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:

$H(s)=\frac{s(s+2\alpha )}{s^2+2\alpha s+{\omega }_n^2}$

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

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:

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

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:

$h(t)*u(t)=V{e}^{-\alpha t}(\cos (\sqrt{{\omega }_n^2-{\alpha }^2}t)+\frac{\alpha }{\sqrt{{\omega }_n^2-{\alpha }^2}}\sin (\sqrt{{\omega }_n^2-{\alpha }^2}t))$

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:

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

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