Tutorial: RLC Circuits and Q

The questions below are due on Monday April 28, 2025; 10:00:00 PM.
 
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This week, we started considering new kinds of circuits involving a resistor, capacitor, and inductor all in one circuit. For this problem, let's consider the following circuits, one with the RLC all in series and another with them all in parallel.

If we wanted to solve these circuits for the values in question using differential equations, the differential equations for the two would have very similar forms. Let's take a look at those.

Series Version: Differential Equation

For the series version of the circuit, we'll have a differential equation like the following (considering only the homogeneous part of the answer):

{\displaystyle {{\rm d}^2 \over {\rm d}t^2}v_o(t) + 2\alpha {{\rm d}\over {\rm d}t}v_o(t) + \omega_o^2v_o(t) = 0}

In order to be consistent with the series circuit above, what is the value of \alpha, in terms of R, L, and C? Enter your answer as a simple Python expression:
\alpha =~

In order to be consistent with the series circuit above, what is the value of \omega_o, in terms of R, L, and C? Enter your answer as a simple Python expression:
\omega_o =~

Parallel Version: Differential Equation

For the parallel version of the circuit, we'll have a differential equation like the following, considering only the homogeneous part of the answer):

{\displaystyle {{\rm d}^2 \over {\rm d}t^2}i_o(t) + 2\alpha {{\rm d}\over {\rm d}t}i_o(t) + \omega_o^2i_o(t) = 0}

In order to be consistent with the parallel circuit above, what is the value of \alpha, in terms of R, L, and C? Enter your answer as a simple Python expression:
\alpha =~

In order to be consistent with the parallel circuit above, what is the value of \omega_o, in terms of R, L, and C? Enter your answer as a simple Python expression:
\omega_o =~

Solutions

OK, so these differential equations (and, in fact, the ones we could write for other voltages and currents in those circuits, too) have very similar forms. Let's go ahead and solve things out. For now, let's focus on the homogeneous part of the solution, which is instead given by the following (with x replaced by i_o or v_o depending on what we care about):

{\displaystyle {{\rm d}^2 \over {\rm d}t^2}x(t) + 2\alpha {{\rm d}\over {\rm d}t}x(t) + \omega_o^2x(t) = 0}

If we assume a form for our answer of x(t) = Ae^{st}, we end up with the following:

s^2 + 2\alpha s + \omega_o^2 = 0

Since this equation is quadratic in s, there will, in general, be two solutions, so our output variable will actually have a form like x(t) = Ae^{s_1t} + Be^{s_2t}, where s_1 and s_2 are given by:

s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_o^2}

Depending on the particular values in question, we can end up with a number of different behaviors:

  • If \alpha > \omega_o, we say that the system is overdamped. In this case, the solutions for s are purely real, and our output looks like the sum of two exponential decays: x(t) = Ae^{s_1t} + Be^{s_2t}

  • If \alpha = \omega_o, we say that the system is critically-damped. In this case, s_1 = s_2 = -\alpha, and our output looks like x(t) = (A + Bt)\left(e^{-\alpha t}\right)

  • Perhaps the most interesting case is when \alpha < \omega_o, in which case we say that the system is underdamped.

The Underdamped Response

If the system is underdamped, s_1 and s_2 are complex: s = -\alpha \pm j\omega_d.

What is \omega_d, in terms of omega_o and alpha?
\omega_d =~

Solving out, we arrive at a solution like the following:

x(t) = Ae^{-\alpha t}\cos(\omega_d + \phi)

where A and \phi depend on initial conditions.

Quality Factor (Q)

In electrical engineering, it's common to use characterize RLC circuits by their "quality factor" Q, which is a single number that tells us how underdamped our system is. Other disciplines sometimes use other related quantities (for example, the damping ratio \zeta is common in mechanical systems), but for EE's, Q is usually the way.

We define the quality factor as follows: Q \equiv {\displaystyle \omega_o\over 2\alpha}. Physically, Q tells us about how much energy is lost per cycle of oscillation.

We can use this parameter Q to characterize whether the system is overdamped, critically-damped, or underdamped.

Which of the following indicates that our system is overdamped?

Which of the following indicates that our system is critically-damped?

Which of the following indicates that our system is underdamped?

Q in terms of Circuit Component Values

In an underdamped system, our output will be of the form x(t) = Ae^{-\alpha t}\cos(\omega_d + \phi), i.e., a sinusoid with an exponential envelope. Q tells us about how quickly the response decays, compared to how quickly it oscillates.

Another way of thinking about this is: the higher the Q value, the closer we are to a perfect LC oscillator like we saw before. Theoretically, as Q\to\infty, our system becomes a perfect oscillator.

Of course, we usually think about Q as a way of thinking about the values in our circuit. So let's go ahead and relate these values back to our original circuit.

In our series RLC circuit, what is Q in terms of R, L, and C?
Q_{\rm series} =~

In our series RLC circuit, which of the following changes, in isolation, lead to a larger Q value?

In our parallel RLC circuit, what is Q in terms of R, L, and C?
Q_{\rm parallel} =~

In our parallel RLC circuit, which of the following changes, in isolation, lead to a larger Q value?

Q in Underdamped Systems: Time Domain

One important physical approximation is that, after Q cycles of oscillation, the amplitude of the sinusoid will be about 4% (1/25) of whatever value it started at, since e^{-\alpha (2\pi / \omega_o) Q} = e^{-\pi} \approx 1/25.

The graph here shows the natural response of an RLC system for some initial conditions. From the graph, what (approximately) is the value of Q? Note that we can use the approximation of the ampilitude being at 4% after Q cycles to also say something about the amplitude after Q/2 cycles, approximately...
Q\approx ~