Richard s Transformations

Transcription

1 4/27/25 Rihard Tranfrmatin.d /7 Rihard Tranfrmatin Reall the put impedane f hrt-iruited and peniruited tranmiin le tub. j tan β, β t β, β Nte that the put impedane are purely reatie jut like lumped element! Hweer, the reatane f lumped dutr and apaitr hae a different mathematial frm t that f tranmiin le tub: j ω C ωc

2 4/27/25 Rihard Tranfrmatin.d 2/7 In ther wrd, the impedane f tranmiin le tub and lumped element (apaitr and dutr) are different funtin with repet t frequeny. Therefre, we an ay general that, fr example: C Hweer, fr a gien lumped element ( r C) and a gien tub (with a gien and length ) the funtin will be equal at preiely ne frequeny! Fr example, there i ne frequeny let all it ω that atifie thi equatin fr a gien, and, : jω j tan β ω jtan p r imilarly atifie thi equatin: tβ ω C ω t p T make thg eaier, let et the length f ur tranmiin le tub t λ 8, where: p 2π λ ω β

3 4/27/25 Rihard Tranfrmatin.d 3/7 Q: Why λ 8? A: Well, fr ne rean, β π 4 and therefre tan ( π 4).! Thi f ure greatly implifie ur earlier reult: π jω j tan 4 j π t ω C 4 Therefre, if we wih t build a hrt-iruited tub with the ame impedane a an dutr at frequeny ω, we et the harateriti impedane f the tub tranmiin le t be ω : jω ω λ 8 ikewie, if we wih t build an pen-iruited tub with the ame impedane a an apaitr C at frequeny ω, we et the harateriti impedane f the tub tranmiin le t be ω C :

4 4/27/25 Rihard Tranfrmatin.d 4/7 j ω C C ω C λ 8 We all thee tw reult Rihard Tranfrmatin. Hweer, it i imprtant t remember that Rihard Tranfrmatin d nt reult perfet replaement fr lumped element the tub d nt behae like apaitr and dutr! Intead, the tranfrmatin i perfet the impedane are equal at nly ne frequeny ( ω ). We an ue Rihard tranfrmatin t replae the dutr and apaitr f a lumped element filter deign. In fat, fr lwpa filter deign, the frequeny ω i the filter utff frequeny. Ug thee tub t replae dutr and apaitr will reult a filter repne imilar t that f the lumped element deign a lw pa filter with utff frequeny ω.

5 4/27/25 Rihard Tranfrmatin.d 5/7 Hweer, the behair f the filter the tpband will be ery different frm the lumped element deign. Fr example, at the (high) frequenie where the tub length beme a multiple f λ 2, the filter repne will be that f ω near perfet tranmiin! Figure 8.37 ge here Q: S why de the filter repne math the lumped element repne well the paband? A: T ee why, we firt nte that the Taylr Serie apprximatin fr tan φ and t φ when φ i mall (i.e., φ ) i: tan φ φ and t φ fr φ φ and φ i expreed radian. The impedane f ur Rihard tranfrmatin hrted tub at me arbitrary frequeny ω i:

6 4/27/25 Rihard Tranfrmatin.d 6/7 ( ω) λ 8 j tan β ω λ j ( ω ) tan p 8 j ( ω ) tan ω 4 Therefre, when ω ω (i.e., frequenie the paband f a lw-pa filter!), we an apprximate thi impedane a: ( ω) j ( ω ) tan ω 4 j ω ω 4 π jω when ω ω 4 Cmpare thi t a lumped dutr impedane: jω Se the alue π 4 i relatiely le t ne, we fd that the Rihard Tranfrmatin hrted tub ha an put impedane ery le t the lumped element dutr fr all frequenie le than ω (i.e., all frequenie f the lw-pa filter paband)! Similarly, we fd that the Rihard tranfrmatin peniruit tub ha an put impedane f apprximately:

7 4/27/25 Rihard Tranfrmatin.d 7/7 ( ω) t ωc ω 4 ω 4 ωc ω π 4 jωc π when ω ω Aga, when mpared t the lumped element apaitr impedane: C jωc we fd that reult are apprximately the ame fr all paband frequenie (i.e., when ω ω ).