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# Microwave Eng I ECE 5320

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This 55 page Class Notes was uploaded by Shyanne Lubowitz on Monday October 26, 2015. The Class Notes belongs to ECE 5320 at University of Utah taught by Staff in Fall. Since its upload, it has received 13 views. For similar materials see /class/230007/ece-5320-university-of-utah in ELECTRICAL AND COMPUTER ENGINEERING at University of Utah.

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ECE 6130 ABCD Parameters and TRL Network Calibration Text Sections 44 and pages 217222 Portfolio Optional Chapter 4 Repeat Problem 418 using ABCD parameters Hint Use table 41 to nd the ABCD parameters of each single transmission line Multiply them together to nd the series con guration ConveIt to S parameters using table 42 Compare with results obtained previously Transmission ABCD Parameters Useful for evaluating cascades of networks Easy to conveIt to and from S parameters V1AV2B12 IlCV2D12 VI A B V2 11 C D 12 For a cascade series of networks VIEWGRAPH FIGURE 411 V1 Al 31 V2 11 C1 D1 12 V2 A2 Bl 1 12 C2 D2 13 SO Ivl A1 Bl A2 Bl Ki 11 C1 D1 C2 D2 13 VIEWGRAPH Table 41 TRL Network Analyzer Calibration Calibration done so far relies on known loads sh01t open matched There are always errors in these loads Matched load for instance is only perfectly matched at a single frequency This is pa1ticularly a problem if you need a holder for a deVice under test DUT and need to calibrate out the effects of the holder Better method TRL Calibration VIEWGRAPH Figures 421 THRU Connect two reference planes exactly Or any line integral of half wavelengths T11 b1 al when a20 s11 22 s1221 3222 T12 bl a2 when al0 s122 1 222 REFLECT Connect two reference planes with any highre ection load open or short R11 blal when a20 811 FL 8122 l S22 FL LINE Connect two reference planes with any length of line OTHER than 05 wavelength multiples Best results occur when the line is quarter wavelength but others are OK too L11 b1 al when a20 s11 322 S122 e39ZYL 1 szzze39ZYL L12 bl a2 when al0 s122 e392yL 1 222 e39ZYL SOLVE for e392yL and FL VIEWGRAPH Equations 478 and 483 Calibration complete Measurement of DUT VIEWGRAPH equations 484 ECE 6130 Impedance Matching Text Seetlons 5e2 Ponfollo questron How do you deslgn a slnglerstub matehmg network for any complex load7 Examples ECE315 tent Chapter 2 avallable on reserve m llbrary or m the IEEE room Problems 353638 slngle stub matehrng IMPEDANCE MATCHING Zln at rnput to matehmg network Zo Onewaytn dn this QnartenWave Transrnrmer Drsaolvantage of14 Wave transforme We must be able to adjustboth L and 20 sueshape ofTL Thls ls not easy Better methnd Single Stub DLatching Use aknown transmrssron lme short or openr both easy to eontrol Haw tn mndel stubs Yin Ys Yd EXAMPLE See transparencies Matching For matched system Yin Yo This is accomplished by Adjusting the line length d until Yin Yo i jX and then adding in a capacitive inductive component shortcircuited stub to remove X SingleStub Matching 1 Plot 2L 2 Re ect it through the origin to nd yL 3 Rotate yL towards the generator until it reaches rL 10 circle there will be two points choose the closest one This is the length d between the load and the stub 4 Read the value of XL 5 For a shortcircuited line Plot zL left hand side and re ect it through the origin to yL right hand side 6 Rotate the shortcircuit towards the generator until it reaches ys 0 j XL This is the length of the stub 7 The new yin l j0 bull39seye EXAMPLE see transparencies Lecture 3 Terminated Lnssless Transmissinn Lines Tux Sectinn zrs Pnrtfnlin 1 39 39 39 Iline39 Slandinngves What isthe re ecu nn cae ieiem and VSWR enmplexlnad impedance Wave Prngzgzunn m Lnssless Transrmssmn 1m VIEWGRAPH Ulaby Table 22 3 y VP do not depend on shape oftransmlsslon hhe forlossless TEMhhes ButZo does zp rs sareuy rea1 for 10551355 hhes no phase ehahge So phee started VI propagate the same way zp ephaols relahve maghrmdes ofVI Re ected Wav W2 Vo equot Vo39 e39 12 Vo zp e39mr Va 20 e 2 traveling Wave comes from the generator 72 travelmg wave rs re ected from the load ZL V1 11 V1 may Vo Vo 11 190 Vo zp eva zp ZL Vo39Vo39Vo39rVo39 20 