# Second-order RLC transfer function

I solved the exercise in the picture with Laplace:

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: I ask you to confirm if it is now correct. Thank you!

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

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\$$.

• Thanks for your reply! I've just updated the post, is is correct? Commented Mar 3 at 14:29