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by: Macy Lowe


Macy Lowe
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This 26 page Class Notes was uploaded by Macy Lowe on Monday October 26, 2015. The Class Notes belongs to ECE1266 at University of Pittsburgh taught by JoelFalk in Fall. Since its upload, it has received 47 views. For similar materials see /class/229451/ece1266-university-of-pittsburgh in Electronics and Computer Technology at University of Pittsburgh.

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Date Created: 10/26/15
ECE 1266 Application ofFields and Waves NOTES for Dielectric Waveguides II This lecture covers Chapter 73 and 74 1 Onedimensional waveguide 2 VV 39 mode wavelength equations Planar Dielectric Waveguides Chapter 7 Dielectric Waveguides II We know that for internal refraction n 1gtn2 in the gure below if the angle 6 is less than 712 BC critical angle we Will have total internal re ection TIR The angle 6 712 6 is not the usual angle of incidence The usual angle of incidence is 6 and is measured With respect to the nonnal to the 1 2 interface The angle 6 is the angle usually considered in discussing Snell s law Now ifnl is sandwiched between two parallel materials of refractive index n2 TIR can take place at both boundaries Which Will trap the light in the n1 layer and guide the light in this layer We call this structure a waveguide In this case see gure below we have 1 dimensional y aXis con nement we call this l D waveguide a e planar gslabl wavegggd ECE 1266 Application of Fields and Waves Now is it true that as long as 9 lt Zi 96 we will have guided light propagating in the waveguide The answer is NO Why The gure below gives a two dimensional picture of the propagation of a wave down the guide Ifthe wave is to travel down the guide the wavefront at point A must be in phase with the wavefront at point C The geometry in the gure above permits us to calculate the angles 9 that permit constructive interference If we follow the wave from A to C we see that it propagates through a distance ABBC and bounces twice off of the sides ofthe guide at point A and point B Chapter 7 Dielectric Waveguides II If the wavefront just before the wave hits the top of the guide at point A is to match that at point C then exp7j llAB 5C 2 l where is the phase shift that the wave suffers on a single totalintemalre ection Thus a waveguide mode will result when the net round trip phase shift is m27r m 12 orwhen 1lAB BC 2 m27r where the minus sign is chosen because near critical angle is near zero and the lhs of the equation is negative It is easy to show that A5 Bo 2a cos 6 so that the requirement that the phase of the wave incident on point A be equal to the phase at point C can be written 12 acos x 27m7r2 The previous notes showed that for the TE mode the phase change resulting from a totalintemalre ection is given by an argon 2mmquot Thus the phase requirement becomes ECE 1266 Application ofFields and Wave tan emu2 a cos 5 2 This is a transcendental eguation that gives the angles of incidence for the TE modes An alternate form ofthis equation is 25 12 m7z s1n sm6 z 1 2 sm 6 The angles in the above equation refer to the angles as measured in the gure on page 2 of these notes These angles are the complements of angles in the equation at the top ofthis page Chapter 7 Dielectric Waveguides II Incidence angle degrees The transcendental equation Eq 1 can be solved graphical by plotting each side of the equation versus 6 or sin 6 and looking for intersections The gure above Text Fig 716a illustrates the process The thicker line plots the right hand side of the phase equation the thinner lines the le hand side f 45GHz a 50min In 520 Hz 10 For TM Waves the solutions follow the same pattern The phase shi on re ection is found from the expressions for the re ectivity for a TM Wave for incidence above the critical angle The re ectivity has a magnitude of 10 and a phase shi that depends on incident angle The phase expression for the TM Wave becomes tan a os 7M sin2 57112n 2 2 n2 r12 cosH ECE 1266 Application ofFields and Waves Figure 717a illustmtes the solutions of this equation As in the previous example the thick line plots