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Total 4 documents matching your query.

1. [Amps] calculations (score: 1): Author: carlseye@tampabay.rr.com (carl seyersdahl); Date: Thu, 21 Mar 2002 14:30:28 -0500; Here's one for the math guru's out there!! given the inductance , input and output capacitances, and knowing at least the input impedance, (50 ohms) can the impedance on the output side of a pi-netwo; /archives//html/Amps/2002-03/msg00476.html (7,170 bytes)
2. [Amps] calculations (score: 1): Author: wc6w@juno.com (Radio WC6W); Date: Thu, 21 Mar 2002 12:34:14 -0800; Hi Carl, Easy, if anyone has a copy of mathcad handy. Just take the Pi-Net equation and solve it for the output impedance. 73, Marv WC6W ______________________________________________________________; /archives//html/Amps/2002-03/msg00477.html (8,132 bytes)
3. [Amps] calculations (score: 1): Author: n8de@thepoint.net (Don Havlicek); Date: Thu, 21 Mar 2002 15:59:51 -0500; Seems to me there is an easier solution: 1. Connect a SWR analyzer [such as the MFJ-259] across the 50-ohm terminals of the PI. 2. Connect a NON-inductive potentiometer across the output terminals of; /archives//html/Amps/2002-03/msg00478.html (7,974 bytes)
4. [Amps] calculations (score: 1): Author: Ian White, G3SEK" <g3sek@ifwtech.co.uk (Ian White, G3SEK); Date: Thu, 21 Mar 2002 21:14:58 +0000; The step-by-step worked example is exactly your pi-network problem. -- 73 from Ian G3SEK Editor, 'The VHF/UHF DX Book' 'In Practice' columnist for RadCom (RSGB) New e-mail: g3sek@ifwtech.co.uk New we; /archives//html/Amps/2002-03/msg00480.html (7,916 bytes)

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