 22 Solving for f and T f N2c2 2Nc1 22 Zc1 f N 222 22 Zc1 The above boundary conditions may need some explanation The boundary conditions state that at z0 the total voltage in region 1 Vi 7 must equal the total voltage in region 2 Nt Likewise the total current in the az direction in region 1 at z0 li lr must equal the current in the az direction at 20 in region 21 Sometimes it seems unclear why the total current in region 1 is l l and not l if The more detailed diagram below may help l 1 lt1 t region 2 ltz x lt2 1 region 1 1 11 r z 0 From this diagram it is clearthat the total voltage in region 1 is Nr and that the total current in region 1 is l lr Lumpedelement loads The reflection coef cient would be unchanged if instead ofa second medium with characteristic impedance 2C2 there were a lumped element with this same impedance zL 26 f NZL 39 2Nc1 ZL Zc1 f 22L ZL Zc1 Consider three important special cases 1 Matched load 2 26 f 0 no re ections 2 Short circuit 2 0 f 1 re ected voltage cancels incident voltage N1 10 N39v V 3 Open circuit ZL 00 f 1 re ected voltage is equal to the incident voltage The total voltage at 20 is twice the incident voltage In general for a passive load the re ection coef cient f exists on or within the unit circle in the complex plane Traveling Waves Constder a vottage putse travehng down a transrntsston hne Vj gt n the frequency domam thts vottage putse conststs of a range of frequenctes f tosses are not a funcuon of frequency that 5 tf as not a funcuon of frequency and e phase vetoctw vp ts not a funcuon offrequency the retattonshtps between the putse s nequenc v shape Losses resutt m the putse decreastng m magmtude put tr vp ts mdependem of frequency the putse s shape M be unchanged as tttravets Gomgfurmer tr thetransrntsston hne ts tosstess r the tosses can be neg ected and tr rnphtu e a hne that M be constdered m regards to Ttrne Domam Re ectometry TDR TransientWaves on Lossless lines n W t a Source reststancet Rs a a s h 2 R5 How does the System evo ve begmmng at t 0 to t aw t At t 0quot the only location involved is at z 0 Att 0quot the system at z 0 can be modeled by two resistors in series Rs and 20 R circuit at 0 z 0 and t 0 2 R 2 5 c Where V 1V The initial traveling step voltage can be described as Z 2 V zt u t 0 Rs Zolt VP This initial step voltage will travel down the transmission line until it encounters the discontinuity at the load with a re ection coef cient FL at which point a re ected wave Vzt traveling in the az direction will be created f RL Zc L RL Zc d z N Vozyt zc FLUti ZC I Lut i R5 Z VP VP Rs Zc Vp VP This re ected wave will then travel in the az direction until it encounters the discontinuity at the source and will itself be reflected with a re ection coef cient 1 Rs Zc 5 R5 Zc vzt 2 fo5 u t a 5 R5 Zc vp vp The wave reflected at the source will then travel to the load where a wave will be re ected to travel to the source etc For the nth wave traveling in the az direction Vn zt 2 mfgquot u t E 3 R5 Zc vp vp For the nth wave traveling in the aZ direction N N N n 2 n1 d vn39zt 2 rLrLrs u t g 1 R5 Zc vp vp The voltage vzt is the sum ofthese waves N Z n 2nd 2 V zt F F u t ERSZ LS VP VP N N N n 2 1 d Zc mm ut l l i quot0 R5 Zc Vp Vp What form does this expression have fort gt oo Z N N n 2nd 2 V F F u 00 210 R Z L s 00 vp Vp w N N N n 2 1 d Z 26 FLFLFS u oo39 n i quot0 R5 Zc Vp Vp wwwi Zc ff 1f Zc 1r fiquot quot 0 R Zc L s L Rs Zc L quot0 L s 1 f Voo 2 N LN s Zc 1 1quotLl 5 Where the formula forthe geometric series 2 aquot 1 for alt1 was utilized a A bit ofalgebra shows that this simplifies as it must to R v L 00 RS RL Bounce Diagrams This process is often analyzed using bounce diagrams A bounce diagram is a plot in the zt plane of the positive and negatively traveling waves A signal travels the length of the transmission line zd in one delay time td As can be seen in the above diagram the slope of the transmission line is inversely proportional to the phase velocity The value ofthe voltage at particular values ofz and t is found by summing the voltages above this point in the bounce diagram Example 8 The voltage at a particular position and time 20 and to in this example is found by adding all the voltages above the point in the bounce diagram plot Nzo to 01 71 2 Example 9 Given the system below 8 vp 10 ms tddlvp100 ns 1509 z0 z10 m i Complete a bound diagram to t 4td ii Find v5 m 25 ns v10 m 150 ns v75 m 350 ns and v25 m 250 ns iii Determine vLt gt 00 iv Sketch v5 m t and v10 m t for O lt t lt 4td 10 m 150 ns ii v5 m25 ns0V v10m150ns12V6V18V v75m350ns12V6V 2V 1V15V v25m250ns12V6V 2V16V iii vLt gtoo18V amp 5154v 25 Q 150 9 iv v5 m t v10 m t 20V 