| Electronics > Beginners |
| Notch active filter transfer function |
| << < (5/5) |
| 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. --- End quote --- 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. |
| Navigation |
| Message Index |
| Previous page |