# NCEES 128 High-Frequency Transmission



## schmidty99 (Feb 24, 2011)

Just wondering if anyone can explain this problem to me. The NCEES answer is using an equation to solve this problem that I haven't come across in the EERM and elsewhere. I can follow the math to get to the answer, but it just seems there would be a better way. Any help would be appreciated. Thanks!


----------



## CincinnatiControlsGuy (Feb 24, 2011)

I'll take a look when I get home from work unless you can post the actual problem. Transmission line problems seem to be relegated to simple compensator circuits that would allow for maximum power transfer (output impedance matches input impedance). There really are only a few points that you need to know and Smith Charts are out. Why? They don't allow any form of compass into the exam, thereby negating that type of problem.


----------



## schmidty99 (Feb 24, 2011)

Sounds great! I think you can get close with a Smith chart by "eyeing" it up, but I'd rather know the analytical way. Thanks!!


----------



## CincinnatiControlsGuy (Feb 24, 2011)

Actually, the equation does exist. It's 57.11. Ask yourself what Zload is on a shorted stub? Clearly, it is 0. This simplifies equation 57.11 to Zin = Zo * j * tan(Bl). Beta can be substituted via equation 57.5.

The really lame part is in the wording. The admittance at point X is 0.2 + 0.4j Siemens. Notice if that pesky 0.4j weren't there then the admittance would be 0.2 and, therefore, the impedance would be 50 ohms. So, the stub needs to be -0.4j Siemens. This is because admittances in parallel add. The -0.4j Siemens admittance from the stub is the same as saying 1/-0.4j = 25j ohm impedance. This is Zin on the stub. Now you have Zin = 25j going into the equation (simplified 57.11 above with Zo = 75 ohms). You can solve for l, which equals 40.9mm.

THAT SAID. There appears to be a typo that I missed when studying for the October exam. The solution contains a (-j) term in the sentence starting with "Analytically, the input...". It would be nice to get a second opinion on that, but their analysis proceeds as if it's positive.

Anyway, I can always write the problem up if need be and post it for clarification.


----------



## schmidty99 (Feb 24, 2011)

Thanks for the explanation. I would have never got that, another weakness. I think I understand the derivation of the equation, mainly that Zload being zero takes out half of the equation and the tan x =sin x/cos x indentity, but I could still use the detailed math explanation if you don't mind. Do you feel that this solution is the only/best way to complete this problem? What text books besides the EERM did you use to study these problems? Thanks!


----------



## CincinnatiControlsGuy (Feb 24, 2011)

No problem. I'll write it out tomorrow morning/afternoon. Unfortunately, this is a pretty solid way of doing these types of problems as Smith Charts aren't super practical for the exam. The EERM and Wikipedia were my only sources for transmission lines and I would be lying if I said they were my strong suit. That said the underlying theme for transmission lines is maximum power transfer, at least as far as practice problems are concerned. Remember that if input impedance matches load impedance then you have maximum power transfer and none is reflected or lost. So, all you are really doing is picking points on the transmission line and adding "compensators" to match the load and generator. What if they had said an open stub? The trick there is knowing that Zload is infinity. Also, stubs don't have to be parallel nor do they have to be strictly shorts or opens. The key is knowing how to handle them. A good place to start is by working with lumped parameter circuits. If you had a circuit that consisted of just a capacitor and a resistor how would you make it appear like just a resistor on the input? You would add a series inductor or a parallel capacitor of sufficient magnitude. I'll throw some pointers on the math explanation to help you out.


----------



## schmidty99 (Feb 24, 2011)

That all sounds reasonable. Again, its hard for me to think about like you do, but as soon as you say it, it makes perfect sense and I understand it. I'll be looking forward to your solution. I really appreciate all this, I hope its not too much trouble.


----------



## CincinnatiControlsGuy (Feb 25, 2011)

schmidty99 said:


> That all sounds reasonable. Again, its hard for me to think about like you do, but as soon as you say it, it makes perfect sense and I understand it. I'll be looking forward to your solution. I really appreciate all this, I hope its not too much trouble.


I'm going to have to hit you up this weekend as I was busy today. Sorry.


----------



## schmidty99 (Feb 25, 2011)

No problem. Whenver you get a chance. Thanks!


----------



## CincinnatiControlsGuy (Feb 26, 2011)

Ok, I'm attaching the problem with an analytical explanation. It also contains the solution had they asked for an open stub instead of a shorted stub. One thing to notice between the result for the shorted stub and the open stub is that they are 0.2 m difference in length. Since the wavelength is 0.8 m, this is the same as saying 1/4 wavelength difference. See results here: http://en.wikipedia.org/wiki/Transmission_...ansmission_line.


----------



## schmidty99 (Feb 28, 2011)

Thanks GB! I'll look it over tonight and let you know if I have any questions.


----------

