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The filter in question takes a mono signal of amplitude between 5 mVp and 200 mVp, and filters the signal based on the frequency. The potentiometer that leads into the inverting input of the op amp determines how this signal is ultimately affected. Depending on which extreme the potentiometer is dialed to, the circuit should amplify either low or high input frequencies, while attenuating the other end of the frequency extrema. When the potentiometer is dialed to the opposite extreme however, its behavior inverts; that is to say that if it amplified high frequencies by a voltage gain of A1 and attenuated low frequencies by A2, it should now amplify the low frequencies by A1 and attenuate the high frequencies by A2. When the potentiometer is set to exactly 50%, the gain should be -1, as no filtering is actually being done.

For this specific purpose, I need A1 to be 3 and A2 to be 1/3, i.e. A2 = 1/A1. I have values that I obtained somewhat coincidentally that achieve this behavior, which are shown in the circuit below. My question, finally arises from my lack of understanding of how to actually approach this circuit. It's almost like a band-pass filter, but not quite as one of the ends will amplify the voltage by the same amount it would have been attenuated. That's pretty much where my confusion lies. In this schematic, the circuit is designed to work towards the gains previously described, and is built to handle frequencies ranging from 20 Hz to 20 kHz.

schematic

simulate this circuit – Schematic created using CircuitLab

I know that: C1 = C2 R1 = R7 R2 = R6 Due to the gain of the circuit being the impedance of the right hand divided by the impedance of the left, and at 50% of the potentiometer, this must equal 1. If the potentiometer is split 10k and 10k for both impedances, then in order for it to equal -1 the above relationships must hold true.

R3 is always a 20k potentiometer. However, as far as the other components go, they can vary so long as they fall in accordance with those relationships. My issue lies in that I'm not quite sure what to do next to get the answers I came up with. I think I maybe have to compare how the expression for gain changes at each extreme with the expression for a gain of -1, but I'm not sure. If anyone can walk me through this, as well as correct any incorrect assumptions I've made, it would be a great help.

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  • \$\begingroup\$ This looks like a 'tilt' control. Are you using it for audio? Your ears are usually a far, far superior tool than equations for designing audio circuits. Have you tried the component values in that schematic? They seem reasonable. No amount of math can make the circuit sound better than trial and error and your ears can. This page shows how you would analyze this kind of circuit on paper: edn.com/design/analog/4368935/… . \$\endgroup\$ – vofa Dec 14 '16 at 22:11
  • \$\begingroup\$ This is actually for a project in school. This is only a single block in the larger circuit. The values work from multisim testing, however I kinda need to know the ins and outs of the circuit in order to have correct derivations for the component values when I turn in the written portion. I did a brief once over of that link and it's exactly what I've been looking for. The jargon and derivations may be a tad too complex for what I actually need at my level, but I'm sure I can simplify it. Thanks! Not knowing what to call a circuit makes it pretty damn hard to find information on it lol. \$\endgroup\$ – The Progenitor Dec 14 '16 at 22:23
  • \$\begingroup\$ If you add a voltage source at Vin you can simulate it using CircuitLab. Add a sweep parameter at R3.K (0 to 1, step 0.25) to see what happens when the pot is adjusted. It is a tilt control. \$\endgroup\$ – Steve G Dec 14 '16 at 22:24
  • \$\begingroup\$ @TheProgenitor - Don't worry about the complexity of that article as long as you understand the concept. Plenty of technical authors care a lot about making you think they're very smart, and they hide every difficulty they had and every mistake they made getting to their conclusion. \$\endgroup\$ – vofa Dec 14 '16 at 22:37
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I did not realize this circuit would be so complicated to analyse : ) when I started to look at it using the FACTs. First off, I redrawn the circuit in which the inputs were no longer floating but ground-referenced. I also considered the open-loop gain \$A_{OL}\$ to later push it to infinity. The newly-redrawn circuit is below. I have split the potentiometer in two distinct resistances linked by a ratio \$k\$ and reflected in the ground-referenced source driving the op amp:

enter image description here

Both dc and ac responses are similar. The next step is to find the dc transfer function for \$s=0\$: open the capacitors and determine the gain in this configuration:

enter image description here

If you do the maths ok, you should get \$H_0=-\frac{R_3(1-k_1)+R_7}{R_1+k_1R_3}\$

