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, 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 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, R
_{F}, and Z, the complex impedance of
the RC branches:
where X indicates the wiper position of potentiometer P
_{1} 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 P
_{1}. All of the other responses, with 0 less than X and X less than P
_{1}, lie between these curves.
The frequency response in
Figure 2 is for the extreme
wiper positions, where X=0 or P
_{1}. All of the other responses,
with 0 less than X and X less than P
_{1}, 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(|T
_{F}|). To get a log/log scale on the graph, substitute
10
^{F} for F on the X axis. Pivot frequency F
_{P} depends on component
value, including the setting of potentiometer P
_{1}, as
it sweeps between an X value of 0 and P
_{1}, where R
_{F} must be
greater than R:
To calculate component values, you first define the maximum
low-boost asymptote as M_{L}, when the frequency goes
to 0 Hz and the potentiometer’s value is also 0Ω. You then
define the maximum high-boost asymptote as M_{H}, when the
input frequency goes to infinity, and set the potentiometer
to its maximum value. This step gives the component values
for R_{F}, R, and C:
For the
equations to work, M
_{L}−1 and (M
_{H}×M
_{L}−1) must
be greater than 0. You can choose any reasonable value of
potentiometer P
_{1}. For example, select a P
_{1} 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 R
_{F} of 16.66 kΩ, an R of 7.14 kΩ, and a C
of 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 in
decibels, so an M
_{L} of 4 is the 12-dB maximum low-frequency
boost, and an M
_{H} 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 M
_{L} and M
_{H},
you can make tilt equalizers that have an asymmetric response
between boost and attenuation.
Figure 4 Voltages V
_{I}, V
_{O}, and V are all referred to ground.
A detailed derivation of the transfer function is included
here. You begin by defining voltages V_{I}, V_{O}, and V, all referred
to ground (Figure 4). In this case, I_{1}, I_{2}, and I_{P} 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. V_{I} and V_{O} 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 I
_{1}
equals I
_{2}. 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 the
unknowns. If you substitute I
_{1} from the third
equation above
into the second
equation, you can find I
_{P}. You then substitute
this 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 is
congruent with the actual numerical value of the examples
in
Reference 1.
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