Chapter 2. Transmission Lines. ECE 130a. otan b. cos sin Z l Z Z l j Z l. b = Z Z j Z l

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1 Chapter Transmissin ines ECE 130a a f cs sin l l j l b b in - csbl j sin bl j l tan b j tan bl

2 Examples f Transmissin ines: (Chapter ) Parallel Strip ine: metal cnductrs w d Caxial ine: metal radius b metal radius a dielectric Parallel Wire ine: Air ine D Dielectric ine ε r Micrstrip ine: Metal cnductr radius a w d Dielectric There is a simple way t iew the guided wae n a transmissin line. Vzt a, f z Izt, Izt af, af

3 The ptential difference (ltage) between the metal cnductrs with equal and ppsite current flwing in them are circuit cncepts, except they depend nt nly n time, but als n the distance z. S we describe the wae as ltage and current waes. itz a, f generatr tz a, f - ength >> π/ β λ lad Other guiding structures: 1. Waeguides -- cnsist f a single hllw metal tube f arius crsssectinal gemetry. An EM wae prpagates lngitudinally inside the hllw structure. The wae prpagatin in waeguides is nt transerse (nt TEM). That is, it has lngitudinal field cmpnent(s). Transerse spatial dependence is fairly cmplicated. The prpagatin cnstant β f a waeguide wae is nt equal t that f plane waes, and elcity f prpagatin thus is nt the same as light.. Optical Fibers -- are used at ptical frequencies, at infrared, and at isible waelengths. An ptical fiber cnsists f a ery thin ( mm) dielectric circular crss sectin cylinder. The material is usually glass r plastic. Inner and uter prtins hae different dielectric cnstants (index f refractin), as shwn belw. cre radius a, index n 1 cladding, index n Optical fibers d nt supprt TEM waes, like the hllw metallic guides. Their prpagatin cnstant and mde structures are een mre cmplicated than fr hllw metallic guides.

4 Deriatin f Transmissin ine Equatins (1-3) et us cnsider a length f a transmissin line at lcatin z. The circuit mdel is clearly a series inductance and resistance (since the whle line with a lad at the end frms a lp) and a shunt capacitance and shunt leakage cnductance (between the gd cnductrs, acrss the dielectric which als has a small cnductance). With the abe circuit parameters being, R, C, and G, the mdel is izt af, R iz a D zt, f zt a, f C G z a D zt, f A straight frward applicatin f Kirchff s p aw gies z zt zt z izt, a D, f - a, f af -D -RDziazt, f, t and Kirchff s Current aw at the upper nde gies z Dz, t i z Dz, t - i z, t - G Dz z Dz, t - C Dz t Diiding thrugh by z and taking the limit z 0, we get: a f a f a f a f i - -Ri (1-5) z t i -C -G (1-6) z t Sinusidal Analysis f Transmissin ines (using phasrs) - arjwfi (1) z i - agjwcf () z Differentiate (1) with respect t z : di - ar jwf z dz.

5 Substitute frm () i/ z : G jwc R jw z Similarly, we get: i ag jwcfar jwfi z The ssless Transmissin ine R 0 (perfect cnductr) G 0 (lssless dielectric) This is nt nly instructie and simple, but als a gd apprximatin f many real lines which are made f ery gd cnductrs (typically, cpper) and nearly lssless dielectrics (e.g., tefln). Fr example, a cmmercial caxial line is a nearly lssless line. jwc jw z -w c Ch z b z - V I - z t The slutins are traelling waes: a fa f e ± jβ z (with e jω t e j wt- bz β ω C ω µε It will be shwn later, when we cnsider transmissin lines frm the Electrmagnetic Field s pint f iew, that these ltage and current waes crrespnd t EM waes, als. It will be shwn that: 1 C µε, where is elcity f light in the medium between the cnductrs. i Als, w i z - C i b i z - I V -C z t already assumed) a f r e jawt bzf r sum f bth. c h

6 We can get the relatinship f V and I by substituting back int the riginal transmissin line equatins. - Ve jawt bzf - i Ie jawt bzf (i.e., nly z ging wae) d di jωi jωc dz dz jβ jωi jβi jωc i ω β β ω C i C β ω C ω µε, This is fr z ging wae nly. is called the characteristic impedance f the line. Fr this lssless line, is real. It means that V and I are in phase. Fr example, pwer is prpagated. Nne is absrbed. (Cntrast this with rdinary circuits, where real impedance means pwer is absrbed and dissipated!) Pwer Prpagated n a ssless ine The pwer prpagated can be calculated either frm the electrmagnetic wae r frm the ltage-current wae. It is usual t use ltage and current fr a transmissin line. The pwer, f curse, is an scillating quantity. We are mstly interested, hweer, in the aerage pwer flw. If (real), then frm the circuit picture: V Pa Re 1 1 i* fr lssless line If we hae nly a z ging wae, the same kind f deriatin we did fr the z ging wae gies: i ( z ging wae) Circuits s. Transmissin ines 1) Q: When des the regular circuit (E17) methd hld crrect; that is, nly e jωt dependence (n prpagatin) and when des the transmissin line methd need t be used?

