Electronics > Beginners
Notch active filter transfer function
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MrAl:

--- Quote from: alex.martinez on April 08, 2019, 11:20:37 am ---
--- Quote from: MrAl on April 08, 2019, 05:05:18 am ---
--- Quote from: The Electrician on April 07, 2019, 06:34:53 pm ---
--- Quote from: kulky64 on April 07, 2019, 01:05:52 pm ---How did you arrive from wo/Qp=1/(R3*C1)+1/(C2*R4) to Qp=wo^(-1)/(C2*R4+R3*C1) ?

--- End quote ---

I assume you're referring to the algebra mistake?



--- End quote ---

Hi,

What that formula actually tested yet though?

--- End quote ---

I think this is getting a bit out of hand... Qp =wo·(C1·C2·R3·R4)/(C2·R4+R3·C1) is algebraically the same as  Qp = 1/[(C2·R4 + C1·R3)· wo], since we define wo = 1/sqrt(C1·C2·R3·R4).

If dimensions are taken into account, both expressions arrive at a dimensionless factor. What I do is I define what is the natural frequency of the system and re-write it in a canonical form:



The I just take the terms that multiply s on the denominator and equate it to wo/Qp, in order to find the expression for Qp.

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Hi,

No i was not referring to the algebra mistake, if there way one, i was referring to teh entire formula as it does not seem right.  If that is meant to calculate the Q of the circuit or the Bandwidth then i dont think it is right.  It is missing two resistors which have an effect on the bandwidth BW.
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