This question is inspired from Andy Aka's question from almost 5 years ago. It is also a nod of my respect to Andy. But what I wrote there is kind of muddled. And it only addresses the under-damped case. I'd like to end my time here with something better than what I provided then to Andy.

I've benefited a lot from some exceptional folks here, appreciating them still more over time. Some I have already missed have left for their own reasons this last decade. I will also sorely miss still more when I also depart in a few days (end of the month.) This is my last question and perhaps, if forced to it, then on December 31st also my last answer.

Assume the following schematic:


simulate this circuit – Schematic created using CircuitLab

The specified initial conditions are \$V_{_0}\$ and \$I_{_0}\$. The values of \$R_1\$, \$C_1\$ and \$L_1\$ are all non-zero and positive and constant over time. They are to be taken as ideal components, too (no parasitics to be added.)

There are three cases to deal with. Critically-damped, under-damped, and over-damped. There's no specification as to which of these apply, so there are potentially three different specific solutions.

I don't want general solutions with unspecified constants. Instead, I want to see the specific solutions for the voltage across the capacitor, using the initial conditions and part values.

The initial conditions apply at \$t=0\$ and the specific solutions for all three cases will be for \$t\ge 0\$.

Finally, the development must start with provided time-domain KCL (see below.) The development should be strictly performed using only time-domain development and must be relatively easy to follow for those familiar with no more than the first six chapters of the 9th edition of Nagle, Saff, and Snider's Fundamentals of Differential Equations. (This excludes Laplace and/or inverse Laplace as that doesn't begin until chapter 7 in that textbook.) Methods such as undetermined coefficients or other more rigorous approaches are fine.

The problem is rather simple in the sense that there are no non-homogeneities (no driving functions.) I'll start the process with the KCL:

$$\begin{align*} \frac{v_{_\text{C}}}{R_1}+C_1\frac{\text{d}}{\text{d}t}v_{_\text{C}}+\frac1{L_1}\int v_{_\text{C}}\:\text{d}t&=0\:\text{A} \\\\ v_{_\text{C}\left(t=0\right)}&=V_{_0} \\\\ i_{_\text{L}\left(t=0\right)}&=I_{_0} \end{align*}$$

Clearly, this is an homogeneous equation. So it should be fairly straight-forward.

Positive currents are downward.

To expand on the above and set the tone for the development:

$$\begin{align*} \frac{v_{_\text{C}}}{R_1}+C_1\frac{\text{d}}{\text{d}t}v_{_\text{C}}+\frac1{L_1}\int v_{_\text{C}}\:\text{d}t&=0\:\text{A} \\\\ \frac{\text{d}}{\text{d}t}\left[\frac{v_{_\text{C}}}{R_1}+C_1\frac{\text{d}}{\text{d}t}v_{_\text{C}}+\frac1{L_1}\int v_{_\text{C}}\:\text{d}t\right.&=\bigg.0\:\text{A}\bigg] \\\\ \frac1{R_1}\frac{\text{d}}{\text{d}t}v_{_\text{C}}+C_1\frac{\text{d}^2}{\text{d}t^2}v_{_\text{C}}+\frac{v_{_\text{C}}}{L_1}&=0\:\text{A} \\\\ \frac{\text{d}^2}{\text{d}t^2}v_{_\text{C}}+\frac1{R_1\,C_1}\frac{\text{d}}{\text{d}t}v_{_\text{C}}+\frac{1}{L_1\,C_1}v_{_\text{C}}&=0\:\text{A} \\\\ \left[\frac{\text{d}^2}{\text{d}t^2}+\frac1{R_1\,C_1}\frac{\text{d}}{\text{d}t}+\frac{1}{L_1\,C_1}\right]v_{_\text{C}}&=0\:\text{A} \\\\ \left[D^2+\frac1{R_1\,C_1}D+\frac{1}{L_1\,C_1}\right]v_{_\text{C}}&=0\:\text{A} \end{align*}$$

To zero out the left side above there are two choices. One is to set \$v_{_\text{C}}=0\$. But given that the problem allows the specification of \$V_0\ne 0\$ and \$I_0\ne 0\$, that option is readily disposed of.

So there's only one other choice: \$\left[D^2+\frac1{R_1\,C_1}D+\frac{1}{L_1\,C_1}\right]=0\$. That's just a matter of solving for its quadratic roots.

Or, it would be if I were permitting complex/imaginary roots to be used in the final solutions. But I'm disallowing those in the final answers. So, while the development may use mathematical properties, including complex domain ones such as Euler's, the final equations must be free of complex/imaginary values and must be entirely symbolic, in form.

I will not accept an answer where the results are just copied out without any development shown. For example, if the critically-damped solution is simply "copied out of a book." At least show some development to get there. There must be a visible step-wise process. I will write an answer if nothing acceptable is provided before the end of the year, as that's when I'm leaving the site and won't be returning. Hopefully, I will accept an earlier answer, though.

