bandpass filter from summed lightly-damped transfer functions

Started by jekain314 2 months ago2 replieslatest reply 2 months ago108 views

I would like to make a bandpass filter for an SDR project by passing the signal through two (or more) transfer functions in parallel. Each transfer function uses a lightly damped denominator so that it picks out a specific frequency and has a bandwidth related to the damping coefficient. The transfer functions add because they are in parallel. If the center frequency of the individual filters are separated, I get a wider bandwidth. By empiracilly adjusting the damping and frequency separation of two filters, I can get a bandpass filter that is flat across the bandwidth. When I compare this bandpass filter to a filter designed using high/low Butterworth filters, I see a much sharper rolloff outside the bandwidth with a lot fewer digital computations. The first example below uses two 2nd order filters with 0.015 damping -- one has center frequency 1.02 and the second has center frequency of 0.98. The bandwidth is about 5% of the center frequency.  The second example below has 2nd order filters with center frequencies 0.999 and 1.001 each with damping coefficient of 0.001.

My question: Is this a useful approach to bandpass filter design? Has anyone studied this in an analytical fashion? Will I encounter numerical issues due to the extreme Q?

image (1)_59656.png

image (2)_47254.png

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Reply by CharlieRaderFebruary 22, 2024

That takes me really far back. In the early 1960s, I knew Robert Lerner at MIT Lincoln Laboratory. I needed to design and build filters (analog) for a filter bank for a vocoder. Lerner suggested an approach where each band-pass filters was realized as a parallel realization of individual resonators.  Suppose one of the resonators has an s-plane frequency response  (B_i(s+a))/((s+a)^2 + b_i^2). With many of those two pole resonators, the b_i are equally spaced. The a is identical for all the filters. The ith and (i+1)th resonator have a crossover frequency and at that crossover frequency we know that the gains of the two resonators are identical and if the phase responses are identical their sum should be the same as the gain of one of the resonators at its center frequency. You get to choose a, which controls the sharpness of the cutoff. The b_i approximately control the center frequency for one of the resonators. The parallel connection of K such resonators, i= i_m,i_(m+1),..., i_(m+K) is the overall bandpass filter and any two consecutive filters in the filter bank share the same resonator. But the constant B_i for the shared resonator at the cross-over frequency b_i is half of what it would be for a resonator not shared.

The Lerner filters were pretty nearly linear phase and nearly equiripple in the pass bands. In your case, you only want one bandpass filter so the "shared resonators" aren't an issue.

This was before I ever heard of digital filtering.  But very soon after that, I wanted to simulate a vocoder on a computer. I began to learn and also invent digital filters using the impulse invariant technique and Lerner filters were the very first I tried on a computer.  

If you can find a copy of Digital Processing of Signals, Gold and Rader, McGraw Hill 1969, pp 52-54. 

The digital "Lerner filters" preserve impulse invariance, but they don't account for aliasing. So be careful. Since you only want to have one band pass filter with a narrow bandwidth, you might be satisfied with this.

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Reply by jekain314February 22, 2024

Charlie .. 

Thanks for the detailed note.

wow ... that goes way back. I found your/Gold's original Lerner filter work at Lincoln Lab circa 1965:


I'll definitly look further at that work. 

I am doing a demo of DirectRF demodulation using the ADI Apollo MxFE chip (ADC to 20GHz) so Direct RF sampling a 100MHz FM signal at 20 samples/cycle is no sweat. The second order lightly damped system I used (transfer function) is: 

s/(s^2+2*psi*wc*s + wc^2 )  

Keeping wc*psi a constant for the two filters results in their peaks being identical and I can solve in closed form for the 3dB down points. Adding the results from the two filters gives the desired flat topped response across the passband -- that is just the separation of the two frequencies. 

I believe the results scale with the carrier. I showed a normalized carrier of 1.0 in my note and a passband of 0.1% of the carrier. An FM channel is ~200KHz for a carrier of ~100MHz so a 0.1% passband should work there. 

My next question is whether I need a massive bit depth on the ADC to treat the internal dynamic range for the processing. 

     Jim Kain