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15th September 2019, 18:45 #1
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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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15th September 2019, 20:06 #2

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26th September 2019, 21:22 #3
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Re: What happens to output power and 3 BW in this circuit
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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27th September 2019, 00:49 #4
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