third order butterworth (ind,cap,ind)

Discussion in 'General Electronics Chat' started by ninjaman, Nov 24, 2013.

  1. ninjaman

    Thread Starter Member

    May 18, 2013
    306
    1
    hello

    please could someone point me in the direction of a site that includes formulas for rlc networks. I have to find current and voltage through three components in a circuit. an inductor in series, a capacitor in parallel and another inductor in series. its a third order butterworth passive filter.
    I have found formulas but only for one inductor in series and one capacitor in parallel.

    does much change when you add another inductor in series?
    I mean when working out the current and voltage?
    do you have to include all three components in the formula or can you do the first two (inductor/capacitor) and then using that result for the second series inductor?

    all the best

    simon
     
  2. shteii01

    AAC Fanatic!

    Feb 19, 2010
    3,387
    497
    Replace inductors and capacitors with impedances.

    Inductor becomes sL.
    Capacitor becomes 1/(sC).

    Now you can treat them like resistors. Do the voltage dividers, current dividers. Resistors in series. Resistors in parallel.
     
    ninjaman likes this.
  3. ninjaman

    Thread Starter Member

    May 18, 2013
    306
    1
    sorry what is sL, series inductor? s is series?

    cheers

    simon
     
  4. MikeML

    AAC Fanatic!

    Oct 2, 2009
    5,450
    1,066
  5. shteii01

    AAC Fanatic!

    Feb 19, 2010
    3,387
    497
    s is complex frequency.
     
  6. LvW

    Active Member

    Jun 13, 2013
    674
    100
    For calculation of voltages and currents (sinusoidal signals) within the network the complex frequency variable is replaced by jw.
     
  7. gootee

    Senior Member

    Apr 24, 2007
    447
    50
    You could use only jω, instead of s = α + jω. But then you would get only the sinusoidal steady-state response and lose the transient response portion. You would basically be downgrading from Laplace to Fourier.

    With Laplace, you are essentially converting the full differential equations of a circuit or system into algebraic equations in terms of s, which are usually much easier to solve than the time-domain differential equations. After the algebraic equations have been solved, you can keep the solution in terms of s, and have the traditional frequency-domain Transfer Function, or directly convert back to the time domain solution of the differential equations.

    For the OP: You can treat each reactive component as a frequency-dependent resistance and then use the standard methods of basic resistive circuit analysis (i.e. writing the node and/or loop equations, using Kirchoff's current and voltage laws, and then solving the set of simultaneous equations).
     
  8. LvW

    Active Member

    Jun 13, 2013
    674
    100
    Don`t forget the phase properties! A frequency-dependent resistance (FDNR) is something else!
     
  9. gootee

    Senior Member

    Apr 24, 2007
    447
    50
    I wasn't forgetting anything!

    The resulting equations would completely characterize not only phase and amplitude in steady state but also the combined transient response and steady-state response solution, if s were used instead of only jω!

    I was NOT suggesting evaluating the impedances at some frequency, first, and then analyzing a resistive network. I was suggesting that the OP should try to realize that the frequency-domain expressions for the components' impedances could be used in the same way that resistances are used, when writing the equations for the circuit network.

    Some beginning EE students experience a significant "Aha!" moment when they first realize that a first-order RC filter can be seen as a frequency-dependent voltage divider, for which they can write the equation very easily.
     
  10. LvW

    Active Member

    Jun 13, 2013
    674
    100
    gootee,

    I know what you mean - however, I don`t think that your explanations are consistent.
    Treating a capacitor/inductor just as a "frequency-dependent resistance" means to write 1/wC and wL, respectively (without the imag. symbol j).
    Now - combining these expressions for calculating a voltage divider provides no phase information at all.
    All I wanted to say in my short remark (post#8) is that the correct expressions 1/jwC or jwL should NOT be called "frequency-dependent resistances" but instead "impedances".
    (Believe me, I also have some experiences (25 years) with EE students)
     
Loading...