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Learn from the explanation here.
Thank you very much!
Everyones graphs are similar, I appreciate the help, but still a bit confused about Cut Off frequency... when calculating by the hand getting around 80Hz, when you guys and other forum people get about 40Hz. Formulas I'm using:
[mod edit: resized the screenshots for better readability]
[mod edit: resized the screenshots for better readability]
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This is what I get for T(s)
Isn't the problem with your
That's the G due to two identical single pole cascaded stages, but your circuit
is not two identical cascaded, there is no buffer of G = 1 between the two stages.
The second stage loads the first stage, so they are not "true" cascaded identical
stages, meeting pole shrinkage factor classic analysis.
Regards, Dana.
Isn't the problem with your
That's the G due to two identical single pole cascaded stages, but your circuit
is not two identical cascaded, there is no buffer of G = 1 between the two stages.
The second stage loads the first stage, so they are not "true" cascaded identical
stages, meeting pole shrinkage factor classic analysis.
Regards, Dana.
Last edited:
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Maybe I am missing something but I thought he had equal L and R values ?
Regards, Dana.
- Joined
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Not missing anything.
The page I linked to gives a bunch of equations and mentions a bit further down the page that the equations given are valid only for unloaded filters i.e. impedance of the 2nd filter stage at least 10 × hihger than the impedance of the first filter stage.
Therefore the simplified equations from this page's beginnings are not suitable for this case.
The page I linked to gives a bunch of equations and mentions a bit further down the page that the equations given are valid only for unloaded filters i.e. impedance of the 2nd filter stage at least 10 × hihger than the impedance of the first filter stage.
Therefore the simplified equations from this page's beginnings are not suitable for this case.
Perhaps maybe you know the equation for this particular case, as my name says I'm really bad at this... people say I need to have Q in my calculations...
You dont "HAVE TO" have Q in your equations, its just another look
at a networks behaviour. That being said a 2 pole filter can exhibit
"peaking" which is where Q is often used as an indicator or the level
of peaking. Example in a LPF :
As poles approach y axis Q rises because energy dissipated in real R drops :
So Q here is meaningful about this peaking effect but is NOT NECESSARY in a T(jw) expression.
Pole Zero locations and effects (exponential growth/decay) on a signal "
Practice makes perfect as the expression goes. You write node and loop
equations, those are fairly simple, and solve for T(jw) or T(s) = Vout / Vin.
Straightforward algebraic manipulation. Note however as the number
pole zeroes rises > 2 that becomes very tedious. Thats why things like
signal flow graphs reduce the pain and tendency to want to jump off a
bridge trying to solve. As was done in post #6.
Regards, Dana.
at a networks behaviour. That being said a 2 pole filter can exhibit
"peaking" which is where Q is often used as an indicator or the level
of peaking. Example in a LPF :
As poles approach y axis Q rises because energy dissipated in real R drops :
So Q here is meaningful about this peaking effect but is NOT NECESSARY in a T(jw) expression.
Pole Zero locations and effects (exponential growth/decay) on a signal "
Practice makes perfect as the expression goes. You write node and loop
equations, those are fairly simple, and solve for T(jw) or T(s) = Vout / Vin.
Straightforward algebraic manipulation. Note however as the number
pole zeroes rises > 2 that becomes very tedious. Thats why things like
signal flow graphs reduce the pain and tendency to want to jump off a
bridge trying to solve. As was done in post #6.
Regards, Dana.
Last edited:
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