V6 zL e Z0 2L 20 V0 antzge Reneeunh Cnef clent propomon ofwave re ected 1quot Vo39 Vo zr Z0 zL 20 raao e uhruess zLzp 71 zLzp 1 r46re 39 Vz Vo e39j z r Vo eJBZ 12 VJ Z0 e39j z r ei z Current Re ection Coef cient Io39IoVo39Vol Example 100ohm lineZo connected to a 50ohm termination RL F 50100 1 50100 1 13 Vo39 Vo For a 1V sine wave 13 V sine wave is re ected back out of phase Draw Vo 1 at open end Example 100ohm line Zo left open RL oo F 00 100 1oo 100 1 1 Vo39 Vo For a 1V sine wave 1 V sine wave is re ected back in phase Draw Vo 1 at open end sum 2V at open end Example 100ohm line Zo with a short RL 0 F 0100 1 0100 1 1 Vo39 Vo For a 1V sine wave 1 V sine wave is re ected back out of phase Draw as function of time Vo 1 at open end sum 0V at open end What about elsewhere Example so how do you get rid of these re ections 100ohm line Zo with RL 100 ohms r 100100 71 100100 1 0 Vo39 Vo For a 1V sine wave 0 V is re ected back Draw as function of time Vo l at open end sum lV at open end Standing Waves We have calculated what happens at one location end of line as a function of time What about the rest of the line Sum of Vo39 and Vo creates a quotstanding wavequot The total wave on the transmission line is the quotinterferencequot of the two waves At any point it is a sine wave same frequency as incident At this point the magnitude and phase is controlled by the sum of the two waves The largest magnitude is when they add in phase max 2V for IV incident The smallest magnitude is when they add out of phase min 0V for IV incident Understanding Standing Wave Diagrams The diagrams in Figure 214 show the ENVELOPE of the wave as a function of distance Inside this envelope the summed wave is still going up and down Simulation Velocity of propagation of a pure standing wave ZERO Wavelength of a standing wave 7 Repetition period 7 2 Volta e Standin Wave Ratio SWR S l Vl maxl Vl min 1 ll l l ll l dimensionless Also F Sl 81 S is large when there is a large reflected wave small when there is a small re ected wave It measures the mismatch Large S is quotbadquot for most things There is a VSWR meter Example Open mrcuxt 520m Example Short cxrcuxt 1quot 0100 71 0100 1 Examp Perfectly matchedlme nu re ectmn 1quot 100100 71 100100 1 01vg Va 5 1 1 Thxsxsthebestwecan aner Wave Trunxfnrmer Text seemm 2 5 A quarter Wavetransfarmens usedta match me REALxmpedance m anather XlA ZL Impedance afthe laad must be stnctly real nu 1magmary pan Za charactensnc 1mpedance afthe 1me yau wanna cannethL m 21 m matchthelme 21 Zn ZL A wavelength afthe 1me mans usedta make me quarterrwave transfarmer Haw dues ths Wark7 R 12 an a Zquot Zthan fw 27rZZ47r2 z fRl2ammeh 5 z1 212 WhatfoL 15 complex7 R Z RnXawlnan 2 10 JXtan3 you reach the REAL lme V Frequency Response of QuaneeWave Matching FIGURE FROM TEXT so me quaneewave seeuon 15 only quaneewave at dunetefreqummes seeuons We mu leam about ths later ECE 61mquot PuwerDividusznd Cuupkrs hummer Haw da yau deag unmcuanpawu amam Wham a uranium and haw dues xivark mmmpmm the mpuu ampumexasjv Puwer Dilidus cm 1m unm 171mm m n General ZrPanSrpuame39zrs 5 5 S 5 5n 522 522 531 5 5 Issues 1 Mawhedhns n n 0 52 5E 517 5n 0 SE 531 52 0 2 Recxpmcalsyxnmemc5 0 Sn 5 S Sn 0 Sn 5 52 o u z Lasslessmmry5 su39sg n SatwaaKSnngu 52339s n sn39sfn 512 2Sj2 Eunhzn39hese can tbe saus ed smyxsa sm H524 2 2 1 1 s s CIRCULATOR Matched mdLassless Eutmtxecxpzacal 001 010 51100 51001 010 100 ECE 6180 Steppel LImpel lance Fillet Design Ponfollo Quesuon How do you deslgn a steppedrlmpedance ltex l Text Sechon 8 6 What IS a stepped nnpedanee lter Made up ofhlgh nnpedanee thln andlownnpedanee thck hnes Wanchlgh Zlow to be as bilge as posslhle so thls IS determined by manufactunng In our case Zhlgh 100 ohms and Zlow 10 ohms Impedanees stay the same but lengths change fol eaeh seouon Each length 15 less than AM Why a steppednnpedanee lter Smallel than stubr ltel Easlel to deslgn no Kmoda ldentmesl why NOT a stepped lmpedance lter Appxoxlmxhons In the deslgn equataons make themless aecmate Amlysls ofsteppedlmpedance lters Note these ate analysls stepsnot deslgn steps You dont do these each lter 15 denved 1 Fox a length of hansmlsslon me see table p 203 Zn BL 2 Convert from ABCD to Zmamx table p 211 