5.6 Binomial Multi-section Matching Transformer
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1 4/14/2010 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 Ji Stiles The Uiv. of Kasas Dept. of EECS
2 4/14/2010 Maxially Flat Trasforer Fuctios preset 1/7 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 = 0 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) dx is o-zero at other values of x. Ji Stiles The Uiv. of Kasas Dept. of EECS
3 4/14/2010 Maxially Flat Trasforer Fuctios preset 2/7 Takig derivates: way too fu to stop! So, we ow have two pieces of iforatio about the fuctio f ( x ). We ca add to this list by cotiuig to take higher-order 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: df ( x) dx x = 1 = 0 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 Ji Stiles The Uiv. of Kasas Dept. of EECS
4 4/14/2010 Maxially Flat Trasforer Fuctios preset 3/7 A ore realistic fuctio 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 ove fro x =1. Ji Stiles The Uiv. of Kasas Dept. of EECS
5 4/14/2010 Maxially Flat Trasforer Fuctios preset 4/7 Copletely flat i ot possible! 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: 3 2 ax bx cx + + = 5 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. Ji Stiles The Uiv. of Kasas Dept. of EECS
6 4/14/2010 Maxially Flat Trasforer Fuctios preset 5/7 How ay derivatives ca be zero? For exaple, we wish to ake the first derivate equal to zero at x =1: 0 = df( x) dx x = 1 2 ( 3ax 2bx c ) = + + = 3a + 2b + c x = 1 Likewise, we wish to ake the secod derivative equal to zero at x =1: 0 = 2 d f ( x) 2 dx x = 1 ( 6ax 2b ) = + = 6a + 2b x = 1 Here we ust stop takig derivatives, as our solutio oly has three degrees of desig freedo (i.e., 3 ukows a, b, c). Ji Stiles The Uiv. of Kasas Dept. of EECS
7 4/14/2010 Maxially Flat Trasforer Fuctios preset 6/7 We re out of degrees of desig freedo 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: 5 = f x = 1 ( ) ( 1) 3 ( 1) 2 ( 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 0 = 3a + 2b + c 0 = 6a + 2b Solvig, we fid: a = 5 b = 15 c = 15 Ji Stiles The Uiv. of Kasas Dept. of EECS
8 4/14/2010 Maxially Flat Trasforer Fuctios preset 7/7 Look! The fuctio is axially flat at x=1! Therefore, the axially flat fuctio (at x =1) is: 3 2 f ( x) = 5x 15x + 15x f(x) x Ji Stiles The Uiv. of Kasas Dept. of EECS
9 4/15/2010 The Bioial Multisectio Matchig Trasforer preset.doc 1/24 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 T = v j 2ωT j 4ωT j2ωt j2ωt e = 0 = Γ 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. Z 0 Z i Z 1 Z 2 Z R L Ji Stiles The Uiv. of Kasas Dept. of EECS
10 4/15/2010 The Bioial Multisectio Matchig Trasforer preset.doc 2/24 Behold the Bioial Fuctio! 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 0, where we wish to achieve a perfect atch: Γ i ( ω = ω 0 ) = 0 That s just oe desig equatio: we eed -1 ore! These additio equatios ca be selected usig ay criteria oe such criterio is to ake the fuctio Γ ( ω) axially flat at the poit ω = ω0. i To accoplish this, we first cosider the Bioial Fuctio: j 2θ ( θ ) A( 1 e ) Γ = + Ji Stiles The Uiv. of Kasas Dept. of EECS
11 4/15/2010 The Bioial Multisectio Matchig Trasforer preset.doc 3/24 What s so special about the Bioial Fuctio! The Bioial Fuctio has the desirable properties that: ad that: jπ ( θ π 2) A( 1 e ) Γ = = + = 0 ( 1 1) = A d Γ ( θ ) d θ θ = π 2 = 0 for = 123,,,, 1 I other words, this Bioial Fuctio is axially flat at the poit θ = π 2, where it has a value of ( θ π ) Γ = 2 = 0. Q: So? What does this have to do with our ulti-sectio atchig etwork? Ji Stiles The Uiv. of Kasas Dept. of EECS
12 4/15/2010 The Bioial Multisectio Matchig Trasforer preset.doc 4/24 A: Plety! 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( C0 + 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 ad it is obvious the two fuctios have idetical fors, provided that: Γ = AC ad ωt = θ Ji Stiles The Uiv. of Kasas Dept. of EECS
