# 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?

 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:

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

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

Image:Example1 underdamped.png