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I solved the exercise in the picture with Laplace:

enter image description here

Not being familiar with systems theory yet, I do not understand one thing (circled in pink in the picture.)

The transfer function found with the black marker (tell me if it's correct) is \$ \frac{s}{s^2 + \frac{s}{RC} + \frac{1}{LC}}\$.

Theoretically it's the same as the transfer function written with the gray marker, \$ \frac{\omega_n^2}{s^2 + 2\varepsilon\omega_n s + \omega_n^2}\$, right?

If yes, then the natural frequency is equal to the square root of 1 or to the square root of 1/LC. Certainly a trivial question, but what I don't understand is just how to interpret the formula written with the gray marker.

After @mond's correction, I update the calculations: enter image description here I ask you to confirm if it is now correct. Thank you!

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  • \$\begingroup\$ I didn't see a grey/gray marker. \$\endgroup\$
    – Andy aka
    Mar 2 at 15:40
  • \$\begingroup\$ @KaleM. Your output seems to be the current. Do you confirm this? \$\endgroup\$ Mar 2 at 18:17

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No, it is not the same function.

You also should use the term "transfer function" which relates input to output.

Your input, it seems you assume a step function where you close the switch at \$t=0\$

This results in a step function

$$ u_{in}(t) = E \cdot \sigma(t) $$

Or

$$ U_{in}(s) = \frac{E}{s} $$

And then the transfer function would be output over input:

$$ \frac{U_{out}(s)}{U_{in}(s)} = \frac{Z_{//}}{Z(s)} $$

In your calculation of \$Z\$ you seem to apply the above right hand side but on the left hand side you write \$I\$ equals. This is an error.

\$U_{out}\$ would be the voltage over \$Z_{//}\$.

With the substitutions

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

and

$$ \epsilon = \frac{1}{\omega R C } = \frac{1}{R} \cdot \sqrt{\frac{L}{C}} $$

$$ \frac{Z_{//}}{Z(s)} = \frac{U_{out}}{U_{in}} = \frac{C L R s^{2}}{C L R s^{2} + L s + R } = \frac{s^{2}}{s^2 + \epsilon \omega s + \omega^2 } $$

So this is NOT equal to your "grey" term.

If you want \$U_{out}\$ (over the \$Z_{//}\$) you multiply with your transfer, then you lose one \$s\$ in the numerator and if you want to lose another \$s\$ you could look at the current through \$L\$.

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  • \$\begingroup\$ Thanks for your reply! I've just updated the post, is is correct? \$\endgroup\$
    – KaleM
    Mar 3 at 14:29

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