Control System Resonance – Is Resonant Frequency for Second Order Transfer Function Same as Frequency at Resonance in Second Order Electrical Circuit?

control systempassive-networksresonancetransfer function

When I studied second order system and thus natural, damped and resonant frequency I wondered if the resonant frequency I find using the formula \$ w _ { r } = w _ { n } \sqrt { 1 – 2 D ^ { 2 } } \$ where D is damping factor, can be applied in any RLC circuit to find the resonant frequency. So I considered a series RLC circuit. Then its impedance is \$ z = R + s L + \frac { 1 } { C s } \$, and I used its reciprocal to form a second order transfer function. By general definition of electrical resonance if I put s = jw in the expression of z and put imaginary part = 0, I get resonant frequency as \$ w = \frac { 1 } { \sqrt { L C } } \$, which is, disappointingly, not equal to the resonant frequency derived from considering 1/Z as a second order transfer function, which is, \$ w = \sqrt { \frac { 1 } { L C } – \frac { R ^ { 2 } } { 2 L ^ { 2 } } } \$ and I don't understand why.
EDIT – I am including derivation for the above formula with relevant diagram:

schematic

^{simulate this circuit – Schematic created using CircuitLab}

Then, the transfer function = Output / Input = \$ \frac { 1 } { z } = \frac { 1 } { R + s L + \frac { 1 } { C s } } = \frac { C S } { L C S ^ { 2 } + R C S + 1 } \$ Now using the standard form of second order transfer function, natural frequency can be derived as \$w _ { n } = \frac { 1 } { \sqrt { L C } }\$ and then, resonant frequency can be given as \$w _ { r } = w _ { n } \sqrt { 1 – 2 D ^ { 2 } } \$, thus \$w _ { r } = \frac { 1 } { \sqrt { L C } } \sqrt { 1 – 2 ( \frac { R } { 2 L } ) ^ { 2 } }\$.

Best Answer

I don't understand why

You appear to be confusing two things; the natural undamped resonant frequency of an RLC circuit and, the damped resonant frequency of the transfer function. There are several notable points in the spectrum of an RLC low-pass filter (for example): -

\$\omega_n\$ is the natural undamped resonant frequency
The damped resonant frequency is \$\omega_n\sqrt{1-\zeta^2}\$
The frequency at which the response amplitude peaks is this: \$\omega_n\sqrt{1-2\zeta^2}\$

Without seeing your derivation or your exact circuit and where you place the input and output nodes, it's a bit of a guess but, I suspect, that your more complicated formula is the latter (above).

Regarding the pole-zero diagram you have this: -

Images from my basic website.

Related Solutions

Electronic – Resonance frequency of an RLC circuit

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);

Electronic – Intuition for resonant, natural, and oscillatory frequencies of RLC circuits

There are quite a few subtle differences between band-pass and low/high filters but, for a simple LCR band-pass filter, damped and un-damped resonant frequencies are the same numerically and formulaically, \$\dfrac{1}{2\pi\sqrt{LC}}\$.

This is also the (un-damped) natural resonant frequency for high/low pass filters.

When damping is added, the natural frequency stays the same but it can (eg for a low pass filter) rotate anticlockwise in the pole zero plane and this leaves (in the jw axis) what is known as the damped resonant frequency, \$\omega_d = \omega_n\sqrt{1 - \zeta^2}\$. See lower part of 1st set of pictures below (it's basically Pythagoras and right-angle triangles).

And for low pass filters, there is the frequency at which peaking occurs and this is slightly different to damped resonant frequency, \$\omega_p = \omega_n\sqrt{1 - 2\zeta^2}\$.

The amplitude that this peaks at is \$\dfrac{1}{2\zeta\sqrt{1-\zeta^2}}\$

The proof of these three different frequencies for a low-pass filter is relatively straightforward but a little long winded. Below is an extract of a design paper for a 2nd order low pass filter where it is shown that a varying Q-factor moved the "peaking" frequency away from the natural frequency (100 Hz in this example) but, as always Q is the value of the peak at the natural resonant frequency: -

Maybe a close up view will be more exciting: -

Best Answer

Related Solutions

Electronic – Resonance frequency of an RLC circuit

Appendix

Electronic – Intuition for resonant, natural, and oscillatory frequencies of RLC circuits

Related Topic