impedance - Resonance frequency of an RLC circuit

^{simulate this circuit – Schematic created using CircuitLab}

Is it safe to consider the resonance frequency of a second order RLC circuit to be alway equal to 1/sqrt(LC) or the transfer function has to be calculated?

When there are three active components such for example for the case of bridged-T RLC network as is shown in the figure, is the resonance frequency still 1/sqrt(LC)? If capacitances are different, would that mean that there are two resonance frequency?

Answer

KCL: (\$v_o\$ over your output, \$v_i\$ over voltage source and \$v\$ over the internal node:

$$ \frac{v_o}{R_3} + \frac{v_o-v}{Z_{C1}}+\frac{v_o-v_i}{R_2} = 0 \\ \frac{v}{Z_L} + \frac{v-v_i}{Z_{C2}} + \frac{v-v_o}{Z_{C1}} = 0 $$

can be solved to get rid of the internal node \$v\$, setting \$Z_{C i} = \frac{1}{sC_i}\$ and \$Z_L = sL\$:

$$ \frac{v_o}{v_i} = \frac{L C_1 C_2 R_2 R_3 s^3 + L (C_1 + C_2) R_3 s^2+R_3}{ L C_1 C_2 R_2 R_3 s^3 + L(R_2+R_3)(C_1+C_2) s^2 + (R_2 R_3 C_1) s + (R_2 + R_3)} $$ which is in the form $$ \frac{s^3+n_2s^2+n_0}{s^3+d_2 s^2 + d_1 s + d_0}$$ with $$ n_2 = \frac{(C_1+C_2)}{C_1 C_2 R_2} = \frac{1}{R_2C_*}$$ $$ n_0 = \frac{1}{LC_1 C_2 R_2} $$ $$ d_2 = \frac{(R_2+R_3)(C_1+C_2)}{C_1 C_2 R_2 R_3} = \frac{1}{R_\| C_*}$$ $$ d_1 = \frac{1}{LC_2} $$ $$ d_0 = \frac{R_2+R_3}{LC_1 C_2 R_2 R_3} =\frac{1}{LC_1 C_2 R_\|}$$ where I introduced the parallel resistance and serial capacitances $$ \frac{1}{R_\|} = \frac{1}{R_2} + \frac{1}{R_3}\\ \frac{1}{C_*} = \frac{1}{C_1} + \frac{1}{C_2} $$

This is a third-order high-pass filter, so general second order LRC intuition does not apply. You have to plug in your values and show them on a Bode-plot in order to find oscillation problems.

Appendix

Maxima script:

sol  : solve([v_o/R3+(v_o-v)*s*C1+(v_o-v_i)/R2=0, 
              v/(s*L)+(v-v_i)*s*C2 + (v-v_o)*s*C1 = 0], [v_i, v_o]) $
trf  : ev(v_o / v_i, sol) $
res  : rat(trf, s);