I started my band-pass designs using the Chebyshev model. I didn’t do my homework. I just found a schematic, chose some values based on someone else’s design using toroids, and totally ignored the “Q” values of components. I built 7 filters and also tested them on a NanoVNA. I didn’t think anything of the 3 or greater dB insertion loss. I was verifying the passband graph shape. Everything looked good. It was not. I’ve been doing lots of homework and here’s some of it.
When you design a 3‑pole Butterworth series‑coupled band‑pass filter, one of the first surprises is that the parallel LC tanks don’t resonate at the same frequency as the filter’s actual passband center. For example, you might calculate:
- Tank resonance: 14.029 MHz
- Filter center frequency: 14.2 MHz
At first glance, this looks contradictory. Why would the tanks be tuned below the desired center frequency?
The answer is simple once you understand how coupled resonators behave.
1. The tanks have a “free” resonance — but the filter has a “forced” resonance
Each LC tank has its own natural resonance frequency:
At this frequency, a parallel LC tank has maximum impedance. If the tank were sitting alone on a bench, it would peak at 14.029 MHz.
But in a real filter, the tanks are not alone. They are:
- coupled to each other,
- loaded by the source and load impedances,
- and constrained by the Butterworth response requirements.
Once you connect everything together, the system develops multiple resonant modes, and the collective behavior shifts the passband center upward.
This is why the filter’s true center ends up at 14.2 MHz, not 14.029 MHz.
2. Coupling “pulls” the resonant modes apart
A 3‑pole filter has three resonant modes. If the tanks were uncoupled, all three would sit at 14.029 MHz.
But coupling splits them into three slightly different frequencies:
- one a bit below the tank resonance,
- one near it,
- one above it.
The Butterworth design equations choose the coupling so that the middle mode becomes the passband center, and that mode ends up at your design frequency (≈14.2 MHz).
So the tanks are intentionally tuned a little low so that, once everything interacts, the system resonance lands exactly where you want it.
3. Think of it like tuning three musical instruments
Imagine three violins, each tuned to 14.029 MHz. When they play together and acoustically couple, the ensemble’s strongest note shifts slightly upward.
The individual strings haven’t changed tuning — but the system resonance has.
Your filter works the same way.
4. What this means for your frequency‑response plot
If you graph insertion loss (dB) vs. frequency:
- At 14.029 MHz Each tank is at maximum impedance, but the filter is not at maximum transmission. You’re on the lower slope of the passband.
- At 14.2 MHz The three resonant modes combine to produce the minimum insertion loss. This is the true center of the Butterworth passband.
- Above and below The response rises smoothly, with the classic maximally‑flat Butterworth shape.
So the filter’s “flat top” is centered at 14.2 MHz even though the tanks themselves peak at 14.029 MHz.
5. The relationship in one sentence
14.029 MHz is the natural resonance of each individual tank. 14.2 MHz is the resonance of the entire coupled system.
The difference is not an error — it’s exactly how a properly designed 3‑pole Butterworth band‑pass filter is supposed to behave.
My About page provides the background of my project, the Freedom7 HF Transceiver.
If this story resonates, comments are welcome. You can also reach me at david [at] kr4bad-dot-communications. no com
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73 KR4BAD David
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