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 $$t\ge0$$. 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: $$ v_n(t)=A_1e^{\big(-\alpha+\sqrt{\alpha^2-\omega_n^2}\big)t}+A_2e^{\big(-\alpha-\sqrt{\alpha^2-\omega_n^2}\big)t} $$

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?

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: $$ i(t)=\frac{dv_c(t)}{dt}C $$

$$ 0=100\cdot10^{-9}\big[4359B_2-1000B_1 \big] $$ Therefore $$B_2\approx-0.229$$. Once $$v_c(t)$$ is known, 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?

Figure 2: Underdamped Resonse File:Example1 underdamped.png