Can a resonance plot be upside down, i.e., inverted?

Question

Given the following RLC circuit with an AC driving source,

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I have found an expression for the amplitude to be

$$ I(t) = \left(\frac{V_{0}}{R}\right)\frac{(\omega \tau)}{\sqrt{(\omega\tau)^{2} + \left(\frac{\omega^{2} / \omega_{0}^{2}}{1 - \omega^{2}/\omega_{0}^{2}}\right)^{2}}}, \tag{Equation 1} $$ which when plotted using Python for three different values of $R$ whilst simultaneously normalizing each plot's amplitude from $0$ to $1$, yields the following result :

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From here I was wondering if the resonance plot is the right hand side figure or the left hand side. Because, wherever I look on the internet or any available literature, the resonance plot is similar to that of the left of this figure. But, in that case the frequency ration $\omega / \omega_{0}^{2}$ is negative , which doesn't make sense to me. Could a resonance plot , be reverted ,i.e., it is the right hand side plot ?

My reasoning is that it is the right hand side plot , i.e the domain $[0, 2]$ ,because a negative frequency ratio is unrealistic for the given circuit. While negative currents are possible I don't see how a negative frequency would be possible as well. I guess that this whole negative frequency thing arises from the square root of the denominator in Equation 1, which in return allows for a negative frequency ratio, but in a real setting I presume that we implicitly assume $\omega >0 \implies \omega/ \omega_{0} > 0 \ \forall \omega.$

Answer 1

The complex impedance $Z$ of the parallel arrangement of the inductor $L$ and capacitor $C$ is $\dfrac 1 Z = j\omega C + \dfrac {1}{j\omega L} \Rightarrow Z = \dfrac{j\omega L}{1-\omega^2LC}$.

You will note that as $\omega \to \sqrt LC$ then $|Z| \to \infty$ so it is not unreasonable that your analysis shows that as resonance is approached the current in the circuit decreases being a minimum at resonance.

In your eqaution for the current it is the $1-\dfrac {\omega^2}{\omega_0^2}$ term which is dominant as resonance is approached.

Answer 2

Yes, it makes sense to have the plot with a minimum. Why would you plot for negative frequencies, anyway? The reason for this is that you plot the total current, the current in the main branch. What you have there is a parallel LC circuit. At resonance the impedance of this circuit becomes very large (ideally infinite) so the current in the main circuit decreases.

Answer 3

The plot has several flaws. At small $\omega \equiv \epsilon$ $I$ should have the following behavior: ($\delta = \epsilon/\omega_0$)

$$ I \sim \frac{\epsilon\tau}{|\epsilon\tau|} \frac{1}{\sqrt{1+ \frac{\delta^4}{((1-\delta^2)\tau\epsilon)^2 }}} \approx \pm 1\cdot C \quad \text{at} \quad |\epsilon|\ll \omega_0 $$

(C is some appropriate constant), so the jump at $\omega=0$ should be symmetric, whereas on the plot it is not (it starts at 0, but should start at -1) In particular the equation (1) as a function of $\omega$ is odd, whereas the function on the plot is not odd.

The circuit is that of a band-stop filter, so it is obvious that it has a negative resonance: At the resonance frequency it should not let any signal through. So the value of the resonance peak should be zero. However, it has a value of $\sim 0.5$ on the plot.