1

are there the poles with positive real part in any function mode of the RLC filter?

enter image description here

The equation to determine the poles is the following:

$$RLCs^2+Ls+R=0$$

$$s_{1,2}=\frac{-L\pm \sqrt {L^2-4R^2LC}}{2RLC}$$

Thank you for your time.

P.s. I posted this question because my professor said that the poles have always real part negative or zero.

Gennaro Arguzzi
  • 482
  • 1
  • 9
  • 26

2 Answers2

2

Do the math for \$s_{1,2}\$, looking at the numerator. Assuming no component is zero, the term \$\sqrt{L^2 - 4R^2 LC}\$ will have a real part no larger than \$L\$. Therefore, the real part of \$-L \pm \sqrt{L^2 - 4R^2 LC}\$ will have a range of \$(0, -L]\$. This gives the result your professor arrived at.

This analysis is valid for the expected linear, positive-valued devices implied by the question and your professor's response. This result does not necessarily hold for non-linear devices that exhibit negative resistances.

user2943160
  • 2,908
  • 1
  • 18
  • 32
  • This doesn't take into account circuits with negative resistance. – Captainj2001 Jun 07 '16 at 18:03
  • Given that this is a linear component analysis problem, this is over-complicating the analysis. A Gunn diode certainly isn't an introductory circuits component. – user2943160 Jun 07 '16 at 18:05
  • The question's author doesn't specify whether or not he is considering purely passive components and negative resistance circuits are used in many designs (especially RF/microwave oscillators) to replace active components when the additional complexity of the full circuit is not required. – Captainj2001 Jun 07 '16 at 18:11
  • @Captainj2001 to continue in the vein of your overly pedantic comment, if you did have a negative resistance in there (and it need not be a gunn diode either - a gyrator would work as well) it is no longer a filter but an oscillator as you pointed out. – placeholder Jun 07 '16 at 21:44
  • @placeholder There is also a large body of research on negative resistance filters. http://ieeexplore.ieee.org/search/searchresult.jsp?newsearch=true&queryText=negative%20resistance%20filter – Captainj2001 Jun 07 '16 at 21:48
  • @Captainj2001 cool !! I didn't realize that such complex systems could be described by $$RLCs^2+Ls+R=0$$ – placeholder Jun 07 '16 at 22:01
  • @placeholder This is not very constructive, enjoy your day. – Captainj2001 Jun 07 '16 at 22:15
0

The only way to have poles in the real half of the complex plane is to have a circuit in which the "resistance" is negative, i.e., the active circuit replacing the load can be modeled as a negative resistance at some operating point. For instance, if the resistor was a Gunn diode operating in it's region of negative resistance, or an amplifier circuit.

With purely passive components the poles will always be in the left-hand plane or on the imaginary axis.

Captainj2001
  • 731
  • 6
  • 13