5.2 Single-Stub Tuning

Transcription

1 3/26/29 5_2 Sgle_Stub Tung.doc 1/1 5.2 Sgle-Stub Tung Readg Assignment: pp Q: If we cannot use lumped elements like ductors or capacitors to build lossless matchg networks, what can we use? A: Recall that a section of lossless transmission le is purely reactive, thus we can build lossless matchg networks usg specific lengths of transmission les. We call these lengths of transmission le distributed elements. The distributed element analogue of the lumped element L- network is the sgle- tuner. Just like the L-network, there are two versions of this design: HO: THE SHUNT-STUB TUNER HO: THE SERIES-STUB TUNER

2 3/26/29 Shunt Stub Tung.doc 1/5 Shunt Stub Tung Consider the follow transmission le structure, with a shunt : Y The two design parameters of this matchg network are lengths and d. An equivalent circuit is: Z, β j B Y z = where of course:

3 3/26/29 Shunt Stub Tung.doc 2/5 Y + j Y tan βd = Y Y L Y + j Y tan β d L and the reactance either: jb of transmission le of length is jb jy tanβ for an open-circuit = jycotβ for an short-circuit Therefore, for a matched circuit, we require: jb + Y = Y Note this complex equation is actually two real equations! i.e., and Re{Y } = Y Im{ jb + Y } = B = B where B Im{Y }

4 3/26/29 Shunt Stub Tung.doc 3/5 Sce Y is dependent on d only, our design procedure is: 1) Set d such that Re{Y } = Y. 2) Then set such that B = B. We have two choices for determg the lengths d and can use the design equations (5.9, 5.1, 5.11) on p. 232,. We OR we can use the Smith Chart to determe the lengths! 1) Rotate clockwise around the Smith Chart from y L until you tersect the g =1 circle. The length of this rotation determes the value d. Recall there are two possible solutions! 2) Rotate clockwise from the short/open circuit pot around the g = circle, until b equals b. The length of this rotation determes the length. For example, your book describes the case where we want to match a load of ZL = 6 j8 (at 2 GHz) to a transmission le of Z = 5Ω.

5 3/26/29 Shunt Stub Tung.doc 4/5 Usg shorted s, we fd two solutions to this problem: Whose length values d and where determed from a Smith Chart: short

6 3/26/29 Shunt Stub Tung.doc 5/5 Q: Two solutions! Which one do we use? A: The one with the shortest lengths of transmission le! Q: Oh, I see! Shorter transmission les provide smaller and (slightly) cheaper matchg networks. A: True! But there is a more fundamental reason why we select the solution with the shortest les the matchg bandwidth is larger! For example, consider the frequency response of the two examples: Clearly, solution 1 provides a wider bandwidth!

7 3/26/29 Series Stub Tung.doc 1/4 Series Stub Tung Consider the followg transmission le structure, with a series : Z Therefore an equivalent circuit is: Z, β j X Z z = where of course: Z + j Z tan βd = Z Z L Z + j Z tan β d L

8 3/26/29 Series Stub Tung.doc 2/4 and the reactance jx is either: jx jz cotβ = jz tanβ for an open-circuit for an short-circuit Therefore, for a matched circuit, we require: i.e., and jx + Z = Z Re{ Z } = Z Im{ jx + Z } = X = X where X Im{ Z } Note the design parameters for this tuner are transmission le lengths d and. More specifically we: 1) Set d such that Re{ Z } = Z. 2) Then set such that X = X. We have two choices for determg the lengths d and. We can use the design equations (5.14, 5.15, 5.16) on pp. 235.

9 3/26/29 Series Stub Tung.doc 3/4 OR we can use the Smith Chart to determe the lengths! 1) Rotate clockwise around the Smith Chart from z L until you tersect the r = 1 circle. The length of this rotation determes the value d. Recall there are two possible solutions! 2) Rotate clockwise from the short/open circuit pot around the r = circle until x equals x. The length of this rotation determes the length. For example, your book describes the case where we want to match a load of ZL = 1 + j8 (at 2 GHz) to a transmission le of Z = 5Ω. Usg open s, we fd two solutions to this problem:

10 3/26/29 Series Stub Tung.doc 4/4 Whose values where determed from a Smith Chart: open Aga, we should use the solution with the shortest transmission les, although this case that distction is a bit ambiguous. As a result, the bandwidth of each design is about the same (dependg on how you defe bandwidth!).