5.6 Binomial Multi-section Matching Transformer

Transcription

1 4/14/21 5_6 Bioial Multisectio Matchig Trasforers 1/1 5.6 Bioial Multi-sectio Matchig Trasforer Readig Assiget: pp Oe way to axiize badwidth is to costruct a ultisectio Γ f that is axially flat. atchig etwork with a fuctio ( ) Q: Maxially flat? What kid of fuctio is axially flat? This fuctio axiizes badwidth by providig a solutio that is axially flat. A: HO: MAXIMALLY FLAT FUCTIOS 1. We ca build a ultisectio atchig etwork such Γ f is a bioial fuctio. that the fuctio ( ) 2. The bioial fuctio is axially flat. Q: Meaig? A: Meaig the fuctio Γ ( f ) is axially flat a widebad solutio! HO: THE BIOMIAL MULTI-SECTIO MATCHIG TRASFORMER

2 4/15/21 Maxially Flat Trasforer Fuctios.doc 1/5 Maxially Flat Fuctios Cosider soe fuctio f ( x ). Say that we kow the value of the fuctio at x =1 is 5: f ( x = 1) = 5 This of course says soethig about the fuctio f ( x ), but it does t tell us uch! We ca additioally deterie the first derivative of this fuctio, ad likewise evaluate this derivative at x =1. Say that this value turs out to be zero: df( x) dx x = 1 = ote that this does ot ea that the derivative of f ( x ) is equal to zero, it erely eas that the derivative of f ( x ) is zero at the value x = 1. Presuably, df( x) at other values of x. dx is o-zero So, we ow have two pieces of iforatio about the fuctio f ( x ). We ca add to this list by cotiuig to take higherorder derivatives ad evaluatig the at the sigle poit x =1. Let s say that the values of all the derivatives (at x =1) tur out to have a zero value:

3 4/15/21 Maxially Flat Trasforer Fuctios.doc 2/5 df ( x) dx x = 1 = for = 123,,,, We say that this fuctio is copletely flat at the poit x=1. Because all the derivatives are zero at x =1, it eas that the fuctio caot chage i value fro that at x =1. I other words, if the fuctio has a value of 5 at x =1, (i.e., f ( x = 1) = 5), the the fuctio ust have a value of 5 at all x! The fuctio f ( x ) thus ust be the costat fuctio: f ( x ) = 5 ow let s cosider the followig proble say soe fuctio f ( x ) has the followig for: 3 2 f ( x) = a x + bx + cx We wish to deterie the values a, b, ad c so that: f ( x = 1) = 5 ad that the value of the fuctio f ( x ) is as close to a value of 5 as possible i the regio where x =1. I other words, we wat the fuctio to have the value of 5 at x =1, ad to chage fro that value as slowly as possible as we

4 4/15/21 Maxially Flat Trasforer Fuctios.doc 3/5 ove fro x =1. Q: Do t we siply wat the copletely flat fuctio f ( x ) = 5? A: That would be the ideal fuctio for this case, but otice that solutio is ot a optio. ote there are o values of a, b, ad c that will ake: + + = ax bx cx for all values x. Q: So what do we do? A: Istead of the copletely flat solutio, we ca fid the axially flat solutio! The axially flat solutio coes fro deteriig the values a, b, ad c so that as ay derivatives as possible are zero at the poit x=1. For exaple, we wish to ake the first derivate equal to zero at x =1: df( x) = dx x = 1 2 ( 3ax 2bx c ) = + + = 3a + 2b + c x = 1

5 4/15/21 Maxially Flat Trasforer Fuctios.doc 4/5 Likewise, we wish to ake the secod derivative equal to zero at x =1: 2 d f ( x) = 2 dx x = 1 ( 6ax 2b ) = + = 6a + 2b Here we ust stop takig derivatives, as our solutio oly has three degrees of desig freedo (i.e., 3 ukows a, b, c). x = 1 Q: But we oly have take two derivatives, ca t we take oe ore? A: o! We already have a third desig equatio: the value of the fuctio ust be 5 at x =1: ( ) ( 1) 3 ( 1) 2 ( 1) 5 = f x = 1 = a + b + c = a + b + c So, we have used the axially flat criterio at x =1 to geerate three equatios ad three ukows: 5 = a + b + c = 3a + 2b + c = 6a + 2b

6 4/15/21 Maxially Flat Trasforer Fuctios.doc 5/5 Solvig, we fid: a = 5 b = 15 c = 15 Therefore, the axially flat fuctio (at x =1) is: 3 2 f ( x) = 5x 15x + 15x f(x) x

