EMRFD Message Archive 2624

2624 2009-01-23 14:32:50 Ivan Rogers Basic Low Pass Filter Design Message Date From Subject I am trying to get my head around different methods of Low pass filter design when using source and load terminations of unequal resistance. First, I understand the use of Bartlet's Bisection Theorem in EMRFD Page 3.6 Fig 3.9 when you first work out the filter design using terminations of equal resistance and then using the theorem to make a simple tranformation on the left hand side of the filter network. This makes sense to me. In other RF design publications another way of calculating LP filters is to use normalised tables that have normalised values based on a source/load ratio when the terminations are unequal. I.E. for 50R terminations at each end Rs/Rl=1 Buterworth values are: .62 1.62 2.00 1.62 .62 For 15R (Rs) termination and 50R (Rl) termination Rs/Rl=.3 Buterworth values are: 1.09 .29 4.84 .54 5.31 (Ref RF Circuit design, Chris Bowick) My question is I would expect when using these normalised values with equations 3.1 and 3.2 (EMRFD page 3.4) is to end up with the same component values as when I used the Bartlet's theorem method above but I don't. Why? Is it because say Bartlet's theorem just transforms the Left hand side of the filter network for impedance trasnformation and that the normalised value ratios use the whole of the circuit for impedence transformation. If so, is there any advantage for one over the other? Just trying to get to grip with the basics first on the 2 approaches above before I start experimenting on the bench. Regards Ivan G0BON Hi Ivan and gang, Very good question. Almost all of the filter work that I've done has been with doubly terminated ladders, although I've done some work with singly terminated ones, which is just the most extreme member of the tables that you mention. I've used Bartlet where modest asymmetry was needed. I elected to look at this situation through study of some design examples. I started with the Bartlet case. I designed a filter with N=5 for a 1 dB Chebyshev shape with 3 dB cutoff of 10 MHz. I used a lot of ripple, for that yields a design with "warts" that are easy to see in graphs. I then duplicated the filter in software, but moved one termination from 50 to 100 Ohms. I did plots in SPICE to get the multicolor display of several waveforms. The Bartlet low pass looked exactly like the original one so far as the output response except that it was displaced in amplitude from the equally terminated case. But the input reflection coefficient was a different story. The equally terminated case was just what we normally see with a Chebyshev  a very good match at those frequencies where the amplitude peaks occur. But the match had degraded severely for the asymmetric termination case. When you think about it though, this makes perfect sense. Consider the behavior at 0 frequency. The reactive components effectively disappear, leaving just the source and load. These resistors are now different, so the match must be poor. Next I went to Zverev and found some low pass tables. The most extreme ripple he had was 0.5 dB, so I selected those. I first did a filter design like the other case, a 10 MHz 3 dB cutoff with equal terminations. The results were as expected. I then shifted to an example with a source resistance that was half the load. This data was used to design a filter with 50 Ohm source and 100 Ohm load. The results for both filters were plotted on the same graph, along with curves showing the magnitude of the reflection coefficient at the 50 Ohm end. Again, the transfer curves were identical with each other except for a displacement in amplitude. But the input impedance match was now severely degraded. The two filters were very different from each other so far as component values go. Now what I need to do is to blend the curves together and come up with one plot that has everything on it. But alas, today I have some painting chores to chase. Incidentally, while chasing these demons, I noticed that my software for the design of the low pass filters generates component values that are identical with those of Zverev with a design based upon the 3 dB cutoff frequencies. However, I noted that the normalized component values