3 Calculate the Zrmamx of a Tjunchon cucult p 195 ZA 232A 211 Zn ZAZC Zn 221 Zc Solve 01211 etc mg 4 Relate 21 24 to Z11 etc ZA 2c 22 77211 125532 4 Axe 21 anch Inductoxs ox capacitoxs l Fox a 1mgth oflme w11hBLlt 71 Z 1 magmary pm mductox 2c r Imagmary pm capautox 2X ZZatan cus i 1 Z 1 32 5 1 1 1 z 7 z asz z 2 C 15 397 SE32 Zacsc z 2am iXZ iXZ In 5 Assume a short length ofhne gm4 accurate a WhenZoIslmge XN 224 Za b When 20 IS small Xo 52 EECYn5L 6 Solve fox lengths ofhnes that me needed fox ltex design LZo z hxgh CZZUW M 7 Note These lengths are given in RADIAN S HPEesof ADS Linecalc eeft is given in DEGREES Multiply these values by 180 7 to get them in degrees Filter Design Steps 1 Design the lumped element filter as before sections 83 and 84 2 Solve for lengths BL of each element Remember to convert to degrees if using Linecalc ECE 6130 7 SPnrnmeters Text Secum 4 3 Desmbe and Campute SrParame39Iers Examples Chapter417mblems 1011 Sal m39nmeters Recall 7 ScaneringDIatrix Spurnmaers W 5n 512 5w W V 7 5n V V1 5m 5m V1 Where 5 v v when Vl 0 far kg 1 Termlnate all pans Exceth wth matchedlaad 2 Dnvepmuwnhv 3 Nleasure re ectedvultage v an pam EXAIVJPLE 3dB attenuator 5 55am 5 55am H 9 e12 4 4F W m Eahms V2 Flnd s11 22 so ahms 21mm 1 8 561418H8 56 50 50 L7th 0 nar ecnan V1 Vl 0 l Flnd 522 Circuit ls symmems 522 7 Sn Flnd 512 S12 V27 V1 when pm 2 15 ter mlnated Since S 11 0 we know that Vl 0 when port 2 is terminated and V2 0 by de nition Then V2 V2 Vmiddle 856 856 50 Vmiddle V1 856 H 1418856 856 H 1418 Return Loss RL 20 logSii dB Insertion Loss TL 20 logSij dB EXAMPLES ECE 6130 Rectangular Waveguides Text Sections 33 Chapter 3 Problem 3 See Appendix I and Derive the TM modes of a rectangular waveguide following the methods described here for TE modes Rectangular Waveguides Recall Method of solution 1 Solve Helmholtz equation for either Hz TE or Ez TM This can be done analytically or numerically In the analytical case you guess the form of the solution which will have several unknown constants like magnitude phase number of cycles 2 Use 319 to 323 to nd transverse components from Ez or Hz 3 Solve for the constants from the boundary conditions In metal boundaries these are that tangential E and normal H 0 on the boundary Now you have Ez or Hz 4 Use Maxwell s equation to nd the other E or H components TE Solution 1 Solve Helmholtz wave equation 62 62 ax 2ay 2kh2xy0 a Use Method of Separation of Variables hxXY XX YY b Substitute into wave equation Helmholtz equation 2 Z d X X d Y abc2 aly2 where k2 kx2 ky2 c Separate the Variables dZX Y k2 0 2 Xd Y ij 0 dy Y k XY 0 2 d Guess the form of the solution 8N V U V O V e hZXy A coskxx B sin kxx C cos kyy D sin kyy The unknown constants are ABCD Also kx and ky Solve for the unknown constants from boundary conditions De ne the boundary conditions Tangential E elds 0 on the metal surfaces walls of the waveguide ex 0 at y0b ey 0 at X0a Obtain appropriate expressions for the boundary condition elds From equations 319 EX 0ij k3 6H2 6y Ey 00 k3 6H2 ax So exXy joau kcz ky A coskxx B sin kxx Csin kyy D cos kyy eyXy imp kcz kx A sinkxx B cos kxx C cos kyy D sin kyy Use boundary conditions to solve for ABCD Substitute ex 0 at y0b into equations above exX0 joau kcz ky A coskxx B sin kxx Csin ky0 D cos ky00 So D 0 Note that a trivial solution also exists if ky 0 exXb joau kcz ky A coskxx B sin kxx Csin kyb 0 cos kyb0 when ky nnb and kx mna Substitute ey 0 at X0a into equations above ey0y imp kcz kx A sinkXO B cos kXO C cos kyy D sin kyy 0 So B0 eyXa Qmu kcz kx A sinkxa B cos kxa C cos kyy D sin kyy0 So kx mn a Now we can simplify the form exXb joau kcz nnb AC cos mnXa sin nnyb exXb Amn cos mnXa sin nnyb Apply constants to H hZXy A coskxx 0 sin kxX C cos kyy 0 sin kyy HZXy AC cos mnXa cos nnyb Am cos mnXa cos nnyb e39jl3Z EX Ey HXHy components are found from equations 319 again TEmn modes Modes are numbered mn indicating how many cosine wave cycles are in the waveguide Cutoff frequency for each mode is different fc kc 2mg 1 mn39 2 717239 2 f5 m 27239 8y 0 b Lowest Order Mode Assuming bgta TE10 mode has the lowest cutoff frequency so will be the first mode in the waveguide ECE 5130 e N crnstripline Filter Implem mtz nn Ponfollo Quest