The Frequency Response of a Quarter-Wave Matching Network

Transcription

1 4/1/29 Th Frquncy Rsons o a Quartr 1/9 Th Frquncy Rsons o a Quartr-Wav Matchg Ntwork Q: You hav onc aga rovidd us with conusg and rhas uslss ormation. Th quartr-wav matchg ntwork has an xact SFG o: a Τ j β b Τ j β Usg our rduction ruls, w can quickly conclud that: b a Τ 2 j 2β = + 1 You could hav lt this siml and rcis analysis alon BUT NOOO!! You had to oist uon us a long, ramblg discussion o th roagation sris and dirct aths and th thory o Jim Stils Th Univ. o Kansas Dt. o EECS

2 4/1/29 Th Frquncy Rsons o a Quartr 2/9 small rlctions, culmatg with th aroximat (i.., lss accurat!) SFG: a j β b j β From which w wr abl to conclud th aroximat (i.., lss accurat!) rsult: b = + a j 2β Th xact rsult was siml and xact! Why did you mak us dtrm this aroximat rsult? A: In a word: rquncy rsons*. Although th xact analysis is about as siml to dtrm as th aroximation rovidd by th thory o small rlctions, th mathmatical orm o th rsult is much simlr to analyz and/or valuat (.g., no ractional trms!). Q: What xactly would w b analyzg and/or valuatg? A: Th rquncy rsons o th matchg ntwork, or on thg. * OK, two words. Jim Stils Th Univ. o Kansas Dt. o EECS

3 4/1/29 Th Frquncy Rsons o a Quartr 3/9 Rmmbr, all matchg ntworks must b losslss, and so must b mad o ractiv lmnts (.g., losslss transmission ls). Th imdanc o vry ractiv lmnt is a unction o rquncy, and so too thn is. Say w wish to dtrm this unction ( ω). Q: Isn t ( ω) = or a quartr wav matchg ntwork? A: Oh my gosh no! A rorly dsignd matchg ntwork will tyically rsult a rct match (i.., = ) at on rquncy (i.., th dsign rquncy). Howvr, i th signal rquncy is dirnt rom this dsign rquncy, thn no match will occur (i.., ). Rcall w discussd this bhavior bor: Figur 5.12 (. 243) Rlction coict magnitud vrsus rquncy or a sglsction quartr-wav matchg transormr with various load mismatchs. Jim Stils Th Univ. o Kansas Dt. o EECS

4 4/1/29 Th Frquncy Rsons o a Quartr 4/9 Q: But why is th rsult: Τ 2 j 2β = + 1 or its aroximat orm: = + j 2β dndnt on rquncy? I don t s rquncy variabl ω anywhr ths rsults! A: ook closr! Rmmbr that th valu o satial rquncy β ( radians/mtr) is dndnt on th rquncy ω o our ign unction (aka th signal ): 1 β = ω v whr you will rcall that v is th roagation vlocity o a wav movg along a transmission l. This vlocity is a constant (i.., v = 1 C ), and so th satial rquncy β is dirctly roortional to th tmoral rquncy ω. Thus, w can rwrit: Jim Stils Th Univ. o Kansas Dt. o EECS

5 4/1/29 Th Frquncy Rsons o a Quartr 5/9 β ω = = ωt v Whr T = v is th tim rquird or th wav to roagat a distanc down a transmission l. As a rsult, w can writ th ut rlction coict as a unction o satial rquncy β : ( β ) = + j 2β Or quivalntly as a unction o tmoral rquncy ω : ( ω) = + j 2ωT Frquntly, th rlction coict is simly writtn trms o th lctrical lngth θ o th transmission l, which is simly th dirnc rlativ has btwn th wav at th bgng and nd o th lngth o th transmission l. β = θ = ωt So that: j 2θ θ = + ( ) Not w can simly srt th valu θ = β to th xrssion abov to gt ( β ), or srt θ = ωt to th xrssion to ω. gt ( ) Jim Stils Th Univ. o Kansas Dt. o EECS

6 4/1/29 Th Frquncy Rsons o a Quartr 6/9 Now, w know that = or a rorly dsignd quartrwav matchg ntwork, so th rlction coict unction can b writtn as: j 2θ θ = 1 + ( ) ( ) Not that: 1 j j( θ θ) jθ + jθ = = = And that: j 2 θ j( θ + θ ) j θ j θ = = And so: j 2θ ( θ ) = ( 1 ) jθ + jθ jθ jθ = ( + ) jθ + jθ jθ = ( + ) + = jθ ( 2cosθ ) Whr w hav usd Eulr s quation to dtrm that: jθ + = 2cosθ + jθ Now, lt s dtrm th magnitud o our rsult: Not that ( ) θ j ( θ ) θ 2cosθ 2 cosθ = = is zro-valud only whn cosθ =. This o cours occurs whn θ = 2 : ( ) = 2 cos = θ θ = 2 2 Jim Stils Th Univ. o Kansas Dt. o EECS

7 4/1/29 Th Frquncy Rsons o a Quartr 7/9 In othr words, a rct match occurs whn θ = 2!! Q: What th hck dos this man? A: Rmmbr, θ = β. Thus i θ = 2: θ λ 2 = = = β 2 λ 4!! As w (should hav) susctd, th match occurs at th rquncy whos wavlngth is qual to our tims th matchg ( Z 1) transmission l lngth, i.. λ = 4. In othr words, a rct match occurs at th rquncy whr = λ 4. Not th hysical lngth o th transmission l dos not chang with rquncy, but th signal wavlngth dos: λ = Q: So, at rcisly what rquncy dos a quartr-wav transormr with lngth rovid a rct match? A: Rcall also that θ = ωt, whr T = v. Thus, or θ = 2: v v 1 θ = = ωt ω = = 2 2T 2 Jim Stils Th Univ. o Kansas Dt. o EECS

8 4/1/29 Th Frquncy Rsons o a Quartr 8/9 This rquncy is calld th dsign rquncy o th matchg ntwork it s th rquncy whr a rct match occurs. W dnot this as rquncy ω, which has wavlngth λ, i..: v v ω 1 ω = = = = = 2T 2 2 4T 4 λ v = = 4vT = 4 Givn this, yt anothr way o xrssg θ = β is: ω v ω θ = β = = = v 2ω 2ω 2 Thus, w conclud: ( ) 2 cos( 2 ) = From this rsult w can dtrm (aroximatly) th bandwidth o th quartr-wav transormr! First, w must d what w man by bandwidth. Say th maximum acctabl lvl o th rlction coict is valu m. This is an arbitrary valu, st by you th microwav ngr (tyical valus o m rang rom.5 to.2). W will dnot th rquncis whr this maximum valu m occurs m. In othr words: m ( ) 2 cos( 2 ) = = = m m Jim Stils Th Univ. o Kansas Dt. o EECS

9 4/1/29 Th Frquncy Rsons o a Quartr 9/9 Thr ar two solutions to this quation, th irst is: 2 1 m m 1 = cos 2 And th scond: 2 1 m m 2 = cos 2 Imortant not! Mak sur -1 cos x is xrssd radians! You will d that < < so, th valus and d m1 m2 m1 m 2 th lowr and ur limits on matchg ntwork bandwidth. m m 1 m 2 All this analysis was brought to you by th siml mathmatical orm o ( ) that rsultd rom th thory o small rlctions! Jim Stils Th Univ. o Kansas Dt. o EECS