the rhs of the equation the thinner lines plot the lhs The frequency and geometry is the same as in the previous gure TE Incidence angle degrees Update October 31 2006 November 2 2006 Typo corrected October 24 2007 Change 01 to e in seveml equations add note about alternate form of Waveguide equation page 3 rst columnAlso typos corrected October 30 2007 Chapter 7 Dielectric Waveguides II Course Notes for ECE 1266 Applications of fields and waves NOTES for Transmission Lines I This lecture covers Chapter 61 and 62 1 Transmission lines Tlines wavelength propagation mode 2 Lumpedelement model 3 Telegrapher s equation and solution Chapter 6 Lecture 1 p t Transmission lines wavelength and TEM wave Transmission lines are used as a carrier to transmit transverse EM waves A Tline connects generator circuits to loads A transverse EM TEM wave is a special EM wave which will be de ned below When do we need to consider a line to be Tline When the wavelength of TEM waves is comparable or smaller than the length of the line we will need to consider the line a Tline Just give a few rough examples 0 A coaxial cable of l m length is just a wire if it transmits signal with frequency of l MHZ Because the wavelength of the TEM wave at this frequency is cf 100300 meter the speed of the light on the coaxial cable is smaller than 3 108 meters However when the signal has frequency of l GHZ this line becomes a Tline because the wavelength of the EM wave is lt03 meter and therefore the voltage along the line varies along the line 0 Conducting wires on a CPU chip can be a few meters long They are just a line when clock frequency is 10 MHZ 20 years ago They must be considered T lines today because the CPU runs at a 23 GHZ clock frequency Below are some examples of Tlines Not all of the lines can be considered as Tlines Strictly speaking only structures that support TEM waves can be considered T lines Because the E eld starts from one conducting plate and ends at the other one which means the voltages 1 Course Notes for ECE 1266 Applications of elds and waves between two conducting plates are welldefined we can use a lumpedelement model to describe the structures below in terms of R L and C circuits mual rlIAl n pim m nudimle m llv lmmurv lmc IInlul m in nIulmml III nnhum nuinu muml Ilzuw llolull In uh sum mm m Mlc mlw mm TEM l39rIIIsuisiIuI Lines 2 Lumpedelement model Because we have a wellde ned voltage between two conducting plates we can use a pair of wires to represent any Tline structures The voltage is measured between the lines and currents ow on both lines shown in the gure below my Mum lh NpImanilww v x x v m u Hmmm eUluiHSAM llAZ mg Kl Ix rm Ix RX 1x Rx LL cm 47 ex x x x ox t m mm mum n mucsmm h m mumlcm CHUH Chapter 6 Lecture 1 Along the length of any Tline structure EM waves will incur loss and phase changes This change depends on the frequency of EM waves All these can effects be well described by R L C circuits It is not hard to imagine that the R L and C along the Tlines depend linearly on the length of the Tlines This means that distributed parameters will be needed to describe voltage and current along the transmission line 0 R The combined resistance of both conductors per unit length Qm o L The V 439 J of both A per unit length Wm G The conductance of the insulation medium per unit length Sm o C The capacitance of the two conductors per unit length Fm It can be proven that for all TEM Tlines the following relationships hold L39C39zyg g C39 8 If the insulating medium between the conductors is air the transmission line is called an air line For an air line 680 g 00 andG 0 Course Notes for ECE 1266 Applications of elds and waves Example Provide a lumpedelement model for the coaXial cable shown below Conductors Mr 2 Insulating mamml 12 u c Chapter 6 Lecture 1 3 Telegrapher s equations and solutions The equations describing the TEM waves propagation on the Tlines are referred to as the Tline equations or telegrapher s equations V 111 N421 tum I Hz quot quot 4 x If the TEM wave carries a wave with a single frequency 03 we can easily find out the relationship between vzAZ w and vz w as well as izAz a2 and iz w Using Kirchoffs laws V2 a2 7 V2 A2 64 iz