# 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 ${\displaystyle v_{f}}$ and a natural response ${\displaystyle v_{n}}$ such that:

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

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

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

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

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

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

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

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

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

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

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

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

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 ${\displaystyle \alpha }$ and ${\displaystyle \omega _{n}}$:

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

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

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

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

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:

${\displaystyle 0=1+B_{1}}$

Therefore ${\displaystyle B_{1}=-1}$. 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 ${\displaystyle v_{c}(t)}$ into the equation for the inductor and solving at the instant the switch was closed (t=0) gives:

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

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

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

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

Figure 2: Underdamped Resonse

Image:Example1 underdamped.png