# How to get accurate coupling coefficient of two microstriipline resonator by HFSS

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#### ermai

##### Member level 1
"Narrow-Band Microwave Filter Design by Daniel G. Swanson, Jr"

The model for extraction of coupling coefficient is as the following picture:

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But I am confused that the two peaks of S12 are depended on the width of gap between probe and resonator and the position of the probe along the resonator,
how do I know how the gap and position is to get the accurate coupling coefficient?

#### Phytech

##### Full Member level 3
You need to get the right In/Out coupling to get an accurate result, i.e. if the In/Out coupling is to strong, you may get smaller k12, and if it is too weak, it is not accurate. Typically, if you get the center Vally of about 20 dB from its peaks, it is about right.

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### hiepsikid007

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#### ermai

##### Member level 1
You need to get the right In/Out coupling to get an accurate result, i.e. if the In/Out coupling is to strong, you may get smaller k12, and if it is too weak, it is not accurate. Typically, if you get the center Vally of about 20 dB from its peaks, it is about right.
Thank for your help very much.
I want to know why it is right when the valley is 20dB from peaks.....thanks

V
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#### Phytech

##### Full Member level 3
Thank for your help very much.
I want to know why it is right when the valley is 20dB from peaks.....thanks

Just by experience. You can also use strong In/Out coupling which results in much lower dB value of the valley. But you may not be able to use the same formula to calculate the k12. However, you can still calculate an accurate k12 value form this simulation. For example, you may get a two pole filter response with the center valley of 1 dB, then you find 1 dB (or 3-dB) bandwidth and deduct the k12 value from there (instead of the distance of the peaks).

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#### ermai

##### Member level 1
Just by experience. You can also use strong In/Out coupling which results in much lower dB value of the valley. But you may not be able to use the same formula to calculate the k12. However, you can still calculate an accurate k12 value form this simulation. For example, you may get a two pole filter response with the center valley of 1 dB, then you find 1 dB (or 3-dB) bandwidth and deduct the k12 value from there (instead of the distance of the peaks).

Thank you...you actually my benefactor.....haha
May I ask you how to deduct k12 from 1dB bandwidth?Thank you very much!:-D

---------- Post added at 03:39 ---------- Previous post was at 03:14 ----------

Just by experience. You can also use strong In/Out coupling which results in much lower dB value of the valley. But you may not be able to use the same formula to calculate the k12. However, you can still calculate an accurate k12 value form this simulation. For example, you may get a two pole filter response with the center valley of 1 dB, then you find 1 dB (or 3-dB) bandwidth and deduct the k12 value from there (instead of the distance of the peaks).

From many artical,the equation of K12 is (W1^2-W2^2)/(W1^2+W2^2)
I don't know what "you may get a two pole filter response with the center valley of 1 dB, then you find 1 dB (or 3-dB) bandwidth and deduct the k12 value from there (instead of the distance of the peaks)" means... thanks

V
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#### Phytech

##### Full Member level 3
Thank you...you actually my benefactor.....haha
May I ask you how to deduct k12 from 1dB bandwidth?Thank you very much!:-D

---------- Post added at 03:39 ---------- Previous post was at 03:14 ----------

From many artical,the equation of K12 is (W1^2-W2^2)/(W1^2+W2^2)
I don't know what "you may get a two pole filter response with the center valley of 1 dB, then you find 1 dB (or 3-dB) bandwidth and deduct the k12 value from there (instead of the distance of the peaks)" means... thanks

It's very simple. You design a two-pole filter with passband ripple of 1-dB with the same 1-dB (or 3-dB) band width from your simulated response. By designing this filter you used a k12 value and this value equals the k12 from your simulation.

ermai

### ermai

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#### ermai

##### Member level 1
It's very simple. You design a two-pole filter with passband ripple of 1-dB with the same 1-dB (or 3-dB) band width from your simulated response. By designing this filter you used a k12 value and this value equals the k12 from your simulation.

you are real smart...... this is a good method..
But I want to extract K12 from HFSS directly.........

My question is that
Following is model for simulation

N is width of gap between resonator and open microstripline,M is probe position along resonator....
I got different peak frequency from HFSS with different M or N

so I want to know the equation K=(W1^2-W2^2)/(W1^2+W2^2) is exact with what M or N..
Thanks.

V
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#### Phytech

##### Full Member level 3
From the picture you showed, it seems to me that you need to reduce the value of N (say to 0.05 mm) to make the In/Out coupling stronger (alternatively, you can increase the width of input/output strips). You can use the above formula to calculate your coupling coefficient after you make you In/Out coupling stronger.

ermai

### ermai

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#### ferdows724

##### Newbie level 2
Hi dear friends...
I simulate two triangle using HFSS for calculate the electric coupling coefficient. I don't see any two peaks in S21!!! I loose coupling in feeds, but I don't see any two peak again! please help me.
View attachment untitled.jpg.bmp

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