RC high pass filter calculation

[sM1101011]

The formula above was used to peform the calculation for the upper section of the filter circuit but i would like to know how would the formula change if i needed to perform a similar calculation to the lower filter circuit. I have outlined both the upper and lower section of the Filter circuit.

 

[BrownOut]

Where did the first equation come from? I might try to make one for the lower circuit, but I just returned from 3 days at the lake, and I'm on my 2nd JD on the rocks. Maybe I'll try later, if I can find the time. BTW, the problem reduces to a fairly simple circuit analysis.

 

[MrAl]

The formula above was used to peform the calculation for the upper section of the filter circuit but i would like to know how would the formula change if i needed to perform a similar calculation to the lower filter circuit. I have outlined both the upper and lower section of the Filter circuit.

Hi again,


Here's your new formula:

Vout=(Vin*R24b*w*C5)/sqrt((1-R24b*w^2*C5*C6*R25-R24a*w^2*C5*C6*R25-R24a*R24b*w^2*C5*C6)^2+(w*C6*R25+R24b*w*C6+R24b*w*C5+R24a*w*C5)^2)

where again w=2*pi*F, and
R24a is the pot upper part resistance, and R24b is the lower part resistance.

I leave it to the reader to simplify more if possible.

I also leave it to the reader to show that for frequencies above 10Hz that the network has a response very very close to another much simpler network made up as follows:

Place a gain block in series with the 10uf cap. Set the gain equal to the ratio R24b/(R24a+R24b).
Eliminate the 10uf cap by shorting it out.
Replace the upper part of the pot (R24a) with a new resistor with a value equal to the parallel parts R24a and R24b, eliminate the lower part of the pot (R24b)...that puts the new resistor in series with R25.
The output is taken from the same place as before, across the capactor C6.
Note that the output is very close to the same as the original network but the formula is greatly simplified:
Vout=G*Vin/sqrt(w^2*C^2*R^2+1)
where
C=C6
R=R24a*R24b/(R24a+R24b)+R25
G=R24b/(R24a+R24b)
w=2*pi*F

 

[The Electrician]

Herewith somewhat simplified formulas. The first is for the case where C5 is replaced with an AC short, so that it is no longer involved in the formula. The second is the full formula.