transfer function
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 18th April 2008, 01:44 PM #11 The Electrician   diyAudio Member   Join Date: Mar 2008 I forgot to mention that there are two coefficients associated with the circuit. Kb is the fraction of rotation of the bass pot, varying from 0 to 1. A value of .5 will give flat response. Likewise, Kt is the treble pot fraction of rotation.
 19th April 2008, 06:08 PM #12 jenks   diyAudio Member   Join Date: Mar 2008 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 :-(
The Electrician
diyAudio Member

Join Date: Mar 2008
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.

The Electrician
diyAudio Member

Join Date: Mar 2008
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.
Attached Images
 tones.jpg (25.2 KB, 138 views)

The Electrician
diyAudio Member

Join Date: Mar 2008
Here is a plot showing the response with the component values shown in post #5.
Attached Images
 bax1.gif (36.6 KB, 136 views)

The Electrician
diyAudio Member

Join Date: Mar 2008
In this plot, the value of R5 has been increased to 47k.
Attached Images
 bax2.gif (49.4 KB, 131 views)

The Electrician
diyAudio Member

Join Date: Mar 2008
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.
Attached Images
 bax3.gif (47.9 KB, 131 views)

The Electrician
diyAudio Member

Join Date: Mar 2008
Now the value of C3 has been decreased to 470p.
Attached Images
 bax4.gif (52.7 KB, 124 views)

The Electrician
diyAudio Member

Join Date: Mar 2008
Here's the bass boost response only, with the treble set flat.
Attached Images
 bax5.gif (17.5 KB, 130 views)

The Electrician
diyAudio Member

Join Date: Mar 2008
Here's the treble response.
Attached Images
 bax6.gif (17.1 KB, 122 views)

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