The specific solutions must cover all three possible cases mentioned above: critical-, under-, and over- damped. A simulation (the only time numeric values are allowed, as that's needed in order to demonstrate that the solutions deduce into specific situations) for each possible case must be included with the answer and at least a few points from each specific solution case equation shown to match up with the simulations of each case. In other words, simulation must confirm the specific solution equations.

For example, here are three curves:

enter image description here

The red curve is for an over-damped case, the green curve is for a critically-damped case, and the dark blue curve is for an under-damped case.

I think this passive formulation, combining one each of the two basic energy storage devices and avoiding complex driving functions, is an essential fundamental skill.

(I will add a bounty to the question on the 27th of December.)

Note: The final equations (there will be three, I believe) cannot use complex or imaginary numbers. They should symbolically use \$R_1\$, \$C_1\$ and \$L_1\$. You may use Euler's in their development, of course. But in the end the final equations must be entirely real-valued in, real-valued out solutions. Exponential functions are fine and encouraged, so long as they are a real-valued in and real-valued out. And don't forget that I'm looking to see their development from start to finish, using the symbols for the three parts and no values specified. No 'skipping over incremental development steps' by using inputs to and resulting outputs from something like Wolfram Alpha, for example. In the end, of course, values will have to be supplied to verify. But no rational-real values may be used during the development of the symbolic solutions.

An answer should make a point of illustrating how we know what we know. The better it does that the more likely I select it.

  • 2
    \$\begingroup\$ To SamGibson you have my special thanks. I can learn technical details from others. And I owe much to them and will miss them. But you are where the better parts of humanity can be found in abundance. And I will still more sorely miss you for that. It's something I find too little of and value very highly in others. You are special in my mind. More, you are in my mind as a kind of unique value in these times in the world where we find ourselves and others, today. Just FYI. Best wishes. \$\endgroup\$
    – jonk
    Dec 25, 2022 at 9:49
  • 2
    \$\begingroup\$ @tobalt I expect nothing. I will answer the question on the 31st if none is provided. And I will depart, regardless. There is sufficient time between now and then. It's not rocket science. This is really pretty basic stuff. And you have my profound respect, by the way. One of the few. I will miss you. \$\endgroup\$
    – jonk
    Dec 25, 2022 at 10:56
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    \$\begingroup\$ Why are you leaving @jonk? \$\endgroup\$
    – Andy aka
    Dec 25, 2022 at 11:04
  • 2
    \$\begingroup\$ Ah well, I guess there is a Joy Of Not Knowing if you don't want to explain. Cheers dude. \$\endgroup\$
    – Andy aka
    Dec 25, 2022 at 11:25
  • 2
    \$\begingroup\$ I have always liked reading your answers (learned a lot!) and your question (although most [all] of your questions are way above my head!). I have always said (to myself) that you should write an autobiography, I think it would be fascinating. Best of luck to you and this site will be less without you. \$\endgroup\$
    – Tyler
    Dec 25, 2022 at 22:19

2 Answers 2


I want to see the specific solutions for the voltage across the capacitor, using the initial conditions and part values.

Obtaining second order differential equation in standard form

$$\frac{V_\text{c}}{R_1} + C_1 \frac{\text{d}V_\text{c}}{\text{d}t} + \frac{1}{L_1}\int V_\text{c} \: \text{d}t = 0. \tag{1}$$

Differentiate both sides with respect to \$t\$

$$\ddot{V_\text{c}}C_1 + \dot{V_\text{c}}\frac{1}{R_1} + V_\text{c}\frac{1}{L_1} = 0.$$

Get the differential equation on standard form

$$\ddot{V_c} + \dot{V_c}\frac{1}{R_1C_1} + V_c\frac{1}{C_1L_1} = 0.$$

There are no driving sources so the zero state response is zero. The only response remaining is the zero input response - aka. the system response to its own internal conditions assuming that the input is zero. Call the damping coefficient \$\beta = \frac{1}{2C_1R_1}\$ and the undamped resonance frequency \$\omega_{_0} = \frac{1}{\sqrt{C_1L_1}}\$ the equation simplifies to

$$\ddot{V_\text{c}} + \dot{V_\text{c}} \cdot 2\beta + V_\text{c}\omega_{_0}^2 = 0 $$

Defining the damping ratio as \$\zeta = \frac{\beta}{\omega_{_0}}\$ the equation becomes

$$\ddot{V_\text{c}} + \dot{V_\text{c}} \cdot 2\zeta\omega_{_0} + V_\text{c}\omega_{_0}^2 = 0 $$

The characteristic polynomial is \$P(\lambda) = \lambda^2 + \lambda \cdot 2\zeta\omega_{_0} + \omega_{_0}^2\$ which has the roots

$$\lambda = \omega_{_0} (-\zeta \pm \sqrt{\zeta^2 -1}) $$

From here, there are 3 possible cases depending on the value of \$\zeta\$:

If \$\zeta > 1\$ the system is overdamped and the zero input response has the form

$$V_\text{c}(t) = K_1e^{\lambda_1 t} + K_2e^{\lambda_2t} $$

where \$\lambda_1, \lambda_2\$ are the roots of the characteristic polynomial.

