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| Notch active filter transfer function |
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| alex.martinez:
--- 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. |
| kulky64:
OK, now I see. --- Quote from: alex.martinez on April 08, 2019, 11:20:37 am ---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 --- But the TF only has this form when the condition (R3*C1-R2/R1*R4*C2)=0 is met, for example if R3=R4=R, C1=C2=C and R1=R2 as suggested in the article you linked in your first post. Then you don't have to bother calculating Q, because it is equal to 1/2. In most general case though, without any restrictions on R1, R2, R3, R4, C1 and C2 values, I think your method for determining Q is no longer valid. Or am I wrong? |
| alex.martinez:
--- Quote from: kulky64 on April 08, 2019, 03:42:44 pm ---But the TF only has this form when the condition (R3*C1-R2/R1*R4*C2)=0 is met, for example if R3=R4=R, C1=C2=C and R1=R2 as suggested in the article you linked in your first post. Then you don't have to bother calculating Q, because it is equal to 1/2. In most general case though, without any restrictions on R1, R2, R3, R4, C1 and C2 values, I think your method for determining Q is no longer valid. Or am I wrong? --- End quote --- I see your point, and it has been bugging me as well because I was getting an s term in the numerator. I really do not know of any second order transfer function that presents a complete quadratic term in the numerator. Regarding (R3*C1-R2/R1*R4*C2)=0, yes the obtained TF and the one in the canonical form coincide just for the aforementioned condition. I do not know how does that relate to the method of obtaining Q. |
| kulky64:
I created Bode diagrams with component values exactly as in the article you mentioned: R3=R4=68kOhm, C1=C2=47nF, but I varied R2/R1 ratio. So w0 is for all curves the same: 312.891113892365 rad/s. F1: R2/R1=1 F2: R2/R1=0.95 F3: R2/R1=0.9 F4: R2/R1=0.8 F5: R2/R1=0.7 F6: R2/R1=0.6 It looks like the Q factor changes according to R2/R1 ratio. Or not? |
| The Electrician:
--- Quote from: kulky64 on April 08, 2019, 09:38:02 pm ---I created Bode diagrams with component values exactly as in the article you mentioned: R3=R4=68kOhm, C1=C2=47nF, but I varied R2/R1 ratio. So w0 is for all curves the same: 312.891113892365 rad/s. F1: R2/R1=1 F2: R2/R1=0.95 F3: R2/R1=0.9 F4: R2/R1=0.8 F5: R2/R1=0.7 F6: R2/R1=0.6 It looks like the Q factor changes according to R2/R1 ratio. Or not? --- End quote --- How then would you calculate the Q factor? |
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