# Electronics/RCL time domain2

Figure 1: RCL circuit

## Example:

Given the following values what is the response of the system when the switch is closed?

 R L C V 0.5H 1kΩ 100nF 1V

$\alpha=\frac{R}{2L}=1000$

$\omega_n=\frac{1}{\sqrt{LC}}\approx 4472$

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

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:

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