Definition of RLC resonance frequency

Question

I have found that a commonly mentioned definition for the resonance frequency in an RLC circuit is defined as the condition where the capacitive and inductive reactances cancel each other, resulting in a purely resistive impedance.

However, though the simplest RLC circuits (e.g RLC all in series or parallel) show maximum/minimum current at the resonance frequency defined above, this is not generally the case. For example, the RL-C series-parallel circuit

has a resonance frequency at $$\omega_0 = \frac{1}{\sqrt{LC}}\sqrt{1-\frac{R^2C}{L}}\tag{1}$$ by the above definition, but shows a minimum current at $$\omega_m = \sqrt{\sqrt{\frac{1}{L^2C^2}+\frac{2R^2}{L^3C}}-\frac{R^2}{L^2}}\tag{2}$$ My question is why is it defined this way? Would it not be more practical to define it as being at the maximum/minimum magnitude of impedance so that the observables (current and voltage) are also at extrema?

Answer 1

For example, the RL-C series-parallel circuit

I'm assuming here you mean a series connection of a resistor and inductor where that series pair is in parallel with a capacitor. The impedance of that circuit can be found to be: -

$$\dfrac{\frac{1}{LC}\cdot\left(R + j\omega L\right)}{\frac{1}{LC} - \omega^2 + j\omega \cdot\frac{R}{L}}$$

Note the \$\frac{1}{LC}\$ terms.

The square root of that term (in these particular types of 2nd order circuits) has a special name. It is called the natural resonant frequency (or pole frequency) and is given the symbol \$\omega_N\$ (or \$\omega_0\$).

Hence, \$\omega_N^2=\frac{1}{LC}\$

That's all it means when we talk about \$\frac{1}{\sqrt{LC}}\$. We find it commonly occurring in these types of equations and, it has enough relevance to be given a meaningful name.

For differing configurations of R, C and L, it can have different connotations but, nevertheless, we still link the name directly with \$\frac{1}{\sqrt{LC}}\$.

In a particular circuit, if we are interested in "other things" (such as maximum impedance or, the frequency at which the impedance is purely resistive) then we have to be specific about what we are talking about. Therefore, it's naïve to talk about a term such as "resonant frequency" with no further definition of what we actually mean.

Definition of RLC resonance frequency

My question is why is it defined this way?

We shouldn't attach a specific meaning to the term "resonance frequency" (or resonant frequency). However, the term "natural resonant frequency" (aka pole frequency) is defined as \$\frac{1}{\sqrt{LC}}\$.

Answer 2

"the resonance frequency in an RLC circuit is defined as the condition where the capacitive and inductive reactances cancel each other, resulting in a purely resistive impedance."

Actually, resonant frequency is a phenomenon that comes from math and it has something to do with the solutions to differential equation. You mention a parallel RLC circuit, so let's have a look at that.

^{simulate this circuit – Schematic created using CircuitLab}

From basic circuit theory, you should know the relationships: \$I_R = \frac{V}{R} \: \: I_C= C \frac{dV(t)}{dt} \: \: I_L=\frac{1}{L} \int V(t) \; dt \$.

Let's write up the relationship between the output voltage, \$V_o \$ and the input current \$I_{in} \$ in form of a node equation. $$\frac{V_o(t)}{R} + C\frac{dV_o(t)}{dt} + \frac{1}{L}\int V_o(t) \; dt = I_{in}(t)$$ We don't want the integral in our equation, so we differentiate every term with respect to \$ t\$. $$C \frac{d^2V_o(t)}{dt^2} + \frac{1}{R}\frac{dV_o(t)}{dt} +\frac{V_o(t)}{L} = \frac{dI_{in}(t)}{dt}$$ This is now a differential equation. We want it in standardform, so we divide with \$C \$ in every term. $$ \frac{d^2V_o(t)}{dt^2} + \frac{1}{RC}\frac{dV_o(t)}{dt} +\frac{V_o(t)}{LC} = \frac{dI_{in}(t)}{dt}$$ And there is our final differential equation. Let's say we wished to find the natural response, that is, the solution to the differential equation when the right hand side is zero. $$ \frac{d^2V_o(t)}{dt^2} + \frac{1}{RC}\frac{dV_o(t)}{dt} +\frac{V_o(t)}{LC} = 0$$ In order to find the solutions, we need the characteristic equation. Because this is a second order differential equation, with the coefficients \$\frac{1}{RC} \$ and \$ \frac{1}{LC} \$ the characteristic equation will look like this. $$\lambda^2 + \frac{1}{RC}\lambda + \frac{1}{LC} = 0 $$ The roots of this equation as it is written right here are not so pretty $$ \lambda = \frac{-L \pm \sqrt{-4R^2CL+L^2}}{2RLC} $$ However, look what happens if we define \$\alpha = \frac{1}{2RC} \$ and \$\omega_0 = \sqrt{\frac{1}{LC}} \$. The charateristic equation now becomes $$\lambda^2 + 2\alpha \lambda + \omega_0 ^2 =0 $$ This equation has the solutions (roots) $$ \lambda = -\alpha \pm \sqrt{\alpha^2 - \omega_0 ^2} $$ The roots are much nicer now. The term \$ \alpha\$ is called the damping coefficient, and \$\omega_0 \$ is called the undamped resonant frequency.