7 A: Simple. When the waelength is large cmpared t the dimensins, we can neglect transmissin line cncepts: π β z z 0 λ That is, wheneer the distance scale (z) is small and/r the waelength is lng; that is, z << λ. ) Fr circuits: If is real (that is, R), Pa 1 I R. This is pwer absrbed in the real impedance (resistance). Fr a lssless transmissin line: is real. If we hae nly ging wae: 1 1 Pa i This is nt absrbed pwer. It is pwer prpagating in the z directin dwn the lssless line. R Pwer is absrbed in the lad t ad where it des smething. If we hae - ging wae: Pa Re i* i The negatie sign means that pwer is prpagating in the negatie directin. Unless therwise stated, we always will mean aerage pwer. Transmissin ine Circuits Transmissin lines are used t carry signals frm ne lcatin t anther. Thus, they usually hae a signal generatr at ne end and a lad at anther end. et s nw examine the behair f a transmissin line with lad. We use phasrs and will assume lssless transmissin line. 1) Transmissin line terminated in its characteristic impedance. This is called matched. et s assume we launched a z ging wae n this transmissin line f characteristic impedance. Then, at any pint z, we knw that / i. At the lad, / i. (This is definitin f.) Thus, if

8 , then ( ) and i i ( ). Thus, there is n reflected wae ( z ging wae). Rather, the wae cntinues int the lad, where it gets dissipated. (In practical situatins, the lad is a whle circuit, fr example, a detectr, a radi, an scillscpe. As lng as its input impedance is, the freging discussin is true.) ) Transmissin line terminated in ther impedance (1-7, text) Assume again we launched a signal at sme frequency. Then, / i. But at the lad, / i. If, cannt be and i cannt be i. The bundary cnditin at the lad can nly be satisfied if there is als a reflected wae, i. This wae prpagates backward in the z directin. Then, at any pint n the line: V e V- e V - - i - i i e jawt bzf V e jawt bzf Nte the sign n the current f the reflected wae! Nw we can satisfy the cnditin at the lad i /. Only ne alue f V will satisfy. et us define: Then: Γ (1) Γ a f a f j wt- bz j wt bz - We sle fr Γ t get: (Prblem: Sle t G ) Γ - Γ, the reflectin cefficient () Nte: Γ is cmplex, in general, because may be cmplex. - 1 G 1 i

9 Fractin f the incident pwer reflected back: Incident Pwer P 1 * N M O V QP 1 N M O QP 1 Re i Re / / Reflected Pwer P r 1 * N M O QP 1 * * Re N M O -i- Re G- G i QP a - 1f 1 Re - G O QP - G i* P i Ttal Pwer Nte: 3) Special Cases f Terminatin i NM N M NM P T Re T T 1 Γ Transmitted pwer (int ad) is purely imaginary. A special case f the general lad is 0, that is, a shrt circuited transmissin line. (A transmissin line may be shrted deliberately in rder t achiee a certain result, r it may be shrted accidentally, smething that has t be repaired.) Fr this case: Γ ~ We can say here that jϕ jπ Γ Γ e 1 e here. - If Ae j wt bz, - jbz jbz z Ae Ae e jw t understd Nte that this time, the z dependence has tw different parts. Thus, the z dependence cannt be lumped tgether with ω t dependence, as it can be O NM a f e j O 1 QP Re i 1 1 G 1- G i e Re G G GG 1 G- G * e j Im G P i * * * * * * a f f i z af - 1 jbz jbz af - Ae Ae c e j j i QP O QP

10 fr a wae traeling in ne directin nly. Nw, the entire V r I, including z dependence, is the phasr: af- z jasinb z A γ iaf c z e jw t understd cs b z The real ltage and current are ( fr A A e j θ ) a f R ST jw tu VW f j Ae i z, t Re q csb ze A csbz csawt q jwt a f m r z t j A e j q, Re - sinb z e A sin bzsin wt q a f V I z Cnductr Bth ltage and current scillate sinusidally in time with different maximum alues (amplitudes) at different lcatins. This is knwn as a standing wae-- standing, because the amplitude remains the same at each lcatin and the scillating pattern is standing. Cntrast this with traelling wae, in which a gien pint n the wae frm prgresses in distance with time. The amplitude f a traelling wae, hweer, is a cnstant alue regardless f lcatin. The instantaneus pwer flw is (must be calculated frm real V, I): a f a fa f Pzt, zt, i zt, 4 A sin bz cs bzsin wt q cs wt q A sin bzsin wt q a a f a f f

