7.6 Coupled-Line Directional Couplers

Transcription

1 4/9/7 7_6 Cupld Lin Dirctinl Cuplrs 1/ 7.6 Cupld-Lin Dirctinl Cuplrs Rding Assignmnt: pp Q: Th Qudrtur Hyrid is db cuplr. Hw d w uild cuplrs with lss cupling, sy 1dB, db, r db? A: Dirctinl cuplrs r typiclly uilt using cupld lins. HO: COUPLED LINE COUPLERS Q: Hw cn w dsign cupld lin cuplrs s tht is n idl dirctinl cuplr with spcific cupling vlu? A: HO: ANALYSIS AND DESIGN OF COUPLED-LINE COUPLERS Q: Lik ll dvics with qurtr-wvlngth sctins, cupld lin cuplr wuld sm t inhrntly nrrw nd. Is thr sm wy t incrs cuplr ndwidth? A: Ys! W cn dd mr cupld-lin sctins, ust lik with multi-sctin mtching trnsfrmrs. HO: MULTI-SECTION COUPLED LINE COUPLERS Q: Hw d w dsign ths multi-sctin cuplrs? Jim Stils Th Univ. f Knss Dpt. f EECS

2 4/9/7 7_6 Cupld Lin Dirctinl Cuplrs / A: All th rquisit dsign xmpls wr prvidd in th lst hndut, nd thr r tw gd dsign xmpls n pgs 45 nd 48 f yur txtk! Jim Stils Th Univ. f Knss Dpt. f EECS

3 4/9/7 Cupld Lin Cuplrs 1/4 Cupld-Lin Cuplrs Tw trnsmissin lins in prximity t ch thr will cupl pwr frm n lin int nthr. This prximity will mdify th lctrmgntic filds (nd thus mdify vltgs nd currnts) f th prpgting wv, nd thrfr ltr th chrctristic impdnc f th trnsmissin lin! Figur 7.6 (p. 7) rius cupld trnsmissin lin gmtris. () Cupld striplin (plnr, r dg-cupld). () Cupld striplin (stckd, r rdsid-cupld). (c) Cupld micrstrip. Gnrlly, spking, w find tht this trnsmissin lins r cpcitivly cupld (i.., it pprs tht thy r cnnctd y cpcitr): Jim Stils Th Univ. f Knss Dpt. f EECS

4 4/9/7 Cupld Lin Cuplrs /4 Figur 7.7 (p. 7) A thr-wir cupld trnsmissin lin nd its quivlnt cpcitnc ntwrk. If th tw trnsmissin lins r idnticl (nd thy typiclly r), thn C11 = C. Likwis, if th tw trnsmissin lins r idnticl, thn pln f circuit symmtry xists. As rsult, w cn nlyz this circuit using dd/vn md nlysis! ε r C 11 C C 1 C 1 Pln f cuplr symmtry Nt w hv dividd th C 1 cpcitr int tw sris cpcitrs, ch with t vlu C 1. Jim Stils Th Univ. f Knss Dpt. f EECS

5 4/9/7 Cupld Lin Cuplrs /4 Odd Md If th incidnt wv lng th tw trnsmissin lins r ppsit (i.., qul mgnitud ut 18 ut f phs), thn virtul grund pln is crtd t th pln f circuit symmtry. + - ε r C 11 C C 1 C 1 irtul grund pln Thus, th cpcitnc pr unit lngth f ch trnsmissin lin, in th dd md, is thus: C = C + C = C + C nd thus its chrctristic impdnc is: = L C Jim Stils Th Univ. f Knss Dpt. f EECS

6 4/9/7 Cupld Lin Cuplrs 4/4 Evn Md If th incidnt wv lng th tw trnsmissin lins r qul (i.., qul mgnitud nd phs), thn virtul pn pln is crtd t th pln f circuit symmtry. + + ε r C 11 C C 1 C 1 irtul pn pln Nt th C 1 cpcitrs hv n discnnctd, nd thus th cpcitnc pr unit lngth f ch trnsmissin lin, in th vn md, is thus: C = C = C 11 nd thus its chrctristic impdnc is: = L C Jim Stils Th Univ. f Knss Dpt. f EECS