If \$\zeta = 1\$ the system is critically damped and the solution is

$$V_\text{c}(t) = e^{\lambda t}(K_1 + K_2t) $$

If \$\zeta < 1\$ the system is underdamped, \$P(\lambda)\$ has complex conjugated roots \$\lambda = \alpha \pm j\omega_{_n}\$ and the solution is

$$V_\text{c}(t) = K_1e^{\alpha t}\cos(\omega_{_n} t) + K_2e^{\alpha t}\sin(\omega_{_n} t).$$

(Source: Electrical Engineering - Principles and Applications by Hambley).

Overdamped case

We know the solution is of the form \$V_\text{c}(t) = K_1e^{\lambda_1 t} + K_2e^{\lambda_2t} \$.

To determine \$K_1\$ and \$K_2\$ the intial conditions \$V_\text{c}(0) = V_{_0}\$ and \$I_\text{L}(0) = I_{_0}\$ are used. We use that \$V_\text{c}(0) = K_1 + K_2 = V_{_0}\$. The current initial conditions can be used knowing the relation between current and voltage for an inductor

$$\begin{align} I_\text{L}(t) = \frac{1}{L_1} \int V_\text{c}(t) \: \text{d}t =\frac{1}{L_1} \int K_1e^{\lambda_1 t} + K_2e^{\lambda_2 t} \: \text{d}t \\ \\ =\frac{1}{L_1} \bigg(\frac{K_1e^{\lambda_1 t}}{\lambda_1} + \frac{K_2e^{\lambda_2 t}}{\lambda_2} \bigg) \end{align}$$

$$I_\text{L}(0) = \frac{1}{L_1} \bigg(\frac{K_1}{\lambda_1} + \frac{K_2}{\lambda_2} \bigg) $$

So now we have

$$\begin{cases} K_1 + K_2 = V_{_0} \\ \\ \frac{1}{L_1} \Big(\frac{K_1}{\lambda_1} + \frac{K_2}{\lambda_2} \Big) = I_{_0} \end{cases}$$

which yields the values for the constants:

$$K_1 = -\frac{\lambda_1 (I_{_0}L_1\lambda_2 - V_{_0})}{\lambda_1 - \lambda_2} \: \: \: \text{and} \: \: \: K_2 = \frac{\lambda_2(I_{_0}L_1\lambda_1 - V_{_0})}{\lambda_1 - \lambda_2} $$

Critically damped case

We know the solution is of the form \$V_\text{c}(t) = e^{\lambda t}(K_1 + K_2t) \$.

We will use the initial conditions to obtain two equations with two unknowns in a similar fashion as before.

$$V_\text{c}(0) = K_1 = V_{_0} $$

$$\begin{align} I_\text{L}(t) = \frac{1}{L_1} \int e^{\lambda t}(K_1 + K_2t) \: \text{d}t \\ \\ = e^{\lambda t} \frac{K_2t\lambda + K_1\lambda - K_2}{L_1\lambda} \\ \end{align}$$

$$I_\text{L}(0) = \frac{K_1\lambda - K_2}{L_1\lambda^2} $$

Inserting \$K_1 = V_{_0}\$ and solving with respect to \$K_1\$ yields

$$K_1 = V_{_0} \: \: \: \text{and} \: \: \: K_2 = -\lambda(I_{_0}L_1\lambda - V_{_0})$$

Underdamped case

We know the solution is of the form \$V_c(t) = K_1e^{\alpha t}\cos(\omega_{_n} t) + K_2e^{\alpha t}\sin(\omega_{_n} t)\$.

The coefficients are determined:

$$V_\text{c}(0) = K_1 =V_{_0}$$

$$\begin{align} I_\text{L}(t) = \frac{1}{L_1} \int K_1e^{\alpha t}\cos(\omega_{_n} t) + K_2e^{\alpha t}\sin(\omega_{_n} t) \\ \\= e^{\alpha t}\frac{((K_1\omega_{_n} + K_2\alpha) \sin(\omega_{_n} t) + (K_1\alpha -K_2\omega_{_n})\cos(\omega_{_n} t)}{L_1(\alpha^2 + \omega_{_n}^2)} \\ \end{align}$$

$$I_\text{L}(0) = \frac{K_1\alpha - K_2\omega_{_n}}{L_1(\alpha^2 + \omega_{_n}^2)} $$

Inserting \$K_1 = V_{_0} \$ and solving with respect to \$K_2\$ gives

$$ K_1 = V_{_0} \: \: \: \text{and} \: \: \: K_2 = -\frac{I_{_0}L_1(\alpha^2 + \omega_{_n}^2) + V_{_0}\alpha}{\omega_{_n}} $$

Plots and simulations

Setting \$C_1 = 10 \: \text{mF}\$, \$L_1 = 40\: \text{mH}\$, \$I_{_0} = 2 \: \text{A}\$ and \$V_{_0} = 10\:\text{V}\$ the value of \$R_1\$ dictates whether the response is overdamped underdamped or critically damped as shown by Maple below:

enter image description here

The following MATLAB script was used to define and calculate the coefficients of the solutions and the plot is also seen below. The coefficients are calculated using the formulas dervied in the sections above.

