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?

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