filter calculations qw

[alfafalfa]

In these examples I would like to know how to calculate the A and B filters



A ; is a high pass filter  but a 6 dB ?  and  is there an easy programme for this calculation ?

B : is a low pass filter ? 

I am familiar with aron's page on filters but A and B i haven't seen there .

Alf

[George Giblet]

A and B are *NOT* filters as they stand.

A is a network which can be used to create a shelf equalizer with a low frequency cut (or  high frequency boost depending on your perspective) - it never really forms a high pass filter in the technically correct sense.  In order to do this you need to have a load resistor to ground - that's what is missing on the schematics.  If R1 is the resistor in parallel with the capacitor and R2 the resistor to ground, the frequency fL= 1/(2*pi * R1 *C) is the lower frequency point where the shelf starts and fH = 1/(2*pi * R12 * C) is the upper frequency point where the shelf flattens off; R12 = R1 in parallel with R2.  If R1 is infinite you get a high pass.

In the same vein B network whch can be used to create a shelf equalizer with a high frequency cut (or low frequency boost).  Again it's not a proper low pass.   B needs a series source resistance between the input terminal and the network.  If R1 is the series resistor and R2 the resistor in parallel with the cap then fL = 1/(2*pi*R2*C) is the point where output starts to drop and fH = 1/(2*pi * R12 * C) is the point where the shelf flattens off again at high frequencies.  If R2 is infinite you get a low pass.

fH and fL are not 3dB points however if fH and fL are far enough appart then both sort of are.

(One thing I can add is calling the *networks* A and B filters is like calling a series resistors a voltage divider.  A series resistor can be used as a divider but without a load resistor the statement doesn't mean much.

[alfafalfa]

Thanks you very much George , for clearing this up .

Your explanation is very clarifying. So now I'm going to do some calculations.
I wanted to calculate the networks in a Mesa Boogie Mk3.

Are there any exelsheets anyone has already devised to do these quickly ?


Thanks , Alf

[PerroGrande]

The thing to keep in mind in both cases (A & B), you're looking at two impedances in parallel.  One is a resistor, contributing straight-out resistance.  The other is a capacitor -- the impedance of which is frequency-dependent. 

To understand "A" fully, the rest of the circuit would have to be in place.  It could be part of a voltage divider or a feedback network.  In either case, as George correctly points out, there needs to be an actual circuit involved.

B is an actual circuit. 

When thinking about B in simple terms, you really have two impedances in parallel to ground.  You have the resistance and the reactive component of the capacitor (which is frequency dependent).  Fortunately, Ohm's law holds for impedances, so you can analyze it the same as one would analyze parallel resistances to ground -- except that the capacitor's "value" (impedance) changes (Xc = 1 / (2 * pi * freq * C).  So -- as frequency increases, Xc drops.  As Xc drops, its lower impedance will dominate the parallel equation.  As Xc rises, the resistor will dominate the parallel equation.

[alfafalfa]

Thanks John , there's more to it than I thought.

It's getting complicated.
I will post part of the circuit I mean.

Alf

[PerroGrande]

Hi Alf,

They're not that bad!  Stick with it.

A "mantra" I have people keep in mind when looking at circuits like this...  "Capacitors pass AC and block DC."  Sure -- it is a bit overly simplistic, but it helps one get the feel for a circuit or a snippet without a lot of math.   When a capacitor shows up, it is going to look like an open circuit to DC -- effectively infinite resistance.  The higher the frequency, the less the capacitor looks like an open circuit (the impedance falls as frequency increases).

Another quick "mantra" is that electricity "takes the path of least resistance."  It should probably read "path of least impedance", but hey, close enough.  So when the signal encounters two possible ways to go, it will divide itself proportionally -- with the "easy" way getting the majority.  What makes one way or another "easy"? Impedance.  Lower = Easier.