Electronics/RCL time domain simple

Figure 1: RCL circuit

When the switch is closed, a voltage step is applied to the RCL circuit. Take the time the switch was closed to be 0s such that the voltage before the switch was closed was 0 volts and the voltage after the switch was closed is a voltage V. The voltage across the capacitor consists of a forced response $v_{f}$ and a natural response $v_{n}$ such that:

$v_{c}=v_{f}+v_{n}$

The forced response is due to the switch being closed, which is the voltage V for $t\geq 0$. The natural response depends on the circuit values and is given below:

Define the pole frequency $\omega _{n}$ and the dampening factor $\alpha$ as:

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

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

Depending on the values of $\alpha$ and $\omega _{n}$ the system can be characterized as:

1. If $\alpha >\omega _{n}$ the system is said to be overdamped. The solution for the system has the form:

$v_{n}(t)=A_{1}e^{{\big (}-\alpha +{\sqrt {\alpha ^{2}-\omega _{n}^{2}}}{\big )}t}+A_{2}e^{{\big (}-\alpha -{\sqrt {\alpha ^{2}-\omega _{n}^{2}}}{\big )}t}$

2. If $\alpha =\omega _{n}$ the system is said to be critically damped The solution for the system has the form:

$v_{n}(t)=Be^{-\alpha t}$

3. If $\alpha <\omega _{n}$ the system is said to be underdamped The solution for the system has the form:

$v_{n}(t)=e^{-\alpha t}{\big [}B_{1}\cos({\sqrt {\omega _{n}^{2}-\alpha ^{2}}}t)+B_{2}\sin({\sqrt {\omega _{n}^{2}-\alpha ^{2}}}t){\big ]}$

How do you calculate these equations?

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

First calculate the values of $\alpha$ and $\omega _{n}$:

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

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

From these values note that $\alpha <\omega _{n}$. The system is therefore underdamped. The equation for the voltage across the capacitor is then:

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

Before the switch was closed assume that the capacitor was fully discharged. This implies that v(t)=0 at the instant the switch was closed (t=0). Substituting t=0 into the previous equation gives:

$0=1+B_{1}$

Therefore $B_{1}=-1$. Similarly at the instant the switch is closed, the current in the inductor must be zero as the current can not instantly change. Substituting the equation for $v_{c}(t)$ into the equation for the inductor and solving at the instant the switch was closed (t=0) gives:

$i(t)={\frac {dv_{c}(t)}{dt}}C$
$0=100\cdot 10^{-9}{\big [}4359B_{2}-1000B_{1}{\big ]}$

Therefore $B_{2}\approx -0.229$. Once $v_{c}(t)$ is known, calculation of the voltage across the inductor and resistor ($V_{out}$) is given by:

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

You have missed a lot of steps, where are they?