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, $i(0)=0$

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

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

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

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:

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

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