7 4/15/21 The Bioial Multisectio Matchig Trasforer.doc 1/17 The Bioial Multi- Sectio Trasforer Recall that a ulti-sectio atchig etwork ca be described usig the theory of sall reflectios as: where: Γ ( ω ) = Γ +Γ e +Γ e + +Γ e i = j 2ωT j 4ωT j2ωt 1 2 = Γ e j2ωt T = v p propagatio tie through 1 sectio ote that for a ulti-sectio trasforer, we have degrees of desig freedo, correspodig to the characteristic ipedace values Z. Q: What should the values of Γ (i.e., Z ) be? A: We eed to defie idepedet desig equatios, which we ca the use to solve for the values of characteristic ipedace Z. First, we start with a sigle desig frequecy ω, where we wish to achieve a perfect atch:

8 4/15/21 The Bioial Multisectio Matchig Trasforer.doc 2/17 Γ i ( ω = ω ) = That s just oe desig equatio: we eed -1 ore! These additio equatios ca be selected usig ay Γ criteria oe such criterio is to ake the fuctio ( ω) axially flat at the poit ω = ω. i To accoplish this, we first cosider the Bioial Fuctio: j 2θ ( θ ) A( 1 e ) Γ = + This fuctio has the desirable properties that: ad that: jπ ( θ π 2) A( 1 e ) Γ = = + = ( 1 1) = A d Γ ( θ ) d θ θ = π 2 = for = 123,,,, 1 I other words, this Bioial Fuctio is axially flat at the poit θ π 2 Γ θ = π 2 =. =, where it has a value of ( ) Q: So? What does this have to do with our ulti-sectio atchig etwork?

9 4/15/21 The Bioial Multisectio Matchig Trasforer.doc 3/17 A: Let s expad (ultiply out the idetical product ters) of the Bioial Fuctio: j 2θ ( θ ) A( 1 e ) j2θ j 4θ j6θ j2θ = A( C + C1 e + C2 e + C3 e + + C e ) Γ = + where: C!!! ( ) Copare this to a -sectio trasforer fuctio: Γ ( ω) = Γ +Γ e +Γ e + +Γ e i j 2ωT j 4ωT j2ωt 1 2 ad it is obvious the two fuctios have idetical fors, provided that: Γ = AC ad ωt = θ Moreover, we fid that this fuctio is very desirable fro the stadpoit of the a atchig etwork. Recall that Γ θ = at θ = π 2--a perfect atch! ( ) Additioally, the fuctio is axially flat at θ = π 2, therefore Γ( θ ) over a wide rage aroud θ = π 2--a wide badwidth!

10 4/15/21 The Bioial Multisectio Matchig Trasforer.doc 4/17 Q: But how does θ = π 2 relate to frequecy ω? A: Reeber that ωt = θ, so the value θ = π 2 correspods to the frequecy: v p 1 π π ω = = T 2 2 This frequecy ( ω ) is therefore our desig frequecy the frequecy where we have a perfect atch. ote that the legth has a iterestig relatioship with this frequecy: v p π 1 π λ π λ = = = = ω 2 β 2 2π 2 4 I other words, a Bioial Multi-sectio atchig etwork will have a perfect atch at the frequecy where the sectio legths are a quarter wavelegth! Thus, we have our first desig rule: Set sectio legths so that they are a quarterwavelegth ( λ 4 ) at the desig frequecy ω. Q: I see! Ad the we select all the values Z such that Γ = AC. But wait! What is the value of A??

11 4/15/21 The Bioial Multisectio Matchig Trasforer.doc 5/17 A: We ca deterie this value by evaluatig a boudary coditio! Specifically, we ca easily deterie the value of Γ ( ω) at ω =. Z Z i Z 1 Z 2 Z R L ote as ω approaches zero, the electrical legth β of each sectio will likewise approach zero. Thus, the iput ipedace Z i will siply be equal to R L as ω. As a result, the iput reflectio coefficiet ( ω ) be: L ( ω ) ( ω ) Zi = Z Γ ( ω = ) = Zi = + Z RL Z = R + Z However, we likewise kow that: ( ) Γ = A 1+ = A 2 j 2 ( ) ( e ) ( 1 1) = A + Γ = ust

