Help me calculate corner frequencies

[lars-musik]

This feels a little stupid, but my web searches didn't come up with a satisfying answer:

If I understand the basics correctly, R1 (and R2 depending on the switch) and C2 form a low pass filter with corner frequencies of 48 Hz (at R=33K) or 120 Hz (R=13.2K).

This seems somehow to be a bit on the dark side soundwise, so I guess the variable resistor (VR1) comes into the equation somewhere. The "standard" RC-filter schematics do not have that resistor following the capacitor. Where would I put it in the equation?

Thanks for answering even trivial questions!!!

[antonis]

Can't see what is puzzling you, Lars..

You have to add VR1 resistance (value according to pot setting) in series with C2..

Think it as a voltage divider (actully, it IS a voltage divider) with lower part consisting of C2 impedance (1/2πfC) in series with VR1 resistance..
(in case of VR1 absence, lower part of voltage divider consists only of C2..)

(same manner with calculation of the HPF on the gain part of the NFB loop of an op-amp..)

[TejfolvonDanone]

I was taught like this.
First just think what kind of result do you expect for different frequencies. At low frequency the cap looks like an open circuit so the output would be the same as the input. As the frequency gets higher the cap's impedance decreases so at high frequencies the cap looks like a short. Then it will be a voltage divider made of R1 and VR1.
Therefore will be two corners: first (f1) when the cap's impedance will be around the same as the resistors and second (f2) when it will be low enough not to make a difference.
As antonis said to calculate the frequencies you have to take the voltage divider equation: Uout/Uin = Z2/(Z1+Z2)
Z1 = R1
Z2 = VR1 + Zc2
This is where in collage we got into the complex numbers and i don't know a way to explain the calculations in any other way. If you don't know what I'm talking about as complex numbers just skip to the last paragraph.
(j is the imaginary unit)
Z2 = VR1 +1/j2πfC
Uout/Uin = (VR1 +1/j2πfC)/(R1+VR1+1/j2πfC) = (VR1*j2πfC+1)/(j2πfC*(R1+VR1)+1)

So the first corner will be at f1 = 1/(2πR1*C) the second at f2 = 1/(2π(R1+VR1)*C)
The damping at high frequencies (above f2) is VR1/(R1+VR1)

[squeezer]

This is what's happening, when the pot is engaged.
The bottom line is the pot with the wiper to lug 1 (1 Ohm, practically the low pass filter without the pot).
The top line is wiper to lug 3 (25k). And some values between.
I don't know formulas to calculate that, but as you see there is more to it than just one corner frequency.

[lars-musik]

Can't see what is puzzling you, Lars..

I think I was puzzled by the fact that I didn't find the formula(s) given by TejfolvonDanone (thanks!)

So the first corner will be at f1 = 1/(2πR1*C) the second at f2 = 1/(2π(R1+VR1)*C)
The damping at high frequencies (above f2) is VR1/(R1+VR1)

or a calculator for a graph as the one so very helpfully visualised by squeezer (thanks again!!!) during my websearch.

If you are used to "electronical thinking" and in-head-visualisation of impedance curves it might all be clear, but that's not the way my brain works - yet!

I am looking for LTSpice (it is a free software, right?) next, so that I might do some curve-drawing myself.

[GibsonGM]

LT Spice is free, just search online and you will find it!  Very helpful for this sort of thing, and modeling this in it will help very much to understand how this works.

The basic formula for first order R-C filters is   Fco = 1 /  2pi R * C       which of course is what TejfolvonDanone described.     

Fco = corner frequency
2pi = 6.28
R in megohms, C in microfarads are usually good units to use

This is for both LPF  and HPF! 

[antonis]

If you are used to "electronical thinking" and in-head-visualisation of impedance curves it might all be clear, but that's not the way my brain works - yet!

You don't have to..

TejfolvonDanone explained you the right way of calculating impedance of any instaneous phase shifted signal (although he should use angular frequency ω instead of 2πf and square root of -1 instead of j tο get you more confused..  ) but practical results can be more simple..

