# Implement an audio-frequency tilt-equalizer filter

Francesco Balena, Electro-Acoustic Design, Conselve (PD), Italy; Edited by Margery Conner and Fran Granville -February 02, 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).

Figure 1 In a tilt audio-tone control, the tilt control tilts the frequency content of the audio signal by simultaneously boosting the treble and cutting the bass frequencies, or vice versa.

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:
where X indicates the wiper position of potentiometer P1 and the values of the resistors and capacitors define Z:

Figure 2 This frequency response is for the extreme wiper positions, where X=0 or P1. All of the other responses, with 0 less than X and X less than P1, lie between these curves.

The frequency response in Figure 2 is for the extreme wiper positions, where X=0 or P1. All of the other responses, with 0 less than X and X less than P1, lie between those curves. To get the frequency responses in decibels, multiply the log of the absolute value of the transfer function by 20: 20log(|TF|). To get a log/log scale on the graph, substitute 10F for F on the X axis. Pivot frequency FP depends on component value, including the setting of potentiometer P1, as it sweeps between an X value of 0 and P1, where RF must be greater than R:
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:
For the equations to work, ML−1 and (MH×ML−1) must be greater than 0. You can choose any reasonable value of potentiometer P1. For example, select a P1 value of 50 kΩ, a desired pivot frequency of 1 kHz, a maximum low-frequency boost of 4, and a maximum high-frequency boost of 2. The equations yield an RF of 16.66 kΩ, an R of 7.14 kΩ, and a C of 12.24 nF (Figure 3).

Figure 3 RF is 16.66 kΩ, R is 7.14 kΩ, and C is 12.24 nF.

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.

Figure 4 Voltages VI, VO, and V are all referred to ground.

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:
Remember that Z is the complex impedance of the RC branches. Now rearrange the equations:
From the first and second equations you can deduce that I1 equals I2. You can now substitute into the last three equations and rearrange them to get the final set:
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.