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From math below it seems no oscillations are possible and the steady state reaches instantly. I know this is wrong but I'm new to differential equations and don't see my mistake.

Summary: For the initial conditions on capacitor

$$v(0) = 0, \qquad i(0) = C v'(0) = 0$$

I'm getting the homogeneous solution constants $c_1 = 0$ and $c_2 = 0$. This means there is no transient response and the response reaches steady-state instantly. How is this possible?

Series RLC circuit


Let $v$ denote voltage across capacitor. Differential equation for series RLC circuit in terms of voltage across the capacitor is

$$Ri + L\frac{di}{dt} + v = V$$

Since $i = C\frac{dv}{dt}$ it follows

$$RC\frac{dv}{dt} + LC\frac{d^2v}{dt^2 } + v = V$$

$$\frac{d^2v}{dt^2} + \frac{R}{L}\frac{dv}{dt} + \frac{1}{LC}v = \frac{V}{LC}$$

With $2p = \frac{R}{L}$ and $\omega_0^2=\frac{1}{LC}$ the differential equation is

$$v''(t) + 2pv'(t) + \omega_0^2 v(t) = \omega_0^2 V$$

The homogeneous solution for the above differential equation is

$$v_h(t) = e^{-pt} ( c_1 \cos(\omega t) + c_2 \sin(\omega t) )$$

where $\omega^2 = \omega_0^2 - p^2$. From the initial conditions it follows

$$v(0) = 0 \implies c_1 = 0$$ $$i(0) = v'(0) = 0 \implies c_2 = 0$$

This means that the differential equation has no transient response! A particular solution is $v_p(t) = V$ and this makes the complete solution! What am I doing wrong?

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2 Answers 2

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TL;DR In the procedure you posted you forgot to include particular solution. The homogeneous solution will always evaluate to $0$ when used as a solution to the general differential equation.


Homogeneous solution

The roots of the differential equation

$$v''(t) + 2p v'(t) + \omega_0^2 v(t) = \omega_0^2 V$$

are $q_{1,2} = -p \pm j \sqrt{\omega_0^2 - p^2}$. For these roots there are three types of homogeneous solutions:

$$v_h(t) = \left\{ \begin{array}{lll} e^{-pt} (A \cos(\omega t) + B \sin(\omega t)), & \omega_0^2 > p^2 & \qquad\text{(Underdamped response)} \\ e^{-pt} (A + B t), & \omega_0^2 = p^2 & \qquad\text{(Critically damped response)} \\ e^{-pt} (A e^{\omega_1 t} + B e^{-\omega_1 t}), & \omega_0^2 < p^2 & \qquad\text{(Overdamped response)} \end{array} \right. $$

where $A$ and $B$ are unknown constants, $\omega^2 = \omega_0^2 - p^2$ and $\omega_1^2 = -\omega^2$. We will now focus only on the underdamped response, as that is the one you analyze in your question.


Particular solution

The particular solution depends on the input (driver) function and the roots of the differential equation. In your case, the particular solution is

$$v_p(t) = K$$

where $K$ is unknown constant.


Solving for unknown constants

The total solution is a linear combination of homogeneous and particular solutions

$$v(t) = v_h(t) + v_p(t) = e^{-pt} \bigl( A \cos(\omega t) + B \sin(\omega t) \bigr) + K$$

If we use this as a solution to the general differential equation we get

$$\underbrace{v_h''(t) + 2p v_h'(t) + \omega_0^2 v_h(t)}_{\text{always } 0} + \omega_0^2 v_p(t) = \omega_0^2 V$$

from which it follows $K = V$. The other two unknown constants are determined from initial conditions $v(0) = v_0$ and $v'(0) = v_0'$

$$A + K = v_0, \qquad -p A + \omega B = v_0'$$

from which it follows $A = v_0 - V$ and $B = \frac{1}{\omega} (v_0' + p v_0 - p V)$.


Final solution

In your special case, $v_0 = 0$ and $v_0' = 0$, and the final solution is

$$v(t) = - V e^{-pt} \Bigl( \cos(\omega t) + \frac{p}{\omega} \sin(\omega t) \Bigr) + V$$

The above solution can be written in a more compact form

$$\boxed{v(t) = V \Bigl( 1 - \frac{\omega_0}{\omega} e^{-pt} \cos\bigl( \omega t - \arctan \frac{p}{\omega} \bigr) \Bigr) }$$


Response overshoot

Note that underdamped (oscillatory) responses naturally have an overshoot. This means that at some point the voltage on the capacitor will be higher than the input voltage.

After the transient response, voltage on the capacitor settles to the input DC voltage

$$v_f = \lim_{t \to \infty} v(t) = V$$

The response overshoot magnitude can be found as

$$\text{PO} = \frac{v(t_m) - v_f}{v_f} \cdot 100\%$$

where $t_m$ is the time of the response maximum. From $v'(t_m) = 0$ it follows that $t_m = k \pi / \omega$ where $k$ is an odd positive number. The response overshoot magnitude is

$$\boxed{\text{PO} = e^{-k \pi p/\omega} \cdot 100\%}$$

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  • $\begingroup$ Ahh looks that is exactly my mistake! So we use the general solution, not just the homogeneous solution while solving constants $c_1, c_2$. But does this mean the input driver function influences the transient solution also? I thought the transient solution depends ONLY on the initial conditions, not on the driver function... is my thinking wrong? $\endgroup$
    – across
    Feb 16 at 16:16
  • $\begingroup$ Wow so $- V e^{-pt} \Bigl( \cos(\omega t) + \frac{p}{\omega} \sin(\omega t) \Bigr)$ is the transient and it oscillates and has a maximum amplitude possible of $V\sqrt{1+(p/\omega)^2}$! Interesting... $\endgroup$
    – across
    Feb 16 at 16:31
  • $\begingroup$ I mean the output voltage can exceed the input by $V\sqrt{1+(p/\omega)^2}$ $\endgroup$
    – across
    Feb 16 at 16:32
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    $\begingroup$ @across You are right that the capacitor voltage could exceed the input voltage. This is commonly known as the (voltage) overshoot and is typical for oscillatory systems. $\endgroup$ Feb 16 at 16:59
  • $\begingroup$ @across Correct, the input (driver) function influences the transient response together with initial conditions. You are also correct about the voltage overshoot, but the overshoot magnitude you mention is not correct. I edited the answer to also include discussion about the overshoot. $\endgroup$ Feb 16 at 22:02
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You have given the general solution to the homogeneous equation

$v'' + 2pv' + \omega_0^2v = 0$

To find the general solution to the inhomogeneous equation you need to add a particular integral that solves the inhomogeneous equation

$v'' + 2pv' + \omega_0^2v = \omega_0^2V$

In this case the constant function $v(t)=V$ works as a particular integral. So the full solution is

$v(t) = V + e^{-pt}(c_1\cos(\omega t) + c_2\sin(\omega t))$

So $c_1=c_2=0$ gives $v(t)=V$ which is indeed the steady state that you expect, but it does not meet your boundary condition $v(0)=0$.

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  • $\begingroup$ Ah so we use the "general solution" to solve the constants $c_1,c_2$. Not the "homogeneous solution"? I thought the homogeneous solution cares only about the initial conditions, not the driver function. Is my understanding wrong? $\endgroup$
    – across
    Feb 16 at 16:11
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    $\begingroup$ @across: The homogeneous solution has to cancel out the general solution at $t = 0$. So the form of the transient will depend both on the form of the general solution and on the initial conditions. $\endgroup$ Feb 16 at 16:29

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