For the next steps, I will determine the time constants \$\tau\$ involving capacitors \$C_1\$ and \$C_3\$. You need to reduce the excitation voltage to 0 V and replace it by a short circuit. You obtain \$\tau_1\$ and \$\tau_3\$. Final step, determine the time constant involving \$C_3\$ while \$C_1\$ is replaced by a short circuit. You obtain \$\tau_{13}\$. Assemble these time constants to form the denominator:

\$D(s)=1+s(\tau_1+\tau_3)+s^2\tau_1\tau_{13}=1+\frac{s}{\omega_0Q}+(\frac{s}{\omega_0})^2\$

Then, calculate and check if the quality factor \$Q\$ is well below 1 to apply the low-\$Q\$ approximation. You can factor \$D\$ as two cascaded poles if \$Q<<1\$: \$D(s)=(1+\frac{s}{\omega_{p1}})(1+\frac{s}{\omega_{p2}})\$ in which \$\omega_{p1}=Q\omega_0\$ and \$\omega_{p2}=\frac{\omega_0}{Q}\$. Unfortunately here, Q is around 0.47 (\$k=0.6\$) so the poles are not far from each other and the low-\$Q\$ approximation does not hold. The resonant frequency in \$D(s)\$ changes from 492 Hz (\$k=0.1\$) to 1160 Hz (\$k=0.9\$).

The numerator can be determined using several approaches: inspect the transformed circuit and see if any impedance combination could lead to a null in \$V_{out}\$ despite the presence of \$V_{in}\$. You could for example check the condition for which \$\epsilon=0\$ and solve the equation. The other option is to apply an Null Double Injection (NDI) and determine the new time constants (as we did for \$D(s)\$) but with \$V_{in}\$ back in place and while \$V_{out}=0\$. These two paths give the simplest expression for \$N(s)\$. I have adopted the generalized form in which I will calculate three gains \$H\$ when the capacitors are alternatively replaced by a short circuit (for \$a_1\$) or when both are shorted (\$a_2\$). This gives a slightly more complex expression than the two other approaches and you will need to rework the final result to simplify it. For a dynamic response analysis like in here, the generalized formula is okay:

\$N(s)=H_0+s(H^1\tau_1+H^2\tau_2)+s^2H^{13}\tau_1\tau_{13}\$

You can also rearrange this expression in a second-order polynomial form as we did for the denominator. Then assemble the transfer function as \$H(s)=H_0\frac{N(s)}{D(s)}\$. The expressions are given in the below Mathcad files:

enter image description here

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I have tweaked these equations so that they depend on \$k\$ as a variable to pass and obtained the following responses:

Magnitude:

enter image description here

Phase:

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I did not do it but I believe you can obtain the frequency at which all the curves cross by differentiating the magnitude expression with respect to \$k\$ and solve the value of \$\omega\$ which cancels the expression. You actually calculate the function sensitivity to \$k\$ and check the frequency at which it is 0.

The cool thing with the FACTs is that you do not need to really bother about the circuitry itself: determine the time constants in certain conditions (\$V_{in}=0\$ for \$D\$ and (\$V_{out}=0\$ for \$N\$) and assemble them as recommended to obtain the final expression. If you are interested by the technique, you can check a seminar taught at APEC in 2016. It is an introduction to show the power of the FACTs and to apply them to simple passive and active circuits.

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I recall this is a variation of the Baxendall Bass/Treble amplifer but combined.

  • Where increasing treble volume reduces bass and visa versa. The unity gain swing point is \$f=\frac{1}{2\pi R1C1}\$

  • R3/R2 controls max differential gain between plateaus for bass and treble, and feedback parts are matched.

It would be like controlling both Bass and Treble band gains in opposite directions simultaneously !! by +/- range = 15dB total

My preference was my old Bogan Stereo Tube filter switch called CONTOUR that unlike LOUDNESS which is useless to me, you run your system Flat at normal levels ( or equalized to sound flat) Then when the phone rings, select CONTOUR and a broad midrange is cut -15dB to ease communication but still sound great.

here ya go https://goo.gl/4KA6ri JAVA sim

If you want a tuneable midrange filter instead with 15 dB cut, make C1/C2=10

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