Electronics/RCL time domain1

Figure 1: RCL circuit

When the switch is closed, a voltage step is applied to the RCL circuit. Take the time the switch was closed to be 0s such that the voltage before the switch was closed was 0 volts and the voltage after the switch was closed is a voltage V. This is a step function given by $V\cdot u(t)$ where V is the magnitude of the step and $u(t)=1$ for $t\ge 0$ and zero otherwise.

To analyse the circuit response using transient analysis, a differential equation which describes the system is formulated. The voltage around the loop is given by:

$Vu(t)=v_c(t)+\frac{di(t)}{dt}L+Ri(t)\text{ (1)}$

where $v_c(t)$ is the voltage across the capacitor, $\frac{di(t)}{dt}L$ is the voltage across the inductor and $Ri(t)$ the voltage across the resistor.

Substituting $i(t)=\frac{d{v}_c(t)}{dt}C$ into equation 1:

$Vu(t)=v_c(t)+\frac{d^2{v}_c(t)}{d{t}^2}LC+R\frac{d{v}_c(t)}{dt}C$

$\frac{d^2{v}_c(t)}{d{t}^2}+\frac{R}{L}\frac{d{v}_c(t)}{dt}+\frac{1}{LC}{v}_c(t)=\frac{Vu(t)}{LC}\text{ (2)}$

The voltage $v_c(t)$ has two components, a natural response $v_n(t)$ and a forced response $v_f(t)$ such that:

$v_c(t)=v_f(t)+v_n(t)\text{ (3)}$

substituting equation 3 into equation 2.

$[\frac{d^2{v}_n(t)}{d{t}^2}+\frac{R}{L}\frac{d{v}_n(t)}{dt}+\frac{1}{LC}{v}_n(t)]+[\frac{d^2{v}_f(t)}{d{t}^2}+\frac{R}{L}\frac{d{v}_f(t)}{dt}+\frac{1}{LC}{v}_f(t)]=0+\frac{Vu(t)}{LC}$

when $t>0s$ then $u(t)=1$:

$[\frac{d^2{v}_n(t)}{d{t}^2}+\frac{R}{L}\frac{d{v}_n(t)}{dt}+\frac{1}{LC}{v}_n(t)]=0\text{ (4)}$

$[\frac{d^2{v}_f(t)}{d{t}^2}+\frac{R}{L}\frac{d{v}_f(t)}{dt}+\frac{1}{LC}{v}_f(t)]=\frac{V}{LC}\text{ (5)}$

The natural response and forced solution are solved separately.

Solve for $v_f(t):$

Since $\frac{V}{LC}$ is a polynomial of degree 0, the solution $v_f(t)$ must be a constant such that:

$v_f(t)=K$

$\frac{d{v}_f(t)}{dt}=0$

$\frac{d^2{v}_f(t)}{dt}=0$

Substituting into equation 5:

$\frac{1}{LC}K=\frac{V}{LC}$

$K=V$

$v_f=V\text{ (6)}$

Solve for $v_n(t)$:

Let:

$\frac{R}{L}=2\alpha$

$\frac{1}{LC}={\omega }_n^2$

$v_n(t)=A{e}^{st}$

Substituting into equation 4 gives:

$\frac{d^{2}A{e}^{st}}{d{t}^2}+2\alpha \frac{dA{e}^{st}}{dt}+{\omega }_n^{2}A{e}^{st}=0$

$s^{2}A{e}^{st}+2\alpha A{e}^{st}+{\omega }_2^{2}A{e}^{st}=0$

$s^2+2\alpha s+{\omega }_n^2=0$

$s=\frac{-2\alpha \pm \sqrt{4{\alpha }^2-4{\omega }_n^2}}{2}=-\alpha \pm \sqrt{{\alpha }^2-{\omega }_n^2}$

Therefore $v_n(t)$ has two solutions $A{e}^{s_{1}t}$ and $A{e}^{s_{2}t}$

where $s_1$ and $s_2$ are given by:

$s_1=-\alpha +\sqrt{{\alpha }^2-{\omega }_n^2}$

$s_2=-\alpha -\sqrt{{\alpha }^2-{\omega }_n^2}$

The general solution is then given by:

$v_n(t)=A_1{e}^{s_{1}t}+A_2{e}^{s_{2}t}$

Depending on the values of the Resistor, inductor or capacitor the solution has three posibilies.

1. If $\alpha >{\omega }_n$ the system is said to be overdamped. The system has two distinct real solutions:

$v_n(t)=A_1{e}^{(-\alpha +\sqrt{{\alpha }^2-{\omega }_n^2})t}+A_2{e}^{(-\alpha -\sqrt{{\alpha }^2-{\omega }_n^2})t}$

2. If $\alpha ={\omega }_n$ the system is said to be critically damped. The system has one real solution:

$v_n(t)=(A_1+A_2)\left(e^{-\alpha t}\right)$

Let $B_1=A_1+A_2$:

$v_n(t)=B{e}^{-\alpha t}$

3. If $\alpha <{\omega }_n$ the system is said to be underdamped. The system has two complex solutions:

$s_1=-\alpha +j\sqrt{{\omega }_n^2-{\alpha }^2}$

$s_2=-\alpha -j\sqrt{{\omega }_n^2-{\alpha }^2}$

$v_n(t)=e^{-\alpha t}[A_1{e}^{j\sqrt{{\omega }_n^2-{\alpha }^2}t}+A_2{e}^{-j\sqrt{{\omega }_n^2-{\alpha }^2}t}]$

By Euler's formula ($e^{j\varphi }=\cos \varphi +j\sin \varphi$):

$v_n(t)=e^{-\alpha t}[(A_1+A_2)\cos (\sqrt{{\omega }_n^2-{\alpha }^2}t)+j(-A_1+A_2)\sin (\sqrt{{\omega }_n^2-{\alpha }^2}t)]$

Let $B=A_1+A_2$ and $B_2=j(-A_1+A_2)$

$v_n(t)=e^{-\alpha t}[B_1\cos (\sqrt{{\omega }_n^2-{\alpha }^2}t)+B_2\sin (\sqrt{{\omega }_n^2-{\alpha }^2}t)]$