11 The aerage pwer flw (we calculate it here directin frm phasrs) is: Re NM This makes physical sense, since n pwer flws int the shrt circuit. Thus, all the pwer must be reflected back, giing net pwer flw equal t zer (aerage). The same result, f curse, culd be deried frm the field pint f iew. The shrt circuit is a perfectly cnducting plate clsing ff the line. N fields penetrate int it. Thus, The Cncept f Input Impedance ( ) Characteristic impedance directin. O N M 1 * 1 Re A VI jasin z QP - b cs bz Re ja sinβzcsβz 0 is the rati f ltage t current f ne wae Since transmissin lines are used in cnjunctin with lumped circuits, we must be able t treat them as regular circuits as well. The input impedance f any circuit is in V/ I at the terminals. Thus, fr a transmissin line, at any pint n it: (purely imaginary) P a ep z ep z 0 in i i * O Q P in T i T, where T and i T are the ttal phasr fields at that pint. Imprtant: Since in T / it, this is nw the same as a regular circuit cncept. S we can always replace a transmissin line circuit with its input impedance! If there is nly a single wae prpagating in ne directin (transmissin line terminated in ), then, f curse, in. But, in general, in, and in is a functin f bth the lad and f the psitin z n the line. et s measure as fllws: 0 at the lad. Therefre, z is negatie n the line z l.

12 Fr the shrt circuited lssless transmissin line: -l ja l ina- lf a f - sin ba a - f i -lf A csb -l a f a f tan tan tan p p f in - l j bl j l j l l The input impedance f a shrt circuited lssless line is purely imaginary (as it shuld be, since it cnsists f distributed capacitance and inductance.) The alues f the input impedance ary with the distance if the frequency is fixed r aries with frequency if the length is cnstant. z z l z 0 (fr a shrted line) f l Reactance can be either capacitie r inductie, and its abslute alue aries frm zer t infinity. This prperty can be utilized in a number f ways. (a) Easily adjustable alues f reactance. (b) catin f an accidental shrt n a line. (Prblem) (c) Matching f a lad nt equal t (will study later) Open Circuited ssless ine In practice, we try t aid pen circuiting a transmissin line. Reasn: Theretically, pen circuit (i.e., pen ends) means infinite lad impedance ( R ). This wuld mean that: Γ 1 That is, all f the wae is reflected, resulting in standing waes. Hweer, in practice, this is nt really true. The pen end is nt a truly infinite impedance. While n current flws beynd the pen end, sme f the electrmagnetic wae ges ut the pen end and becmes a free space

13 wae. Only a small fractin des, since the pen circuited transmissin line is prly matched t space (i.e., it is a pr antenna). Yet, if nt all the pwer is reflected, we biusly d nt hae infinite impedance. ssless transmissin line terminated in a real lad R π 01 A 01 0 R R 01 A The abe tw cases are identical as far as T.. 1 is cncerned, since the impedance at pint A is fr bth. R We can always replace any transmissin line by its equialent input impedance. The fields up t that input pint will be unchanged. Fr the abe, since : 0 01 Γ R R Γ is real and its alue aries frm 1 t 1. 1 A 0 R R ine 1 has bth z and z ging wae. ine has nly z ging wae, since it is prperly terminated. Just t the left f junctin A: 1 1 Just t the right f junctin A: These are biusly equal. S we replace at A by R : R in A τ Γ 1

14 τ is called the transmissin cefficient. Thus: τ 1 τ 1 Γ τ can be greater than unity, but that des nt mean mre than 100% f pwer transmitted. Pwer balance: Pr Pt Pi 0 t 1 1 Pr Γ Γ Pi τ 1 1 Pt τ Substituting fr Γ and τ and writing the sum f the reflected and transmitted pwer incident: Example: 1 MF HG I NM KJ QP b g b Cmmercial caxial line (Radi Shack) is 50Ω, flat antenna line is 300Ω. The abe situatin wuld happen if yu cnnect them tgether withut a matching transfrmer. (We shall learn later what a matching transfrmer is.) At the junctin Az a 0f: in Γ in τ 1 Γ The fractin f pwer reflected is Γ The fractin f pwer transmitted t 300Ω line and n t the lad is g F HG IO KJ P 1 1 Q.E.D. in 300Ω z 0 50Ω 300Ω 300Ω A

