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?

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

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

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:

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