7 4/9/7 Anlysis nd Dsign f Cupld Lin Cuplrs 1/11 Anlysis nd Dsign f Cupld-Lin Cuplrs A pir f cupld lins frms 4-prt dvic with tw plns f rflctin symmtry it xhiits D symmtry. prt β prt4 prt1 β C c prt As rsult, w knw tht th scttring mtrix f this furprt dvic hs ust 4 indpndnt lmnts: S S11 S1 S1 S41 S1 S11 S41 S 1 = S1 S41 S11 S1 S41 S1 S1 S 11 Jim Stils Th Univ. f Knss Dpt. f EECS

8 4/9/7 Anlysis nd Dsign f Cupld Lin Cuplrs /11 T dtrmin ths fur lmnts, w cn pply surc t prt 1 nd thn trmint ll thr prts: β s + - β C c Typiclly, cupld-lin cuplr schmtic is drwn withut xplicitly shwing th grund cnductrs (i.., withut th grund pln): β s + - β C c Jim Stils Th Univ. f Knss Dpt. f EECS

9 4/9/7 Anlysis nd Dsign f Cupld Lin Cuplrs /11 T nlyz this circuit, w must pply dd/vn md nlysis. Th tw circuit nlysis mds r: Evn Md Circuit s - + β 4 C c I= C c s β Odd Md Circuit s - + β 4 C c = C c s β Jim Stils Th Univ. f Knss Dpt. f EECS

10 4/9/7 Anlysis nd Dsign f Cupld Lin Cuplrs 4/11 Nt tht th cpcitiv cupling sscitd with ths mds r diffrnt, rsulting in diffrnt chrctristic impdnc f th lins fr th tw css (i..,, ). Q: S wht? A: Cnsidr wht wuld hppn if th chrctristic impdnc f ch lin whr idnticl fr ch md: = = Fr this situtin w wuld find tht: = nd = 4 4 nd thus whn pplying suprpsitin: = + = nd = + = indicting tht n pwr is cupld frm th nrgizd trnsmissin lin nt th pssiv trnsmissin lin. β = 4 = s + β Jim Stils Th Univ. f Knss Dpt. f EECS

11 4/9/7 Anlysis nd Dsign f Cupld Lin Cuplrs 5/11 This mks sns! Aftr ll, if n cupling ccurs, thn th chrctristic impdnc f ch lin is unltrd y th prsnc f th thr thir chrctristic impdnc is rgrdlss f md. Hwvr, if cupling ds ccur, thn, mning in gnrl: nd 4 4 nd thus in gnrl: = + nd = Th dd/vn md nlysis thus rvls th munt f cupling frm th nrgizd sctin nt th pssiv sctin! β 4 s + β Nw, ur first stp in prfrming th dd/vn md nlysis will t dtrmin scttring prmtr S 11. T ccmplish this, w will nd t dtrmin vltg 1 : = Jim Stils Th Univ. f Knss Dpt. f EECS

12 4/9/7 Anlysis nd Dsign f Cupld Lin Cuplrs 6/11 Th rsult is it cmplictd, s it wn t prsntd hr. Hwvr, qustin w might sk is, wht vlu shuld S 11? Q: Wht vlu shuld S 11? A: Fr th dvic t mtchd dvic, it must zr! Frm th vlu f S 11 drivd frm ur dd/vn nlysis, ICBST (it cn shwn tht) S 11 will qul t zr if th dd nd vn md chrctristic impdncs r rltd s: = In thr wrds, w shuld dsign ur cupld lin cuplr such tht th gmtric mn f th vn nd dd md impdncs is qul t. Nw, ssuming this dsign rul hs n implmntd, w ls find (frm dd/vn md nlysis) tht th scttring prmtr S 1 is: ( ) S1 = ct β + + ( ) Thus, w find tht unlss =, pwr must cupld frm prt 1 t prt! Q: But wht is th vlu f lin lctricl lngth β? Jim Stils Th Univ. f Knss Dpt. f EECS

