Resonate Frequency of Bandpass Filter

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Discussion Overview

The discussion revolves around the resonant frequency of a bandpass filter, specifically examining the calculation methods and the underlying transfer function. Participants explore theoretical aspects, mathematical reasoning, and potential shortcuts in deriving the resonant frequency.

Discussion Character

  • Homework-related
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions the validity of calculating the resonant frequency using the method of setting imaginary Z(s) to zero, expressing uncertainty about the outcome.
  • Another participant suggests examining the transfer function H(s) and relating it to the general second order system to describe the natural frequency (Ѡn) and quality factor (Q), while expressing caution about shortcuts in calculations.
  • A later reply reiterates the importance of understanding the second-order filter denominator and its coefficients, suggesting that it is beneficial to memorize the expression and its meaning.
  • One participant expresses gratitude for the clarification provided, indicating that it simplifies their understanding of the topic.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the validity of the initial calculation method for resonant frequency. Multiple viewpoints regarding the analysis of the transfer function and the use of shortcuts remain present.

Contextual Notes

Some participants note the need for clarity on how Z(s) was transformed to include imaginary terms, indicating that this step requires further examination. There is also a mention of different expressions for the second-order filter, which may depend on the context of control theory versus filter theory.

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Homework Statement



See attached.

Can somebody explain how the resonant frequency w0=1/RC. I worked it out by setting imaginary Z(s)=0. The answer I get is j√2 which is obviously wrong. Is it wrong to calculate the resonant frequency in this manner in this case?


Homework Equations





The Attempt at a Solution



See Attached
 

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For a bandpass function, as with any filter, we are interested in Vout/Vin. So they derived this transfer function as H(s) near the bottom of the solution. Are you able to examine that denominator and by relating it to the general second order system describe Ѡn and Q here? (Or the damping ratio, ζ zeta)?

As for your approach, I'm cautious about endorsing shortcuts. But I think it might be valid, thereby allowing you to determine Ѡn at least. But you haven't indicated how you changed Z(s) to something with imaginary terms, so that needs to be checked.
 
NascentOxygen said:
For a bandpass function, as with any filter, we are interested in Vout/Vin. So they derived this transfer function as H(s) near the bottom of the solution. Are you able to examine that denominator and by relating it to the general second order system describe Ѡn and Q here? (Or the damping ratio, ζ zeta)?

As for your approach, I'm cautious about endorsing shortcuts. But I think it might be valid, thereby allowing you to determine Ѡn at least. But you haven't indicated how you changed Z(s) to something with imaginary terms, so that needs to be checked.

Are you able to examine that denominator and by relating it to the general second order system describe Ѡn and Q here? (Or the damping ratio, ζ zeta)?

No. Are you aware of a good internet resource which explains it?
Thanks for your help
 
Towards the end of this article: http://www.swarthmore.edu/NatSci/echeeve1/Ref/FilterBkgrnd/Filters.html

they state that the second-order filter denominator takes the form:

https://www.physicsforums.com/images/icons/icon2.gif s² + (Ѡn/Q)s + Ѡ²n

where (Ѡn/Q) is the bandwidth, with Q being the "Q-factor" of the system.

It's well worth memorizing this expression, and what the co-efficients mean.

If you prefer the corresponding one from control theory, it's: s² + 2ζѠns + Ѡ²n
where zeta is the co-efficient of damping and you can see ζ=1/(2Q)
 
Last edited by a moderator:
Thanks for that. Makes things somewhat easier than how I was doing it.
 

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