mXR39Az my A2 KVL iz a2 7 iz A2 64 V2 A2 wG Az ij Az KCL Dividing both equations by AZ and letting AZ gt0 we derive the telegrapher s equations in the frequency domain dVz 72 L G jaC V2 dz R jaL z Course Notes for ECE 1266 Applications of fields and waves Of course we can translate these equations into the time domain easily by replacing ja by ath To solve the telegrapher s equation we take derivative of the rst equation respect to 2 then substitute the 2nd equation into the rst We get y2Vz 0 12 d 1Z y21z 0 61122 y a m The parameter y is called a complex propagation constant or is an attenuation constant N pm and 5 as phase propagation constant radm The equations for Vz and 2 can be easily solved 72 7 72 VZ V0 e V0 e 2 gem Ige Both Vz and 2 contain two terms the forward propagation wave lSt term and the backward term 2rld term Now we have four variables VJ V0 10 and 10 We can reduce the number of variable to two by observing the relationship between voltage and current waves We rst substitute Vz into the voltage telegrapher equation We have 7 7 7 7 7 Iz7Ve Ve le le RWLL 0 l 0 0 We compare two exponential terms and nd Z R39jaL R jaL VI E 0 y G jaC39 10 1039 Chapter 6 Lecture 1 The parameter Z0 is a very important parameter It is called the characteristics impedance of the transmission line With the de nition of the Z0 we now rewrite the solutions for current and voltage VZ VJe Vo e V V 2 2 0 e 72 0 e 20 Z The solution represents a forward propagating and a backward propagating solution In general the constant y is complex and can be written in terms of its real and imaginary parts y orjB The propagation phase velocity is determined from B Example A air Tline is a Tline for which air is the dielectric material present between the two metal wires which renders G 0 If the conductors are super good so R0 If this Tline has a characteristics Z050 Q and the phase constant 520 radm at 700 MHz Find L and C File updated September 12 2006 minor change September 25 2006 File updated September 12 2007 Course Notes for ECE 1266 Applications of elds and Waves 1Impedance matching NOTES for Transmission Lines V A load is connected through a Tline to a generator Maximum power is transferred from the Tline to the load when the characteristic impedance of the line Z0ZL This ensures that This lecture covers Chapter 65 and 67 there is no re ection from the load Of course in general Z0ZL In the event that Z0 ZL we need a matching network to eliminate a re ection back to the enerator With reference to the dia ram 1 39 Impedance matchlng below we wish to makegZn Z0 so that the load seen at WZ0 2 Transwnts on Tllnes or1zinYin1ZOY0 j A Matching netuur39k Generator Transmission line Load The usual way to match the load to the Tline is to use a single stub matching network as shown below The load admittance YL shown in the diagram and the load impedance are related by YLlZL lvmlllllr VIP l I muim I Chapter 6 Lecture 5 l Course Notes for ECE 1266 Applications of fields and waves The two transmission lines length d and l are use to transform the load admittanceYL1ZL into an admittance lZo The admittances looking into the two Tlines are Yd and YS respectively The line of length d is used to transform the real part of the load admittance into lZo and the shorted line of length l is used to compensate for any reactive imaginary impedance resulting from the Tline of length d terminated in ZL This matching means that Yd YS 2Y1 Y0 Example Construct the shorted shuntstub matching network for a 509 line terminated in a load ZL 20 7j55 or zL 04 7j11 text example 67 Chapter 6 Lecture 5 Course Notes for ECE 1266 Applications of elds and waves Fiure 628 p 302 a The generic layout ofthe shorted shuntstub matching network b Adding shunt admittances c Using the Smith Chaitto find through line and stub lengths Values on the chart apply to Example 67 Wenlwo Copyright 2005 by John Wiley amp Sons Inc All rights reserved Chapter 6 Lecture 5 Course Notes for ECE 1266 Applications of elds and Waves Transients 0n Tlines The transient response of a voltage pulse on a transmission line is a time record of its back and forth travel between the sending and receiving ends of the line taking into account