on How do you rmplement a lter deslgn uslng stubs7 Inelude lnformanon on applylng the Rlchardson Transformatlon and Kuroda rdenhhes Richardsnn Tranxfnrmztinns Frlter deslgns are developed as ladders oflnductors and eapaertors These ean be rmplemented m mrerosmplrne as open or short erreurted stubs The equrvalent erreurts are R lchavdsun Tyanstmmallnns Ml at a L shun clteull l 0 mm Open clteull To venfy these see Smth Chart examples Use Lhistn implement arrlter Example Deslgn a LP lterforfabncatlon uslng mrerostnplrnes The spees are cutoff frequency of2 GHz attenuahon of at least 30 dB at4 GHz rmpedanee of 50 ohms 3 dB equal npple eharaetenshe Step 1 Frndthe order othe fllter See Flgure 8 27b lm m4 71 GHz 2 GHz7110 The attenuauon of 30 dB requlres afourth order lter N t Step 2 Frndthe lter eoemerents and draw the LC lter See Table 8 4b for 3dB equalrnpple Step 3 Convert to stub network This would work but for microstrip con guration we need all parallel stubs Series stubs can t be built in microstrip Kuroda Identities Used to convert for buildability Convert to parallel stubs See notes ECE 6130 LECTURE 5 Text Secnun 2 56 mmth Far a terminated lassless kansmxssmn lme 1 Desenbe hawta nd me ynuage at any paint anthelme as afuncnan afnme 2 Desenbe hawta nd me ynuage alang the hue at any pman time TRANSIENT S 0N TRANSMISSION LINES Equivalent mun an 21 time t x c n o The much 5 1asednayauage has maved dawn the txansmxssmn lme All the generator 5225quot 5 Zn 2L is we far away I g kg Za means pasmvertxavelmg wave 1quot means rstwave I Za Vg Za Rg Za vm Fwst Wave waves dawn n Velaertv afwave vp 1 swat Wave maves dawn TL vvrth nu re ecnans until rt reaches the laad Trmeta reachlaad TL vp Wm Wave re ects afflaad and starts back V1 y r 21 2a 22a1quotr 13 Thrs 15 real whathappens rfrtrs complex7 Get phase e as well as re eetmr W1 2v Wave ve eds mt genevmm v2 v1 W w W Luau m Wave Vr nUW re ects uffgerreratar New rtxavelmg wave 15 pmduced V 1 rs Zr Za zg2a Nance that re eetmns are gemng pragressrvely smaller Waves keep re eetmg until STEADY srATEts reached addmonal re ections are neglxgbly small VwVg zL kg zL Voltage on hhe at ste dy state This is the SAME as we would have observedm the DC case IwVwZyVgRgZy BOUNCE DIAGRAMS z ISL T T ZT 1T 31 31 1 Axes Time and dASLance 2T ZT 3T 3T 1 W 1quot 1 V1 gv rgrhvt Ve39rhve rg1quoth2vr Howto nd Vzt for any 2 ahdt 1 Fmdthe pomt zt on the bounce dAagram eg z L2 t 3T 2 Draw alme backto T0 3 Add up all othe V traces you cross v T W ZT 7 W N V2 31 7 31 VL23T V Vy v Example Handout ECE 6130 Impedance and Multiport networks Text Sections 41 Portfolio Question Explain the concept of quotimpedancequot as it relates to microwave transmission lines and waveguides Microwave Network Theory Circuit Network Low Frequency Elements can be modeled as lumped elements Given a set of RLC components at low frequencies you can nd voltage and current and impedance at any points BETWEEN these elements using circuit theory KVL KCL etc You do not nd the values of voltage eld etc in points INSIDE the elements They are usually not of interest Microwave Network High Frequency Elements cannot be modeled as lumped elements Elements are large compared to wavelength In Microwave network analysis a Field theory Maxwell s equations is used to nd the complete eld distribution in each element b Determine equivalent circuit properties of each element Like Z0 and Zin for transmission lines from complete elds c Equivalent circuit models are then analyzed with Circuit Theory Example Coaxial Transmission line analysis a Use Maxwell s equations and boundary conditions to nd E and H everywhere in the line Text section 35 b Use E and H to nd RLGC parameters text section 22 c Analyze RLGC parameters using circuit theory text section 2 l d One step fu1ther Combine RLGC parameters to give VIZ relations Use these to analyze multiple lines So what we need for any case is a method to determine VIZ values so we can combine effects of multiple elements using circuit theory Impedance For 2conductor transmission line V is voltage between the conductors V 506 IbLodr Vlnba V a rlnba lnba I is current owing on the positive conductor 2 I Hod L0 rd I I Characteristic impedance of the line