# Resonance condition in transmission line

In this document, a transmission line terminated at both ends with purely reactive loads is considered as an example for the Transverse Resonance Method.

In order for this circuit to be resonant, these conditions must be satisfied:

$$V^r = V^l$$

$$I^r = I^l$$

and being $\overleftarrow{Z}_{in} = - V^l / I^l$ and $\vec{Z}_{in} = V^r / I^r$, this implies

$$\overleftarrow{Z}_{in} = -\vec{Z}_{in}$$

How can the conditions $V^r = V^l$ and $I^r = I^l$ be related to resonance? They rather seem continuity equations, which should trivially be satisfied by every transmission line, even without resonance.

So, how can the above conditions instead represent the ability of the circuit to maintain nonzero voltage/currents even with zero sources, that is the resonance?

I am not asking what is the meaning of the above formulas in the linked document. I am asking about what they physically represent. A resonant transmission line is a line where a standing wave exists, even without a source, and which is ideally terminated on reactive, mismatched loads. How can this condition be represented by $V^r = V^l$ and $I^r = I^l$ at any position $x_0$? I can't see any link between these two equations and the resonance condition.

• As for how the equations might not be satisfied, give us some more context about what's written around that figure in your book. It looks to me like they're trying to figure out at which frequency the line is resonant...but it's hard to be sure. – The Photon Jul 4 '17 at 14:25
• @ThePhoton It was a typo, sorry, I fixed it. Unfortunately, apart from the linked document itself, I have no more context. There, the purpose is to obtain the equation $\overleftarrow{Z}_{in} = -\vec{Z}_{in}$, which can be used as a constraint to finally obtain the resonance frequency of a cavity, or wavenumbers for dielectric guides. But my question is about the physics of this problem: how can such conditions on voltages and currents at the reference plane $x_0$ mathematically imply the resonance? – BowPark Jul 4 '17 at 14:46
• Not having Office I can't read your linked document. My guess is that it's exactly the point --- if the voltages and currents aren't continuous, then that's not a solution for the equations describing the system. So to find resonance, you need to find a frequency where the voltages and currents can be continuous. – The Photon Jul 4 '17 at 17:30
• @ThePhoton Sorry for being late. Maybe you got the point, but I still can't understand. The only equations I know about this system are the Telegrapher equations and the corresponding wave equations $\displaystyle \frac{\partial^2 i(z,t)}{\partial z^2} = LC \frac{\partial^2 i(z,t)}{\partial t^2}$ (and the one for $v$). This kind of equation can accept any solution such as $f(t - z/v)$, where $1/v^2 = LC$. I can't see instead any link between this and the continuity that you mentioned. – BowPark Jul 10 '17 at 10:50
• @ThePhoton The fact is that any wave must have continuous voltages and currents: this is IMHO not a special condition for standing waves. So, can you please rephrase your statement? – BowPark Jul 10 '17 at 10:56

Here's an alternate way of looking at it.

Resonance in this structure happens when the signal propagates through the structure back to the starting point, and when it gets there it is in the same phase it was when it started. This means the wave is essentially constructively interfering with the delayed version of itself.

Another way to say this is that if it propagates round trip through the structure and doesn't end up with the same phase it had when launched, you wouldn't produce a standing wave, you'd produce a transient that would die out quickly due to losses in the system (which we otherwise ignore).

The cleverness of the transverse resonance method (as far as I can tell) is recognizing that it's mathematically equivalent to say "the input impedance looking in the two directions is equal" or "the round trip phase change is a multiple of $2\pi$."

• Even if after a long time, would you like to talk about this in chat? Or would you prefer here with comments? I'd like to discuss with you a little bit more. – BowPark Oct 4 '17 at 20:02
• (If you want or can, I'll be here for a couple hours from now on). – BowPark Oct 4 '17 at 20:08
• @BowPark, I'm in the main chat but it might be better to start a new room. – The Photon Oct 4 '17 at 20:19
• New room here: chat.stackexchange.com/rooms/66608/… – The Photon Oct 4 '17 at 20:20