Electronics/RCL frequency domain

Figure 1: RCL circuit

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

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

${\displaystyle {\frac {1}{LC}}=\omega _{n}^{2}}$

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

${\displaystyle H(s)={\frac {s{\big (}s+2\alpha {\big )}}{s^{2}+2\alpha s+\omega _{n}^{2}}}}$

${\displaystyle H(s)={\frac {s{\big (}s+2\alpha {\big )}}{{\big (}s+\alpha +j{\sqrt {\omega _{n}^{2}-\alpha ^{2}}}{\big )}{\big (}s+\alpha -j{\sqrt {\omega _{n}^{2}-\alpha ^{2}}}{\big )}}}}$

When the switch is closed, this applies a step waveform to the RCL circuit. The step is given by ${\displaystyle Vu(t)}$. 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 ${\displaystyle Vu(t)}$. Therefore the output is given by multiplication in the s-domain H(s)U(s), where ${\displaystyle U(s)=V{\frac {1}{s}}}$ is given by the Laplace Transform available in the appendix.

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

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

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

3. If ${\displaystyle \alpha <\omega _{n}}$ the system is said to be underdamped The solution for h(t)*u(t) is given by:

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

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 ${\displaystyle \alpha }$  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:

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

Figure 2: Underdamped Resonse