13 4/9/7 Anlysis nd Dsign f Cupld Lin Cuplrs 7/11 A: Th lctricl lngth f th cupld trnsmissin lins is ls dsign prmtr. Assuming tht w wnt t mximiz th cupling nt prt, w find frm th xprssin v tht this is ccmplishd if w st β such tht: ct β = Which ccurs whn th lin lngth is st t: β = π λ = 4 Onc gin, ur dsign rul is t st th trnsmissin lin lngth t vlu qul t n-qurtr wvlngth (t th dsign frquncy). = λ 4 Implmnting ths tw dsign ruls, w find tht t th dsign frquncy: S1 = + This vlu is vry imprtnt n with rspct t cuplr prfrmnc. Spcificlly, it is th cupling cfficint c! c = + Jim Stils Th Univ. f Knss Dpt. f EECS

14 4/9/7 Anlysis nd Dsign f Cupld Lin Cuplrs 8/11 Givn this dfinitin, w cn rwrit th scttring prmtr S 1 s: S 1 = 1 c tnβ c + tn β Cntinuing with ur dd/vn md nlysis, w find (givn tht = : S 1 = 1 c 1 c cs β + sinβ nd s t ur dsign frquncy, whr β = π, w find: 1 c S = = 1 c 1 ( ) ( ) 1 c + 1 Finlly, ur dd/vn nlysis rvls tht t ur dsign frquncy: S 41 = Cmining ths rsults, w find tht t ur dsign frquncy, th scttring mtrix f ur cupld-lin cuplr is: Jim Stils Th Univ. f Knss Dpt. f EECS

15 4/9/7 Anlysis nd Dsign f Cupld Lin Cuplrs 9/11 S 1 c c 1 c c = c 1 c c 1 c Q: Hy! Isn t this th sm scttring mtrix s th idl symmtric dirctinl cuplr w studid in th first sctin f this chptr? A: Th vry sm! Th cupld-lin cuplr if ur dsign ruls r fllwd rsults in n idl dirctinl cuplr. If th input is prt 1, thn th thrugh prt is prt, th cupld prt is prt, nd th isltin prt is prt 4! prt (cupld) β prt4 (isltin) prt1 (input) β C c prt (thrugh) Jim Stils Th Univ. f Knss Dpt. f EECS

16 4/9/7 Anlysis nd Dsign f Cupld Lin Cuplrs 1/11 Q: But, hw d w dsign cupld-lin cuplr with spcific cupling cfficint c? A: Givn ur tw dsign cnstrints, w knw tht: = nd c = + W cn rrrng ths tw xprssins t find slutins fr ur dd nd vn md impdncs: 1+ c 1 c = = 1 c 1+ c Thus, givn th dsird vlus nd c, w cn dtrmin th prpr vlus f nd fr n idl dirctinl cuplr. Q: Ys, ut th dd nd vn md impdnc dpnds n th physicl structur f th cupld lins, such s sustrt dilctric ε r, sustrt thicknss (d r ), cnductr width W, nd sprtin distnc S. Hw d w dtrmin ths physicl dsign prmtrs fr dsird vlus f nd?? Jim Stils Th Univ. f Knss Dpt. f EECS

