# Line Level EQ insertion loss



## Anthony (Oct 5, 2006)

I'm not sure where to put this, so I'll start here and see where it goes  It's about EQ of a subwoofer, albeit passive.

To compensate for my dipole subwoofer's 6dB/oct rolloff I want to insert a line level low-pass filter at 6dB/oct at 15 Hz. It's a simple R-C circuit with the resistor in series and the capacitor in parallel.

What I'm having trouble understanding is the insertion loss and matching impedance.

Linkwitz's site has some pages regarding a line-level crossover and filter for his dipole sub. I found some others that have good info, and the math is simple. 2*pi*f = 1/(R*C) where f is the frequency you want the "knee" to be at.

My subwoofer amplifier has an input impedance of 12k Ohms. What values of R should I use to minimize the insertion loss? I've seen a lot of recommendations, but no explanations.

The plan is to use a 7.5k Ohm resistor and 1.4 uF Capacitor -- both items I have parts for already, but I wanted to verify the theory.

Thanks.


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## Wayne A. Pflughaupt (Apr 13, 2006)

By "insertion loss," do you mean a reduction in signal strength? I don't know much about passive filters, but I would think the desired rate of attenuation and frequency knee would determine the resistor values - no?

Regards,
Wayne


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## Anthony (Oct 5, 2006)

I keep reading about these filters and they all mention to be very careful about choice in resistor values. You can add too much resistance and lower the sensitivity or you can add too little and strain the preamplifier.

I'm just trying to get my head around the output impedance, filter, and input impedance interactions. I know you want input to be ~10x output so that it behaves as a pure "voltage reader". But how the filter changes the impedance the preamp and amplifier sees is what I'm having a hard time with. Not one website has a clear explanation or circuit diagram, but many have recommendations and such for resistor values.

so I know 7k Ohms is not a bad choice, but is it the best choice? I believe you get 6dB insertion loss if R1 = Ramp -- but there's a limit to how low you can go. Those are the kind of trade-offs I'm trying to get my head around.


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## Wayne A. Pflughaupt (Apr 13, 2006)

Well, if it's just a signal level reduction issue, then the sub amp's gain control can compensate (up to a point of course). Not sure why it would have a bad effect on a pre-amp. Brucek knows more about these things than I do; since he hasn't replied it probably means he's not sure what to tell you either.

Maybe we should move this to another Forum so you can get the info you need, but I'm not sure which one...

Regards,
Wayne


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## SturmMD (Dec 30, 2008)

The low-pass and line level input impedance will combine to have the two following effects.
1. In the pass-band (under 15 hz) the 7.5kohms and 12kohms will form a voltage divider reducing the voltage amplitude to a factor of 0.61. This gives you an "insertion loss" of 4.2 dB.
2. The low impedance (compared to the 7.5K) of the input will affect the frequency of the filter. Using the normal equation fc = 1/(2*pi*R*C) The R value is equal to the parallel combination of the RC filters' resistance and the output impedance. In your case this this will raise the corner frequency to 25Hz. This is why it is a general rule of thumb that the output of a RC filter should have a load at least 10x larger than the resistance.

In your case (12kohms input impedance) you should just lower your RC filter to 1 to 2kohms and adjust your capacitor accordingly. This will also reduce your insertion loss.


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## SturmMD (Dec 30, 2008)

another comment, with regard to having too low of a resistor value...
After a couple of octaves above the cutoff frequency the capacitor can be considered a short to ground. This means that filter's resistor is equal to the load placed on your preamplifier or whatever is providing the signal.

Whether too low will actually causes damage is unlikely. A low value will load the output and cause the voltage to drop as a voltage divider with the output impedance is created. Even if the preamp was designed for a load of 10kohms as low as 1kohms should pose no problem.


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## Anthony (Oct 5, 2006)

That's what I was trying to understand, thanks.

I understood the simple voltage divider stuff (done that before), but how the filter affected the preamp (i.e. too low of a load) and changing the frequency was what I kept going in circles around.

The good news is, my preamp seems up to the task. Everything sounds pretty good with 6dB of gain (old filter, made with a 2k Ohm resistor), and the sub amp at 12 o'clock. I didn't expect too many problems, but wanted to make sure I understood everything.