15 General Transmissin ine Nw let s return t a general case and find the ttal ltage and current n the line as a functin f distance: - jb z jb z - jb z jb z 1 V 1 e GV 1 e V e c1 Ge h (et V 1 V ) V 1 1 i e jb z V e jb z V e jb z e jb z G c G h e jω t ariatin understd! We hae drpped The real fields, f curse, ary sinusidally in time. 1 and i 1 are the cmplex phasr amplitudes. Since, fr a cmplex ltage Ae jω t, the maximum alue (r sinusidal amplitude) is A, and the sinusidal amplitude f V 1 alng the line is: - jbz jbz 1 V e 1 Ge V 1 G csbz jsin bz (fr G real) Remember that Γ may be psitie r negatie. 4π βz z λ The minimum alue f this is (if Γ is psitie): a e jω t V 1 Gcsbz G sin bz V 1 G Gcsbz a f V 1 1 G - G V 1 - min G Fr a general Γ : V min b G Similarly: V 1 1 V 1 max G G G The lcatin f cnsecutie minima and maxima are λ/ apart. At the lad, we hae either a maximum r a minimum (max. fr Γ psitie). The shape f the functin s. z is nt a pure sinusid. The minima are nt infinitely sharp (as they are fr a shrt circuit). a f f b. g g ad z

16 An A.C. ltmeter, f curse, des nt measure the time ariatin, but nly the amplitude (tp slid line). (Actually the RMS alue) This is called a partial standing wae. An imprtant measurable quantity is the Standing Wae Rati (SWR) S. S can range frm 1 t. S V 1 max 1 V min Example: Fr the preius example, S We nte als that fr Γ psitie ( >, real), 1 S 1 Γ Γ and fr Γ negatie ( <, real), 1 S 1 This als agrees with ur example: If a transmissin line is cnnected t an unknwn lad, the SWR can be measured and, frm this, the lad alue fund. But t fully understand hw, we must cnsider a mre general lad. ssless Transmissin ine Terminated in a General Cmplex ad Mst practical circuits/deices (whether the deice be a transmitting antenna, a TV, a receier, a micrwae amplifier, an scillscpe, etc.) hae cmplex input impedance. Thus, if such a deice is fed by a transmissin line, the transmissin line sees a cmplex lad. Withut smething special being dne, the line is nt matched t the lad. There is a reflected wae / 50

17 Γ Assume fr simplicity that V 0 and that the lad is at 0. Therefre, n the transmissin line, z l. Then, i V j l j l T, e β V e β Γ where l is the distance measured backwards frm the lad. The input impedance at a pint l distance back frm the lad is: Substituting the alue f When, we get. Nw returning t the ltage and current: jβz jβz T Ve ΓVe V V i j z j z T e β e β Γ Γ jβl T Ve (cmplex, in general) ΓVe e β Γ e in jβl e Γ e and rearranging, we get the alternate frms: This is the same expressin as befre ( z l), except nw Γ is cmplex Γ Γ θ r. It is easiest t see hw this changes the standing wae by cnsidering phasr diagrams. The amplitude f the quantity in square brackets determines the amplitude f the standing wae, since e jβ l 1. jβl j l jβl 1 Γ e 1 Γ e a f cs sin jβl jβl jβl l l j l b b in - cs bl j sin bl in jβl Ve 1 Γ e V i e jβl 1 Γ e jβl jβl j l tan b j tan b l 1 Γ e in 1 Γ e anther frm jβl jβ l

18 Definitin f Generalized Reflectin Cefficient j l V () l V e G() l G e V () l - b - -( 0) - b ( 0) jbl V ( 0) e af G G 0 In my ntatin, reflectin cefficient at the lad (cmplex quantity ~ in general) is equialent t the bk s Γr. Nte: 1) Γ is real if lad is real and line is lssless. Γ is cmplex if lad is cmplex. ) Gaf l is always cmplex, except at certain pints. These pints hae special significance! 3) Galf G, that is, ming abut n the line nly changes the phase angle f Gaf l. Im At l 0 (ming back, away frm lad) 1Γ V min. Γ V max. Re Max. Γ 1 Γ j l z -l It is bius frm this phasr diagram that the cases f real lad and cmplex lad are nt fundamentally different. θ r 0 r 180 fr a real lad, but can be anything fr a cmplex lad. The phasr rtates as l increases, making a cmplete cycle fr βl π ; that is, 4πl / λ π, r l λ /. V max is still 1Γ and V min 1 Γ, where Γ Γ. Obiusly, the shape f the standing wae is als the same as fr a real lad, except the maximum r minimum n lnger ccurs at the lad. catin f maxima and minima: Maximum ccurs at: θr βlmax 0, ± π, ± 4π, etc. Similarly, θr βlmin ± π, ± 3π, etc.