13 4/15/2010 The Bioial Multisectio Matchig Trasforer preset.doc 5/24 See, the Bioial Fuctio is very useful! Moreover, we fid that this fuctio is very desirable fro the stadpoit of a atchig etwork. Recall that Γ ( θ ) = 0 at θ = π 2--a perfect atch! Additioally, the fuctio is axially flat at θ π 2 rage aroud θ = π 2--a wide badwidth! =, therefore ( θ ) 0 Γ over a wide 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 0 = = T 2 2 This frequecy ( ω 0 ) is therefore our desig frequecy the frequecy where we have a perfect atch. Ji Stiles The Uiv. of Kasas Dept. of EECS
14 4/15/2010 The Bioial Multisectio Matchig Trasforer preset.doc 6/24 What about the legth of each sectio? ote that the sectio-legth has a iterestig relatioship with this frequecy: v p π 1 π λ0 π λ0 = = = = ω 2 β 2 2π 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 quarter-wavelegth ( λ 0 4 ) at the desig frequecy ω 0. Ji Stiles The Uiv. of Kasas Dept. of EECS
15 4/15/2010 The Bioial Multisectio Matchig Trasforer preset.doc 7/24 Ad that pesky costat A? Q: I see! Ad the we select all the values Z such that is the value of A?? Γ = AC. But wait! What A: We ca deterie this value by evaluatig a boudary coditio! Specifically, we ca easily deterie the value of Γ ( ω) at ω = 0. Z 0 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 0. As a result, the iput reflectio coefficiet ( ω 0) i Γ = ust be: ( ω 0) ( ω 0) Z = Z i R Z Γ ( ω = 0) = = Z = + Z R + Z 0 L 0 0 L 0 Ji Stiles The Uiv. of Kasas Dept. of EECS
16 4/15/2010 The Bioial Multisectio Matchig Trasforer preset.doc 8/24 Are t boudary coditios great? However, we likewise kow that: ( ) Γ 0 = A 1+ = A 2 j 20 ( ) ( e ) ( 1 1) = A + Equatig the two expressios: Ad therefore: R Z0 Γ ( 0) = A 2 = L RL + Z0 A = 2 L RL R Z + Z 0 0 (A ca be egative!) We ow have a for to calculate the required argial reflectio coefficiets Γ : Γ = AC = A! ( )!! Ji Stiles The Uiv. of Kasas Dept. of EECS
17 4/15/2010 The Bioial Multisectio Matchig Trasforer preset.doc 9/24 How do I deterie characteristic ipedace? 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 ote this is a iterative result we deterie Z 1 fro Z 0, 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. Ji Stiles The Uiv. of Kasas Dept. of EECS
18 4/15/2010 The Bioial Multisectio Matchig Trasforer preset.doc 10/24 The BIG proble with this result! ote that there are +1 coefficiets 0,1,, ) 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 Ji Stiles The Uiv. of Kasas Dept. of EECS
19 4/15/2010 The Bioial Multisectio Matchig Trasforer preset.doc 11/24 Our degrees of freedo have ru out! But there is oe ore atheatical requireet! The last argial reflectio coefficiet ust likewise satisfy: R Z L 0 Γ = AC = 2 R + Z L 0 where we have used the fact that C = 1. But, we just selected Z to satisfy the requireet for Γ 1, we have o physical desig paraeter to satisfy this 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! Ji Stiles The Uiv. of Kasas Dept. of EECS
20 4/15/2010 The Bioial Multisectio Matchig Trasforer preset.doc 12/24 Q: You big Why did you waste all y tie by discussig a overcostraied desig proble that ca t be built? A: Relax; there is a solutio to our dilea albeit a approxiate oe. 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 -1.0 Ji Stiles The Uiv. of Kasas Dept. of EECS
21 4/15/2010 The Bioial Multisectio Matchig Trasforer preset.doc 13/24 Use this approxiatio for value A! 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. + 1 Thus, we use the approxiatio: Z Z 1 Z Γ = l Z Z 2 Z + 1 Likewise, we ca also apply this approxiatio (although ot as accurately) to the value of A : A R Z L 0 ( 1) R L = l R + Z Z L 0 0 Ji Stiles The Uiv. of Kasas Dept. of EECS
