# Electronics/RCL frequency domain

Figure 1: RCL circuit

Define the pole frequency $\omega_n$ and the dampening factor $\alpha$ as:

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

$\frac{1}{LC}=\omega_n^2$

To analyze the circuit first calculate the transfer function in the s-domain H(s). For the RCL circuit in figure 1 this gives:

$H(s)=\frac{s\big(s+2\alpha\big)}{s^2+2\alpha s+\omega_n^2}$

$H(s)=\frac{s\big(s+2\alpha\big)}{\big(s+\alpha+j\sqrt{\omega_n^2-\alpha^2}\big)\big(s+\alpha-j\sqrt{\omega_n^2-\alpha^2}\big)}$

When the switch is closed, this applies a step waveform to the RCL circuit. The step is given by $Vu(t)$. Where V is the voltage of the step and u(t) the unit step function. The response of the circuit is given by the convolution of the impulse response h(t) and the step function $Vu(t)$. Therefore the output is given by multiplication in the s-domain H(s)U(s), where $U(s)=V\frac{1}{s}$ is given by the Laplace Transform available in the appendix.

The convolution of u(t) and h(t) is given by:

$H(s)U(s)=\frac{V\big(s+2\alpha\big)}{\big(s+\alpha+j\sqrt{\omega_n^2-\alpha^2}\big)\big(s+\alpha-j\sqrt{\omega_n^2-\alpha^2}\big)}$

Depending on the values of $\alpha$ and $\omega_n$ the system can be characterized as:

3. If $\alpha < \omega_n$ the system is said to be underdamped The solution for h(t)*u(t) is given by:

$h(t)*u(t)=Ve^{-\alpha t}\big(\cos(\sqrt{\omega_n^2-\alpha^2}t)+\frac{\alpha}{\sqrt{\omega_n^2-\alpha^2}}\sin(\sqrt{\omega_n^2-\alpha^2}t)\big)$

## 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:

$h(t)*u(t)=e^{-1000 t}\big(\cos(4359 t)+0.229\sin(4359t)\big)$

Figure 2: Underdamped Resonse