I have a circuit of the following form.

Circuit diagram:

enter image description here

How do I decide the resonant frequency ?

What is the criteria here to select the resonant frequency ?

Background: I am testing the transfer function for the system, and I need to know what is the proper value of "w" (omega) I can use.

Please let me know if anyone has the knowledge on this.


  • \$\begingroup\$ Have spice make you a bode plot? Run a signal generator into the input and measure the output as a function of frequency. I trust the inductors aren't coupled. \$\endgroup\$ – George Herold Aug 30 '16 at 16:20
  • \$\begingroup\$ @GeorgeHerold...Please let me know how did you think it is not coupled. I want to know. Regards \$\endgroup\$ – cppiscute Aug 30 '16 at 16:22

First find the equivalent impedance of your circuit using, e.g., Thevenin theorem. This impedance will depend on \$\omega\$. The resonant frequency(ies) is(are) the value(s) of \$\omega\$ that makes the reactive part of the impedance zero.

  • \$\begingroup\$ Is there any tool which does the calculation of thevenin equivalent voltage and imedance of the circuit?..I really can't believe my calculations, so I am keep looking for some verification methods. \$\endgroup\$ – cppiscute Sep 5 '16 at 17:51
  • 1
    \$\begingroup\$ I use a computer algebra system to help doing the calculations. In my case, it's maxima, but I think Mathematica and even Matlab also do the same thing. Basically I write the impedance for each element and combine them manually in series or parallel. Although it's not automated it does help a lot in the calculations. For example, for your example, it took me less than 5 mins to find that the resonant frequencies are \$\omega=0\$ and \$\omega=1/C_1(L_1+L_3)\$. \$\endgroup\$ – hcabral Sep 6 '16 at 13:49
  • \$\begingroup\$ The full denominator of the Thevenin impedance is \$j\omega^2C_1C_2R_1 + \omega C_2(1 -\omega^2C_1(L_1 + L_3))\$. \$\endgroup\$ – hcabral Sep 6 '16 at 13:52
  • \$\begingroup\$ Please scrape off the \$\omega=0\$ resonant frequency. \$\omega=0\$ actually zeroes the full denominator, which means the impedance is infinite and the load will not get anything. \$\endgroup\$ – hcabral Sep 6 '16 at 14:00
  • \$\begingroup\$ OK, hold on because in my hurry I forgot \$R_L\$. \$\endgroup\$ – hcabral Sep 6 '16 at 14:22

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