all the multiple re ections at both ends We use a unit step to describe this problem Vg t VgUt Um 7 1 z 2 0 7 0 z lt 0 Since a unit step propagates at a nite speed at different times the wave sees different impedances Let s get this into detail We denote the time for the pulse to travel from generator to load as T At time t0 the wave only can see the impedance of the line 20 and cannot see the load it takes time for the pulse to travel to the load The initial z 0 current and voltage on the line are therefore calculated by voltage division as If Vg Rg ZU 7 ngu Rg Z D For a step input the effective wavelength is in nite and the transmission line in steadystate looks like a wire This means that i 1 Chapter 6 Lecture 5 for a step mction the nal steady state currents and voltages at the load will be 1 Vi L ZL Rg ZL V 7 ngL Rg Z L We now discuss the approach to the steady state solution in detail The initial pulse not only experiences a re ection at the load re ection coef cient FL but experiences a re ection when it travels back to the generator The re ection coef cient for the wave traveling backwards toward the generator is given by Pg 7 Rg39zo RgZo Our textbook treats a nearly identical problem shown in Figure 633 At t 0 a switch is closed connecting a source Thevenin equivalent VS in series with ZS to a load ZL via a transmission line of characteristic impedance Zo Each time the pulse reaches the load it is partially re ected each time it reaches the generator it is also partially re ected We use a bounce diagram to represent this situation Course Notes for ECE 1266 Applications of elds and Waves t 0 I o t m 2t P b 85 V0 3t Steve M Chapter 6 Lecture 5 Z 7 Z At the load the re ection coef cient is given by PL L U at ZL Zn ZS 7 Zn the genemtor the re ection coef cient is given by F ZS Zn We plot distance on the horizontal scale and time on the Vertical scale We can keep tmck of What pulses are present at any time and any position on the transmission line by locating the position between the source and load and the time on the Vertical scale and adding up all of the pulses that have reached a particular position at a particular time Example 610 page 314 illustrates this process for a 6 cm long transmission line The transit time for a pulse to propagate from source to load is calculated as 2ns Course Notes for ECE 1266 Applications of elds and waves V3cm t The problem is someone more dif th is the input to the line is a The re ection coef cient at the load is 1257512575 025 pulse rather than a step function d at the source 25752575 05 The bounce diagram illustrates the re ections at the load and source We can look at 39ddle of the diagram dashed line to identify the pulses that Consider the situation shown in the text Figure 636 shown below pass the exact middle of the line 2 3 cm and can plot the voltage at the middle of the line That plot is given below Chapter 6 Lecture 5 Course Notes for ECE 1266 Applications of elds and waves pulse that occurs at the middle of the transmission line VJ t T V0 T V0 3 t 2t 3t The opening and closing of the switches produces the pulsed waveform shown n the top gure The pulse travels down the line to a load not shown The effect of the pulse can be calculated by 4t considering the pulse as a sum of two unit steps V V0 Ut 7 V0 UtT and using a bounce diagram to calculate the effect of each unit step The bounce diagrams are show below as is the resulting a V0 t t 2t 3t 13 Chapter 6 Lecture 5 Course Notes for ECE 1266 Applications of fields and waves To give a practical illustrationsuppose the 4V step Voltage in example 610 these notes page 5 is replaced by a 4V pulse of 3 ns duration The appropriate bounce diagram and the pulse that exists at the middle of the line are shown below V3cm t Additional Example A timedomain re ectometer TDR is an instrument used to locate faults on a Tline The fault is located at Chapter 6 Lecture 5 Course Notes for ECE 1266 Applications of elds and Waves 1n n vi I v L A m The mm A 11m Cblshlnu cpmemed b u Minor change September 25 2005 mm mt min 17 m min 17 mm m mlm1n 7nn Minor correction October 3 2005 Reviewed typos corrected September 25 2007 Major changes September 25 2008 Chapter 