19 Example: An RF signal is sent n a 300Ω parallel wire transmissin line, which has ε r 5, t a receier. A VSWR f S 3 is measured n the line. The distance between tw minima is 067. m. and the nearest minimum t the lad minimum is 045. m. away. Find the frequency f the signal and the The nenin equialent circuit f the receier input m m. 300Ω S 3 Slutin: λ m / λ m 134. λ λm εr m. c f MHz λ 3 π π 1 β 149. π m λ m 134. βlmin θr π 98. πl m π θ r. 34π θ 1 S r 0. 34π 61. Γ 1 Γ Γ S 1 S Γ j l 300 Γ Gf 300a1 Gf a j0. 44f 300a1. 4 j0. 44f 14. j j j34 Ω R jω h 8 π 10 ad is R 9Ω 054. µ h ANS.

20 in and at the rtating phasr diagram f the pre- king at the definitin f ius page, we als see that: At the lcatin f a ltage maximum: 1 Γ in S 1 Γ At the lcatin f a ltage minimum: 1 - G in 1 G S in S in at ltage minimum S This is an imprtant bseratin. It will be used sn fr matching. Matching a transmissin line t an unmatched lad Reflectins n a transmissin line are undesirable fr a number f reasns. Fr example, digital signals frm PCM reflected pulses can bunce back again frm the sending end if that end is als unmatched. Then false signals can be detected. Q: Hw d we match an unmatched lad? A: There are seeral methds. We nw study the first ne. I. Quarter Wae Matching ( λ /4 Transfrmer ) at ltage maximum Suppse we hae a transmissin line terminated in a real lad in 0 If the length f this line is exactly l λ/4, then in, since 0 π λ π sin βl sin sin 1 λ 4 and csβl 0 0 is intrinsic impedance. b R g

21 This then ffers a methd t match this lad t any lssless transmissin line 01, intrinsic impendance 0 t be matched t. real br g A λ 4 We insert a transmissin line f λ /4, length ( ), and haing Then, at ( A ): 0 01 in 01! Since in A 01, we are matched. N reflectin, since at pint A, we culd replace the whle cmbinatin by in A. 01 in 01 Hw is this pssible physically? We wn t g int a detailed analysis f what really happens t the right f pint A. But briefly, we d hae (in actual physical fact) reflectin f waes at pint A and at. The tw reflectins, hweer, cancel each ther ut n the line 1. Q: This biusly wrks if the lad is real, since 01 is als real, resulting in real 0. But if the lad is cmplex, then what t d? A: Recall that the input impedance at a ltage minimum r maximum is real S and / S. Thus, these are the steps fr matching a cmplex lad t a line 01. 1) Add a sectin f 01 line t the lad. The length shuld be l max r l min f the standing wae that results. Say, we use : l max l max This is equialent t a lad f S. 01

22 ) Nw use λ /4, matching n this equialent lad. Fr example, insert a line: S S, length λ / B λ /4 A 01 l max The impedance at pint B then is: S S 01 in B 01 S 1 Thus we are matched. N reflectin ging back n line 1. Same cmment applies as befre. Actually, there are reflectins at pints A and B, but they are 180 ut f phase and cancel ut. The 50W - 300W transfrmer supplied fr TV and ide cmpnents is a lumped circuit simulatin f a λ /4 transfrmer! π λ π Imprtant Nte: l λ /4 means βl λ 4 π f π The matching is nly alid at ne frequency! Since β, if f is ν λ changed, the waelength is changed and βl π/ fr the same l. Example: (a) Match a 50 caxial cable t a 300W real input impedance deice at a frequency f 180 MHz. Assume all transmissin lines used hae ε r 6. (b) Find the reflectin cefficient f the matched circuit at 90 MHz

23 (a) At 180 MHz: 8 c 3 10 λ 167. m. 8 f λ λ n the T.., λ m 068. m. 6 T.. impedance needed Ω ans length needed m. ans. 50Ω 1. 5Ω 300Ω We cannt readily buy a 1.5W transmissin line. Hweer, any transmissin line at these R.F. frequencies can be simulated by a prper lumped circuit netwrk. (In this class, we wn t study hw.) (b) If the frequency is changed t 90 mhz: We hae: λ 017. m. 4 λ m π π βl ina 50Ω 1. 5Ω l λ 8 300Ω 300 j j j Reflectin cefficient at A is: j j Γ j j ans Fractin f pwer reflected back at A is Γ π βl 4 tan βl 1