Thanks again.


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## brucek (Apr 11, 2006)

> he hasn't replied it probably means he's not sure what to tell you either


hehe, no, I've just been kinda busy lately, sorry Anthony.

Yeah, SturmMD covered what you need to know. Even a simple design requires lots of trade offs, as nothing in engineering is ever perfect. A passive filter requires many compromises, and you'll never get ideal conditions through both the stop and pass bands.

Your best bet in analyzing this is to examine the extremes and see what problems will exist. In your case, you take the upper frequency in the stop band of 20KHz and see that the capacitor is now basically a short. Redraw your circuit with that knowledge and examine the problems with attention to the input and output impedance. Then redraw the circuit with an Xc in the pass band, and examine the problems. Use ~100 ohms as a standard preamp Zout.

As SturmMd pointed out, you've broken the 10:1 rule of thumb with your capacitor (of 1.4uf), and so the fc of your filter will suffer (because you would need to include the input amplifiers impedance in parallel with the Xc of the capacitor). If you honour the 10:1 rule, then you can ignore the amplifiers input impedance. This is why SturmMD suggested you lower the filters Xc output to about 1-2Kohm. This would demand a larger capacitor in about the 8-10uF range and the resulting series Resistor to about 1000ohms. 
At 15Hz, the Xc of the 10uf capacitor would be about 1000ohms. This is a suitable output impedance for this stage to be able to ignore any effect from the next amplifiers input impedance of 12Kohm. The fc would not be compromised. As he said, this would reduce your insertion loss considerably, but may offer new concern when the frequency was high and the capacitor acts as a short. I wouldn't be too concerned about that as most op amps today offer full time DC short protection.

So from an insertion loss and output impedance view, a 15Hz RC filter of 10uf and 1000ohms looks quite good. From an input impedance view, it may not be that good. I suppose if your preamp offered an output impedance of 100ohms (fairly typical), then it may be fine, but I would probably want to raise the input impedance of the filter a bit higher. Values of about 5uf and 2Kohms would offer a fairly good compromise.

Anyway, you must have an idea now about why values are good or not.....

brucek


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## Anthony (Oct 5, 2006)

Yeah, thanks.

thankfully I have lots of crossover caps in the 0.33 uF to 100 uF range, so I have a wide range to play with.

The amp itself is a subwoofer amp and it has almost zero output by 2kHz (unloaded). 

Linkwitz in his example of these filters used 2kOhm series and an appropriate capacitor, so I figured that was ballpark close. I just wanted to know why. He also mentioned scaling if your amp did not have a 10kOhm input impedance (mine is 12k) -- he showed the math, but didn't explain why (which is what I really wanted to know).

Thanks again guys. I am planning on doing some measurements this weekend of different filters and effect on frequency response. I already have a few, but the values weren't selected very scientifically. Hopefully I will have some charts this weekend.


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## aNomad (Feb 17, 2009)

The advice given was good. I can add to it in a minute way by pointing out that most preamps have a resistor in series with the output jack. That will usually be around 560 ohms, but can be as high as 1K. If you have the manual, it might say, but a good method of determining it is to put a steady state signal out (a sine wave), load it with a known load, and use the resistor divider formula to calculate the output impedance.

However, once known, and if you know the sub's input impedance (also almost always resistive), you can simply take those values into account when calculating the cap value. As Sturm said, the impedance (to the cap) is the parallel combo of the sub's input impedance and the series resistance (preamp output impedance plus series resistor value). Calculate the final parallel impedance as follows (I'll use 7.5K for series resistor, 560 ohm as preamp output impedance, and 12K as sub input impedance):

R1 is total series resistance (560 + 7.5K or 8060 ohms). R2 is sub's input impedance (12K). Impedance to the cap is 1/ (1/R1 + 1/R2). 1/R1 = 1.24^-4, 1/R2 = 8.333^-5. The sum of those is 2.074^-4. 1 over that is 4821 ohms.

Also, be sure to use a film cap, never electrolytic or ceramic. film caps will not cause distortion.

Have you considered an active filter?


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