2.5 The Quarter-Wave Transformer

Transcription

1 /3/5 _5 The Quate Wave Tansfome /.5 The Quate-Wave Tansfome Reading Assignment: pp By now you ve noticed that a quate-wave length of tansmission line ( λ 4, β π ) appeas often in micowave engineeing poblems. HO: The Quate-Wave Tansfome Q: Why does the quate-wave matching netwok wok afte all, the quate-wave line is mismatched at both ends? A: HO: Multiple Reflection Viewpoint

2 /3/5 The Quate Wave Tansfome /5 The Quate-Wave Tansfome Say the end of a tansmission line with chaacteistic impedance is teminated with a esistive (i.e., eal) load. We typically would like all powe taveling down the line to be absobed by the load. But ifr, the line is unmatched and some of the incident powe will be eflected. Q: Can all incident powe be deliveed to a esistive load if R?? A: Yes! We can inset a matching netwok between the tansmission line and the load. Matching Netwok

3 /3/5 The Quate Wave Tansfome /5 A matching netwok is a lossless, -pot device. Its job is to tansfom the load ( o even ) to a value. In othe wods, we want the input impedance of the matching netwok to be in, so that Γ in --no eflection! Since none of the incident powe is eflected, and none is absobed by the lossless matching netwok, it all must be absobed by the load! Q: These matching netwoks sound too good to be tue. Exactly how do we build them? A: Thee ae many methods and ways, but pehaps the easiest is the quate-wave tansfome. Fist, inset a tansmission line with chaacteistic impedance and length λ 4 (i.e., a quate-wave line) between the load and the tansmission line. in λ 4 The λ 4 line is the matching netwok!

4 /3/5 The Quate Wave Tansfome 3/5 Q: But what about the chaacteistic impedance ; what should its value be?? A: Remembe, the quate wavelength case is one of the special cases that we studied. We know that the input impedance of the quate wavelength line is: in ( ) ( ) R Thus, if we wish fo in to be numeically equal to, we find: in ( ) R Solving fo, we find its equied value to be: ( ) ( ) R R R In othe wods, the chaacteistic impedance of the quate wave line is the geometic aveage and!

5 /3/5 The Quate Wave Tansfome 4/5 Theefoe, a λ 4 line with chaacteistic impedance R will match a tansmission line with chaacteistic impedance to a esistive load. in R λ 4 Thus, all powe is deliveed to load! Impotant Note: We find that in only if the matching if the quate-wave tansmission line is exactly one-quate wavelength in length λ 4. The poblem with this, of couse, is that a physical length of tansmission line is exactly one-quate wavelength at only one fequency f! Remembe, wavelength is elated to fequency as: v p λ f f C whee v p is the popagation velocity of the wave.

6 /3/5 The Quate Wave Tansfome 5/5 Fo example, assuming that v p c (c the speed of light in a vacuum), one wavelength at GHz is 3 cm ( λ.3 m ), while one wavelength at 3 GHz is cm ( λ. m ). As a esult, a tansmission line length 7.5 cm is a quate wavelength fo a signal at GHz only. Thus, a quate-wave tansfome povides a pefect match ( Γ in ) at one and only one signal fequency!

7 /3/5 Multiple Reflection Viewpoint /7 Multiple Reflection Viewpoint The quate-wave tansfome bings up an inteesting question in µ-wave engineeing. Γ in R λ 4 Q: Why is thee no eflection at z? It appeas that the line is mismatched at both z and z. A: In fact thee ae eflections at these mismatched intefaces an infinite numbe of them! Fist, lets define a few tems:

8 /3/5 Multiple Reflection Viewpoint /7 Τ Τ i j z β j z + β Γ Γ R Γ 3 λ 4 Γ patial eflection coefficient of a wave incident on the z inteface fom the line: Γ + Γ R Γ patial eflection coefficient of a wave incident on the z inteface fom the line: Γ Γ + Γ R Γ 3 patial eflection coefficient of a wave incident on the z inteface fom the line: R Γ 3 R + R Γ 3

9 /3/5 Multiple Reflection Viewpoint 3/7 T patial tansmission coefficient of a wave incident on the z inteface fom the line: Τ + Τ R T patial tansmission coefficient of a wave incident on the z inteface fom the line: Τ + Τ R Now let s ty to intempeate what physically happens when the incident voltage wave: i j z β R eaches the inteface at z.

10 /3/5 Multiple Reflection Viewpoint 4/7. At z, the chaacteistic impedance of the tansmission line changes fom to. This mismatch ceates a eflected wave: i j z β + j β z R i whee V Γ V.. Howeve, a potion of the incident wave is tansmitted ( T ) acoss the inteface at z, this wave tavels a distance of β 9 to the load at z, whee a potion of it is eflected ( Γ 3 ). This wave tavels back β 9 to the inteface at z, whee a potion is again tansmitted ( T ) acoss into the tansmission line anothe eflected wave ( V )! i j z β + j β z R whee we have found that taveling β 8 has poduced a minus sign in ou esult: V Τ e Γ e ΤV j9 j9 3 Τ Τ Γ 3 V i i

11 /3/5 Multiple Reflection Viewpoint 5/7 3. Howeve, a potion of this second wave is also eflected ( Γ ) back into the tansmission line at z, whee it again tavels to β 9 the load, is patially eflected ( Γ 3 ), tavels β 9 back to z, and is patially tansmitted into ( T ) ou thid eflected wave! i j z β 3 + j β z R whee: V Τ e Γ e Γ e Γ e ΤV j9 j9 j9 j ( ) 3 Τ Τ Γ Γ V i i n. We can see that this bouncing back and foth can go on foeve, with each tip launching a new eflected wave into the tansmission line. Note howeve, that the powe associated with each successive eflected wave is smalle than the pevious, and so eventually, the powe associated with the eflected waves will diminish to insignificance! Q: But, why then is Γ?

12 /3/5 Multiple Reflection Viewpoint 6/7 A: Each eflected wave Vn is a coheent wave. That is, they all oscillate at same fequency ω ; the eflected waves diffe only in tems of thei magnitude and phase. Theefoe, to detemine the total eflected wave, we must pefom a coheent summation of each eflected wave a opeation easily pefomed since we have expessed ou waves with complex notation: + j βz + jβz Vn e n It can be shown that this infinite seies conveges, with the esult: Γ +ΓΓΓ3 ΤΤΓ 3 i V V +ΓΓ3 Thus, the total eflection coefficient is: V Γ i V Γ +Γ Γ Γ Τ Τ Γ 3 3 +Γ Γ 3 Using ou definitions, it can likewise be shown that the numeato of the above expession is: Γ +Γ Γ Γ Τ Τ Γ 3 3 ( R ) ( + )( R + )

13 /3/5 Multiple Reflection Viewpoint 7/7 It is evident that the numeato (and theefoe Γ) will be zeo if: R R Just as we expected! Physically, this esults insues that all the eflected waves add coheently togethe to poduce a zeo value! A simple example of this phenomenon is the addition of two waves with equal magnitude and opposite phase (i.e., thei phase diffeence is 8 ). ( ω ) ( ω ) ( ω ) ( ω ) cos t + cos t + 8 cos t cos t Note all of ou tansmission line analysis has been steady-state analysis. We assume ou signals ae sinusoidal, of the fom exp( jω t ). Note this signal exists fo all time t the signal is assumed to have been on foeve, and assumed to continue on foeve. In othe wods, in steady-state analysis, all the multiple eflections have long since occued, and thus have eached a steady state the eflected wave is zeo!