is de ned Zo VI ohms BUT for a singleconductor transmission line V and I are not wellde ned Start with a rectangular waveguide Table 32 shows the elds of a rectangular waveguide Sketch the TElO and TMlO modes TE10 mode of rectangular waveguide Ey Xyz imua TE A sin TE X a eXp sz HXXyz GBa TE A sin TE X a eXp sz Voltage between top and bottom of waveguide is V integral E odl jwua pi A sin pi X a integral dy jwua pi A sin pi X a b This depends on X which value of X should we take There is no unique answer to this Uses of impedance 0 Wave impedance ZW E1 Ht Depends on size and shape of guide frequency type of wave ZTEM ZTE ZTM o Intrinsic Impedance of the material 11 We 0 Characteristic Impedance Zo V 1 unique for TEM waves nonunique for TE and TM waves So how do we decide on de nitions of impedance for TE and TM modes 1 Voltage and current are de ned for a particular mode Choose voltage proportional to Etmsvme and current proportional to Htmsvme 2 Choose V and I so that VI power ow ofthe mode 3 For a single traveling wave or V I Zo which is usually chosen to wave impedance of line or normalized to 1 Example Rectangular Waveguide 1 Voltage and current are de ned for a particular mode Choose voltage proportional to 5mm 2xyAe quotZ A eW Z 2xyCiVe quotZ V eW Z 1 and current proportional to Hxyz Zapata 32 A ew z ZxyCiIe quotZ I M 2 where 2x y transverseE sinmc 61 Rx y transverseH 1 sinmc 61 ZTE also 2 x2xy h x y ZW and Zn V I V I E1 and current proportional to H1 2 Choose V and I so that VI power ow ofthe mode This is going to partially de ne C1 and C2 Power owing in positivetraveling eld wave only 2 13 1J39EXH dS1J39 EnydxdyM 2 S 2 S 4ZTE Power owing in positivetraveling voltage and current wave only Pt 12 VT Setting these two powers equal gives 2 ale 21W 4ZTE 2 Re call from above Vl AC1 and 1 AC2 3 Set ratio of V I for single traveling wave to be wave impedancefor that mode V C1 1 C2 TE Solving Clab C221 ab 2 Z 2 ECE 6130 Quadrature Caupler Text Section 7 5 What re a quadrature caupler and haw dues rt Wark7 Quadrature lepler 90 degree phase shx between autput parts Pawer Evenly dvrded ads between dutput pans All pans matched Na cauplmg ta part 4 camplete xsalauon 0110 41001 515100 0110 zosmmz Evmrodd Made Analyer Nmmalxzedlmpedance Crreurt Made EV39ENMODE 112 9 129 K T g 1 lSSmbHLM 7Z18ampub 1 0 ma 120mg 1 0 Y 1 fZasm 05 Y 1 A3CD39 rl71J 7 2 2 71 77 71 ABCD inmind J5lt J 1 121 g 1 1 8 Stub1 4 TL1 8 Stub l 0 cos jZosin l 0 Y 1quotjZosin cos Y l 100 39J51011 lJ llijxE c0 li f ll lf ll 2F AB C D1j39 j39 1J 0 2 ABCD 1jj1 2 2 1 ABCD139391J 71j 212 COMBINE EVEN and ODD MODES Incident voltage at node 1 V1e V10 12 Incident voltage at node 4 V4e V40 12 Amplitudes of waves exiting each node B1 V1e Fe V10 F0 Fe 2 FoZ 0 Port 1 is matched B2 V16 T6 V10 T0 Te 2 T0 2 j sq1t2 Half power 90 degree phase shift from port 1 to 2 B3 V46 T6 V40 T0 T6 2 T0 2 lsqrt2 Halfpower 180 degree phase shift from port 1 to 3 B4 V4e Fe V40 To Fe 2 F0 2 0 Port 4 is isolated This is the rst row of the S matrix 0110 1j001 SJ 100 0110 Other rows are found by transposition Row 2 for example 1920 291139 3941 493 0 EcE mu LUMPED ELEMENT IMPEDANCE MATCHING Reference anxck RF Cuculs pp m7 14mm Am Hmdvut sauces faxSMT up mductm39s The LeN eomk me afferent farms uf me Lesmpea N eomk LawPass AH and page P 55 cm Purpase quthlemen39s SHUNT Element Transfm39m lug ZL m mane wluewnhxeal pm equal m me 1251 pm ust SERIES Element Resma z wnhcmce me magnume athe mpeamee Deagq mp5 X X5 maxp cmbe anther capacxuve ax mducuve bunnme af appame type Example 5n X 91R 06 X 7RQ1IEIl n szHz c abvve ng1 Pass canfxguahm D wauld waka equa ywzll XnLeLXn159nl1 PROELEMSLIMITATKONS wah39hsme39had 2 T111515 ampr my mtbmhmhr m hndswxlltake can arms 2 Cmnmmawhxmpedanceslessthan39helmeunpedmceQbecamesxmagmy 4 ma jmaya became Wham MATCHING COMPLEX IMPEDANCES Twamz hndscmbeusedmxemwe39helaadxeacunce 1 Absmpummang the laadreactmce aspm arm demedmnlchmg urcut 2 ResamncewngmLtacmczl AC a AC tau151 ml Example Same Example as mm mum mahm areachve campmznl zLmn125 nhms Th2 X 725mm 15 equivalenlm a capsular m senesw hthz laadxeastarwe R The Wm quL lle A pF This gves um mm m cucu39s Shawn belaw L CL T RL 7 HPVC AESORPTION 0F cL cL can mw be absurbed