Replace cap with it's equivalent resistance (at any specific frequency of interest) and use Ohm's law on a purely resistive divider..

P.S.
Don't apply the above in case of some cascaded filters.. 

[lars-musik]

.. he should use angular frequency ω instead of 2πf and square root of -1 instead of j....

I spotted that one at once!   

[GibsonGM]

.. he should use angular frequency ω instead of 2πf and square root of -1 instead of j....

I spotted that one at once!   

Square root of - 1?   Get real...    

[lars-musik]

So all in all: The tone control of the OCD is a high cut. More tone-shaping is not possible by simply changing the values of the parts.

[GibsonGM]

So all in all: The tone control of the OCD is a high cut. More tone-shaping is not possible by simply changing the values of the parts.

Right, like your guitar.  You CAN move the corner around by changing values.  You'd have to add things to get a different shape, tho. 

[antonis]

Square root of - 1?   Get real...   

I can call it "j" if you think is closer to reality.. 

[EBK]

Square root of - 1?   Get real...   

I can call it "j" if you think is closer to reality.. 

[rant]
That j notation is one of the great mysteries of electrical engineering (to me, anyway).
Why do we use j for the square root of negative one when everyone else uses i?
The usual answer is simply because we use i for current.  But why do we use i for current?  Apparently, it originated with the French phrase intensité de courant, which Ampère abbreviated to i.  But if the French word for current is courant, then why didn't Ampère just abbreviate it to c? 
[/rant]

[antonis]

Maybe the guy who enriched M.K.S. system with A. wasn't drunk enough.. 

[EBK]

Maybe the original use of j was a typo, and no one wanted to admit it. 

[GibsonGM]

C is the speed of light.  But I'd imagine that "I" was chosen before Einstein's Theory, so yes - WHY NOT?? 

I think we should petition to re-do these abbreviations in a way that makes more sense!

And "E" for voltage....eeew*    I used to write "I = V/R" before I better trained myself!  It was just so easy to do so.

*Yes, I know, electromotive force...still...

[anotherjim]

While being able to target a particular frequency is all very cool, in guitar land I don't find it so straightforward.
Guitar amplification usually adds some compression or clipping, and that flattening tends to widen the pass band of a simple 3dB/oct filter, moving the cutoff -3dB corner downward at the bass end or upward at the treble.

You may discover that your filter corner frequency can need to be as much as an octave or more away from target, so it is at least changed by 6db where you want it to be noticeably different.

Often, I find it comes down to changing the calculated value of caps by a factor of 2 one way or the other, depending on LP or HP, gets the filter working something like I intended.

[ElectricDruid]

Others have given you all the nitty-gritty details, so I won't go there.

Instead, I'd just like to point out that what you've got there is a "shelving filter", in this case a shelving lowpass (or hi-cut, same thing).
This kind of arrangement of an RC filter with an extra resistor in is handy if you want the RC action, but you need to limit its effect in some way.

There's some more about the different configurations (and active ones too) at the Linkwitz labs site:

http://www.linkwitzlab.com/filters.htm

HTH,
Tom

[antonis]

You may discover that your filter corner frequency can need to be as much as an octave or more away from target, so it is at least changed by 6db where you want it to be noticeably different.

An "elementary" case is IN & OUT HPFs of an AC coupled Emitter follower..

Making same (or overlap) 3db points of two cascaded filters ensure larger (at least 6 db) than predicted signal attenuation at the lowest frequency of interest..
(a greater than doubling capacitor values is required for actuall cut-off point remedy..)

[TejfolvonDanone]

(although he should use angular frequency ω instead of 2πf and square root of -1 instead of j tο get you more confused..  )

The angular frequency (for me at least) never had that real life meaning as frequency. And because ω = 2πf they are almost the same. Also it can be easily understandable of the OP.
The square root of -1 is not the same as the imaginary unit (i or j) because the square root of -1 is j and -j.

Apparently, it originated with the French phrase intensité de courant, which Ampère abbreviated to i.  But if the French word for current is courant, then why didn't Ampère just abbreviate it to c? 

Because he didn't abbreviate courant he abbreviated intensité de courant, didn't he?