transfer function
 User Name Stay logged in? Password
 Home Forums Rules Articles diyAudio Store Blogs Gallery Wiki Register Donations FAQ Calendar Search Today's Posts Mark Forums Read Search

 Solid State Talk all about solid state amplification.

 Please consider donating to help us continue to serve you. Ads on/off / Custom Title / More PMs / More album space / Advanced printing & mass image saving
jenks
diyAudio Member

Join Date: Mar 2008
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.
Attached Images
 bass pot 100%.jpg (60.6 KB, 296 views)

jenks
diyAudio Member

Join Date: Mar 2008
bass pot simplified

see below
Attached Images
 bass pot 100% (next step).jpg (59.5 KB, 271 views)

 17th April 2008, 07:29 PM #3 KSTR   diyAudio Member     Join Date: Jul 2007 Location: Central Berlin, Germany 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: http://www.schmarder.com/radios/tech/tone.htm - Klaus
 17th April 2008, 07:32 PM #4 Limhes   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 + 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
jenks
diyAudio Member

Join Date: Mar 2008
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.
Attached Images
 baxandall_scan.jpg (20.2 KB, 280 views)

 17th April 2008, 07:40 PM #6 jenks   diyAudio Member   Join Date: Mar 2008 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, 07:43 PM #7 KSTR   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, 07:44 PM #8 Limhes   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...
KSTR
diyAudio Member

Join Date: Jul 2007
Location: Central Berlin, Germany
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
diyAudio Member

Join Date: Mar 2008
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).
Attached Images
 tonexferfunction.gif (15.1 KB, 235 views)

 Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is Off Forum Rules
 Forum Jump User Control Panel Private Messages Subscriptions Who's Online Search Forums Forums Home Site     Site Announcements     Forum Problems Amplifiers     Solid State     Pass Labs     Tubes / Valves     Chip Amps     Class D     Power Supplies     Headphone Systems Source & Line     Analogue Source     Analog Line Level     Digital Source     Digital Line Level     PC Based Loudspeakers     Multi-Way     Full Range     Subwoofers     Planars & Exotics Live Sound     PA Systems     Instruments and Amps Design & Build     Parts     Equipment & Tools     Construction Tips     Software Tools General Interest     Car Audio     diyAudio.com Articles     Music     Everything Else Member Areas     Introductions     The Lounge     Clubs & Events     In Memoriam The Moving Image Commercial Sector     Swap Meet     Group Buys     The diyAudio Store     Vendor Forums         Vendor's Bazaar         Sonic Craft         Apex Jr         Audio Sector         Acoustic Fun         Chipamp         DIY HiFi Supply         Elekit         Elektor         Mains Cables R Us         Parts Connexion         Planet 10 hifi         Quanghao Audio Design         Siliconray Online Electronics Store         Tubelab     Manufacturers         AKSA         Audio Poutine         Musicaltech         Aussie Amplifiers         CSS         exaDevices         Feastrex         GedLee         Head 'n' HiFi - Walter         Heatsink USA         miniDSP         SITO Audio         Twin Audio         Twisted Pear         Wild Burro Audio

 Similar Threads Thread Thread Starter Forum Replies Last Post NIC1138 Digital Source 9 15th July 2007 12:15 AM Limhes Solid State 3 26th April 2007 05:17 AM jclouse Multi-Way 22 1st July 2006 01:19 PM cm961 Multi-Way 1 31st August 2004 06:42 PM Thomas B Multi-Way 4 12th March 2003 04:04 AM

 New To Site? Need Help?

All times are GMT. The time now is 09:14 PM.