What happens to output power and 3 BW in this circuit

Hi, I calcaulated L and C values for a butterworth filter, to get 3 dB BW of 50 GHz and attenuation of 6 dB at 60 GHz. Next, I gave 10 dBm input power and I got the expected result.
Again I took two sections of the same T network and gave input at different points as shown in figure, I am not sure what is its expected behavior. can someone explain how this circuit should work? Thanks

circuitslave said:

Again I took two sections of the same T network
and gave input at different points as shown in figure,
I am not sure what is its expected behavior.

Click to expand...

Why do you set three ports.
Where is driving point ?

What do you expect ?

circuitking said:

Again I took two sections of the same T network and gave input at different points as shown in figure

Click to expand...

The ports connected to the capacitors providing input power of 10 dBm each (loaded LC ladder section). I kept ports to find the ZP and SP at those points. I expect to see more output power at the last port(Port 3 in the figure). But I am not sure its working technicalities and analytical equations yet.

circuitslave said:

But I am not sure its working technicalities and analytical equations yet.

Click to expand...

Can you understand filter ladder circuit ?
Learn very basic things before EDA Tool Play.