20V 13v 15V 15V 10 V39 10 V39 nal 5V39 5V I I I ns 100 200 300 400 q dt C C Clearly charge cannot accumulate instantaneously which then prevents an instantaneous change in capacitance voltage This is important here because when the positively traveling wave rst impinges on the capacitive load the capacitor will remain unchanged and the voltage across it will remain zero At that instant the capacitor will act as a short circuit so will require the sum ofthe incident wave and re ected wave at that instant to be zero For ttd1 L 1 As time elapses and the capacitance becomes fully charged its current will become zero and the capacitance will act as an open circuit For large t 1 Fort gt td the re ection coef cient changes from 1 to 1 The rate of change is that of a rstorder circuit with a time constant rZCC Analysis Vzt ut i 3 Vdt ut 1 VP VP d s nthe sdomain vds e d d e i 26 e Vds FL 510 which using partial fraction expansion becomes s 7 Z sC 15 15 V d s e p 2 s 1 s Z C c The inverse transform for this sdomain expression is readily found t dvp Vdt 12e39 ZcC uti VP The reflected voltage and the total voltage at the load can be sketched V 39dt V dt V 39dt FL Consider for the same system the voltage at the source Note that the source is matched to the transmission line so that I 0 V 0t VD39Ot 2 v 1V 2dlv in Note Due to losses curves measured in the laboratory will not have as sharp as the ideal model predicts An actual curve may look something like V 0t VD39Ot 2V quot quot 1ch 1V 2dlvp Example 10 As a second example of reactive loads consider a RL load an inductance in series with a resistance as shown below z0 zd As with the previous example the source is matched to the line so that there will be no V1 wave or 7139 72 V239 etc 1 c RL for t t3 the inductance acts as an open circuit check Vtt gt oo RL Z fort gt co the inductance becomes a short circuit At the load zd for the reflected wave left graph and for the total voltage right graph Va39dt Vadt V 39dt 1 LIZ RL r 1LIZCRL t 1 V Tota vouage at me want to me tranSmwSSwon hne 20 wuwno 1V My Note frequencwes mm My Example 11 A 10 von pu se 5 mtroduced mo 5 750 transmwsswon hne havmg vp 105 ms M 3 n5 aler mm the dwscontmmty and us ocauom vans Vn 5 5 unknown V zc mad gtz U Suumn From v at me want to me transmwsswon me me re ecuon coef cwent can be deterrmned From me measured 1 me oad nnpedance can be ca c at r V Zc 2 2L 50 o ZL Zc From the time of the re ected pulse the distance to the load 2 d can be determined Note that the time is the sum of the time required forthe incident pulse to reach the load and the time required for the re ected wave to reach the input Further discussion Note that a pulse can be considered as the sum oftwo step functions and that likewise the re ection can be considered as the sum oftwo step functions v volts v volts 0 1o 1 b t HS 2 t1 HHS 1 This viewpoint is especially valuable when considering re ections from reactive loads For example consider the results when a pulse is applied to the above 75 Q line with a load consisting of a capacitance in parallel with a 50 Q resistance In this case the P will initially be 1 and will decay to 02 as the capacitance charges v volts v volts 1O Example 12 Consider the system below 10utut1psV Forthe input voltages shown below determine 20 d and the load Vin 0 av 8I3V t its 1 3 4 From v ZC 200 Q From the time of re ected waves tr 3 us d 300 m From the shape of the re ected waves the load is a resistance combination From its amplitude R 2200 Q ii 5V 4 5L trus l 1 V 5v From v ZC 50 Q From the time of re ected waves tr 4 us d 400 m From the shape of the re ected waves the load is a parallel RC combination Since F gt 0 as t grows larger R ZC 50 Q From this measurement C could also be determined from the time constant of the reflected waves From these curves an accurate measure of c is difficult but approximating c 01 ps ill rRC 2 0120391 S2nF R 50 Q III In thIs case d Is known to 3335 be 400 m WAS From v Zc 25 Q 1 10 Since there are no re ections ZL ZC 25 Q Example 13 Given the ossess transmission line below connecting the 25 Q source to ZL Find i 20 ii zL iii vp iv e v 1 vquot volls 25 0 5E 125ulul5nsV 9 RL 100 z0 z5m uns 51015 20 25 30 35 40 45 50 Solution i ZC can be calculated from the voltage of the initial pulse v1OOV125VL 2 Zc1OOQ Zc 25 Q ii ZL can be determined from the reflection coefficient From the shape ofthe re ected pulse it can be seen that the load must be real Also since the reflected wave is positive ZL gt 20 r1rs r1 06 008 r02 2 z150g2 ZLZc 2d 25m 8 III v 210 ms P tr 5010399s iv vp 41 6 and 26 3 45 and 321 V vp cvp e 50pFm
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