12 4/15/21 The Bioial Multisectio Matchig Trasforer.doc 6/17 Equatig the two expressios: Ad therefore: R Z Γ ( ) = A 2 = L RL + Z A = 2 L RL R Z + Z (A ca be egative!) We ow have a for to calculate the required argial reflectio coefficiets Γ : Γ = AC = A! ( )!! Of course, we also kow that these argial reflectio coefficiets are physically related to the characteristic ipedaces of each sectio as: Γ = Z Z Z + Z Equatig the two ad solvig, we fid that that the sectio characteristic ipedaces ust satisfy: 1+Γ 1+ AC Z Z Z = = Γ 1 AC

13 4/15/21 The Bioial Multisectio Matchig Trasforer.doc 7/17 ote this is a iterative result we deterie Z 1 fro Z, Z 2 fro Z 1, ad so forth. Q: This result appears to be our secod desig equatio. Is there soe reaso why you did t draw a big blue box aroud it? A: Alas, there is a big proble with this result. ote that there are +1 coefficiets,1,, ) i the Bioial series, yet there are oly desig degrees of freedo (i.e., there are oly trasissio lie sectios!). Γ (i.e., { } Thus, our desig is a bit over costraied, a result that aifests itself the fially argial reflectio coefficiet Γ. ote fro the iterative solutio above, the last trasissio lie ipedace Z is selected to satisfy the atheatical requireet of the peultiate reflectio coefficiet Γ : 1 Z Z Γ = = AC Z + Z 1 Thus the last ipedace ust be: Z 1 + AC 1 = Z 1 1 AC 1

14 4/15/21 The Bioial Multisectio Matchig Trasforer.doc 8/17 But there is oe ore atheatical requireet! The last argial reflectio coefficiet ust likewise satisfy: Γ = AC = 2 L RL R Z + Z where we have used the fact that C = 1. But, we just selected Z to satisfy the requireet for Γ, we have o physical desig paraeter to satisfy this 1 last atheatical requireet! As a result, we fid to our great costeratio that the last requireet is ot satisfied: R Z Γ = L R + Z AC L!!!!!! Q: Yikes! Does this ea that the resultig atchig etwork will ot have the desired Bioial frequecy respose? A: That s exactly what it eas! Q: You big #%@#$%&!!!! Why did you waste all y tie by discussig a over-costraied desig proble that ca t be built? A: Relax; there is a solutio to our dilea albeit a approxiate oe.

15 4/15/21 The Bioial Multisectio Matchig Trasforer.doc 9/17 You udoubtedly have previously used the approxiatio: y x 1 y l y + x 2 x A approxiatio that is especially accurate whe y x is sall (i.e., whe 1). y x y l 2 x y x y + x y x ow, we kow that the values of Z + 1 ad Z i a ulti-sectio atchig etwork are typically very close, such that Z Z is sall. Thus, we use the approxiatio: + 1 Z Z 1 Z Γ = l Z Z 2 Z + 1

16 4/15/21 The Bioial Multisectio Matchig Trasforer.doc 1/17 Likewise, we ca also apply this approxiatio (although ot as accurately) to the value of A : A R Z L ( 1) R L = l R + Z Z L So, let s start over, oly this tie we ll use these approxiatios. First, deterie A : ( 1) 2 + R L A l Z (A ca be egative!) ow use this result to calculate the atheatically required argial reflectio coefficiets Γ : Γ = AC = A! ( )!! Of course, we also kow that these argial reflectio coefficiets are physically related to the characteristic ipedaces of each sectio as: 1 Z l + 1 Γ 2 Z

17 4/15/21 The Bioial Multisectio Matchig Trasforer.doc 11/17 Equatig the two ad solvig, we fid that that the sectio characteristic ipedaces ust satisfy: Z = Z exp Γ ow this is our secod desig rule. ote it is a iterative rule we deterie Z 1 fro Z, Z 2 fro Z 1, ad so forth. Q: Huh? How is this ay better? How does applyig approxiate ath lead to a better desig result?? A: Applyig these approxiatios help resolve our overcostraied proble. Recall that the over-costrait resulted i: R Z L Γ = R + Z AC L But, as it turs out, these approxiatios leads to the happy situatio where: 1 R L Γ l A C = 2 Z Saity check!! provided that the value A is likewise the approxiatio give above.