It turns out that if \$\alpha > \omega_0 \$ the solution to the homogenous differential equation is $$V_o(t) = K_1 e^{\lambda_1 t} + K_2 e^{\lambda_2 t} $$ If \$\alpha = \omega_0 \$ the solution is $$V_o(t) = K_1 e^{\lambda_1 t} + tK_2 e^{\lambda_2 t} $$ If \$\alpha < \omega_0 \$ the solution is $$V_o(t) = K_1 e^{-\alpha t} \cos \bigg(\sqrt{\omega_0^2 - \alpha^2} \; t \bigg) + K_2 e^{-\alpha t} \sin \bigg(\sqrt{\omega_0^2 - \alpha^2} \; t \bigg) $$ If you have experience with solving 2nd order differential equations this will make sense to you.

As you can see, the resonant frequency \$\omega_0 \$ is saying a lot more than just "where the resulting impedance is purely resistive", and it is defined in a different way than what you initially thought.

It just so happens that in both a series and parallel RLC circuit (with only 1 of each component) at resonant frequency the equivalent impedance is purely resistive. Note, however, that this would not be the case for a, let's say, RRLC circuit, for example.

Answer 3

Seeking for a maximum or a minimum impedance only makes sense for pure series or parallel RLC circuits; what you have there is a mix, which means you can no longer apply the same reasoning.

The circuit you're showing has the property that the parallel C cancels the series L, thus leaving R. It's a common circuit in QVar compensation, where the series RL would be some motor, and C the reactive compensator. This means that the imaginary part of the impedance is zero, which means a purely resistive load.

With this in mind, the imaginary part of the impedance and the roots are:

$$\begin{align} Z&=-\dfrac{Lj\omega+R}{LC\omega^2-RCj\omega-1}\tag{1} \\ \Im{(Z)}&=-\dfrac{L^2C\omega^3+(R^2C-L)\omega}{L^2C^2\omega^4+(R^2C^2-2LC)\omega^2+1}\tag{2} \\ 0&=L^2C\omega^3+(R^2C-L)\omega\tag{3} \\ &\begin{array}{x} \left\{ \begin{aligned} \omega_1&=0 \\ \omega_{1,2}&=\pm\sqrt{\dfrac{L-R^2C}{L^2C}}=\pm\sqrt{\dfrac{1}{LC}-\dfrac{R^2}{L^2}} \end{aligned} \right. \end{array} \end{align}$$

If you plot this for random values of RLC, you'll see that it often leads to imaginary numbers. This is because the C is meant to compensate the RL, which means only for certain values of all RLC will the numbers end up right. Otherwise, it may not even be a resonant peak (think of the three cases: over-, under-, and critically damped). Here is a plot for the absolute value of \$\omega_{1,2}\$ with R=1, L=2, C=3, with varying R (log-log axis):

R goes to zero when \$\omega=0\$, or at \$\sqrt{L/C}\$.

Answer 4

Resonance is a mathematical condition and not a mathematical definition.

Resonance can be found in many fields and not only in electronics.

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In electronics, resonance comes in when a linear circuit made of R, L, and C components is excited by a sine wave where the frequency f is though as a parameter.

v(t) = Asin (2pi*f + phi1)

Let's call i(t) the current sourced by v(t)

i(t) = Bsin (2pi*f + phi2)

The frequency fr at which

phi1 = phi2

is called resonance frequency fr.

When voltage and current are in phase the load is purely resistive.

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Resonance is a condition experienced by the sources of a circuit.

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Non linear systems can experience resonance conditions as well.