22 4/15/2010 The Bioial Multisectio Matchig Trasforer preset.doc 14/24 Let s try this agai with approxiatios! So, let s start over, oly this tie we ll use these approxiatios. First, deterie A : ( 1) 2 + R L 0 A l Z (A ca be egative!) ow use this result to calculate the atheatically required argial reflectio coefficiets Γ : Γ = AC = A! ( )!! Ji Stiles The Uiv. of Kasas Dept. of EECS
23 4/15/2010 The Bioial Multisectio Matchig Trasforer preset.doc 15/24 Here s (fially) our secod desig rule! 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 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 0, Z 2 fro Z 1, ad so forth. Ji Stiles The Uiv. of Kasas Dept. of EECS
24 4/15/2010 The Bioial Multisectio Matchig Trasforer preset.doc 16/24 I do t uderstad what just happeed 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 over-costraied proble. Recall that the over-costrait resulted i: R Z L Γ = R + Z L AC But, as it turs out, these approxiatios leads to the happy situatio where: 1 R L Γ l 2 Z = A C A Saity check!! provided that the value A is likewise the approxiatio give above. Ji Stiles The Uiv. of Kasas Dept. of EECS
25 4/15/2010 The Bioial Multisectio Matchig Trasforer preset.doc 17/24 I still do t uderstad what just happeed 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 Γ ω! reflectio coefficiet fuctio ( ) ote that as we icrease the uber of sectios, the atchig badwidth icreases. Ji Stiles The Uiv. of Kasas Dept. of EECS
26 4/15/2010 The Bioial Multisectio Matchig Trasforer preset.doc 18/24 Badwidth: How do we defie it? Q: Ca we deterie the value of this badwidth? bad width (bad widt h, -witt h ) - ou 1. the rage of frequecies withi a. A: Sure! But we first ust defie what we ea by badwidth. As we ove fro the desig (perfect atch) frequecy f 0 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 0 f 2 Ji Stiles The Uiv. of Kasas Dept. of EECS
27 4/15/2010 The Bioial Multisectio Matchig Trasforer preset.doc 19/24 Badwidth: How do we calculate it? ote there are two values of frequecy f oe value less tha desig frequecy f 0, ad oe value greater tha desig frequecy f 0. These two values defie the badwidth Δ f of the atchig etwork: ( ) ( ) Δ f = f f = 2 f f = 2 f f 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 0.2 or less. Ji Stiles The Uiv. of Kasas Dept. of EECS
28 4/15/2010 The Bioial Multisectio Matchig Trasforer preset.doc 20/24 We get to perfor soe Algebra!! 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 j θ Acosθ cosθ Ji Stiles The Uiv. of Kasas Dept. of EECS
29 4/15/2010 The Bioial Multisectio Matchig Trasforer preset.doc 21/24 It gets better eve ore algebra!! ow, we defie the values θ where Γ ( θ ) = 2 =Γ as θ. I.E., : ( θ θ ) Γ = Γ = Acosθ 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. Ji Stiles The Uiv. of Kasas Dept. of EECS
30 4/15/2010 The Bioial Multisectio Matchig Trasforer preset.doc 22/24 Covertig θ to f Recall thatωt = θ, therefore: v ω = θ = θ T 1 p But recall also that = λ 4, where 0 λ 0 is the wavelegth at the desig frequecy f 0 (ot f!), ad where λ 0 = v f0. p Thus we ca coclude: v 4v ω θ θ θ ( 4f ) p p = = = 0 λ0 or: ( 4 ) θ ( 2 ) 1 v f f θ f = = = 2π 2π π p 0 0 θ where θ is expressed i radias. Ji Stiles The Uiv. of Kasas Dept. of EECS
31 4/15/2010 The Bioial Multisectio Matchig Trasforer preset.doc 23/24 Therefore: Ad thus the badwidth is f 1 1 2f Γ = cos + π 2 A f 2 1 2f Γ = cos π 2 A Thus, the badwidth of the bioial atchig etwork ca be deteried as: ( ) Δ f = 2 f f f Γ = 2f0 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! Ji Stiles The Uiv. of Kasas Dept. of EECS
32 4/15/2010 The Bioial Multisectio Matchig Trasforer preset.doc 24/24 I suary, our desig steps 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 Z0, 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 0 0. Ji Stiles The Uiv. of Kasas Dept. of EECS
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