[600MRG] LPF - Pi vs Tee form and Constant K impedance match

[Roger Graves]

With my Pi form Low Pass Filter between the tapped bucket coil 630 mtr tuning inductor and my RX which has a 50 Ohm input, I looked at suppression of BC stns near 950 kHz as a check on potential 2nd harmonic suppression. The difference between no LPF and w/ the filter was only 12 dB. When driven by a 50 Ohm generator the suppression was 26 dB, just as predicted by the design program. I built a 2nd filter, with identical parameters but Tee configuration. This one also showed 26 dB suppression with a 50 Ohm source but now the BC stns were 32 dB down when driven by the tapped inductor, which must be wildly reactive and way different from 50 Ohm resistive at 950 kHz. So, with my antenna system, the Tee configuration LPF provided 20 dB better suppression of 950 kHz. For my SDR RX, 20 dB btr suppression of BC stns is important for eliminating IMD spurs. 

The btr suppression with the Tee form should also be seen going the other way if the filter were driven by a 50 Ohm source TX. TX output circuits will likely not be 50 Ohm resistive, especially on the harmonics, so the actual harmonic suppression of a LPF should be measured, not assumed based on the LPF design plot shown for pure 50 Ohm input and output.

BTW, I discovered that besides the “Voltage Effective Gain” plot, the “Input Impedance” plot on the filter design programs is very important. The nice sharp cut-off filters I was designing (with 50 Ohm termination) turned out to have non 50 Ohm input Z at 475 kHz, and the Z curve was very non flat around 475 kHz. I found that a “Constant K” design with 50 Ohm reactance of the L and C elements provided a nice flat 50 Ohm match. The lower Q would also reduce voltage on the L and C (which would be of interest to those running more than my 1 Watt output!).

73,
Roger, VE7VV