Implement an audio-frequency tilt-equalizer filter
Developing the transfer functions lets you customize the filter.
Francesco Balena, Electro-Acoustic Design, Conselve (PD), Italy; Edited by Margery Conner and Fran Granville -- EDN, February 2, 2012
In the 1970s, Quad Ltd developed a “tilt” audio-tone
control, which first appeared on the company’s model
34 preamplifier. The tilt control tilts the frequency content
of the audio signal by simultaneously boosting the treble and
cutting the bass frequencies, or vice versa (Figure 1). Only
one knob is needed to tilt the frequency response around a
pivot frequency, FP (Figure 2).Quad Ltd never published a transfer function for the filter. You need a Spice simulation and many trial-and-error cycles to tune it to your desired response. By deriving the transfer function, you can easily select the component values. Surprisingly, the transfer function also shows how you can make the tilt response asymmetric, with different amounts of boost and cut. You begin deriving the transfer function by expressing the input versus the output as a function of dc-feedback resistor, RF, and Z, the complex impedance of the RC branches:
To calculate component values, you first define the maximum
low-boost asymptote as ML, when the frequency goes
to 0 Hz and the potentiometer’s value is also 0Ω. You then
define the maximum high-boost asymptote as MH, when the
input frequency goes to infinity, and set the potentiometer
to its maximum value. This step gives the component values
for RF, R, and C:
You take 20 times the log of ML to get the response in
decibels, so an ML of 4 is the 12-dB maximum low-frequency
boost, and an MH of 2 represents the 6-dB maximum high-frequency
boost. When you normalize the resistor and capacitor
values to standard values, you get only a minor error in
your desired response. By defining the variables ML and MH,
you can make tilt equalizers that have an asymmetric response
between boost and attenuation.A detailed derivation of the transfer function is included
here. You begin by defining voltages VI, VO, and V, all referred
to ground (Figure 4). In this case, I1, I2, and IP are the minimal
number of unknown currents. Because an op amp servos the
output to keep the input pins at the same voltage, the potentiometer
wiper is at 0V, a virtual ground. Further assume the
infinite input impedance of the op-amp input pins so that the
current at the inverting pin is 0A. VI and VO are unknown,
letting you write a set of equations for the conditions:
The goal is to find VO/VI; you need not solve all of the
unknowns. If you substitute I1 from the third equation above
into the second equation, you can find IP. You then substitute
this IP into the fourth equation and find the ratio of VO/VI,
yielding the first equation in this Design Idea. This result is
congruent with the actual numerical value of the examples
in Reference 1.|
Reference |
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Talkback
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Interesting circuit and application, but I'd be wary about having a pot wiper connected directly to an op amp inverting input, unless it's a trimpot very close to the IC. The issue would be capacitance to ground or noise sources as well as leaving the input floating if the pot is scratchy.
Steve Trapp - 2012-13-2 09:12:27 PST -
Unfortunately, as usual the "download PDF" does not work so I can't store the article for later reference.
Cor - 2012-10-2 12:53:47 PST
