i have a set of issues,
1) i'm using AWR design environment. In case of the basic (conventional) power divider i'm getting the output. But in case of the optimized power divider in this software if i'm adding the capacitors at the schematic level i'm getting the output (i.e.) my Scattering parameters are showing high isolation , high return loss at ports and 3.3 dB insertion loss. but when i'm trying to do the EM simulation using the AXIEM simulator in the software by using the method of extraction. the EM structure is being generated but im not getting the desired results and the results that i had got in the schematic are also modified and it is same as that of the EM structure
This is a common issue that new designers face. Simulations with closed-form component models (i.e. MLIN, MTEE, MBEND, etc) do not take into account all of the coupling between adjacent metal structures. When you port the design from schematic to layout and simulate it with a true EM solver, you are seeing a more realistic answer. If you fabricated the circuit, it should work very similar to the AXIEM results. These are the steps most designers follow in practice:
1) Create basic model with schematic simulation to prove the basic concept and get a starting point
2) Tune the schematic-based model
3) Port the schematic model to layout and simulate with you favorite EM solver (AXIEM, Momentum, etc).
4) Tweak/tune your layout using the EM solver (iterate until it looks right)
5) Fabricate, test circuit, compare results to simulation from step 4.... retune/repeat as necessary
2) I'm using thin film resistor for the isolation resistor in case of Software simulation but for physically testing the fabricated layout using a network analyzer if i solder a SMD resistor in the place of the thin film resistor will i have any problem
At 1 GHz, you shouldn't have any problem. Watch out for your resistor's impact to the signals' phase, so you maintain good cancellation inside the structure, and thus maintain good isolation in your Wilkinson design.
For a Wilkinson divider, you want the length of the resistor to be small portion of a wavelength. Follow this little exercise to see how some of the signals cancel each other inside the structure. A signal reflected from output 1 will bounce back and split at the junction for the resistor. Some of the energy will go around the two quarterwave sections (i.e. winds up 180 degs out-of-phase) and will meet up with the in-phase signal that went straight across the resistor. If you make the resistor too long, you won't get the 180 degree-delayed and in-phase signals to meet up perfectly, because the "in-phase" signal will acquire some delay by traversing a long resistor (no longer perfect cancellation, less than perfect port-to-port isolation).
A good rule of thumb is < 5 degrees. This length varies depending on your frequency (1 GHz?) and dielectric constant of the substrate. Assuming an Er = 4, then 5 degs @ 1 GHz would be 82 mils (0.082"). So, and 0805 resistor should be the largest you'd want to go, initially. In case you didn't know the convention, an 0805 resistor/capacitor has .08" x .05" package dimensions. Small resistors (shorter length, less phase delay, better cancellation) is better. So go as small as you can, but remember the power handling of the part. If one port is a perfect reflection (open or short), then the resistor will dissipate 1/2 of the incident power. The other half of the power will go the other port (if it's terminated in a good matched load). If both output ports are matched, then the resistor sees no power dissipation.