Variable Low Frequency Droop Formula?

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Ufdah

Member
Joined
Jul 19, 2005
Messages
8
Hi guys, I found a white paper on equalization that has a "variable low frequency droop" EQ that I'd like to implement but I dont know how to figure the values. Is it the same as a standard RC High-Pass?
Fc=1/(2*pi*R*C)
Or does the addition of R1 change things? If it's similar how do I size R1?

I'm looking for an Fc around 250Hz and don't want much less than 10Kohms to ground.

Here's the circuit:
98tves.gif
 
Qualitatively, look at what is going on: at high frequencies the impedance of the cap gets small, so the input from the left gets to the output without attentuation.

At low frequencies/"d.c." the impedance of the cap is high so the circuit just looks like a voltage divider with the output a fraction of the input:

E out = Ein (R2/(R1 + R2)).

I'm assuming zero source impedance and no output loading.

The precise expression for the magnitude of the transfer function |Eout/Ein| as a function of frequency is fairly hairy* and always involves both R's and the C. Note that the shape is going to be a function of the amount of attenuation, and that again is given by the simple voltage divider equation.



*

|Eout/Ein| =

square root of {[R2^2 + (2pi f R1 R2 C)^2]/[(R1+R2)^2 + (2pi f R1 R2 C)^2]}

This is unity as f goes to infinity and R2/(R1 + R2) for f = 0.
 
Thanks bcarso!

Your methodical approach makes perfect sense to me. So by graphing the equation with x=F and inputing static values for the resistors I'm able to see the frequency response in terms of signal gain (the y-axis).

Thanks again.
 
> Is it the same as a standard RC High-Pass?

To a first approximation: yes, the point where it starts drooping can be estimated with the hi-pass solution.

> Or does the addition of R1 change things?

Yes, if R1 is similar to R2. Then you get a better approximation if you figure the parallel equivalent of R1||R2 and work your lo-pass calculation.

> If it's similar how do I size R1?

Their ratio sets the gain at super-low frequencies. If R2=10K (which solves your "don't want much less than 10Kohms to ground" criteria) and R1 is 1Meg, then the gain at super-low frequency will be about 10K/1Meg= 1/100= -40dB and the lower corner will be about 1/100th of the frequency of the top corner and the slope will be almost 6dB/octave. I don't know any audio use for a -40dB shelf, you might have use for 10K and 100K or about 1:10 or about -20dB. The two corners will be almost 10:1 apart in frequency and the slope won't be over 5dB/octave. For 10dB or 6dB shelves, the math gets very messy.... but the result is always gentle, simple to the ear, and we have no need to compute the exact response.

> by graphing the equation with x=F

GRAPH???? You mean, with slide-rule and graph-paper? And hard sweat? Man, we didn't do that in the Old Days, not unless we HAD to.

To some degree, you rough-up the math, build it, and listen. Pick a resistor ratio and values, pick a cap to put you in the right part of the spectrum, and try it.

Changing the cap moves everything up/down in frequency. Changing R1 changes the height of the lower shelf. Make big changes, like double or half, and you quickly find what you want (or prove you need a more sophisticated filter).

Radiotron has some practical schemes; see Chapter 15 in the 4th edition. Note that some of these assume triode voltage amplifiers: source resistance is low but far from zero. A few use odd effects of pentodes and other schemes not readily applied to modern work. Also because 78s and AM radio had good bass and awful treble, much of the focus is on bass-boost. But the chart in Fig 15.4 is general: you can turn it upside down for your bass-drop circuit, or left-right for treble boost/cut filters. But knowing that response is -2.345dB at 1.234 times nominal frequency does not tell you what it sounds like: "the ear is the only judge".
 
Thanks for the link to the Radiotron stuff, looks like some good info.

When I talk about graphing I mean on a graphing calculator... I just plug in the equation:

y=sqr.{[(B^2+(2 pi x A B C)^2]/[(A+B)^2+(2 pi x A B C)^2]}

Then I can say A=100K, B=20K, and C=.033uF, press GRAPH and it shows me a frequency response curve! Then I just tweak my values until the .707 is where I want it and I have the load that I'm looking for and whalla. It really is an easy way to do it. I guess that's one of the perks to growing up in day and age that I have.
 
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