RLC DC Transient (Part 2)

It isn't really necessary to have a closed form solution, so I won't bother deriving one. Instead, I will employ a very useful concept; that of the "state space". The "state space" formulation expresses the time-derivatives of the things you want to solve for, in terms of the present values of the things you want to solve for (and any forcing functions). That is useful because knowing the rate of change allows us to predict the state a very short time later. The system dynamics can thus be simulated in small, sequential, steps.

$\dot{x} = f(x,u)$

In the above equation, $x$ represents a state variable to be solved for (like the voltage across a capacitor or the current through an inductor). The quantity $u$ is the system forcing function (the voltage source, for example). Below is a more general representation of the circuit for $t \geq 0$ .

Let the current and voltage for the inductor be $I_L, V_L$ , and let the current and voltage for the capacitor be $I_C, V_C$ . $I_L$ and $V_C$ are the state variables that we solve for. Note that in circuit analysis, the state variables are associated with energy storage. The fundamental equations are:

$V_L = L \, \dot{I_L} \\ I_C = C \, \dot{V_C}$

The right-hand sides already resemble one side of the state-space equation. But we need to get the left-hand sides expressed in terms of state variables $I_L$ and $V_C$ . Start by writing an expression for the source current.

$I_S = I_L + I_C \\ I_S = I_L + \frac{V_S - R_S I_S - V_C}{R2}$

Re-arranging gives:

$I_S = \frac{I_L + V_S/R_2 - V_C / R_2}{1 + R_S/R_2}$

Now we can write the state-space equation for the inductor current (recall that $I_S$ contains both state variables):

$V_L = L \, \dot{I_L} \\ V_S - R_S I_S - R_1 I_L = L \, \dot{I_L} \\ \dot{I_L} = \frac{V_S - R_S I_S - R_1 I_L}{L}$

And for the capacitor voltage (recall that $I_S$ contains both state variables):

$I_C = C \, \dot{V_C} \\ I_S - I_L = C \, \dot{V_C} \\ \dot{V_C} = \frac{I_S - I_L}{C}$

Numerically integrating the derivatives (from appropriate starting conditions) results in the following plot. I have plotted all the way to $t = 10$ for illustration purposes. By inspection, we know that the source current should start at $5$ and approach $\frac{10}{3}$ as time goes to infinity. The plot shows this behavior. The desired integral can be included as part of the simulation.

We are going to use s-domain analysis and Laplace transformation. (Roughly speaking, $s$ corresponds to $j\omega$ , where $j=\sqrt{-1}$ .) The impedance of the circuit is $Z=1+\frac{(2+s)(1+1/s)}{3+s+1/s}$ . The admittance is $Y=1/Z$ . The Laplace transform of the unit step function is $1/s$ ; for a 10V source it is $10/s$ . The time dependent current is the inverse Laplace transform of

$\frac{10}{s}Y=\frac{10}{s\left(1+\frac{(2+s)(1+1/s)}{3+s+1/s}\right)}$

Feeding this function to an inverse Laplace transformator (e.g. here ) we get

$i(t)=10\left[ \frac{1}{3} + \frac {\sqrt{3}+1}{12} e^{-\alpha t} -\frac{\sqrt{3}-1}{12} e^{-\beta t} \right]$

where $\alpha=\frac {3-\sqrt 3}{2}$ and $\beta=\frac {3+\sqrt 3}{2}$

Feeding $i(t)-\frac{10}{3}$ to an integral calculator (like this ) yields $Q=3.04897$