m the HPVD can gunuan Thms c cmm z xxpp Cmatchm seneswnhCL CmatchCL CL Cmmm OR Cmmh c CLCLV C 3 xzpnas APP 6 A 12pm 5 32 pF LIMITATION Ths mlywmks xfCL Ls genie um cum Yambank has a mum example whzxe the laadxsmadeledthh apua el cR cm gu39ahan TO CONVERT FROM smas TO PARALLEL LoAD ADMITTANCE Wm up lmpedan e mm mm cm gunums 2 RP UPC RPHJ CP RP MHGCP RP ZsRslJmCs 2 25 zp Tm W m mm mm mums The mmwns m up ms orqu Rpcp OR Rpcp orqu Ms mm W Rs XPszXPZ Rp X5 RpRs X where 71 z oR mL 0 R52 232 Rs RP X R5 Fax aux casewhzreRFl El mamas nhms we getRyFlEl 25 ahms dep 7425 nhms ThsxsashmtcapanlmceXp 7 mac whexec nmpp RESONANCE ach w M u u v mmductance Thsxseamxwhznthz c mdL arempan el 11 an 3ch 772a Pmceedthhnamml mmhmg as mm rm Example 5 alsa EXAMPLE A 3 mm mam Lp RP I CD E ca 51806180 Mimwzve Putzer Lectues meea element lms alsn apphes hm hm freq Jany mus sum mm suppeaxmpem mm c We Lm mm Lulnped Elnnznl Finn39s 1 m SemnnX 3 menlm Quzsnm wa an Maggi map element lt usng39helnsemansstd Puwer Lass Ra P Paw Avaxlable from Source 39 Power Delxwred to Load 39 Insen39nn Lass u m IngPu Ipmtnetwmk V1 Fm Wtched swam 2m he Fwy12 Plaad V IS Irm 1 maul m Dewbym mmmmm gamma in mm 9 st and mm mm F tzrpzmnetns Passband W frequmues that are passed by ltex Stwpbandn equznmesmat axerzjened Inmm lass W hww muchpwwex xs Lransfen39ed nu ma m passband Ammanm W hww muuh pawn xs rejecu mtuansfened m the ma mm stapband Cutoff rate or attenuation rate how quickly the filter transitions from passtostop or stopto passban s Phase response Linear phase response in the passband means that signal will not be distorted Classes of Filters Determined by F We have not proven this yet but a useful mathematical proof section 41 shows that lFoJl2 is an even function of m So lFoJl2 can be written as a polynomial in OJ 2 Ma2 Imml Ma2Na2 PLR 1L Na The class of filter is controlled by the type of polynomial used Polynomials M and N can be I Binomial Butterworth Maximally Flat I Chebyshev Equal Ripple I Elliptic Specified Minimum Stopband Attenuation faster cutoff I Linear Phase Binomial Butterworth Maximally Flat Low Pass Filter Design 2N a PLR 1k2m 5 N Filter Order OJ frequency of interest we cutoff frequency At we PLR 1k2 If the 3dB point is defined to be the cutoff point common kl For 0Jgtgtoac then PLR mkz OJme2N which means Insertion Loss increases at a rate of 20N dB decade This allows us to increase the steepness of the cutoff by adding more sections Chebyshev Equal Ripple Filters a PL 1k2T1 w 5 Where TN are Chebyshev polynomials Ripples are equalinsize 1k2 Cutoff Rate is 20N dBdecade same as binomial Insertion loss in the stopband is ZZN4 greater than binomial Elliptic and Linear Phase Filters Other options see textbook Filter Design Method 1 Design a LP filter for normalized ZOJ 2 Scale Z z ommmam LP m HP a a as amed A ommmam mpeam nimbuud elements as awed 1 Binnm39nl mag quP Finn m Numzl39ned z n Datevmmt how many elemmt m maid N Fmd mat and mm at m gue m ammuhanm the impband Fng 26 p 450 Ex mp1 Haw many elements are xequnred m deag a maxxmallyr at lm wah a culufffxequency uf 2 3111 m m mm mmtpmwdg 2 dB af 311mman AM GH17 FarthmcaseJmmgrl 7 42H 7 Khanam ms Fmlemzan ltexthahsAEOVELhedemed an 4 attenuah b Emvesummeovconmcmmevnluesfvom 71151283 n 1n g4l c ChooseLFEImvon oype Why chums am aver the v1th Amman campanznls pranses Ufme are xdenncal gg 1 L2g2 RLgn1l C1g1 C3g3 1 Away start and and with c fwd L1g1 Start and endwith L 1 7 n11 RL C2g2 0 my any h h pan 2 see hmdau39 Thsxs effemvelymatchmg mammg smu unevusly z Deag SD n has canadexednan nahzedxmpedancestRL andmmahzed equencymk 1 Us impedance and frequency scahngxhhzy Aren t 1 Chehyshw Equal 1239quot I mag quP Finn m Numzliud 2 same guys asfm bmmmal There are an y afew differences a Dmmun number af 49mm N Ths wm always be 155am a equal m bmmmal Us Fxgure 2 21mm chance af x ufnpple b Us mm 2 A wah same chance af pp e as used m M a c Same asbmmmal 2 lmpedanceznd qumy Scalhgnnmz zzl39nn Ta mam same lm fax Zn Rg RL and a gven culufffxequency a U5 m same mm 9mm va scale the values a Top l zvpvotoype R ZDSD c1 g Za mg 5 mm m Eu 3 a i Lzaawj m mam g4 Emmmal momma are u same hm exceptfax an umce RL faxbxmmxal 5 always mum w 5mg W 4984 r r Qumexwave mam