Your GUESS? YOUR GUESS? Via the Smith Chart, one can easily see that traveling on a constanct VSWR circle from the load clockwise to the source that 1/4 wavelength trip will put you at the 25 Ohm poi
Correct, Pete. As you move along the transmission line (assuming a 50 Ohm line in a 50 Ohm system), the VSWR remains constant. But the reactances definitely change and at some points, you do have zer
How can one calculate it mathematically? Gee, that's tough! Use a computer! :-) What do you mean by most "extreme" impedance? OK, with a normal Smitch Chart NORMALIZED to 50 Ohms: 100 Ohms is on the
Amps aren't shielded with metal shielding to protect your eyes! They are made of metal primarly for the purposes of strenght and RF grounding. The filaments in a tube are dimmer than a lightbulb. You
No, that's not correct necessarily. A 100 Ohm termination used in a 50 Ohm system has no reactive components. Adding a 1/8 wavelength of 50 Ohm transmission line will make that 100 Ohm real load look
Approximately 176.7 Ohms. It's not necessarily exact since I took 10 seconds to complete it using the Smith Chart normalized to 93 Ohms. 73, Jon KE9NA -- The Second Amendment is NOT about duck huntin
OK. I understand what you are saying now. But I think a more accurate way to describe it would be to say minimum reactance (obviously zero reactance is a minimum). If you have a fixed antenna and a f
This isn't correct, Vic. In a 93 Ohm system, the SWR is unchanged. However, we are attempting to measure a 50 Ohm load in a system where 50 Ohms is our base (the SWR meter has been specified as being
How is that Weird Sceince, O Great One? Or do you say that the Smith Chart is not correct and that all the people throughout the world who use it are wrong? If I have a 50 Ohm piece of coax in a 50 O
Obviously, Vic, you've missed the whole discussion I have participated in. If my transmission line impedance is the same as the impedance that I am operating in, then you move along a constant VSWR c
But Vic, Source impedance is 50 Ohms. Load impedance is 50 Ohms. The transmission line is a length where even you agree that the transformed impedance at the end of it is 50 Ohms. So where is the imp
It has zero REAL resistance. However, when calculating complex impedance it has plenty of Ohms. The complex (or imaginary) part has plenty of resistance in the case of a 100 Ohm load at 1/8 wavelengt
Correct. I wrote the statement above yours and in my entire post specified the system as a 50 Ohm system. Not exactly, Rich. If the load impedance is 100 Ohms and the transmission line impedance is 9
No. Not correct. If so, then please explain the operation of a quarter wave transformer. The VSWR is only dependent on the load/cable mismatch PROVIDED THE CABLE IS THE SAME IMPEDANCE AS THE SOURCE!
IF THE SOURCE WAS 93 OHMS, YES! But you are dealing with a 50 OHM source! -- -- -- -- You have to think of the entire system. The 93 Ohm line acts as an impedance transformer. The impedance at the in
Oops! Roger, you are correct. Mea Culpa! When I was reading the chart this afternoon, I noticed that but ignored it since I was thinking in terms of an X-Y axis. At 1/8 lambda, you have no X componen
No, RIch, they don't if your coax and your exciter are of the same characteristic impedance. The impedance of the load will appear to vary as you move down the line (if it is different than the chara
Not if the impedance of your transmission line is the same is the impedance of your source. Draw it out on a Smith Chart, Rich. You move along a constant VSWR circle with 50 Ohms as the center of tha
I admitted to this mistake. At 1/8 wavelength a 100 Ohm load on a 50 Ohm line will appear as 40-j30 Ohms. I wasn't careful to read my Smith Chart and my foot tastes really good to me right now! 73, J
Rich, MUST you debate everything? Even when somebody agrees with you? It IS a 50 Ohm system because you are defining the measuring device as 50 Ohms. Therefore, you want to see what the SWR with a 50