18 4/15/21 The Bioial Multisectio Matchig Trasforer.doc 12/17 Effectively, these approxiatios couple the results, such that each value of characteristic ipedace Z approxiately satisfies both Γ ad Γ + 1. Suarizig: * If you use the exact desig equatios to deterie the characteristic ipedaces Z, the last value Γ will exhibit a sigificat ueric error, ad your desig will ot appear to be axially flat. * If you istead use the approxiate desig equatios to deterie the characteristic ipedaces Z, all values Γ will exhibit a slight error, but the resultig desig will appear to be axially flat, Bioial reflectio coefficiet fuctio Γ ω! ( ) Figure 5.15 (p. 25) Reflectio coefficiet agitude versus frequecy for ultisectio bioial atchig trasforers of Exaple 5.6 Z L = 5Ω ad Z = 1Ω.

19 4/15/21 The Bioial Multisectio Matchig Trasforer.doc 13/17 ote that as we icrease the uber of sectios, the atchig badwidth icreases. Q: Ca we deterie the value of this badwidth? A: Sure! But we first ust defie what we ea by badwidth. As we ove fro the desig (perfect atch) frequecy f the value Γ ( f ) will icrease. At soe frequecy (f, say) the agitude of the reflectio coefficiet will icrease to soe uacceptably high value ( Γ, say). At that poit, we o loger cosider the device to be atched. Γ ( f ) Γ Δ f f f 1 f f 2 ote there are two values of frequecy f oe value less tha desig frequecy f, ad oe value greater tha desig frequecy f. These two values defie the badwidth Δ f of the atchig etwork: ( ) ( ) Δ f = f f = 2 f f = 2 f f

20 4/15/21 The Bioial Multisectio Matchig Trasforer.doc 14/17 Q: So what is the uerical value of Γ? A: I do t kow it s up to you to decide! Every egieer ust deterie what they cosider to be a acceptable atch (i.e., decide Γ ). This decisio depeds o the applicatio ivolved, ad the specificatios of the overall icrowave syste beig desiged. However, we typically set Γ to be.2 or less. Q: OK, after we have selected Γ, ca we deterie the two frequecies f? A: Sure! We just have to do a little algebra. We start by rewritig the Bioial fuctio: j 2θ ( θ ) A( 1 e ) j θ + jθ jθ = Ae ( e + e ) j θ + jθ jθ = Ae ( e + e ) Γ = + = Ae j θ ( 2cos θ ) ow, we take the agitude of this fuctio: ( θ ) 2 Γ = = 2 Ae jθ Acosθ cosθ

21 4/15/21 The Bioial Multisectio Matchig Trasforer.doc 15/17 ow, we defie the values θ where Γ ( θ ) = 2 ( θ θ ) Γ = Γ = Acosθ =Γ as θ. I.E., : We ca ow solve for θ (i radias!) i ters of Γ : θ Γ = cos 2 A θ Γ = cos 2 A ote that there are two solutios to the above equatio (oe less that π 2 ad oe greater tha π 2)! ow, we ca covert the values of θ ito specific frequecies. Recall thatωt = θ, therefore: v ω = θ = θ T 1 p But recall also that = λ 4, where λ is the wavelegth at the desig frequecy f (ot f!), ad where λ = vp f. Thus we ca coclude: v 4v ω θ θ θ ( 4f ) p p = = = λ

22 4/15/21 The Bioial Multisectio Matchig Trasforer.doc 16/17 or: ( 4 ) θ ( 2 ) 1 v f f θ f = = = 2π 2π π p θ where θ is expressed i radias. Therefore: f 1 1 2f 1 1 Γ = cos + π 2 A f 2 1 2f 1 1 Γ = cos π 2 A Thus, the badwidth of the bioial atchig etwork ca be deteried as: ( ) Δ f = 2 f f 1 1 4f 1 1 Γ = 2f cos + π 2 A ote that this equatio ca be used to deterie the badwidth of a bioial atchig etwork, give Γ ad uber of sectios. However, it ca likewise be used to deterie the uber of sectios required to eet a specific badwidth requireet!

23 4/15/21 The Bioial Multisectio Matchig Trasforer.doc 17/17 Fially, we ca list the desig steps for a bioial atchig etwork: 1. Deterie the value required to eet the badwidth ( Δ f ad Γ ) requireets. 2. Deterie the approxiate value A fro Z, R L ad. 3. Deterie the argial reflectio coefficiets Γ = AC required by the bioial fuctio. 4. Deterie the characteristic ipedace of each sectio usig the iterative approxiatio: Z = Z exp Γ 5. Perfor the saity check: 1 R L Γ l 2 Z = A C 6. Deterie sectio legth = λ 4 for desig frequecy f.