walid Posted August 2, 2009 Report Share Posted August 2, 2009 Hellow,I build the following pi filter to match 100 ohm source to 1000 ohm load, I did the calculations based on RF Circuit Design BookAs u can see at the center freq fc = 30MHz & Q=15 I had 3.771dB, I was surprised by where this value? this is amplification!!!!The second question is: If i want to match 1000 ohm source to 100 ohm load, Simply replace the capacitors, one must place the other, when i did so and simulate i had the following:[img width=680 height=478]Look!! -9.8dB at the center freq.can you please comment, how to match big source to small load without that bad attenuationthanks alot Quote Link to comment Share on other sites More sharing options...
Kevin Weddle Posted August 4, 2009 Report Share Posted August 4, 2009 A pi filter can be used to match loads? Most impedance matching that I've seen match the source and load impedance. Although depending on the application, pi filters can play a role in impedance matching. Quote Link to comment Share on other sites More sharing options...
walid Posted August 4, 2009 Author Report Share Posted August 4, 2009 Thank you KevinIV for replymy question is why and how filter amplify signals?? Quote Link to comment Share on other sites More sharing options...
Kevin Weddle Posted August 4, 2009 Report Share Posted August 4, 2009 I'm surpised at the results. I know that mismatching impedances can cause standing waves, and it may depend on where you probe on the transmission line. But I think there would still be enough signal loss, and it would seem hard to get that good of a measurement to make a comparison. Quote Link to comment Share on other sites More sharing options...
audioguru Posted August 4, 2009 Report Share Posted August 4, 2009 A series LC circuit is a low impedance at resonance. The signal current is high. The voltage across its capacitor and the voltage across its inductor is higher than the input signal voltage. Quote Link to comment Share on other sites More sharing options...
indulis Posted August 5, 2009 Report Share Posted August 5, 2009 This is a "second order" circuit. You took a bode plot over a frequency range. Things are going to change over frequency. Have you solved the differential equation for this circuit to find the poles and zeros? It's also what is called a "passive circuit"... so, unless you hit a resonant frequency (as AG pointed out) in your sweep, this thing can only attenuate. Quote Link to comment Share on other sites More sharing options...
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