17 4/9/7 Anlysis nd Dsign f Cupld Lin Cuplrs 11/11 A: Tht s much mr difficult qustin t nswr! Rcll tht thr is n dirct frmultin rlting micrstrip nd striplin prmtrs t chrctristic impdnc (w nly hv numriclly drivd pprximtins). * S it s n surpris tht thr is likwis n dirct frmultin rlting dd nd vn md chrctristic impdncs t th spcific physicl prmtrs f micrstrip nd striplin cupld lins. * Instd, w gin hv numriclly drivd pprximtins tht llw us t dtrmin (pprximtly) th rquird micrstrip nd striplin prmtrs, r w cn us micrwv CAD pckgs (lik ADS!). * Fr xmpl, figurs 7.9 nd 7. prvid chrts fr slcting th rquird vlus f W nd S, givn sm ε r nd (r d). * Likwis, xmpl 7.7 n pg 45 prvids gd dmnstrtin f th singl-sctin cupld-lin cuplr dsign synthsis. Jim Stils Th Univ. f Knss Dpt. f EECS

18 4/9/7 MultiSctin Cupld Lin Cuplrs 1/8 Multi-Sctin Cupld Lin Cuplrs W cn dd multipl cupld lins in sris t incrs cuplr ndwidth. Figur 7.5 (p. 46) An N-sctin cupld lin l W typiclly dsign th cuplr such tht it is symmtric, i..: c = c, c = c, c = c, tc. 1 N N 1 N whr N is dd. Q: Wht is th cupling f this dvic s functin f frquncy? A: T nlyz this structur, w mk n pprximtin similr t tht f th thry f smll rflctins. Jim Stils Th Univ. f Knss Dpt. f EECS

19 4/9/7 MultiSctin Cupld Lin Cuplrs /8 First, if c is smll (i.., lss thn.), thn w cn mk th pprximtin: Likwis: S ( θ ) = 1 1 S ( θ ) = c tnθ 1 tn c + c tnθ 1 + tnθ = c sinθ θ 1 c θ 1 cs sin 1 csθ + sinθ = c θ + θ whr f curs θ = β = ωt, nd T = v p. W cn us ths pprximtins t cnstruct signl flw grph f singl-sctin cuplr: 1 1 c sinθ θ c sinθ θ c sinθ θ 4 c sinθ θ 4 Jim Stils Th Univ. f Knss Dpt. f EECS

20 4/9/7 MultiSctin Cupld Lin Cuplrs /8 Nw, sy w cscd thr cupld lin pirs, t frm thr sctin cupld lin cuplr. Th signl flw grph wuld thus : 1 c1 θ θ c θ θ c θ θ c1 θ θ c θ θ c θ θ 1 c1 sinθ θ c1 θ θ c sinθ θ c θ θ c sinθ θ c θ θ 4 4 Nt tht this signl flw grph dcupls int tw sprt nd grphs (i.., th lu grph nd th grn grph). 1 c1 θ θ c θ θ c θ θ c1 sinθ θ c θ θ c θ θ 4 4 c1 θ θ c sinθ θ c θ θ c1 θ θ c θ θ c θ θ 1 Jim Stils Th Univ. f Knss Dpt. f EECS

21 4/9/7 MultiSctin Cupld Lin Cuplrs 4/8 Nt tht ths tw grphs r ssntilly idnticl, nd mphsiz th symmtric structur f th cupld-lin cuplr. Nw, w r intrstd in dscriing th cupld utput (i.., ) in trms f th incidnt wv (i.., 1 ). Assuming prts, nd 4 is mtchd (i.., 4 = ), w cn rduc th grph t simply: 1 c1 θ θ c sinθ θ c θ θ c1 θ θ c θ θ Nw, w culd rduc this signl flw grph vn furthr r w cn us th multipl rflctin viwpint t xplicitly ch prpgtin trm! Q: Multipl rflctin viwpint! I thught yu sid tht this ws prticulrly d wy t prfrm ntwrk nlysis? A: Gnrlly spking it is, s w wuld hv t ccunt fr n infinit numr f trms. Hwvr, in crtin cnditins, ust fw trms dmint this infinit sris. If w cn crrctly idntify ths fw trms, w cn writ n xcllnt pprximtin t th xct slutin! An xmpl f tht ws th thry f smll rflctins, whr w nly cnsidrd trms invlving singl rflctin. Jim Stils Th Univ. f Knss Dpt. f EECS

