5.4 The Quarter-Wave Transformer

Transcription

1 3/4/7 _4 The Qurter Wve Trnsformer /.4 The Qurter-Wve Trnsformer Redg Assignment: pp , 4-43 By now you ve noticed tht qurter-wve length of trnsmission le ( = λ 4, β = π ) ppers often microwve engeerg prolems. Another ppliction of the = λ 4 trnsmission le is s n impednce mtchg network. HO: THE QUARTER-WAVE TRANSFORMER HO: THE SIGNA-FOW GRAPH OF A QUARTER-WAVE TRANSFORMER Q: Why does the qurter-wve mtchg network work fter ll, the qurter-wve le is mismtched t oth ends? A: HO: MUTIPE REFECTION VIEWPOINT Jim Stiles The Univ. of Knss Dept. of EECS

2 3/3/7 The Qurter Wve Trnsformer /7 The Qurter-Wve Trnsformer Sy the end of trnsmission le with chrcteristic impednce is termted with resistive (i.e., rel) lod. R Unless R =, the resistor is mismtched to the le, nd thus some of the cident power will e reflected. We cn of course correct this sitution y plcg mtchg network etween the le nd the lod: Mtchg Network R In ddition to the designs we hve just studied (e.g., - networks, stu tuners), one of the simplest mtchg network designs is the qurter-wve trnsformer. Jim Stiles The Univ. of Knss Dept. of EECS

3 3/3/7 The Qurter Wve Trnsformer /7 The qurter-wve trnsformer is simply trnsmission le with chrcteristic impednce nd length = λ 4 (i.e., qurterwve le). R = λ 4 The λ 4 le is the mtchg network! Q: But wht out the chrcteristic impednce ; wht should its vlue e?? A: Rememer, the qurter wvelength cse is one of the specil cses tht we studied. We know tht the put impednce of the qurter wvelength le is: ( ) ( ) = = R Thus, if we wish for to e numericlly equl to, we fd: ( ) = = R Jim Stiles The Univ. of Knss Dept. of EECS

4 3/3/7 The Qurter Wve Trnsformer 3/7 Solvg for, we fd its required vlue to e: ( ) ( ) R = = = R R In other words, the chrcteristic impednce of the qurter wve le is the geometric verge of nd R! Therefore, λ 4 le with chrcteristic impednce = R will mtch trnsmission le with chrcteristic impednce to resistive lod R. = = R R = λ 4 Thus, ll power is delivered to lod R! Als, the qurter-wve trnsformer (like ll our designs) hs few prolems! Jim Stiles The Univ. of Knss Dept. of EECS

5 3/3/7 The Qurter Wve Trnsformer 4/7 Prolem # The mtchg ndwidth is nrrow! In other words, we ot perfect mtch t precisely the frequency where the length of the mtchg trnsmission le is qurter-wvelength. But rememer, this length cn e qurter-wvelength t just one frequency! Rememer, wvelength is relted to frequency s: v p λ = = f f C where v p is the propgtion velocity of the wve. For exmple, ssumg tht v p = c (c = the speed of light vcuum), one wvelength t GHz is 3 cm ( λ =.3 m ), while one wvelength t 3 GHz is cm ( λ =. m ). As result, trnsmission le length = 7. cm is qurter wvelength for signl t GHz only. Thus, qurter-wve trnsformer provides perfect mtch ( = ) t one nd only one signl frequency! Jim Stiles The Univ. of Knss Dept. of EECS

6 3/3/7 The Qurter Wve Trnsformer /7 As the signl frequency (i.e., wvelength) chnges, the electricl length of the mtchg trnsmission le chnges. It will no longer e qurter wvelength, nd thus we no longer will hve perfect mtch. We fd tht the closer R (R ) is to chrcteristic impednce, the wider the ndwidth of the qurter wvelength trnsformer. Figure. (p. 43) Reflection coefficient mgnitude versus frequency for sgle-section qurter-wve mtchg trnsformer with vrious lod mismtches. We will fd tht the ndwidth cn e cresed y ddg multiple 4 λ sections! Jim Stiles The Univ. of Knss Dept. of EECS

