Multi inductor or capacitor RLC circuits

[ricc]

So I know that the equation for the natural frequency of an RLC circuit is:
ω₀=(LC)^-1/2
I'm just wondering how this would change for a circuit with more than one inductor or capacitor. Say for instance an inductor in parallel with a capacitor, both connected in series to another inductor.

Cheers in advance for any help you can give me.

 

[DragonPetter]

Do you know how to perform circuit analysis on a circuit like you described?

 

[the_emi_guy]

ricc,
With multiple LRCs you don't get a single natural frequency. One LC tank can be resonating at 1GHz while another is resonating at 1MHz. Your step or impulse response will likely be a more complicated waveform consisting of a superposition of multiple damped frequencies. Have you studied Laplace transforms yet?

 

[ricc]

Up until now I though I could. I can find total impedance and everything, it's just getting the frequency for complicated circuits like this. I can't find any literature on it either.

 

[ricc]

the_emi_guy
No, I haven't done laplace transforms. I'm asking this because I'm doing a computer project and want my programme to be able to give me the natural frequency. Laplace transforms sound a bit too out of scope for this.

 

[the_emi_guy]

ricc
Unfortunately there may not be a single natural frequency.
Total impedance is relatively easy to compute because it involves steady-state behavior of the circuit. Transient behavior of the circuit, which would include natural frequencies, requires a more advanced set of tools such as Laplace transforms.

 

[DragonPetter]

ricc, did you learn how to use phasors in AC circuit analysis?

The actual circuit analysis is very similar to what you might have done with resistors, but now you have complex impedances instead of just resistance.

You can use a phasors approach or laplace transform to consider single frequencies or a transfer function that will tell you how the circuit behaves over the frequency spectrum.

 

[ricc]

the_emi_guy
Ah well, thanks for your help. I can just put a limiting factor in the programme to stop it from trying to compute this type of problem.