wauldhave znfm 9241 Zn 3 Cunven cm H m HP ngmPassCan gumms Rg c2 gm gm RL C1 C3 H f F m m bothprofogpe c am a L Zum a For foppvo ogpe R a ga a gm 02m mam Ry Fbvb R 7 Zaga Za g 3 Convert from LP to Bandpass or Bandstop Normalized bandwidth A m 0 031 lower limit 032 upper limit 0 Noaloaz See Table 86 p 461 for conversions To use for unnormalized filters L9LZo C9CZo ECE 6130 TEM TE TM Fields Text Sections 31 Portfolio Question What are TEM TE and TM waves Do Chapter 3 Problem 1 TEM Transverse Electromagnetic Waves In transmission lines we considered TEM propagation This meant E and H were both perpendicular transverse to direction of propagation Now we will generalize this and consider TE and TM modes also Fields in a waveguide closed cavity or transmission line are generalized to the SUM of combinations of TEM TE TM modes At the lowest resonant frequency only one mode will resonate As the frequencies become higher the lowest modes and additional higherorder modes MAY be excited Which modes will actually be excited depends on the source location and type General Wave propagating in lossless material oc0 the z direction E Exx Ey y Ez z e39lBZ Ett Ez z e39JBZ transverse direction X and or y H Hxx Hy y Hz z e39iBZ Htt sz e39jBZ Maxwell s Equations in sourcefree region no external charges or currents V x E joa u H V x H joag E For zdependence e39jBZ ddz j3 Apply to Maxwell s Equations and equate vector components VIEWGRAPH and handouts of equations 3334 We can solve homework problem for the transverse X and y components as functions of the z components VIEWGRAPH of equations 35 GENERAL eld solutions Wave propagation Number k 27 9 0 Vug Previously we called this 3 NOWletk27t7toa lug And B2k2 kf Cutoff wavenumber kc depends on geometry of wave guide This will be the wavenumber kc 03c VHS no is the cutoff frequency of the waveguide in the mode TEMTETM mode of interest The cutoff frequency is the LOWEST frequency where a given mode will propagate The mode MAY propagate at frequencies of no or higher TEM field solutions The TEM fields are defined when Ez Hz 0 Substitute into equations 35 BUT we get the result ExEyHxHy0 UNLESS kc 0 SO kc 0 for TEM waves This is why we never saw kc in the transmission line section TEM waves will propagate from DC to high frequencies To find TEM field solutions Return to Maxwell s Equations and derive Helmholtz wave equation VIEWGRAPH p l 6 Apply to EX and By and HX and Hy components Examine TEM modes For 2conductor lines there can be a TEM mode For a closedconductor line like a waveguide there cannot be a TEM mode since V0 metal all around then ExEy0 and Ez already 0 so there cannot be TEM modes Wave Impedance ZTEM EX Hy lug n 377 This is NOT the Characteristic Impedance of the transmission line Zo Vo Io Method of Solution i Solve Laplace s Equation This can be done analytically or numerically Analytical methods you guess the form of the solution which will have several unknown constants Example waveguide will have a sine function but we don t know the amplitude or number of periods N Use boundary conditions CD or dCDdN on surface to find unknown constants Now know potential distribution CD 3 Use CD and V12 q31 q2 ZEW to findE FindHfrom VXEjoauH E Modes Transverse Electric Ez 0 Hz i 0 VIEWGRAPH eqns 319 Now kc i 0 so 32 k2 kc2 function of frequency and guide geometry TE waves cannot propagate at DC TE waves CAN propagate in wave guides transmission lines open environment Method of solution Solve Helmholtz equation for either Hz TE or Ez TM This can be done analytically or numerically In the analytical case you guess the form of the solution which will have several unknown constants like magnitude phase number of cycles i N Use 319 o 323 to find transverse components from Ez or Hz U Solve for the constants from the boundary conditions In metal boundaries these are that tangential E and normal H 0 on the boundary Now you have Ez or Hz 4 Use Maxwell s equation to find the other E or H components Wave Impedance ZTE EX Hy Ey HX k n 3 TM waves Same as TE except Transverse Magnetic Hz0 Ez i 0 ZTMEXHyEyHxB nk General method of solution Given the physical dimensions materials etc Of a guide or line determine the modes that MAY be excited Then