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Old 18th April 2008, 02:44 PM   #11
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Quote:
Originally posted by Limhes
Just apply Kirchhof's and Ohm's laws for all resistors and impedances and you can work it out, knowing that the opamp's negative input terminal is at 0V...
Here's the transfer function for the whole thing. I made a few changes in the designators. The two resistors on either side of the bass pot are R1. The bass pot itself is R2. The two resistors on either side of the treble pot are R3, and the treble pot is R4. The two bass capacitors are C1 and C2, with C1 being the furthest from the opamp. The capacitor in series with the treble pot wiper is C3 and the resistor in series with the bass pot wiper is R5. The opamp gain is assumed to be infinite (ideal opamp).
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Old 19th April 2008, 07:08 PM   #12
jenks is offline jenks  United Kingdom
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Hey there Electrician! That transfer function looks like a beast. How did you come up with it? I'm doing it in bits atm and it's taking ages with a pen and paper :-(
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Old 19th April 2008, 10:57 PM   #13
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Quote:
Originally posted by jenks
Hey there Electrician! That transfer function looks like a beast. How did you come up with it? I'm doing it in bits atm and it's taking ages with a pen and paper :-(
I did it with Mathematica. A person could probably use any of the other modern mathematical programs that can do symbolic algebra, such as Maple or possibly Matlab. I don't think it would be possible to get it right by hand!

I posted it somewhat as a joke. You can see how complicated it is, and I'm not sure that it's much help for understanding the circuit operation.

But, it is in fact correct, and if you substitute in the values for the various components and then plot it as a function of frequency, you get plots that are what you expect. You have to substitute 2*Pi*j*f for the Laplace variable s, and take the absolute value of the function (square root of the sum of the squares of the real and imaginary parts) for plotting.

One nice thing about using the full transfer function is that the effect of the bass components on the operation of the treble control, and vice versa, is correctly calculated.

One thing I did notice while doing some plots is that the R5 resistor (you called it R4 in your circuit in another thread; the one with the mid-band control) seems too low by at least an order of magnitude. As given, it limits the treble boost and cut to amounts much less than the corresponding bass boost and cut.

Sometimes I have found that if you substitute numbers for all the components but one, leaving that one symbolic, the transfer function simplifies enough to give some insight.

You can probably get more insight into the circuit operation by using a simulator and plotting families of response curves as you vary one component.

At any rate, you are right, the transfer function IS a beast. It's amazing how fast they get out of hand as soon as you have 4 or more nodes in your circuit.
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Old 20th April 2008, 03:20 AM   #14
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Here's the schematic from which I got the transfer function.

I'll show plots of the gain with the treble and base pots going from flat to full up in increments of .1 for the coefficients Kb and Kt (fractional pot rotation) in subsequent posts. I don't know how to attach more than one image; maybe one of the moderators can tell me how to do this.
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Old 20th April 2008, 03:20 AM   #15
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Here is a plot showing the response with the component values shown in post #5.
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Old 20th April 2008, 03:20 AM   #16
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In this plot, the value of R5 has been increased to 47k.
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Old 20th April 2008, 03:20 AM   #17
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In this plot, the value of R5 has been increased to 470k.

Now we get full boost of the treble, but the corner frequencies are wrong.
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Old 20th April 2008, 03:20 AM   #18
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Now the value of C3 has been decreased to 470p.
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Old 20th April 2008, 03:20 AM   #19
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Here's the treble response.
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Old 20th April 2008, 03:20 AM   #20
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Here's the full boost and cut response of the circuit with both treble and bass controls varied independently over their full ranges in .1 increments of the Kb and Kt coefficients.
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