Electronic – RLC circuit: unexpected behaviour

To establish the damping calculate ζ (damping ratio).

\$\zeta = \dfrac{R_{series}}{2}\sqrt{\dfrac{C}{L}} \$

If ζ is <1 then it is underdamped. If ζ=1 it is critically damped etc..

Q (quality factor) is defined as the peak energy stored in the circuit divided by the average energy dissipated in it per cycle at resonance and, Q is \$\dfrac{1}{2\zeta}\$.

You decide what's the best method for you but for me it's easier to use the ζ formula providing you know the values for R, L and C. If you don't know them then there are other methods that involve looking at the step transient response or the frequency response.

Impedance is \$\dfrac{R + j\omega L}{R + j\omega L +\frac{1}{j\omega C}}\cdot\dfrac{1}{j\omega C}\$ = \$\dfrac{R + j\omega L}{j\omega RC + j^2\omega^2 LC +1}\$ = \$\dfrac{R + j\omega L}{j\omega RC - \omega^2 LC +1}\$

If you were to multiply both the numerator and the denominator by the denominator's complex conjugate you'd get a complex equation on the top and a non-complex (real) equation on the bottom. Resonance is when the imaginary part of the new numerator equals zero so, working only on the new numerator we get: -

New Numerator = \$(R+j\omega L)\cdot (-j\omega RC - \omega^2 LC +1)\$

The imaginary part boils down to \$j(\omega L - \omega^3 L^2 C - \omega C R^2)\$

Equating this to zero we get, \$\omega L - \omega^3 L^2 C - \omega C R^2 = 0\$

Therefore, \$\omega^2 L^2 C = L - CR^2\$ and \$\omega^2 = \dfrac{1}{LC} - \dfrac{R^2}{L^2}\$

EDIT section that describes other "resonances"

It's also interesting to note that there is another resonance at play. Firstly, the "resonance" above is defined by the impedance looking into the LCR network being purely resistive - this is slightly different to the "natural" resonant frequency (also purely resistive) when R is ignored.

In this situation \$\omega = \sqrt{\dfrac{1}{LC}}\$

However, the real peak of the response (as would be seen on a spectrum analyser and not purely resistive) is found by equating the denominator to zero and solving for s where s = j\$\omega\$.

\$s^2LC +sRC +1=0\$ or

\$s^2 + s\frac{R}{L} + \frac{1}{LC} = 0\$

Using the general solution to a quadratic, we get: -

\$s= \dfrac{-\frac{R}{L} +/- \sqrt{\frac{R^2}{L^2} - \frac{4}{LC}}}{2}\$

If we use a trick of reversing the signs inside the square root part and bring the square root of -1 (j) outside we get: -

\$s = \dfrac{-\frac{R}{L} +/- j\sqrt{\frac{4}{LC} - \frac{R^2}{L^2}}}{2}\$

Now, divide thru by two and we get: -

\$s = -\frac{R}{2L} +/- j\sqrt{\frac{1}{LC} - \frac{R^2}{4L^2}}\$

As we should know the "jw" part is what we see in a bode plot so the peak magnitude of the response (not when the impedance is purely resistive) is when: -

\$\omega = \sqrt{\dfrac{1}{LC} - \dfrac{R^2}{4L^2}}\$

If you drew a pole zero diagram, the imaginary parts of the poles would be at

+/-\$\sqrt{\dfrac{1}{LC} - \dfrac{R^2}{L^2}}\$ i.e. slightly different to +/-\$\sqrt{\dfrac{1}{LC} - \dfrac{R^2}{4L^2}}\$