Diplexer Design and Optimization Problem

[ali_cmi]

Hello!

I was given a task to optimize a contiguous LPF-HPF suspended-substrate diplexer (doubly-terminated filters). The modelling was done in HFSS. I re-shaped/designed the model to make it almost fully relative, that is, it accepts the variations of length and width in one of the filter elements.
The problem is, when I try to optimize it, all four optimization algorithms of HFSS just don't reduce the Error-Factor (EF) below 100. But when I do it myself, the EF reduces to around 28 and the response is still not-acceptable. There are 7 variables in total which control the pole-positions of LPF (2-3 poles) and HPF (4 poles) and I don't understand there relation effectively. When one of the filters responds as it should, the other one's totally messed up.
View attachment dip_208G_fab_test_12_LPF_HPF_combined_10_start-8_HPF_mods_16.rar
As a susceptance-annulling stub, a physical screw is being simulated in HFSS: changing its height at the tee-junction cancels the imaginary part of impedance.

Kindly guide me to a better understanding of such diplexers by sharing some design information/material/data. Attached is my design. Take a look at it.

Waiting for your response.

Muhammad Ali

 

[biff44]

you have been given an impossible task then. to make a continguous diplexer filter with matching at all 3 ports, you need singly terminated prototype filters.

In other words, you need a lowpass that has an admitance at the junction of Y= 1 + jB(f) in the lowpass passband. The highpass has Y=0-jB(f) in its the lowpass passband. and vice versa.

That way the junction admitance is 1 (since the reactive admittance of the HP cancels the reactive admittance of the LP).

There are some papers by Rhodes that are useful to understand the concept.

You might have to jigger the HP and LP cutoff frequencies to get G=.5 real part right at the crossover frequency. at the crossover you want junction admittance = admittance of LP + admittance of HP = 0.5 +Jb' + 0.5 -Jb' = 1, i.e. matched.

Obviously, the lowpass by itself will have very poor performance. It only works when the singly terminated LP end is attached to the singly terminated HP port.

 

[ali_cmi]

@Biff44

Thanks a ton for replying.

BIff, I understand that reactive parts should cancel each other, leaving us with the beauty: 50 ohms. But how do I do it practically and with the given design? A pragmatic example would do me great favor.

Ali