Improving matching by adding a stub

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yefj
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Hello, I have a gap transmission line resonator the gaps are sighned in arrows.The matching is -5dB.
Is there a way to increase the matching by adding an opencircuit stube to the transmission line resonator?
It will be great to have some physical intuition.
Thanks.
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We live in a world of mismatched impedance. Where there is no change in impedance, we cannot see the boundary. If you match the resonator to the line, it will no longer resonate. The greater the end mismatch, the greater the Q of the resonator.

Do you want a deep rejection notch or a BPF?
You are plotting S22, so what do you connect to port 1 on the line passing through the resonator? How will the be used in a circuit?


Why use gaps when you can overlap λ/2 dipoles, en echelon, which reduces the area of the PCB. There are several filters etched on PCBs shown here.
https://en.wikipedia.org/wiki/Distributed-element_filter
 
Hello Baluncore, Its not a filter but a resonator, I am supposed to connect a capacitor in one gap to shift the resonance frequency as shown below, however that dip of S22 gets small ,I thought to use a stube to incread the dip is it possible?
Thanks.
"Do you want a deep rejection notch"
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I cannot see how a stub can help you.

You have a dipole resonator in series with a transmission line. That resonator will act as a deep notch filter for signals travelling in the line. The signal voltage will appear across the resonator, not at the output port.

If the dipole resonator was connected between one port, or a signal node, and ground, then the resonator would enhance signals of that frequency, while all others would be attenuated.

I cannot guess what you are trying to do, so it is difficult for me to suggest how or where to change the system. Can you produce a block diagram of the wanted circuit?
 
The Smith Chart shows a reasonable conductance value but a large negative susceptance. If you shunt a positive susceptance across the line at the generator side that will cancel this out. An open circuit line shorter than a quarter of a wavelength will provide the correct compensation. X = Zo / tan (theta), or Y = tan (theta) / Zo.
 
yefj said:
I am supposed to connect a capacitor in one gap to shift the resonance frequency as shown below
Wait, is this question for your schoolwork?
 
Hello Tech99,I am trying to understand the words you said.

I have a microstrip with two gaps ,I am having problem visualising where do I need to put the open circuit stub.
Could you please tell me on drawing below where should the stub be located?
Thanks.
The Smith Chart shows a reasonable conductance value but a large negative susceptance. If you shunt a positive susceptance across the line at the generator side that will cancel this out. An open circuit line shorter than a quarter of a wavelength will provide the correct compensation. X = Zo / tan (theta), or Y = tan (theta) / Zo.
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yefj said:
Hello Tech99,I am trying to understand the words you said.

I have a microstrip with two gaps ,I am having problem visualising where do I need to put the open circuit stub.
Could you please tell me on drawing below where should the stub be located?
Thanks.
The Smith Chart shows a reasonable conductance value but a large negative susceptance. If you shunt a positive susceptance across the line at the generator side that will cancel this out. An open circuit line shorter than a quarter of a wavelength will provide the correct compensation. X = Zo / tan (theta), or Y = tan (theta) / Zo.
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The stub needs to be positioned on the transmission line not on the resonator. It needs to go on the sending end. It should be placed at the point where you measured your Smith Chart.
 
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