Electronics/RCL time domain

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 $$t\geq0$$ 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:

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

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 $$i(t)=\frac{dv_c(t)}{dt}$$ into equation 1:

$$ Vu(t)=v_c(t)+\frac{d^2v_c(t)}{dt^2}LC+R\frac{dv_c(t)}{dt}C $$ $$ \frac{d^2v_c(t)}{dt^2}+\frac{R}{L}\frac{dv_c(t)}{dt}+\frac{1}{LC}v_c(t)=\frac{Vu(t)}{LC} \mbox{ (2)} $$

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

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

substituting equation 3 into equation 2.

$$ \bigg[\frac{d^2v_n(t)}{dt^2}+\frac{R}{L}\frac{dv_n(t)}{dt}+\frac{1}{LC}v_n(t)\bigg]+\bigg[\frac{d^2v_f(t)}{dt^2}+\frac{R}{L}\frac{dv_f(t)}{dt}+\frac{1}{LC}v_f(t)\bigg]=0+\frac{Vu(t)}{LC} $$

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

$$ \bigg[\frac{d^2v_n(t)}{dt^2}+\frac{R}{L}\frac{dv_n(t)}{dt}+\frac{1}{LC}v_n(t)\bigg]=0 \mbox{ (4)} $$ $$ \bigg[\frac{d^2v_f(t)}{dt^2}+\frac{R}{L}\frac{dv_f(t)}{dt}+\frac{1}{LC}v_f(t)\bigg]=\frac{V}{LC} \mbox{ (5)} $$

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:

$$ v_f(t)=K $$ $$ \frac{dv_f(t)}{dt}=0 $$ $$ \frac{d^2v_f(t)}{dt}=0 $$

Substituting into equation 5:

$$ \frac{1}{LC}K=\frac{V}{LC} $$ $$ K=V $$ $$ v_f=V \mbox{ (6)} $$

Solve for $$v_n(t)$$:

Let: $$ \frac{R}{L}=2\alpha $$ $$ \frac{1}{LC}=\omega_n^2 $$ $$ v_n(t)=Ae^{st} $$

Substituting into equation 4 gives:

$$ \frac{d^2Ae^{st}}{dt^2}+2\alpha\frac{dAe^{st}}{dt}+\omega_n^2Ae^{st}=0 $$ $$ s^2Ae^{st}+2\alpha Ae^{st}+\omega_2^2Ae^{st}=0 $$ $$ s^2+2\alpha s+\omega_n^2=0 $$

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

Therefore $$v_n(t)$$ has two solutions $$Ae^{s_1t}$$ and $$Ae^{s_2t}$$

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:

$$ v_n(t)=A_1e^{s_1t}+A_2e^{s_2t} $$

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

2. If $$\alpha = \omega_n$$ the system is said to be critically damped

3. If $$\alpha < \omega_n$$ the system is said to be underdamped

Example
Given the general solution

$$ \alpha=\frac{R}{2L}=1000 $$ $$ \omega_n=\frac{1}{\sqrt{LC}}\approx 4472 $$

$$ s_1=-1000-4359j $$ $$ s_2=-1000+4359j $$

$$ v_n(t)=A_1e^{(-1000-4359j)t}+A_2e^{(-1000+4359j)t} $$

Thus by Euler's formula ($$e^{j\phi}=\cos{\phi}+j\sin{\phi}$$):

$$ v_n(t)=e^{-1000}\big[A_1\big(\cos(-4359t)+j\sin(-4359t)\big)+A_2\big(\cos(4359t)+j\sin(4359t)\big)\big] $$ $$ v_n(t)=e^{-1000t}\big[(A_1+A_2)\cos(4359t)+j(-A_1+A_2)\sin(4359t)\big] $$

Let $$B_1=A_1+A_2$$ and $$B_2=j(-A_1+A_2)$$

$$ v_n(t)=e^{-1000t}\big[B_1\cos(4359t)+B_2\sin(4359t)\big] $$

Solve for $$B_1$$ and $$B_2$$:

From equation \ref{eq:vf}, $$v_f=1$$ for a unit step of magnitude 1V. Therefore substitution of $$v_f$$ and $$v_n(t)$$ into equation \ref{eq:nonhomogeneous} gives:

$$ v_c(t)=1+e^{-1000t}\big[B_1\cos(4359t)+B_2\sin(4359t)\big] $$

for $$t=0$$ the voltage across the capacitor is zero, $$v_c(t)=0$$

$$ 0=1+B_1\cos(0)+B_2\sin(0) $$ $$ B_1=-1\mbox{ (7)} $$

for $$t=0$$, the current in the inductor must be zero, $$i(0)=0$$

$$ i(t)=\frac{dv_c(t)}{dt}C $$ $$ i(0)=100\cdot10^{-9}\big[e^{-1000t}\big(-4359B_1\sin(4359t)+4359B_2\cos(4359t)\big)-1000e^{-1000t}\big(B_1\cos(4359t)+B_2\sin(4359t)\big)\big] $$ $$ 0=100\cdot10^{-9}\big[4359B_2-1000B_1 \big] $$

substituting $$B_1$$ from equation \ref{eq:B1} gives

$$ B_2\approx-0.229 $$

For $$t>0$$, $$v_c(t)$$ is given by:

$$ v_c(t)=1-e^{-1000t}\big[\cos(4359t)+0.229\sin(4359t)\big] $$

$$v_{out}$$ is given by:

$$ v_{out}=V_{in}-v_c(t) $$ $$ v_{out}=Vu(t)-v_c(t) $$

For $$t>0$$, $$v_{out}$$ is given by:

$$ v_{out}=e^{-1000t}\big[\cos(4359t)+0.229\sin(4359t)\big] $$