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17th April 2008, 08:17 PM  #1 
diyAudio Member

transfer function
Hey, I'm trying to work out how the bass pot of a baxandall tone circuit works. Atm I'm looking at the pot when it's turned fully to the right. That's the image I've attached to this post.
From some basic circuit electronics I know I can combine C1 & R2 to get Z(RC) = 1 / (s + 1/R2C1) but I'm a bit stuck then because you get three impedances 'looking' into the same node (I'll attach a drawing of this below) and that R3 really throws me off. If it wasn't there I'd know what to do but I'm really confused with it. If anybody out there can offer me a helping hand then please do! :) Cheers, Jenks. 
17th April 2008, 08:21 PM  #2 
diyAudio Member

bass pot simplified
see below

17th April 2008, 08:29 PM  #3 
diyAudio Member
Join Date: Jul 2007
Location: Central Berlin, Germany

If the output is unloaded (highZ) then R3 doesn't matter. In the passive Baxandall, R3 is only there to isolate the bass section from the treble section.
Here a nice article which deals with the details: http://www.schmarder.com/radios/tech/tone.htm  Klaus 
17th April 2008, 08:32 PM  #4 
diyAudio Member
Join Date: Apr 2005
Location: Nijmegen!

The transfer function you want to calculate doesn't take input/output impedances in account. So, if you imagine a source output impedance of 0 and an infinite input impedance, you can simply ignore R3, leading to:
Vout/Vin = R4 / (R4 + R1 + R2ZC1) If you know your input/output impedances, then ofcourse you can calculate everything (I assume?). So, in most cases I assume Rin >> R3 and Rin >> R4. Then, effectively, R4 becomes R4(R3 + Rin), leading to R4 again. If Rout << R1, then you can ignore that one also. Hope this helps... .edit: too late 
17th April 2008, 08:36 PM  #5 
diyAudio Member

Cheers for that Klaus, however the circuit I'm using is a little different to that featured at the link you've included. I'm using the Baxandall circuit but with and opamp incorporated and the circuit isn't grounded the same as it is in the link you provided either so I suspect the same rules don't apply.
I've attached a screenshot of my entire circuit. If you can help me out in trying to discover how the component values affect the frequency response of my input signal I'd really appreciate it. Thanks. 
17th April 2008, 08:40 PM  #6 
diyAudio Member

This circuit will be manipulating line level signal so does the input imedance match the value of the 4.7kohm (R3) resistor?
What then? 
17th April 2008, 08:43 PM  #7 
diyAudio Member
Join Date: Jul 2007
Location: Central Berlin, Germany

Look at the references given at the end. The spanish site (actually peru) deals with the opamp variant, which is indeed completely different.
Another nice way to figure out, especially if you're not into tedious math (well, who is?), is to use simulation. The free LTspice/SwCADIII (from linear.com) is quite fine, if not to say ideal for such tasks.  Klaus 
17th April 2008, 08:44 PM  #8 
diyAudio Member
Join Date: Apr 2005
Location: Nijmegen!

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...

17th April 2008, 08:47 PM  #9  
diyAudio Member
Join Date: Jul 2007
Location: Central Berlin, Germany

Quote:
 Klaus 

18th April 2008, 02:44 PM  #10 
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.

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