L1 = 40e-3;
I0 = 2;
t = (0:0.001:0.1)';

lambda1 = 50*(-2+sqrt(3));
lambda2 = -50*(2+sqrt(3));
K1 = -lambda1 *(I0*L1*lambda2-V0)/(lambda1-lambda2);
K2 = lambda2 * (I0*L1*lambda1-V0)/(lambda1-lambda2);
y1 = K1*exp(lambda1*t) + K2*exp(lambda2*t);

%Critically damped
lambda = -50;
C1 = V0;
C2 = -lambda *(I0*L1*lambda-V0);
y2 = exp(lambda*t).*(C1+C2*t);

alpha = -33.333;
wn = 37.2678;
A1 = V0;
A2 = (-I0*L1*(alpha^2+wn^2)+V0*alpha)/wn;
y3 = A1*exp(alpha*t).* cos(wn*t) + A2*exp(alpha*t).*sin(wn*t);

hold on
plot(t,y2, 'color','blue')
plot(t,y3, 'color', 'red')
legend('Overdamped', 'Critically damped', 'Underdamped')
grid on
xlabel('Time [sec]');
ylabel('Capacitor Voltage [V]');
hold off

enter image description here

Finally, an LTspice simulation of the exact three cases investigated above is shown below. We obtain the exact same waveforms as we did in the MATLAB plot for all three cases:

enter image description here

  • \$\begingroup\$ Not what I'm looking for. I'd like to see resulting equations that only incorporate the values for R, C, and L and logic related to selecting which equation, as well as their development. The final equations cannot use any complex numbers. They must only use real numbers. I'll add that notation to my question. I assumed and then forgot about it. See my added note at the end. The entire development must be doable written in sand with fingers on a deserted island and certainly cannot use a numeric solver. This is to be paper and pencil stuff and real numbers only, in the end. \$\endgroup\$
    – jonk
    Dec 25, 2022 at 21:35
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    \$\begingroup\$ @jonk Well, I tried. I think I will leave my answer is it currently is (if that is okay) because I’m interested in other people’s answers to this problem - especially your own answer to this. Thank you for all the help you have given me through the years, Jonk. \$\endgroup\$
    – Carl
    Dec 25, 2022 at 21:45
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    \$\begingroup\$ I think you should leave it here, too. It's my failure in not being as clear as I might have wished. Plus, your answer and my added note as well as comments work to clarify what I'm asking for, too. So I completely agree, Keep it! And thanks! I don't mean to be, in any way, 'down' on your answer. It's just not what I was looking to see. There are some foundational math tools I'd like to see being used. I'm also curious. \$\endgroup\$
    – jonk
    Dec 25, 2022 at 21:51
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    \$\begingroup\$ Oh, and Carl. You seem to have simply "took from a book" the standard answer for a critical damping case (\$\zeta=1\$), for example. I need to see its development. Not left with the feeling that it's "taken from this book" thing. How and why does that particular equation exist? (It's just a step or two's work to show why. But I want to see those steps.) This is about how we know what we know. \$\endgroup\$
    – jonk
    Dec 25, 2022 at 22:08
  • 1
    \$\begingroup\$ And thanks for the kind words, Carl. I don't deserve them. But they are nice to hear! :) \$\endgroup\$
    – jonk
    Dec 25, 2022 at 22:19

Question Recap

I'd like to start with a quick recap taken directly from my question:

$$\begin{align*} \frac{v_{_\text{C}}}{R_1}+C_1\frac{\text{d}}{\text{d}t}v_{_\text{C}}+\frac1{L_1}\int v_{_\text{C}}\:\text{d}t&=0\:\text{A} \\\\ \frac{\text{d}}{\text{d}t}\left[\frac{v_{_\text{C}}}{R_1}+C_1\frac{\text{d}}{\text{d}t}v_{_\text{C}}+\frac1{L_1}\int v_{_\text{C}}\:\text{d}t\right.&=\bigg.0\:\text{A}\bigg] \\\\ \frac1{R_1}\frac{\text{d}}{\text{d}t}v_{_\text{C}}+C_1\frac{\text{d}^2}{\text{d}t^2}v_{_\text{C}}+\frac{v_{_\text{C}}}{L_1}&=0\:\text{A} \\\\ \frac{\text{d}^2}{\text{d}t^2}v_{_\text{C}}+\frac1{R_1\,C_1}\frac{\text{d}}{\text{d}t}v_{_\text{C}}+\frac{1}{L_1\,C_1}v_{_\text{C}}&=0\:\text{A} \\\\ \left[\frac{\text{d}^2}{\text{d}t^2}+\frac1{R_1\,C_1}\frac{\text{d}}{\text{d}t}+\frac{1}{L_1\,C_1}\right]v_{_\text{C}}&=0\:\text{A} \\\\ \left[D^2+\frac1{R_1\,C_1}D+\frac{1}{L_1\,C_1}\right]v_{_\text{C}}&=0\:\text{A} \end{align*}$$

I had snuck in that \$D=\frac{\text{d}}{\text{d}t}\$ in the last line above. It's just short-hand to save on scribbling.