7 3/3/7 The Qurter Wve Trnsformer 6/7 Prolem # Recll the mtchg solution ws limited to lods tht were purely rel! I.E.: = R + j Of course, this is BIG prolem, s most lods will hve rective component! Fortuntely, we hve reltively esy solution to this prolem, s we cn lwys dd some length of trnsmission le to the lod to mke the impednce completely rel: z R, β r r possile solutions! However, rememer tht the put impednce will e purely rel t only one frequency! We cn then uild qurter-wve trnsformer to mtch the le to resistnce R : Jim Stiles The Univ. of Knss Dept. of EECS

8 3/3/7 The Qurter Wve Trnsformer 7/7 = = R R λ 4 Ag, sce the trnsmission les re lossless, ll of the cident power is delivered to the lod. Jim Stiles The Univ. of Knss Dept. of EECS

9 3/4/7 The Signl Flow Grph of QurterWve Trnsformer /9 The Signl Flow Grph of Qurter-Wve Trnsformer First, let s consider the sctterg mtrix of perfect connector n electriclly very smll two-port device tht llows us to connect the ends of different trnsmission les together. I I + + V V Port Port If the connector is idel, then it will exhiit no series ductnce nor shunt cpcitnce, nd thus: V = V I = I The sctterg mtrix for such this idel connector is therefore: S= Jim Stiles The Univ. of Knss Dept. of EECS

10 3/4/7 The Signl Flow Grph of QurterWve Trnsformer /9 As result, the perfect connector llows two trnsmission les of identicl chrcteristic impednce to e connected together to one semless trnsmission le. Now, however, consider the cse where the trnsmission les connected together hve dissimilr chrcteristic impednces (i.e., ): I I + + V V Port Port Q: Won t the sctterg mtrix of this idel connector rem the sme? After ll, the device itself hs not chnged! A: The impednce, dmittnce, nd trnsmission mtrix will remed unchnged these mtrix quntities do not depend on the chrcteristics of the trnsmission les connected to the device. Jim Stiles The Univ. of Knss Dept. of EECS

11 3/4/7 The Signl Flow Grph of QurterWve Trnsformer 3/9 But rememer, the sctterg mtrix depends on oth the device nd the chrcteristic impednce of the trnsmission les ttched to it. After ll, the cident nd exitg wves re trvelg on these trnsmission les! The idel connector this cse estlishes semless terfce etween two dissimilr trnsmission les. Rememer, this is the sme structure tht we evluted n erlier hndout! From the results of tht nlysis we cn conclude tht the sctterg mtrix of the idel connector (when connectg dissimilr trnsmission les) is: + + S = + + Jim Stiles The Univ. of Knss Dept. of EECS

12 3/4/7 The Signl Flow Grph of QurterWve Trnsformer 4/9 Or more compctly stted: S= where + + = = For qurter wve trnsformer, we set such tht: = R = R Insertg this to the expressions ove, we fd: R = = R R + R + Sce the device is lossless, we cn conclude (nd likewise show) tht: = + = + where this lst expression is (only) true ecuse nd re rel vlued. Jim Stiles The Univ. of Knss Dept. of EECS

13 3/4/7 The Signl Flow Grph of QurterWve Trnsformer /9 Jim Stiles The Univ. of Knss Dept. of EECS Now, qurter wvelength of trnsmission le hs the sctterg mtrix: j j = S While lod hs sctterg mtrix of: R R = = + S Note tht =!!! 4 = λ 3 j j R

14 3/4/7 The Signl Flow Grph of QurterWve Trnsformer 6/9 Jim Stiles The Univ. of Knss Dept. of EECS Of course, if we connect the idel connector to qurter wvelength of trnsmission le, nd termte the whole thg with lod R, we hve formed qurter wve mtchg network! We cn likewise put the signl-flow grph pieces together to form the signl-flow grph of the qurter wve network: And simplifyg: 4 λ = R 3 j j j j

15 3/4/7 The Signl Flow Grph of QurterWve Trnsformer 7/9 Now, let s see if we cn reduce this grph to determe: From the series rule: From the splittg rule: From the self-loop rule: Jim Stiles The Univ. of Knss Dept. of EECS

16 3/4/7 The Signl Flow Grph of QurterWve Trnsformer 8/9 Ag with the series rule: And flly with the prllel rule: So tht: = Q: Hey wit! If the qurter-wve trnsformer is mtchg network, shouldn t =?? A: Who sys it doesn t! Recll tht for the qurter-wve trnsformer, we found tht =, thus: Jim Stiles The Univ. of Knss Dept. of EECS

