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 $t\ge 0$. 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:

$v_n(t)=A_1{e}^{(-\alpha +\sqrt{{\alpha }^2-{\omega }_n^2})t}+A_2{e}^{(-\alpha -\sqrt{{\alpha }^2-{\omega }_n^2})t}$

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

$v_n(t)=B{e}^{-\alpha t}$

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

$v_n(t)=e^{-\alpha t}[B_1\cos (\sqrt{{\omega }_n^2-{\alpha }^2}t)+B_2\sin (\sqrt{{\omega }_n^2-{\alpha }^2}t)]$

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
1kΩ	0.5H	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:

$v_c(t)=1+e^{-1000t}[B_1\cos (4359t)+B_2\sin (4359t)]$

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:

$i(t)=\frac{d{v}_c(t)}{dt}C$
$0=100\cdot 10^{-9}[4359B_2-1000B_1]$

Therefore $B_2\approx -0.229$. Once $v_c(t)$ is known, the voltage across the inductor and resistor ($V_{out}$) is given by:

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

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

Figure 2: Example 1 Underdamped Response
Figure 2: Underdamped Resonse

File:Example1 underdamped.png