# 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

Oct 2, 2009
5,450
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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)