Calculating resonant frequency

Question

Here is my attempt:

Can someone explain to me how to proceed? How do I deal with the negative sign here? If I get the resonant frequency now, it will be complex.

Am I not supposed to use the negative sign for the capacitive reactance?

the resonant frequency is 447.2... rad/s (minimum impedance of all circuit). Max value of voltage on R occurs @ 441.6 rad/s. EE&O ... — Antonio51, Jun 11 '22 at 21:35

Andy aka · Accepted Answer · 2022-06-11T16:05:58.113

2

Their reactances are equal magnitude but opposing sign: -

$$j\omega L = -\dfrac{1}{j\omega C}$$

The minus sign allows us to mathematically equate them as equal reactances: -

$$j^2\omega^2 = -\dfrac{1}{LC}$$

And, because \$j^2 = -1\$: -

$$\omega^2 = \dfrac{1}{LC}$$

edited Jun 11 '22 at 16:05

answered Jun 11 '22 at 15:36

Andy aka

Isn't the reactance the wL and 1/wC parts? In other words, it's the coefficients of the complex term j? Or am I mistaken? I thought it's Impedance = Resistance + (j)*(Reactance). So I am equation reactance of the inductor wL to the reactance of the capacitor -1/(wC) – alayoiskgfbfqhxjiw Jun 11 '22 at 16:04
1

The complex reactance includes the "j". If you are talking about reactance magnitudes then you can equate them without "j" and without the -1. Both ways are acceptable. – Andy aka Jun 11 '22 at 16:07
That clears it up. Thanks! – alayoiskgfbfqhxjiw Jun 11 '22 at 16:09
In this case then, the resonant frequency is 447.2 rads/s? Book says 435 – alayoiskgfbfqhxjiw Jun 11 '22 at 16:10
@alayoiskgfbfqhxjiw perhaps the book's solution first converter parallel RC to its series RC equivalent...and only then calculated resonance? – glen_geek Jun 11 '22 at 17:28
parallel RLC and series RLC should have the same equation for resonant frequency. – alayoiskgfbfqhxjiw Jun 11 '22 at 17:39
I think they have miscalculated; it should be 447.2 rad/sec: http://www.stades.co.uk/RLC%20filters/RLC%20LPF.html#RLC_low_pass_filter_type_2 – Andy aka Jun 11 '22 at 18:07

Chu · Answer 2 · 2022-06-12T18:11:27.317

Using the Laplace representation for R, L and C: $$\small R\rightarrow R ;\:\: C\rightarrow \frac{1}{sC}\:;\: L\rightarrow sL$$

First derive a Laplace transfer function (TF) where the output signal is across R//C.

Output impedance is R in parallel with L, $$z_o(s)\small=\frac{R/sC}{R+1/sC}=\frac{R}{1+RCs}$$

Input impedance is \$\small sL\$ in series with \$ z_o(s)\$,

Thus,

$$ z_i(s)\small=sL+\frac{R}{1+RCs}=\frac{RCLs^2 +sL+R}{1+RCs}$$

The transfer function is then, $$\frac{z_o(s)}{z_i(s)}\small=\frac{R}{RCLs^2 +sL+R}= \frac{1/LC}{s^2 +s/RC+1/LC}$$

By inspection, the natural frequency is, $$\omega_n \small=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{5\times10^{-6}}}=447.21 \:rad/sec$$

and the damping coefficient is, $$\zeta=\small\frac{1}{2R}\sqrt{\frac{L}{C}}=0.11 $$

Finally, the resonant frequency (or 'damped natural frequency') is then, $$\omega_r =\omega_n \small\sqrt{1-2\zeta ^2} = 441.59\: rad/sec$$

2 Answers2