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

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

${\omega }_n=\frac{1}{\sqrt{LC}}\approx 4472$

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

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:

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

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

$0=1+B_1\cos (0)+B_2\sin (0)$

$B_1=-1\text{ (7)}$

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

$i(t)=\frac{d{v}_c(t)}{dt}C$

$i(0)=100\cdot 10^{-9}\left[e^{-1000t}\right(-4359B_1\sin (4359t)+4359B_2\cos (4359t))-1000e^{-1000t}(B_1\cos (4359t)+B_2\sin (4359t)\left)\right]$

$0=100\cdot 10^{-9}[4359B_2-1000B_1]$

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

$B_2\approx -0.229$

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

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

$v_{out}$ 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:

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