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I have the following network (page 805, example 14.5, Agarwal & Lang, Foundations of Analog and Digital Circuits): given circuit

The associated transfer function for \$i_z, v_z\$ is \$ H(s) = \frac{I_z (s)}{V_z (s)} = \frac{s^2 + (R/L) s + 1/(LC)}{s/C + R/(LC)}\$. I can see how to get this from the impedance model of the circuit.

The example says that the numerator of this transfer function is the characteristic equation. The textbook example concludes that "the roots of the characteristic equation [numerator of transfer function] are complex and therefore the circuit is resonant", brackets comment mine.

I'm a bit confused by this since I thought that generally the characteristic equation is the denominator of the transfer function.

I think that the denominator of the transfer function is the characteristic equation of the output variable, so if you took the reciprocal of this transfer function, then \$s^2 + (R/L) s + 1/(LC)\$ would become the denominator and would be the characteristic equation of the new output variable \$v_z\$, so then you could analyze \$s^2 + (R/L) s + 1/(LC)\$ to find that voltage's oscillation frequency \$\omega_0\$, damping frequency \$\alpha \$, quality factor Q, etc. However, these would not directly correspond to \$i_z\$ or the magnitude/phase plots of the original transfer function \$ H(s) \$. Is this correct?

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    \$\begingroup\$ You are showing an admittance transfer function Y=I / V so it's inverted.poles are specified by the denominator of the transfer function. However, the zeros of the transfer function are evaluated using the numerator. \$\endgroup\$ – Tony Stewart EE75 Jul 21 '20 at 0:56
  • \$\begingroup\$ @TonyStewartSunnyskyguyEE75 I suppose inverted if you always expect a voltage to be the output but I don't think that's a requirement in general, right? A transfer function is just some output and some input. Is it true in general that the denominator of a transfer function is the characteristic equation of the output variable? \$\endgroup\$ – knzy Jul 21 '20 at 1:02
  • \$\begingroup\$ Yes, the denominator of the transfer function of the system, when equated to 0, provides the characteristic equation of that particular system. \$\endgroup\$ – Tony Stewart EE75 Jul 21 '20 at 1:04
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I'm a bit confused by this since I thought that generally the characteristic equation is the denominator of the transfer function.

If the example question had a current source (instead of a voltage source) and asked you to find the terminal voltage (as the output) then the TF would be inverted because it's V/I = impedance. In this case it would be the denominator that is the characteristic equation.

So, the answer depends on what you are trying to solve.

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