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    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
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    Re: What happens to output power and 3 BW in this circuit

    Quote Originally Posted by circuitslave View Post
    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.
    Why do you set three ports.
    Where is driving point ?

    What do you expect ?



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    Re: What happens to output power and 3 BW in this circuit

    Quote Originally Posted by circuitking View Post
    Again I took two sections of the same T network and gave input at different points as shown in figure
    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.



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    Re: What happens to output power and 3 BW in this circuit

    Quote Originally Posted by circuitslave View Post
    But I am not sure its working technicalities and analytical equations yet.
    Can you understand filter ladder circuit ?
    Learn very basic things before EDA Tool Play.



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