17 3/4/7 The Signl Flow Grph of QurterWve Trnsformer 9/9 = = And likewise sce = : = = = = A perfect mtch! Jim Stiles The Univ. of Knss Dept. of EECS

18 3/4/7 Multiple Reflection Viewpot /7 Multiple Reflection Viewpot The qurter-wve trnsformer rgs up n terestg question µ-wve engeerg. z = z = = = R R = λ 4 Q: Why is there no reflection t z =? It ppers tht the le is mismtched t oth z = nd z =. A: In fct there re reflections t these mismtched terfces n fite numer of them! We cn use our signl flow grph to fct determe ll the propgtion pths through the qurter-wve trnsformer. Jim Stiles The Univ. of Knss Dept. of EECS

19 3/4/7 Multiple Reflection Viewpot /7 + V ( z ) V ( z ) = R R + = V ( z = ) = V ( z = ) = λ 4 j j Now let s try to terpret wht physiclly hppens when the cident voltge wve: + V ( z ) R R reches the terfce t z =. Jim Stiles The Univ. of Knss Dept. of EECS

20 3/4/7 Multiple Reflection Viewpot 3/7. At z =, the chrcteristic impednce of the trnsmission le chnges from to. This mismtch cretes reflected V z : wve, wve tht we shll cll ( ) + V ( z ) V ( z ) R R + so V ( z = ) = V ( z = ). j. However, portion of the cident wve is trnsmitted () cross the terfce t z =, this wve trvels distnce of β = 9 to the lod t z =, where portion of it is reflected ( ). This wve trvels ck β = 9 to the terfce t z =, where portion is g trnsmitted () cross to the V z )! trnsmission le nother reflected wve ( ( ) j + V ( z ) V ( z ) R R Jim Stiles The Univ. of Knss Dept. of EECS

21 3/4/7 Multiple Reflection Viewpot 4/7 where we hve found tht trvelg β = 8 hs produced mus sign our result: j9 j9 + V ( z = ) = e e V ( z =) + V ( z ) = = j j 3. However, portion of this second wve is lso reflected () ck to the trnsmission le t z =, where it g trvels to β = 9 the lod, is prtilly reflected ( ), trvels β = 9 ck to z =, nd is prtilly trnsmitted to () our third reflected wve! + V ( z ) V ( z ) 3 R R where: j9 j9 j9 j9 V ( z = ) =e e ( ) e e V + ( z =) 3 + ( ) V ( z ) = = Jim Stiles The Univ. of Knss Dept. of EECS

22 3/4/7 Multiple Reflection Viewpot /7 j j n. We cn see tht this ouncg ck nd forth cn go on forever, with ech trip lunchg new reflected wve to the trnsmission le. Note however, tht the power ssocited with ech successive reflected wve is smller thn the previous, nd so eventully, the power ssocited with the reflected wves will dimish to significnce! Q: But, why then is =? A: Ech reflected wve V ( z ) n is coherent wve. Tht is, they ll oscillte t sme frequency ω ; the reflected wves differ only terms of their mgnitude nd phse. Therefore, to determe the totl reflected wve, we must perform coherent summtion of ech reflected wve opertion esily performed sce we hve expressed our wves with complex nottion: V ( z) V ( z ) = n = n Jim Stiles The Univ. of Knss Dept. of EECS

23 3/4/7 Multiple Reflection Viewpot 6/7 It cn e shown tht this fite series converges, with the result: + V ( z = ) = V ( z = ) Thus, the totl reflection coefficient is: = Usg our defitions, it cn likewise e shown tht the numertor of the ove expression is: = ( R ) ( + )( R + ) It is evident tht the numertor (nd therefore ) will e zero if: R = R Just s we expected! Physiclly, this results sures tht ll the reflected wves dd coherently together to produce zero vlue! Note ll of our trnsmission le nlysis hs een stedy-stte nlysis. We ssume our signls re susoidl, of the form exp( jω t ). Note this signl exists for ll time t the signl is ssumed to hve een on forever, nd ssumed to contue on forever. Jim Stiles The Univ. of Knss Dept. of EECS

24 3/4/7 Multiple Reflection Viewpot 7/7 In other words, stedy-stte nlysis, ll the multiple reflections hve long sce occurred, nd thus hve reched stedy stte the reflected wve is zero! Jim Stiles The Univ. of Knss Dept. of EECS