design a feed system to get those modes 1 Find the TEM TE and TM modes individually Each will have a different cutoff frequency N For a given frequency determine all modes that MAY be present For a given feed system find the modes that are compatible or design a feed system that is compatible with the desired modes Total modes in a given guide at a given frequency is the sum of modes with cutoff frequencies below the frequency of interest that also are compatible with the feed system designed U 4 EXAMPLE Please read the example of the parallelplate waveguide EXAMPLE Rectangular wave guide ECE 5130 Lecture 4 SMITH CHARTS Text Seem 2 4 Ponfoho Quesuon 1 Desenbe andDemonstxate howto use a Smnh Chan to nd1mpedance me Vmax SW R re ection coef c1entetc See Chap1e121gtmb1ems7e12 Smith Chan Circles Imaginary F Reneeuon coe ment 1 015 alw ysON 01mm eee1e Real F Tam F am 111 Chart fur Renemm Cnemeiem and and Impedance Re ecuon Coef ment and Load z are duectly related 1quot 2L 20 e 1 zLz 1 2L 7 1 zL 1 OR zLz zL 11quot1e1quotlt 39I39h1515NORNLALIZEDload1mpedance JxL 21 n are functions FT on the same chart Remember 1n 1 Capacmve top half XL 712 Induetme bottomhalf tFU L 12 rL 1 rL2 Example Gwen Z nd1quotusmg smtth Chan See txansparenmes Copxes to be made avalable m eopy room How to mar 1 Fmd Normahzedloadlm mee zL zLZo 11 1 XL er e modes for n and XL and PLOT 21 2 Draw lme from center of smtth ehaet to or through A 4 Read angle o rftom outstde of smtth e an wtt aprotractor and compare to lme on bottom ofsmxth ehaet or r39 1abe1ed Ref Coeff Zo 100 ohms ZL open circuit 1 ZL 00 00 0 2 PLOT far right 3 Draw Line through zL Read 40 4 Measure using a protractor or this one is obviously 1 l P l 1 F l A 0 which is what we expect for an open circuit ZL open circuit 1 2L 0 0 j 0 2 PLOT far left 3 Draw Line through zL Read 4180quot 4 Measure using a protractor or this one is obviously 1 l P l 1 F l 4 180 l which is what we expect for an short circuit ZL 100 j 0 ohms l ZLZLZ0lj0 2 PLOT center of smith chart 3 Draw Line through zL Not so easy A 4 Measure using a protractor or this one is obviously 0 l P l 0 F 0 A which is what we expect for a matched load ZL 100 j 100 ohms l ZLZLZ0ljl 2 PLOT top right quadrant 3 Draw Line through zL about 463 4 Measure using a protractor l P l 045 F 045 463 F zL l zL l 0jl 2jl 1490quot 2236 42656 045 46343quot How do you nd load impedance if given F 1 Plot F 2 Read zL zL j 2L 3 Unnormalize ZL zL Zo Admittance vs Impedance Admittance yL l zL F zL lzL 1lyL llyL l yL lyL l 180 out ofphase Steps to nd F from yL Find normalized yL Zo ZL g ij Plot it Using same curves gr and bX Transform it through the origin Rotate 180 degrees draw a line of equal length through the origin Now you have found zL 4 Read F as before WNH VVV EXAMPLE See transparencies Input Impedance ZinZ0 1 Fe391l31 1 Fe391l31 2zinzO 1Fe39j2l31 lFe39j2l31 Define re ection coefficient at the input NOT Pg as the re ection coefficient looking into the load frm the input location F1 FL AZ l This represents moving 2E1 radians towards the generator You can convert this distance to degrees and read it off the outer circles on the Smith Chart notice DIRECTION to the generator is marked OR Z l 2211 7t 1 411 l 7 This has been normalized for you on the outside circle around the Smith Chart Observe that if 1 this represents 2 complete rotations around the Smith Chart L M 2 represents one complete rotation Does this make sense For a Transmission line of length L M2 traveling from generator to the load and back would represent a phase shift of 360 degrees one complete rotation Thenzin11 1 lF1 How to find Zin 1 Normalize zL ZL Z0 2 Plot zL This also gives you FL 3 Rotate F distance 1 given in wavelengths TOWARDS the generator 4 Read zin and F1 5 Zin zin Zo EXAMPLE see transparencies Standing Wave Ratio To read SWR from the Smith Chart 1 PLOT 2L 2 Draw a circle through it 3 Read SWR from real aXis to right SWR Z l EXAMPLE See transparencies Voltage Minima and Maxima To read Voltage maXima off Smith Chart 1 PLOT 2L 2 First Voltage maximum occurs on right side of real axis First Voltage minimum occurs on left side of real aXis EXAMPLE See transparencies

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