22 4/9/7 MultiSctin Cupld Lin Cuplrs 5/8 Hr w n pply similr mthdlgy, pplying thry f smll cuplings. In thr wrds, w cnsidr nly th prpgtin pths whr n cupling is invlvd th signl prpgts crss cupld-lin pir nly nc! Nt frm th signl flw grph tht thr r thr such mchnisms, crrspnding t th cupling crss ch f th thr sprt cupld lin pirs: 1 c1 θ θ c sinθ θ c θ θ c1 sinθ θ c θ θ θ θ θ θ θ θ θ ( c1 sin θ c sinθ c sinθ ) θ θ 5θ ( c1 sin θ c sin θ c sin θ ) = + + Nt tht ll thr trms f th infinit sris wuld invlv t lst thr cuplings (i.., thr crssings), nd thus ths trms wuld xcding smll (i.., c ). Thrfr, ccrding t this pprximtin: θ θ S 1 ( θ) = ( θ ) = ( θ) = c1 sn i θ + c sin θ + c sin θ θ Mrvr, fr multi-sctin cuplr with N sctins, w find: Jim Stils Th Univ. f Knss Dpt. f EECS

23 4/9/7 MultiSctin Cupld Lin Cuplrs 6/8 S ( θ) c sinθ c sinθ c sinθ = θ θ 5θ 1 1 c sinθ N ( N 1) θ And fr symmtric cuplrs with n dd vlu N, w find: Nθ ( ) = sin cs( 1) + cs( ) S1 θ θ c1 N θ c N θ + c cs ( N 5) θ c M whr M ( N ) numr. = + 1. Nt M is n vn intgr, s N is n dd Thus, w find th cupling mgnitud s functin f frquncy is: c( θ) = S ( θ) 1 ( ) ( ) ( ) = c sinθ cs N 1 θ + c sinθ cs N θ 1 + c sinθ cs N 5 θ + + c sinθ M And thus th cupling in db is: C ( θ ) = c ( θ ) 1lg 1 Nw, ur dsign gls r t slct th cupling vlus c 1, c, cn such tht: Jim Stils Th Univ. f Knss Dpt. f EECS

24 4/9/7 MultiSctin Cupld Lin Cuplrs 7/8 1. Th cupling vlu C ( θ ) is spcific, dsird vlu t ur dsign frquncy.. Th cupling ndwidth is s lrg s pssil. Fr th first cnditin, rcll tht th t th dsign frquncy: θ = β = π I.E., th sctin lngths r qurtr-wvlngth t ur dsign frquncy. Thus, w find ur first dsign qutin: ( ) = ( ) + ( ) c θ c cs N 1 π c cs N π θ = π 1 ( ) + c cs N 5 π + + cm whr w hv usd th fct tht sin ( π ) = 1. Nt th vlu c ( θ ) θ = π is st t th vlu ncssry t chiv th dsird cupling vlu. This qutin thus prvids n dsign cnstrint w hv M-1 dgrs f dsign frdm lft t ccmplish ur scnd gl! T mximiz ndwidth, w typiclly imps th mximlly flt cnditin: Jim Stils Th Univ. f Knss Dpt. f EECS

25 4/9/7 MultiSctin Cupld Lin Cuplrs 8/8 d m c( θ ) d θ m θ = π = m = 1,,, M 1 B crful! Rmmr t prfrm th drivtiv first, nd thn vlut th rsult t θ = π. On finl nt, yu my find tht this trig xprssin is hlpful in simplifying yur drivtivs: Fr xmpl, w find tht: 1 1 sinφ csψ = sin φ + ψ + sin φ ψ ( ) ( ) ( ) ( ) ( θ) sin ( θ) ( θ) sin ( θ) sinθ csθ = sin θ + θ + sin θ θ = sin + = sin Jim Stils Th Univ. f Knss Dpt. f EECS