| Vikash |
I've been pulling my hair out trying to calculate the the values for two parallel notch filters to attenuate two frequency peaks. I've checked out out the various calculators around the net but the values don't seem to work when plugged into modelling software.
Concentrating on the worst one, the notch is required for:
f = 6.5kHz (5db peak)
f1 = 5.8kHz (-3db)
f2 = 7kHz (-3db)
Using this calculator for example gives the following LC values but plugged into the software it addresses the 900Hz region instead.
L = 0.12mH
C = 4.62uf
I've tried in both Speaker Workshop and LspCAD (on different occassions) and neither models it even close to what is expected.
These are the values that work from trial and error in LspCAD:
L = 6.8uH (yes, microhenries!)
C = 1.5uf
I'd appreciated any input... |
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| Ron E |
| Post your circuit, please. |
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| Vikash |
Hi Ron,
After much hair pulling I've found a problem with my mates version of LspCAD which seems to do weird things from saving/reloading. The same filter values on the same curves seem to give different results from before/after a restart.
Anyway, after learning not to cheat on SW so easily, I've come up with the following. The curve looks ok, but are the values reasonable? I'm looking for a cheap meter that will read L to make sure the values are close and will get them by unwinding standard 0.1mH air core. |
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| soongsc |
| Looks very nice, let us know how it measures.:) |
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| salas |
| Looks nice, let us know how it sounds in a stereo pair. In case it sounds heavy, its gonna be the power response and then the resistors must be half of what they are now. |
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| phase_accurate |
You can always use Thomson's equation to check the center (= resonance) frequency of your notch. At resonance the parallel impedance of coil and cap gets infinite (at least theoretically) so you'll have a voltage divider given by the impedance of your driver and the resistor of the correction network. This way you can check if your values aren't way off.
Regards
Charles |
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| consort_ee_um |
| Vikash, wouldn't a series resonant circuit be better to eliminate the 6.5kHz peak? Interesting that you have gone for big C's and small L's whereas my active notch filter posted on the FR125s thread uses small C's and big simulated L |
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| Vikash |
From what I've read and brielfy modelled with series circuits, they're used to flatten the impedence at driver resonance or the rising impedence due to VC inductance. I'm not sure how it would tame a peak in frequency.
Those calculators actually gave a reasonable starting point once I plugged them into Speaker Workshop.
As I understand it the C x L product dictates the resonant freq. in which case as long as it's kept the same you can use any values. However the different values affect the bandwidth of the notch. So fine tuning required increasing C and decreasing L to contain the notch in the affected bandwidth area of the peak. i.e. not sure if could be done with larger L and smaller C. |
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| Vikash |
Here's the measured FR with the notches in place. I've been very impressed with ARTA, and with the accuracy of designing filters with SW (first time I've done this!).
There is an improvement in the output but still sharp in the top end as is obvious from the FR in the 10-15k region. Any more filter work to this speaker and it's just not worth it, so I'll call it a day here. Will make for a superb everyday HT speaker.
(edit: the enclosure is only lightly stuffed with polyfill. I wonder if the FR will be tamed when the wall foam goes in...)
V
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| soongsc |
| Vikash, you are using ARTA for measurements, was the Speaker Workshop measurement capability not good? |
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| Vikash |
ARTA has some serious usability advantages. Plus it's not limited to MLS signals.
For FR measurements I haven't come across anything better.
V |
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| EKIZER |
I too searched for a decent calculator for parallel notch filters without any luck. The few that I found were simply wrong. The impedance of the driver at the frequency notch must be known to make the calculations of the C, L and R. So it boils down to a relationship between not only C and L but how C and L react with R and Z.
Not knowing your Z at 6.5KHz try modeling one of these for a 5db loss.
If Z = 7, model C=20.3uF, L=0.026mH, R=5.6
If Z = 8, model C=17.76uF, L=0.03mH, R=6.4
If Z = 9, model C=15.79uF, L=0.034mH, R=7.2
If Z = 10, model C=14.21, L=0.038mH, R=8
have fun |
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