6 Lecture 5 Course Notes for ECE 1266 Applications of fields and waves NOTES for Antennas III This lecture covers 85 1 Antenna arrays Chapter 8 Wentworth In the last two lectures we studied a number of antennas including Hertzian M2 and a dipole of arbitrary length A problem with all of these antennas is that their directivities are not large and their directions of maximum intensity are not controllable For radar or high bandwidth free space communication we need to deliver energy in a much con ned fashion This can be accomplished by an antenna array ie a collection of properly oriented and properly driven antennas Antenna array For an isolated antenna element the radiation field is detemiined by two things the spherical propagation factor e39j rr and directionally dependent element the normalized radiation intensity Pn63 The field within a multiplicative constant is given by Eltna gt mm The directional component Pn62 is detemiined by the intrinsic characteristics of the antenna However to improve the radiation directionality or steer the beam change the radiation direction we can use an antenna array arranged in a formation shown below Phase Aiupli ci 5 shitters or nlteuumm s Al m m elements I 7 7 Element All I I 7 Element N Z u t l39 El I Mud Element 139 T n I i 17 Elementllqe 7 7 ElementO v I b Array geometry relative to ml Anny elemenu vnh mdwtdual Observation pojm mnplmlde and plmxe cnmml QR0 e 4 Course Notes for ECE 1255 Applications of elds and waves Typically all antenna elements have identical radiation characteris 39cs However we can control the composite antenna pattern by can controlling the amplitude and phase shi of the current driving each antenna element We illustrate this principle by considering N antennas spaced ev d meters along the z axis Each antenna s radiation characteristic is described by W E r 11 A ER 11094 Where R is the distance from each antenna to the point of observation Q A is proportional to the drive current amplitude and phase to each antenna 3 Element N l Element N 2 Element 1 1d Element 1 7 Element 0 We proceed in a manner similar to that used to analyze a long dipole The variation of R with antenna is important only in the exponential term expj BR where Chapter 8 Wentworth R E R0 7 icos e Z r icos e where R0 r is the distance from antenna element 0 i is the antenna element number and e is the angle between the antenna array and the line from the antenna to the point of observation The total eld at the observation point Qr 19 W is enj r N71 Er 6 r 6 7 x 24M W 7 10 The corresponding power density Poynting vector is gt 2 N4 2 gt Pr 6 5 i LAG 2 A e wcosg a39 0 10 gtlt r Where 4 1121 measure the current driving each antenna This discussion looks at multiple antennas oriented along the z axis The rst term l 19 rl2 is called a unit factor the N71 second Extewmg is called an array factor Our text Wu treaw two slightly different examples Example 1 An array of two vertical Hertzian dipoles oriented along the x axis text page 427 Figure 830 p 428 39 of zoriented A pair Hertzian dipole antennas 2 Course Notes for ECE 1266 Applications of elds and waves separated by a distance d on the xaxis The text looks at the electric field at a point of observation that is located at 612 ie the point of observation is in the xy plane The evaluation of the electric field proceeds in a fashion that is identical to the previous example ie the fields from the two dipoles are added Thus Eatam1 Em Em j77u a j77u as Sl ear3r a Sl ear3392 a 177 7 4 47 7 l 72 The subscript 0 refers to the point of observation the subscript S refers to a phasor quantity If the currenm driving each antenna are equal but out of phase with one another ie if 15 ID and 151 Ine we can find the total electric field using assumptions similar to those employed when we summed the contributions of Hertzian dipoles to find the electric field due to a long dipole These assumptions are a where distances appear in a denominator they can be assumed equal r1 r2 r Chapter 8 Wentworth b angles between each Hertzian dipole individual antenna and the point of observation are the same For this example this means C131 C132 CID c Differences between r1 and r are important