To zero out the left side above there are two choices. One is to set \$v_{_\text{C}}=0\$. But given that the problem allows the specification of \$V_0\ne 0\$ and \$I_0\ne 0\$, that option is readily disposed of.

The Quadratic Solution

There's only one other choice. \$D^2+\frac1{R_1\,C_1}D+\frac{1}{L_1\,C_1}\$ must somehow always be zero.

Given the quadratic form of \$ax^2+bx+c=0\$ the roots are found at \$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\$.

But we have something more like \$x^2+\textsf{a}\,x+\textsf{b}=0\$, with roots at: \$x=-\frac12\textsf{b}\pm\frac12\sqrt{\textsf{a}^2-4\textsf{b}}\$.

By setting \$\beta=\sqrt{\textsf{b}}\$ and \$\alpha=\frac{\textsf{a}}{2\sqrt{\textsf{b}}}\$ (or setting \$\beta=\sqrt{\frac{c}{a}}\$ and \$\alpha=\frac{b}{2\sqrt{a\,c}}\$) then the roots (of \$x^2+2\alpha\beta\,x+\beta^2=0\$) are at \$x=\beta\left(-\alpha\pm\sqrt{\left(-\alpha\right)^2-1}\right)\$ and where \$\alpha\ge 0\$. This separates things out so that \$-\alpha\$ is the relative axis of symmetry and \$\beta\$ is a scaling factor. Now, a simple inspection informs us immediately whether or not roots are complex conjugates. If \$\alpha\lt 1\$ then they are complex conjugates. Otherwise, they are real and different or else real and the same.

In electronic transfer functions where the homogeneous response is 2nd order, \$\omega_{_0}\$ (or \$\omega_{_\text{c}}\$ for crossover or \$\omega_{_\text{p}}\$ for pole) is a special angular frequency used instead of \$\beta\$ and \$\zeta=\frac1{2\,Q}\$ is the damping factor and is used instead of \$\alpha\$.


If you knew that \$\frac{\text{d}}{\text{d}t}\,f\left(t\right)=0\$, then you'd be pretty sure that \$f\left(t\right)\$ wasn't a function of time, but instead was constant. So you'd guess that the general solution is \$f\left(t\right)=A_1\$, where \$A_1\$ was some constant determined by some initial condition. You could get to the specific form if you knew the initial condition.

In similar fashion, it follows that if we knew that \$\left[\frac{\text{d}}{\text{d}t}+\alpha\right]f\left(t\right)=0\$, we could guess that the general solution is \$f\left(t\right)=A_1\,e^{^{-\alpha\,t}}\$. We could test this:

$$\begin{align*} \left[\frac{\text{d}}{\text{d}t}+\alpha\right]f\left(t\right)&=0 \\\\ \left[\frac{\text{d}}{\text{d}t}+\alpha\right]A_1\,e^{^{-\alpha\,t}}&= 0 \\\\ \frac{\text{d}}{\text{d}t}A_1\,e^{^{-\alpha\,t}}+\alpha\,A_1\,e^{^{-\alpha\,t}}&= 0 \\\\ -\alpha\,A_1\,e^{^{-\alpha\,t}}+\alpha\,A_1\,e^{^{-\alpha\,t}}&= 0 \\\\ 0&=0 \end{align*}$$

So the operator \$\left[D+\alpha\right]\$ (just snuck \$D\$ back in) annihilates functions of the general form of \$f\left(t\right)=A_1\,e^{^{-\alpha\,t}}\$. (It's also conversely true that the operator \$\left[D-\alpha\right]\$ annihilates functions of the general form of \$f\left(t\right)=A_1\,e^{^{\alpha\,t}}\$.)

What does the operator \$\left[D+\alpha\right]^2\$ annihilate? In general, any operator of the form \$\left[D+\alpha\right]^m\$ annihilates any solution of the general form \$f\left(t\right)=\sum_{k=0}^{m-1} A_k\,t^k\,e^{^{-\alpha\,t}}\$. When \$m=2\$, this means it annihilates any function of the general form, \$f\left(t\right)=A_0\,e^{^{-\alpha\,t}}+A_1\,t\,e^{^{-\alpha\,t}}\$. Let's see:

$$\begin{align*} \left[D+\alpha\right]^2 f\left(t\right)&=0 \\\\ \left[D+\alpha\right]\left[D+\alpha\right]\left[ A_0\,e^{^{-\alpha\,t}}+A_1\,t\,e^{^{-\alpha\,t}}\right]&= 0 \\\\ \left[D+\alpha\right]\left\{D\left[ A_0\,e^{^{-\alpha\,t}}+A_1\,t\,e^{^{-\alpha\,t}}\right]+\alpha\left[ A_0\,e^{^{-\alpha\,t}}+A_1\,t\,e^{^{-\alpha\,t}}\right]\right\}&= 0 \\\\ \left[D+\alpha\right]\left\{ -\alpha\,A_0\,e^{^{-\alpha\,t}}+D\left[A_1\,t\,e^{^{-\alpha\,t}}\right]+\alpha\,A_0\,e^{^{-\alpha\,t}}+\alpha\,A_1\,t\,e^{^{-\alpha\,t}}\right\}&= 0 \\\\ \left[D+\alpha\right]\left\{D\left[A_1\,t\,e^{^{-\alpha\,t}}\right]+\alpha\,A_1\,t\,e^{^{-\alpha\,t}}\right\}&=0 \\\\ \left[D+\alpha\right]\left\{A_1\,e^{^{-\alpha\,t}}-\alpha\,A_1\,t\,e^{^{-\alpha\,t}}+\alpha\,A_1\,t\,e^{^{-\alpha\,t}}\right\}&=0 \\\\ \left[D+\alpha\right]A_1\,e^{^{-\alpha\,t}}&=0 \end{align*}$$

And we already know where that leads to. So \$\left[D+\alpha\right]^2\$ annihilates \$f\left(t\right)=A_0\,e^{^{-\alpha\,t}}+A_1\,t\,e^{^{-\alpha\,t}}\$!

The following are perhaps the two more important general annihilator forms (in electronics) to have in an annihilator toolbox:

  • \$\left[D-\alpha\right]^m\$ annihilates any solution of the general form:

    \$\quad\quad f\left(t\right)=\sum_{k=0}^{m-1} A_k\,t^k\,e^{^{\alpha\,t}}\$.

  • \$\left[\left(D-\alpha\right)^2+\beta^2\right]^m\$ annihilates any solution of the general form:

    \$\quad\quad f\left(t\right)=\sum_{k=0}^{m-1} A_{2k}\,t^k\,e^{^{\alpha\,t}}\cos\left(\beta\,t\right) + A_{2k+1}\,t^k\,e^{^{\alpha\,t}}\sin\left(\beta\,t\right)\$

(When I need to know how to annihilate some function, \$f\left(t\right)\$, then I look at the denominator of the Laplace transform, \$F\left(s\right)=\mathcal{L}\, f\left(t\right)\$. Let's call that denominator \$X\left(s\right)\$. If \$X\left(s\right)\$ is a polynomial then \$X\left(D\right)\$ will be an annihilator of \$f\left(t\right)\$. I don't have the time before I leave to find or create a proof, or a counter-factual. But that's been my experience.)

Annihilators are rigorous and powerful tools.

They can be used in cases where we do not have an homogeneous equation -- the left side isn't zero and instead includes a non-homogeneity. All we need to do is to annihilate the non-homogeneity on the right side by applying the appropriate annihilator to both sides, equally.

Doing so causes the non-homogeneity to become zero (annihilated), turning a non-homogeneous equation into an homogeneous one (read: easier to solve.)

It's a way of zeroing out the driving function. Had this circuit been one that included an input source through a resistor for example, we'd then have had a non-homogeneous equation to mess around with. But by annihilating the input source, we get back a simpler problem (even if some more zeroes to worry over.) Nice.

When you squash the problem down until there's only zeroes left, it's just easier!

In real-estate it's all about "location, location, & location." Perhaps in electronics math it's all about "annihilation, annihilation, & annihilation."

Critically Damped

Returning to the problem at hand, \$\left[D^2+\frac1{R_1\,C_1}D+\frac{1}{L_1\,C_1}\right]v_{_\text{C}}=0\$, the critically damped case is where \$\zeta=1\$. This happens when \$L_1=4\, C_1 R_1^2\$. (Seeing a \$\text{Farad}\cdot\text{Ohm}^2\$ don't we all just think, "Oh, that's a \$\text{Henry}\$!")

Substituting, we find \$\left[D^2+2\frac1{2\,R_1\,C_1}D+\left(\frac1{2\,R_1\,C_1}\right)^2\right]v_{_\text{C}}=0\$. If we set \$\alpha=-\frac1{2\,R_1\,C_1}\$ then this is:

$$\left[D^2-2\alpha D+\alpha^2\right]v_{_\text{C}}=\left[D-\alpha\right]^2\:v_{_\text{C}}=0$$

We now easily recognize the \$\left[D-\alpha\right]^{m=2}\$ annihilator! And clearly, the roots happen when \$D=\alpha\$.

As \$m=2\$, the general solution is:

$$\begin{align*} v_{_\text{C}}&=\sum_{k=0}^{m-1} A_k\,t^k\,e^{^{\alpha\,t}} \\\\ &= A_{_0}\,e^{^{\alpha\,t}}+A_{_1}\,t\,e^{^{\alpha\,t}} \\\\ &=e^{^{\alpha\,t}}\left(A_{_0}+A_{_1}\,t\right) \end{align*}$$

At \$t=0\$ we know that \$v_{_\text{C}}=V_{_0}\$. Therefore, \$A_{_0}=V_{_0}\$.

