Implement an audio-frequency tilt-equalizer filter

In the 1970s, Quad Ltd developed a “tilt” audio-tonecontrol, which first appeared on the company’s model34 preamplifier. The tilt control tilts the frequency contentof the audio signal by simultaneously boosting the treble andcutting the bass frequencies, or vice versa (Figure 1 ). Onlyone knob is needed to tilt the frequency response around apivot frequency, F_P (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 thefilter. You need a Spice simulation and many trial-and-errorcycles 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 canmake the tilt response asymmetric, with different amountsof boost and cut. You begin deriving the transfer functionby expressing the input versus the output as a function ofdc-feedback resistor, R_F , and Z, the complex impedance ofthe RC branches:

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where X indicates the wiper position of potentiometer P₁ andthe 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 extremewiper positions, where X=0 or P₁ . All of the other responses,with 0 less than X and X less than P₁ , lie between thosecurves. To get the frequency responses in decibels, multiplythe log of the absolute value of the transfer function by 20:20log(|T_F |). To get a log/log scale on the graph, substitute10^F for F on the X axis. Pivot frequency F_P depends on componentvalue, including the setting of potentiometer P₁ , asit sweeps between an X value of 0 and P₁ , where R_F must begreater than R:

For the equations to work, M_L −1 and (M_H ×M_L −1) mustbe greater than 0. You can choose any reasonable value ofpotentiometer P₁ . For example, select a P₁ value of 50 kΩ, adesired pivot frequency of 1 kHz, a maximum low-frequencyboost of 4, and a maximum high-frequency boost of 2. Theequations yield an R_F of 16.66 kΩ, an R of 7.14 kΩ, and a Cof 12.24 nF (Figure 3 ).

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

You take 20 times the log of M_L to get the response indecibels, so an M_L of 4 is the 12-dB maximum low-frequencyboost, and an M_H of 2 represents the 6-dB maximum high-frequencyboost. When you normalize the resistor and capacitorvalues to standard values, you get only a minor error inyour desired response. By defining the variables M_L and M_H ,you can make tilt equalizers that have an asymmetric responsebetween boost and attenuation.

Figure 4 Voltages V_I , V_O , and V are all referred to ground.

Remember that Z is the complex impedance of the RCbranches. 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 V_O /V_I ; you need not solve all of theunknowns. If you substitute I₁ from the third equation aboveinto the second equation , you can find I_P . You then substitutethis I_P into the fourth equation and find the ratio of V_O /V_I ,yielding the first equation in this Design Idea. This result iscongruent with the actual numerical value of the examplesin Reference 1 .