only when these terms appear in phase terms ie in terms exp jBrl or exp jBrz In those terms the differences between r1 and r2 must be explicitly considered Simple geometry see text Figure 831 shows rr7cos andr riicos 1 2 1P z 2 Figure 831 p 429 a Parameters used to evaluate the far fields in the x 7y plane for a pair of Hertzian dipole antennas b Expanded geometry near the dipoles With some geometry it then becomes easy to show Course Notes for ECE 1266 Applications of elds and waves Il e 39 15 d a 9 EMU y 27 T2e Z cos acos g aa The total Poynting vector can be calculated as PM Rein Xng With several lines of arithmetic this becomes 2 z z a PM cosZ gcos a The total Poynting vector is the product of the Poynting vector due to one antenna an an array factor The array factor is the factor located between the square brackem Example 2 NElement arrays A similar approach can be used to calculate the field in the x y plane due to N Hertzian dipole antennas spaced a distance d apart along the X axis The situation is show schematically in text figure 833 The text should note but does not that each of the antennas is a Hertzian dipole Chapter 8 Wentvvorth Again we assume that the currents driving each antenna have equal amplitudes but are not in phase with one another For simplicity we assume that the currents are spaced equally in phase from one another ie 1g 1 152 Ina rS Ind 15 Ind etc The total field at an observation point in the xy plane is found by superposition Summing a geometric series we find I 9175 liex N Eamm1 J77 i 47 r 17 exp j w Where 11 d cos a Again the total field is the product of the field from one antenna and an array factor The Poynting vector can be calculation from the expression for electric field With some algebra we can show that the normalized power pattern PH lt15 that resulm from an N element array is given by Course Notes for ECE 1266 Applications of elds and waves N 1ij PM 727 N sinZ Z 2 The maximum of the normalized power pattern occurs at 1 0 The control of 1p 3d costp a ie control of the phase shift among the current drives to each antenna controls the direction of maximum normalized power The following figure shows the radiation pattern text example 89 for an antenna with five Hertzian dipole elements spaced M4 apart from each other and driven by equal amplitude currenm separated in phase from each other by steps of 30 degrees Figure 834 p 434 Fundamentals DIEleclmmignehcs m Engmeermg Appncaemwsmm M Wemvwnh capvngm 2uu5 bv JahnWilev a Sans lru All mam vesevved Chapter 8 Wentworth Revision December 5 2006 minor typo corrections Dec 7 2006 revisions November 26 and 27 2007 Corrections to page 2 column 2 made on November 27 2007 Figure for example 89 and typo corrected December 4 2007 Course Notes for ECE 1266 Applications of fields and waves NOTES for Waveguide I This lecture covers 71 and 72 l Rectangular waveguides 2 Transverse magnetic TM modes Waveguides Lecture 1 A waveguide is used to transmit highfrequency EM signals Waveguides are better than Tlines at higher frequencies 3 300GHz Transmission lines become inefficient at those frequencies due to the skin effect and due to dielectric losses More generally waveguide analysis is important when wavelengths are comparable to the cross sectional dimension of the waveguide In that case a single voltage or current is not adequate to describe the guide s signal Another difference between waveguides and T lines is that Tlines support TEM waves at any frequency A waveguide in general has a cutoiT frequency a minimum frequency that will propagate 1 Rectangular waveguides The simplest waveguide is a metal rectangular waveguide We begin by assuming that the waveguide is lled with a lossless material eg air as shown below m Mquot a 2m We will eventually assume that the waveguide carries energy in the z direction The Efreld and the Hfreld each have three components Xy and z and we will use Maxwell s equations to describe each of these components separately Course Notes for ECE 1266 Applications of fields and waves We will use the phasor form of Maxwell s equations in a chargefree and currentfree environment V X E jaJIIS V X 1 S 10812 The subscript 5 indicates a phasor If we write