The other initial condition is the current in the inductor. Let's recall the KCL again, replace that initial current into the equation, and follow through:

$$\begin{align*} \frac{v_{_\text{C}}}{R_1}+C_1\frac{\text{d}}{\text{d}t}v_{_\text{C}}+\frac1{L_1}\int v_{_\text{C}}\:\text{d}t&=0\:\text{A} \\\\ \frac{v_{_\text{C}}}{R_1}+C_1\frac{\text{d}}{\text{d}t}v_{_\text{C}}+I_{_0}&=0\:\text{A} \\\\ \frac{v_{_\text{C}}}{R_1}+C_1\frac{\text{d}}{\text{d}t}\left[e^{^{\alpha\,t}}\left(A_{_0}+A_{_1}\,t\right)\right]+I_{_0}&=0\:\text{A} \\\\ \frac{v_{_\text{C}}}{R_1}+C_1\left[\alpha\,e^{^{\alpha\,t}}\left(A_{_0}+A_{_1}\,t\right)+e^{^{\alpha\,t}}\left(A_{_1}\right)\right]+I_{_0}&=0\:\text{A} \end{align*}$$

At \$t=0\$, which is when we care about finding \$A_{_1}\$, this collapses down to:

$$\begin{align*} \frac{v_{_\text{C}}}{R_1}+C_1\bigg[\alpha\,A_{_0}+A_{_1}\bigg]+I_{_0}&=0\:\text{A} \end{align*}$$

From there, and substituting in for \$\alpha=-\frac1{2\,R_1\,C_1}\$, \$v_{_\text{C}}=V_{_0}\$ at \$t=0\$ and \$A_{_0}=V_{_0}\$, we find that \$A_{_1}=\frac{-1}{C_1}\left(\frac{V_{_0}}{2\,R_1}+I_{_0}\right)\$.

And now we can finally lay out the specific solution for the critically damped case:

$$v_{_\text{C}}=V_{_0}\cdot e^{^{\frac{-t}{2\,R_1\,C_1}}}\cdot \left[1-\frac1{C_1}\cdot \left(\frac1{2\,R_1}+\frac{I_{_0}}{V_{_0}}\right)\cdot t\right]$$

This incorporates the initial conditions.

Now, keep in mind something. If one of the initial conditions is \$V_{_0}=0\:\text{V}\$, then propagate \$V_{_0}\$ through and into the right side, first, before setting it to zero. Then the above solution reduces to:

$$v_{_\text{C}}= -\frac{I_{_0}}{C_1}\cdot t\cdot e^{^{\frac{-t}{2\,R_1\,C_1}}}$$

Obviously, if both \$V_{_0}=0\:\text{V}\$ and \$I_{_0}=0\:\text{A}\$ then the whole thing collapses, as it should, to \$v_{_\text{C}}=0\:\text{V}\$.

Note that \$L_1\$ is nowhere to be seen in the specific solution. The reason is simple. If we know \$R_1\$ and \$C_1\$ then we know \$L_1\$ as the situation is critically damped and there's only one possible value for this unique situation. We do have to know two of the three values, but it doesn't have to be the same two I picked above.


In the under-damped case (\$\zeta\lt 1\$), we have \$L_1\lt 4\, C_1 R_1^2\$. We are still solving for \$\left[D^2+\frac1{R_1\,C_1}D+\frac{1}{L_1\,C_1}\right]v_{_\text{C}}=0\$ and we will still set \$\alpha=-\frac1{2\,R_1\,C_1}\$.

Expanding \$\left[D-\alpha\right]^2\$ gives \$D^2+\frac1{R_1\,C_1}D+\frac{1}{4\,R_1^2\,C_1^2}\$. Everything is good except for the last term.

To get to the \$\left(D-\alpha\right)^2+\beta^2\$ form, we now also set \$\beta=\sqrt{\frac1{L_1\,C_1}-\frac1{4\, R_1^2\,C_1^2}}\$.

The general solution is:

$$\begin{align*} v_{_\text{C}}&=\sum_{k=0}^{m-1} A_{2k}\,t^k\,e^{^{\alpha\,t}}\cos\left(\beta\,t\right) + A_{2k+1}\,t^k\,e^{^{\alpha\,t}}\sin\left(\beta\,t\right) \\\\ &= A_{_0}\,e^{^{\alpha\,t}}\cos\left(\beta\,t\right)+A_{_1}\,e^{^{\alpha\,t}}\sin\left(\beta\,t\right) \\\\ &=e^{^{\alpha\,t}}\left[A_{_0}\cos\left(\beta\,t\right)+A_{_1}\sin\left(\beta\,t\right)\right] \end{align*}$$

Again, at \$t=0\$ we know that \$v_{_\text{C}}=V_{_0}\$. And just as before, \$A_{_0}=V_{_0}\$.

