Resonance definition through dip in S-parameters

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SUMMARY

Resonance in transmission line stubs is characterized by dips in S-parameters, specifically S11 or S21, depending on the network configuration. An open-ended λ/4 transmission line stub creates a virtual short circuit at the junction, resulting in a dip in S21 due to energy reflection and cancellation at port 2, while a dip in S11 typically indicates resonance in a one-port resonator where energy is trapped and reflected. The discussion clarifies that the concept of "trapped energy" applies differently in high-Q resonators versus transmission line stubs, where standing waves form from the superposition of forward and reflected waves along the stub. Resonance here is defined by the formation of standing waves and impedance inversion on the Smith chart, not by continuous energy circulation as in lumped LC tanks.

PREREQUISITES

  • Understanding of S-parameters (S11, S21) in RF networks
  • Transmission line theory, specifically λ/4 stub behavior and impedance transformation
  • Smith chart interpretation for impedance inversion
  • Concept of standing waves and resonance in transmission lines

NEXT STEPS

  • Study the impact of stub termination (open vs short) on S-parameter responses
  • Analyze high-Q resonator behavior versus transmission line stub resonance
  • Explore standing wave formation and measurement techniques on transmission lines
  • Apply Smith chart tools to visualize impedance transformations in multi-port networks

USEFUL FOR

RF engineers, microwave circuit designers, and students studying transmission line resonators and S-parameter analysis will benefit from this discussion. It clarifies the interpretation of resonance phenomena in multi-port networks and the physical meaning of dips in S11 and S21 parameters related to energy reflection and standing wave formation.

yefj
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Hello , I see things from Electro magnetic point of view thats why I said that energy is traped and goes from electric to magnetic withing the stucture.
I am use to see resosnace as a dip in S11 why in the phot the say that resonance in a dip in S21?
Thanks.
 

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An open-ended λ/4 transmission line stub, appears as a short circuit in the network.
λ/4 is halfway around the Smith chart, so the impedance is inverted about Zo.

S11 = 0dB, says at the resonant frequency, all the incident energy is reflected by the virtual short circuit to ground.
S21 = -40dB, says at resonant frequency, energy does not pass through the network, because port 2 is virtually shorted to ground.

Where does the energy go?
It is reflected by the open-ended λ/4 transmission line stub resonator.

If the stub was instead shorted to ground, the network would appear to be an open circuit, which is sometimes called a metal insulator.
 
The basic definition of resonance is "energy trapped within a cavity and it going from electric to magnetic type"
So whenrgy is trapped in a covity and doesnt go back thats why I understand we have a DIP in S11 of one port resonator.

but how can we understand the dip of S21 using same "trapped energy definition of resonance"

or there is some other basic definition of resonance which allows DIP in S11 and another case Dip in S21(as shown in this two port case)?

I have a conflict in definitions regarding what is resonance because a resonator has dip in S11 unline in this case, so why they both considered the same "resonance" where is the trapped energy in the two port case?
Thanks.
 
The "trapped" energy is not relevant. Once a high-Q resonator is oscillating, no more energy is required to keep it oscillating. S-parameters assume a steady state. Maybe it is your concept of trapped energy circulating continuously in the resonator, like an LC tank circuit, that confuses the issue.

Avoid high-Q resonance, and look instead at the length of the stub. The input signal travels along the stub from the junction, then is reflected back to the junction. With an open-ended line, the phase of the incident wave cancels the reflected wave at the junction, making it look like a ground. If the stub were shorted at the end, the phase of the reflected wave would reinforce the incident wave.

The signal only propagates along the stub, once each way. You can see that it is not a high-Q resonator, because the dip is very wide, not narrow, as it would be with a high-Q resonant circuit.
 
yefj said:
the reflected wave cancels the incoming wave at port 2 so why its a standing wave situation?
The cancellation occurs where the stub joins the line between port 1 and port 2. The forward wave in the stub is reflected from the impedance mismatch at the open-end of the stub, to become the reflected wave. The forward and reflected waves propagate independently. The standing wave is formed from the sum of the forward and reflected waves in the stub.
 
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yefj said:
Hello baluncore , from reading at the following link resonance is a situation when we have standing waves.
is this what happens here?
the reflected wave cancels the incoming wave at port 2 so why its a standing wave situation?
Thanks.

https://www.allaboutcircuits.com/textbook/alternating-current/chpt-14/standing-waves-and-resonance/
In practice, the standing wave will not be 'perfect'. The range of amplitude with time (and position) inside the resonator will be due to the basic resonance and the power flow through the resonator.

The term 'standing wave' refers to a resonator with dimensions greater than a wavelength. A 'lumped component 'resonator may not have identifiable maxes and mins in the same way as a transmission line. This is another instance where a commonly used term may add confusion rather than enlightenment.
 

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