Cannot figure out resonance of an LRC circuit

Question

this is my first time here, nice to meet you all.

In one of my classes we are studying the resonance of an LRC Circuit where the C branch is in parallel with an RL branch. This circuit is driven by a function generator made 1Vpp sine wave, and we are looking for the natural frequency, and two corner frequencies (where Vo/Vin = 0.707)

The setup of the circuit is like this: the signal goes through a 1k resistor, then at the opposite side of the resistor, the node splits into 2 parallel branches, where 1 has a capacitor of 0.011uF and the other branch has a 51 ohm resistor in series with a 9.8mH inductor. The circuit can be seen here:

http://jmp.sh/v/C8E5a0YXEAAXY2VPxwP7

I set the transfer function equal to 0.707 for Vo/Vin but I ended up getting s=jw=16060+-93637i. I do not understand this result, mainly because we have barely even touched on transfer functions and solving for resonance this way. How can I get from my result (if it is even correct) to the angular frequency and thus the actual frequency?

You don't need to solve it for me, just a gentle nudge in the right direction would help tons. I asked the professor but I think I'm on my own for this one.

Thank you all

C

Answer 1

we are looking for the natural frequency, and two corner frequencies (where Vo/Vin = 0.707)

The natural resonant frequency is purely determined by L and C and is when the impedance of L equals the impedance of C. Why not write those two formula down and equate then solve for frequency. This part is quite simple.

The two corner frequencies I take as meaning the -3dB points of the band-pass filter - are you aware that what you have described IS a BPF? Anyway this is more onerous mathematical solution but you should end up with a formula that generally is: -

TF = \$\dfrac{2\zeta\omega_n s}{s^2 + 2\zeta\omega_n s + \omega_n^2}\$

I use the term "general" because any 2nd order filter be it mechanical or electronic uses the same general terms although for an electronic circuit the terms R, L and C will be part of the formula.

Where s = jw and \$\zeta\$ = 1/2Q (do you understand Q?).

\$\omega_n\$ is the natural resonant frequency in radians per second.

Q (or inversely \$\zeta\$) dictate how "fat" the response is in the frequency domain. For instance a very high Q gives a very sharp response with the -3dB points very close in frequency. Another definition of Q is that Q = natural resonant frequency divided by -3dB bandwidth.

Hope this helps.