these elds in terms of their components siX equations result After some vector calculus we nd 515 615 25 390 62 J Hm 6E 6E xs zs 390 62 ax J Hy 6E 6E y5 xs39a ax 6y J st 6H 6H ys j39ngm 32 32 6H 6H xs zs ng S 32 3x y 6H 4 aHxs j39ngzs 3x 5y We can then solve for each of the four transverse components Ex Ey Hx and Hy in terms of the two zdirected components EZ and HZ For instance solve the fth equation for Ey and substitute into the rst equation This will result in an equation for Hx in terms of HZ and El Waveguides Lecture 1 If the direction of EM wave propagation is assumed along the z direction it is fair to assume that all eld components follow Eisxyaz Eitxyeij z Hisxyz Hitxyeij z The subscript i can be X y or z Againthe subscript 5 indicates phasors Similar to the approach we took analyzing Tlines and UPWs we refer to 3 as a propagation constant whose value is to be determined The value of 3 will not have the same value as that of a wave propagating in an unbounded medium This substitution leads to four equations for the transverse X and y components of the electric and magnetic elds E L6 Wai x uz z 3x uz z 3y E J 3152 Jaw 0H2 jaw 6Ez j 6H Hz z xii J32 6y xii J32 6x jaw 6Ez j 6H2 Hy2 z 2 z 6x 5 3 6y Aw 8 In general we can classify EM waves from the above expressions 1 TEM waves where EZHZ0 These do not eXist in a metal rectangular waveguide Transverse electric TE modes where Ez0 Hz 0 Transverse magnetic TM modes where Hz0 Ez 0 WN Course Notes for ECE 1266 Applications of fields and waves 4 HE modes where Hz 0 Ez 0 We will be concerned with 2 and 3 TE and TM waves The phasor expressions for the elds within the waveguide satisfy the wave equations we will follow the text and continue to use the subscript s to denote phasors In phasor form the wave equations are V25xyz 312566 y z 0 V21 xayaz 3121306 y 2 0 3 718 You should compare these wave equations to those that result for elds propagating in an unbounded medium The equations are nearly identical Here we are assuming lossless materials so that the propagation constant 7 described in Chapter 5 is imaginary and the role of 12 in the wave equations given in chapter 5 is played by 7Bu2 We will nd solutions to these wave equation for TM waves and for TE waves For TM waves we need to solve for EZ and all the other eld components will be determined from the equations on the previous page for TE wave we need to solve for HZ Solution of TM modes Hz0l Eli0 Separation of variables The text page 3523 shows how to solve for El We will look at this solution in detail in class Waveguides Lecture 1 General results TM waves In summary the solutions for TM waves are E25 E0 sin mjsinje39mZ a E 7j M E cos 7mm sin L eijpz x 2 0 u 2 a a b E 7j M sin 7mm cos L e j aZ y 2 0 u 2 b a b H 17 ME Sin M cos L e lt XS 2 0 u 2 b a b H 7jw8 M E cos 7mm sin Lily eijpz 5 Z 0 u 2 a a b 2 Z x y f y2 Mj nit a a b m 123 71 123 The propagation characteristics depend on the propagation constant The propagation constant can be written as 2 2 f 55 f a b From the above expression we can see that 3 must be real for propagation and that since 3H2 mzus there is a minimum value of 03 or a minimum value of frequency f JD27E for propagation We refer to that frequency as the cutoff frequency fc Course Notes for ECE 1266 Applications of fields and waves 0 Cutoff frequency for a given mode m n the cutoff frequency is found from the requirement that there is a minimum value of 3JLL for propagation ie if 3JLL 0 then f fc This means that for propagation 2 2 m7 mt u 2 l7r 5 Z and the cut off frequency is found from Z 2 f 2 where c 71 andtherefore a2u wu 1 fc f2 Z M1 fc f2 c Below the cutoff frequency fc 3 is imaginary or ygt0 so the wave decays with z and is known as an evanescent wave 0 The phase velocity of the guide for a given mode m n is C Q p 2 1H f u o The intrinsic wave impedance of the mode is de ned by the ratio ExHy 234 ExH JE 1 L 77 1 L y mg g f u f Update October 12 2006 1020 am8 Minor typo correction October 18 2006 Corrections October 19 2006 November 2 2006 November 8 2006 October 10 2007 Correction October 18 2007 minor typos corrected October 1 2008 Waveguides Lecture 1


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