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Two-Pole Passive RL Low Pass filter

Efexynoob

Sep 5, 2023
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A bit new to passive filters, so wondering what is the formula of frequency cutoff for the two pole Low Pass RL filter? And how Gain/Phase graphs would look like? Thank you!RDT_20230905_0806246720477943089937920.png
 

Efexynoob

Sep 5, 2023
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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:

Screenshot_20230906_080110.jpg

Screenshot_20230906_080124.jpg

[mod edit: resized the screenshots for better readability]
 
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danadak

Feb 19, 2021
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This is what I get for T(s)

1693997609301.png

1693997992530.png



Isn't the problem with your 1693998390185.png
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.
 
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Harald Kapp

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Nov 17, 2011
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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.
 

Efexynoob

Sep 5, 2023
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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.
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...
 

danadak

Feb 19, 2021
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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 :


1694083716263.png

1694083450333.png



As poles approach y axis Q rises because energy dissipated in real R drops :


1694083959100.png


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 "

1694085892898.png
I'm really bad at this

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.
 
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