Electronics/RCL frequency domain

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Figure 1: RCL circuit
Figure 1: RCL circuit

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



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}


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:


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)


Given the following values what is the response of the system when the switch is closed?

0.5H 1kΩ 100nF 1V

First calculate the values of \alpha and \omega_n:


\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: Example 1 Underdamped Response
Figure 2: Underdamped Resonse