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 jenks 17th April 2008 08:17 PM

transfer function

2 Attachment(s)
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.

 jenks 17th April 2008 08:21 PM

bass pot simplified

2 Attachment(s)
see below

 KSTR 17th April 2008 08:29 PM

If the output is unloaded (high-Z) 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:

- Klaus

 Limhes 17th April 2008 08:32 PM

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 + R2||ZC1)

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 :whazzat:

 jenks 17th April 2008 08:36 PM

2 Attachment(s)
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 op-amp 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.

 jenks 17th April 2008 08:40 PM

This circuit will be manipulating line level signal so does the input imedance match the value of the 4.7kohm (R3) resistor?

What then?

 KSTR 17th April 2008 08:43 PM

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

 Limhes 17th April 2008 08:44 PM

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...

 KSTR 17th April 2008 08:47 PM

Quote:
 Originally posted by jenks This circuit will be manipulating line level signal so does the input imedance match the value of the 4.7kohm (R3) resistor? What then?
You need a buffer ahead of it, otherwise it doesn't work as expected. Or a redesign with very high resistor/impedance values, but that's not really a good idea.

- Klaus

 The Electrician 18th April 2008 02:44 PM

2 Attachment(s)
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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