education
are there the poles with positive real part in any function mode of the RLC filter?
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.
It can amplify the voltage (near to the resonant frequency) at the expense of load current. In other words you can't get power amplification. The frequency response will look like this: -
For a high damping (R is low in value, zeta = 0.5) the frequency response will be the lower blue curve but, with a high value resistor several tens of dB voltage amplification can occur. That amplification is not constant with frequency (as you can see) but can be significant from half resonance to a little over resonance.
A problem of significance occurs if you try and get too much amplification. Consider the situation when damping is really low i.e. resistance really high. For now the resistor can be ignored and what you are left with is a series resonant LC tuned circuit to ground. As you may know, a series resonant circuit (with zero or very low damping) will act as a short circuit - if you try and apply a voltage input at the resonant frequency it will be shorted by the LC. Because the inductor will never be perfect the input impedance won't be zero but the DC resistance of the inductor (maybe an ohm).
I know that the circuits which employ negative feedback are stable
Any feedback system is potentially unstable. What might seem like negative feedback at low frequencies can easily turn to positive feedback at higher frequencies.
whereas those which employ positive feedback are not stable.
Also, incorrect. A comparotor that uses hysteresis can be regarded as stable when one or the other threshold has been exceeded.
How can I prove that a voltage buffer (ideal op-amp) which uses positive feedback is not stable?
You can derive a stable condition for an op-amp with positive feedback but, if you introduce the slightest bit of noise as an influence, the circuit becomes a massive noise amplifier. If you then model input offset voltage and bias currents and how those change with temperature you get an unpredictable circuit.
Best Answer
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.