I would like to suggest that use the ends of the split rings to access the antenna electrically. From the electrical Fields distribution, one may see that the outer terminals are suitable to feed the antenna as the electric field is emanating from one and enter in the other.

You can study the suitability of this solution by calculating S11 of this port.

Please it would be good when you display S11 calculations.

Best wishes

Dear Abdelhalim abdelnaby Zekry thank you very much. I have simulate the design in cst by exciting through discrete port on outer ring. Unfortunately S11 parameters doesn't match with original results. My design pic are given below.

Dear Muhammad,

It is required to see S11 as a function of frequency to judge on your design.

Is there is a ground plane for this antenna? and so you need only one point on the ring to feed the antenna. I could not access the paper which you reproduce its results. The main point is that one search for a feed point matching the 50 ohm source.

Best wishes

Dear Abdelhalim abdelnaby Zekry yes it has ground plane on the back side. I have simulated the design in CST. but result doesn't match. My cst file and paper is attached below.

Dear Muhammad,

welcome.

Th reference paper indicates that you can excite the outer ring at the point x making an angle phip with horizontal axis. Please refer to Fig. 3a in the paper.

This point can be changed to match the ring to the output impedance of the tank.

Best wishes

Deaar Abdelhalim abdelnaby Zekry I understand your point. As you know single discrete port has real power 50 ohm. on the other hand CHIP has complex power. It means a+jb. therefore, for perfect matching antenna must has a-jb impedance. My point is that how discrete port match this conjugate impedance.

Dear Muhammad,

welcome,

I think the discussion will be fruitful and finds a solution ISA.

Yes You must choose a point where the two resistances are eqaul where and the two reactances cancel each other. Since the the ASIC chip has cpacitive reactance the resonance circuit of the ring must be operated to the left of the resonance to be inductive. So, it is so that you search a point for the equality of the resistance and shift the operating frequency down ward resonance. This is the trick!

Best wishes

Dear Abdelhalim abdelnaby Zekry thank you very much for you time and helpful suggestions. I will pray for your success, carrier and hereafter.

thanks a lot.