Home |
Search |
Today's Posts |
#11
![]() |
|||
|
|||
![]()
On 16 jun, 20:50, K1TTT wrote:
On Jun 16, 11:16*pm, Cecil Moore wrote: On Jun 16, 6:04*pm, K1TTT wrote: would you care to provide us with the general equation for A given a complex load impedance and a line length other than a multiple of 90 degrees? It's simply the relative phase angle between the forward voltage and reflected voltage at the source resistor. I trust that you can manage that without my help. -- 73, Cecil, w5dxp.com ah, so you do have to calculate the voltages, which would requires two transformations of the length of the coax plus calculating the magnitude and angle of the complex reflection coefficient then adding them all up. *still sounds like more work than necessary. Hello all If I do not make any mistake, my numbers agree with yours Cecil, it is a pleasure (my procedure was the old plain and simple standard) :) I know, 1000 examples probe nothing, only one popperianists man can bring a black swan at any time and falsify the induction :). but now seems more clear to me you have a predictive model that render identical numbers of my classical one, at least in basic tests, it is a step forward to better considerate your efforts and with goodwill begin to establish basic points of agreement. I recognize it is a more simple approach the classic method (as said K1TTT) to me too, but also I think not always one good method or model it is more convenient to understand another useful view of phenomena. Phasorial solutions are good, practical and likely complete electric solutions but in my opinion they do not married very well with other more general electromagnetic and physics models (wave guides perhaps? I have not experience); unification has its advantages too (However my favourite answer was definitely the Roy's one in the fourth post of the thread, hi hi...) 73 - Miguel - LU6ETJ PS: Owen do not be upset with me :) most of my available newsgroup time it is spent in translate to english without flaws that may induce to misinterpretations (all of you are very demanding with precise wording and exact definitions), one mistake and I will need three or four painly translations more to clarify :D Today I tested your interesting formula with a Half wave 50 ohms TL, loaded with 100 and 25 ohms respectively. Vs=100 Vrms, Rs=50 ohm give 2:1 VSWR. In both cases then Pf 50 W, Pr=5.5 W, Pnet=44.4 W. Giving Vf=50 V and Vr=16,6 V aprox. for both loads. I am using (Pf=Pnet / (1-Rho^2) and Pr=Pnet / ((1/1/Rho^2) -1) formula which not gives different sign to Vr, in such case applying to PRs=(((Vs/2)-Vr)^2)/Rs = ((50/2)-16.66)^2/50. I get PRs=22,2 with both loads because the sign of Vf (always +) from simple Pr and Pf formulas, changing the sign of Pr render Rs=88,8 W for Prs (OK). I have in my disk a very descriptive, advisable and friendly article downloaded from: http://www.ittc.ku.edu/~jstiles/622/...es_package.pdf In page number 88 there is a agreement with your formula in "Is zo of aa HF ham tx typycally 50+j0?". |
Thread Tools | Search this Thread |
Display Modes | |
|
|
![]() |
||||
Thread | Forum | |||
Mismatched Zo Connectors | Antenna | |||
Calculating loss on a mismatched line | Antenna | |||
Collins R390 power cord and power line filter | Boatanchors | |||
Collins R390 power cord and power line filter | Boatanchors | |||
Astron RS-20A Power Supply Great Condition - used to power a VHF radio | Swap |