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Analog Filter Circuits

  1. Oct 8, 2014 #1
    3nSCvlR.png
    NCEES Reference Handbook p.206


    Hi,

    My question is: In Fig.2, how can R1 and R2 be parallel? If anything, it should be in series because capacitor in the long run is an open.

    Thanks ahead.
     
    Last edited: Oct 8, 2014
  2. jcsd
  3. Oct 8, 2014 #2

    sophiecentaur

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    Gold Member

    Hi and welcome.
    "Series" and "parallel" are just names. They are not relevant to the analysis of circuits - except for the most elementary cases. But, if you bear in mind that a voltage source (v1) has zero impedance and if you replace it with a piece of wire, then, from the point of view of the C, they are in parallel and the equivalent resistance can be calculated with that formula.
     
  4. Oct 8, 2014 #3
    Consider form of the transfer function in the s-domain for that circuit. Introducing Rp gives a shorter way to write it down and calculate.
     
  5. Oct 8, 2014 #4
    thanks for that answers (and others too )

    how did they come to the equation H(s)?
    I tried to do the circuit analysis but it didn't come out the same, will put it up here when i find a decent equation editor
     
  6. Oct 8, 2014 #5
    Let i1 be current through R1, ic current through C, and i2 current through R2.
    For sine wave voltage input it holds:

    i1 = ic+ i2
    v1 = i1·R1 + ic/(jωC)
    0 = i2·R2 - ic/(jωC)

    From this you find (express) i2.
    Then, H(s)=H(jω)=v2/v1= (i2·R2)/v1.
    Finally, when you introduce Rp=R1·R2/(R1+R2), you'll get the result.
     
  7. Oct 8, 2014 #6
    Is eq.2 missing: v1 = i1·R1 + ic/(jωC) + i2·R2?
    Is eq.3 suppose to be: 0 = i1·R1 - i2·R2 - ic/(jωC)?
    Also, how does eq.5 implement onto eq.4?
     
  8. Oct 9, 2014 #7
    Nothing is missing. 2nd eq. is for the first closed loop: v1 - R1 - C and 3rd eq. is for second closed loop: C-R2.
    Do you understand this?
     
  9. Oct 11, 2014 #8
    Okay, I understand eq.1-3, 2&3 are from KVL. i2 should express from which eq.? Why v1 become a short?
     
    Last edited: Oct 11, 2014
  10. Oct 11, 2014 #9
    i2 to be expressed as one of the solutions to the given system of 3 linear equations (other two solutions are i1 and ic but you don't need them). What "v1 become a short"? I don't understand you
     
  11. Oct 11, 2014 #10

    NascentOxygen

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    Eqn 3 comes from saying "because C and R2 are connected in parallel, then the voltage across C equals the voltage across R2."

    Eqn 1 comes from saying "the voltage across C equals V1 minus the voltage lost across R1".
     
  12. Oct 12, 2014 #11
    That is how they get the RP to be R1 and R2 parallel right?

    As for finding H(s), the path using system of equation is kind of tedious. My i2 came out to be pretty ugly, maybe the algebra is wrong somewhere or it just simply ugly. I try to use the voltage divider approach, where Rx is C and R2 parrallel.

    vf7y5Dl.png
     
    Last edited: Oct 12, 2014
  13. Oct 12, 2014 #12

    NascentOxygen

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    H(s) tells you Vout/Vin. I'm wondering why you are needing to find I2 ?

    For a complex impedance, you should be using Zx not Rx. Rx should be used only where the element is known to be a pure resistance.
     
    Last edited: Oct 12, 2014
  14. Oct 12, 2014 #13
    I was trying out the method suggested by zokia85 in post #5
     
  15. Oct 13, 2014 #14
    Oho, looks like you know how to use a voltage divider principle in complex impedance domain:)
    Then, it is obvious you obtained what you wanted:

    cdot%20\frac{R_{1}R_{2}}{R_{1}+R_{2}}}%3D\frac{R_{2}}{R_{1}%281+sCR_{2}%29+R_{2}}.gif

    I don't know what the heck you were doing with the system of equations, but i2 turns up quickly and quite nicely in a form:

    gif.latex?i_{2}%3D\frac{v_{1}}{R_{1}+R_{2}+j\omega%20\cdot%20CR_{1}R_{2}}.gif

    He doesn't need to, but when someone asks questions wether R1 and R2 are in series or parallel it is more pedagogical to show approach with fundamentals of circuit analysis.
     
  16. Oct 13, 2014 #15
    Okay, so far the algebra checked off their final equation against my. Though I still don't understand how they see R1 and R2 as being parallel, the v1 is in the way. If you say, turn v1 into a short, why?
     
  17. Oct 13, 2014 #16
    Not 100% sure, but my guess is they prefer approach to the problem by the method of Thevenin's theorem
    In that case the filter circuit can be replaced by equivalent Thevenin's circuit with:
    gif.latex?V_{th}%3D\frac{R_{1}}{R_{1}+R_{2}}\cdot%20v_{1}.gif
    gif.latex?Z_{th}%3D\frac{R_{1}R_{2}}{R_{1}+R_{2}}%3D%20R_{p}.gif
    in order to find current and voltage of capacitor's impedance Xc :
    gif.latex?i_{c}%3D\frac{V_{th}}{Z_{th}+X_{c}}.gif
    gif.gif
     
  18. Oct 13, 2014 #17
    Ugh, I see. They want to be fancy schmancy. :rolleyes:

    Okay, one last thing. That omega, the way they set it up seem like RP will ALWAYs be equal to capacitor impedance. Is that the case?
     
    Last edited: Oct 13, 2014
  19. Oct 13, 2014 #18
    Are you drunk?
     
  20. Oct 13, 2014 #19
    No, I'm sober, are you?
    Av6Bxz1.png
     
  21. Oct 14, 2014 #20
    ωc is the critical angular frequency. ω can vary between 0 and rad/s . Just have a look at Fig.1 . And stop trolling.
     
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