Like before, the other initial condition is still the current in the inductor. So let's recall the KCL again, replace that initial current into the equation, and follow through:

$$\begin{align*} \frac{v_{_\text{C}}}{R_1}+C_1\frac{\text{d}}{\text{d}t}v_{_\text{C}}+\frac1{L_1}\int v_{_\text{C}}\:\text{d}t&=0\:\text{A} \\\\ \frac{v_{_\text{C}}}{R_1}+C_1\frac{\text{d}}{\text{d}t}v_{_\text{C}}+I_{_0}&=0\:\text{A} \\\\ \frac{v_{_\text{C}}}{R_1}+C_1\frac{\text{d}}{\text{d}t}\left[e^{^{\alpha\,t}}\left[A_{_0}\cos\left(\beta\,t\right)+A_{_1}\sin\left(\beta\,t\right)\right]\right]+I_{_0}&=0\:\text{A} \\\\ \frac{v_{_\text{C}}}{R_1}+C_1\,e^{^{\alpha\,t}}\left[\left(\alpha\,A_{_0}+\beta\,A_{_1}\right)\cos\left(\beta\,t\right)+\left(\alpha\,A_{_1}-\beta\,A_{_0}\right)\sin\left(\beta\,t\right)\vphantom{e^{^{\alpha\,t}}}\right]+I_{_0}&=0\:\text{A} \end{align*}$$

At \$t=0\$, which is when we care about finding \$A_{_1}\$, this collapses down to:

$$\begin{align*} \frac{v_{_\text{C}}}{R_1}+C_1\left(\alpha\,A_{_0}+\beta\,A_{_1}\right)+I_{_0}&=0\:\text{A} \end{align*}$$

From there, and substituting in for \$\alpha=-\frac1{2\,R_1\,C_1}\$, \$v_{_\text{C}}=V_{_0}\$ at \$t=0\$ and \$A_{_0}=V_{_0}\$, we find that \$A_{_1}=\frac{-1}{\beta\,C_1}\left(\frac{V_{_0}}{2\,R_1}+I_{_0}\right)\$. (Yeah, I didn't want to expand \$\beta\$ into that.)

And now we can finally lay out the specific solution for the critically damped case:

$$\begin{align*} v_{_\text{C}}&=V_{_0}\cdot e^{^{\frac{-t}{2\,R_1\,C_1}}}\cdot \left[\cos\left(\beta\,t\right)-\frac1{\beta\,C_1}\cdot \left(\frac1{2\,R_1}+\frac{I_{_0}}{V_{_0}}\right)\sin\left(\beta\,t\right)\right] \end{align*}$$

(You can incorporate the sine and cosine terms, using a phase computed from the constants. If you want to see where I do that, see near the bottom of my answer to Andy at the link given in the first sentence of the question.)


In the over-damped case (\$\zeta\gt 1\$), we have \$L_1\gt 4\, C_1 R_1^2\$.

In this case, we also set \$\beta=\sqrt{\frac1{L_1\,C_1}-\frac1{4\, R_1^2\,C_1^2}}\$.

The general solution follows the same process as for the under-damped case and that at \$t=0\$ then \$A_{_0}=V_{_0}\$ and \$A_{_1}=\frac{-1}{\beta\,C_1}\left(\frac{V_{_0}}{2\,R_1}+I_{_0}\right)\$. Same as before.

But note here that \$\beta\$ will be imaginary!

So let's set \$\beta^{'}=\sqrt{\frac1{4\, R_1^2\,C_1^2}-\frac1{L_1\,C_1}}\$, which is real, and let \$\beta=j\,\beta^{'}\$.

And now we can finally lay out the specific solution for the critically damped case:

$$\begin{align*} v_{_\text{C}}&=V_{_0}\cdot e^{^{\frac{-t}{2\,R_1\,C_1}}}\cdot \left[\cos\left(\beta\,t\right)-\frac1{\beta\,C_1}\cdot \left(\frac1{2\,R_1}+\frac{I_{_0}}{V_{_0}}\right)\sin\left(\beta\,t\right)\right] \\\\ &=V_{_0}\cdot e^{^{\frac{-t}{2\,R_1\,C_1}}}\cdot \left[\cosh\left(\beta^{'}\,t\right)-\frac{1}{\beta^{'}\,C_1}\cdot \left(\frac1{2\,R_1}+\frac{I_{_0}}{V_{_0}}\right)\sinh\left(\beta^{'}\,t\right)\right] \end{align*}$$

All real values, now. The \$j\$ is all gone.


Let's use \$R_1=1\:\text{k}\Omega\$ and \$C_1=10\:\mu\text{F}\$ and supply \$\zeta\$ to vary \$L_1\$. This means using the following equation: \$L_1=\zeta^2\left( 4\,C_1\,R_1^{\,2}\right)\$.

I'll do a run with enough lines to make up a reasonable demonstration (16 lines) going from \$\zeta=0.25\$ to \$\zeta=2\$ (three octaves) with 5 lines per octave.

enter image description here

Good match!


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