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:


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


  • \$\begingroup\$ Your link leads to a blank page. You can insert pictures directly into your question. \$\endgroup\$ – JRE Apr 28 '16 at 7:52
  • \$\begingroup\$ user108391, Vo/Vin is a complex transfer function. How could you equalize it with 0.707? For finding the bandwidth you must find the magnitude of Vo/Vin and set it equal to Amax*0.707. \$\endgroup\$ – LvW Apr 28 '16 at 9:52
  • \$\begingroup\$ take at look at the equation at the bottom of this page electronics-tutorials.ws/accircuits/parallel-resonance.html \$\endgroup\$ – JIm Dearden Apr 28 '16 at 10:06
  • \$\begingroup\$ Thanks for all the replies everybody. JRE, I just accessed the link on a different computer than I used to upload the picture and also to post this thread last night, and the picture works for me, but here's another link: !RLC Circuit LvW yes, it is complex but I set the magnitude of H(s) = 0.707 because it is v0/vin where the magnitude of vin is 1 volt peak to peak and JImDearden will that equation work for my case? \$\endgroup\$ – user108391 Apr 29 '16 at 0:38

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.

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  • \$\begingroup\$ Thank you Andy, I understand the concept of Q although at the outset of this lab, we had not covered barely any of this material, and had only done simple transfer functions using LaPlace transforms, and that looks to be the idea here. I tried to math my way through it and got a weird equation with omega to the fourth power etc, but I know it should be quadratic like you have. The text shows finding the magnitude of s=jw by taking the root of the sum of the squares etc, but since we have s^2 in there, it messes things up when I try to solve it. Maybe I should switch my major to psychology? \$\endgroup\$ – user108391 Apr 29 '16 at 0:44
  • \$\begingroup\$ Is that the formula when equating reactances or the more difficult 2nd formula? \$\endgroup\$ – Andy aka Apr 29 '16 at 8:10

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