• We’re currently investigating an issue related to the forum theme and styling that is impacting page layout and visual formatting. The problem has been identified, and we are actively working on a resolution. There is no impact to user data or functionality, this is strictly a front-end display issue. We’ll post an update once the fix has been deployed. Thanks for your patience while we get this sorted.

Transmission Line Gurus...

H

hypn0tik

Diamond Member
I have the following question from a problem set of mine.

Clicky

I found the impedances looking into the open and the shorted segments (Z1 and Z2 as indicated in red) and found them to be:

jZ0Bd and Z0/jBd.

This seems go suggest that you have an inductor in series with a capacitor.

However, when I add them up to find the equivalent impedance:

Z1 + Z2 = Z0(jBd + 1/jBd) = Z0(-(Bd)^2 + 1)/jBd.

However, using the approximation, we find that -(Bd)^2 + 1 approx.= 1
=> Ztotal = Z0/jBd

When you look at part two of the question, they ask you for an expression for the resonant frequency. This seems to suggest that you cannot make the approximation to get rid of the inductor.

Any ideas?

Thanks in advance.

<Obligatory> Do your own homework </Obligatory>

Edit:
<Obligatory> Head explodes </Obligatory>
 
I was horrible at electromagnetics. Sorry. But it sounds like they want the equation that will generate your frequency response of the system.
 
Then don't make use of the small angle approximation. Keep it in exact terms and you can solve for a resonance such that \beta d < 1. I would not describe it as being << 1 but obviously making the small angle approximation requires you to break that relationship to achieve resonance.
 
Anyone?

I decided to say 'screw it' to part a and left them as they are without employing the approximations again when adding them.

For part (b), the series impedance of a LC circuit can be written as:

[1-(w/w0)^2]/jwC, where w is the frequency of the excitation signal and w0 =1/sqrt(LC)

Any ideas on what I do from there?
 
For part a), it seems to me that you should be taking the parallel combination, rather than the series, from the given geometry.

As for b), I'm tempted to guess that for frequencies much smaller than w0, it won't work. If you are able to define some sort of Q(Quality factor) for this resonanace- here it's infinite becasue of ideal elements, then maybe the width of the resonance would be a good approximation.....

BTW, I could be completly wrong, as I have yet to take any transmission line theory. Ask me the same question next semester, and I'll give you a more accurate answer 😉
 
For part a), it seems to me that you should be taking the parallel combination, rather than the series, from the given geometry.

As for b), I'm tempted to guess that for frequencies much smaller than w0, it won't work. If you are able to define some sort of Q(Quality factor) for this resonanace- here it's infinite becasue of ideal elements, then maybe the width of the resonance would be a good approximation.....

BTW, I could be completly wrong, as I have yet to take any transmission line theory. Ask me the same question next semester, and I'll give you a more accurate answer
Click to expand...

You would take the elements in series here. You would only consider them being in parallel, or better yet look at the admittance, if both of the elements were tuning stubs. But the first element is in series with the signal line. If it was in parallel, its return would be on the larger circuit's return, not its signal line. I would say that for b I would think about it as a Q factor as well but I'm not going to spend anymore time looking into this. I know from a quick calculation that you can acheive resonance in part a if you do not do any approximations. You can simply use the \beta d << 1 relationship to pin it down to a single \omega_0. Otherwise, since \beta d is periodic, you could get any number of resonant frequencies depending on what \beta d you choose.
 
When I woke up this morning, I was hoping that I'd feel vastly stupid at one point today.

Thanks for granting my wish.
 
You must log in or register to reply here.

TRENDING THREADS

Back
Top