# 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

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

${\displaystyle \omega _{n}={\frac {1}{\sqrt {LC}}}\approx 4472}$

${\displaystyle v_{n}(t)=e^{-1000t}{\big [}B_{1}\cos(4359t)+B_{2}\sin(4359t){\big ]}}$

Solve for ${\displaystyle B_{1}}$ and ${\displaystyle B_{2}}$:

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

${\displaystyle v_{c}(t)=1+e^{-1000t}{\big [}B_{1}\cos(4359t)+B_{2}\sin(4359t){\big ]}}$

for ${\displaystyle t=0}$ the voltage across the capacitor is zero, ${\displaystyle v_{c}(t)=0}$

${\displaystyle 0=1+B_{1}\cos(0)+B_{2}\sin(0)}$

${\displaystyle B_{1}=-1{\mbox{ (7)}}}$

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

${\displaystyle i(t)={\frac {dv_{c}(t)}{dt}}C}$

${\displaystyle i(0)=100\cdot 10^{-9}{\big [}e^{-1000t}{\big (}-4359B_{1}\sin(4359t)+4359B_{2}\cos(4359t){\big )}-1000e^{-1000t}{\big (}B_{1}\cos(4359t)+B_{2}\sin(4359t){\big )}{\big ]}}$

${\displaystyle 0=100\cdot 10^{-9}{\big [}4359B_{2}-1000B_{1}{\big ]}}$

substituting ${\displaystyle B_{1}}$ from equation \ref{eq:B1} gives

${\displaystyle B_{2}\approx -0.229}$

For ${\displaystyle t>0}$, ${\displaystyle v_{c}(t)}$ is given by:

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

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

${\displaystyle v_{out}=V_{in}-v_{c}(t)}$

${\displaystyle v_{out}=Vu(t)-v_{c}(t)}$

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

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