Electronics/RCL time domain simple

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Figure 1: RCL circuit
Figure 1: RCL circuit

When the switch is closed, a voltage step is applied to the RCL circuit. Take the time the switch was closed to be 0s such that the voltage before the switch was closed was 0 volts and the voltage after the switch was closed is a voltage V. The voltage across the capacitor consists of a forced response v_f and a natural response v_n such that:


v_c=v_f+v_n

The forced response is due to the switch being closed, which is the voltage V for t\ge0. The natural response depends on the circuit values and is given below:

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


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


\omega_n=\frac{1}{\sqrt{LC}}

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

1. If \alpha > \omega_n the system is said to be overdamped. The solution for the system has the form:


v_n(t)=A_1e^{\big(-\alpha+\sqrt{\alpha^2-\omega_n^2}\big)t}+A_2e^{\big(-\alpha-\sqrt{\alpha^2-\omega_n^2}\big)t}

2. If \alpha = \omega_n the system is said to be critically damped The solution for the system has the form:


v_n(t)=Be^{-\alpha t}

3. If \alpha < \omega_n the system is said to be underdamped The solution for the system has the form:


v_n(t)=e^{-\alpha t}\big[B_1\cos(\sqrt{\omega_n^2-\alpha^2} t)+B_2\sin(\sqrt{\omega_n^2-\alpha^2} t)\big]

How do you calculate these equations?


Example:[edit]

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

R L C V
1kΩ 0.5H 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:


v_c(t)=1+e^{-1000t}\big[B_1\cos(4359t)+B_2\sin(4359t)\big]

Before the switch was closed assume that the capacitor was fully discharged. This implies that v(t)=0 at the instant the switch was closed (t=0). Substituting t=0 into the previous equation gives:


0=1+B_1

Therefore B_1=-1. Similarly at the instant the switch is closed, the current in the inductor must be zero as the current can not instantly change. Substituting the equation for v_c(t) into the equation for the inductor and solving at the instant the switch was closed (t=0) gives:


i(t)=\frac{dv_c(t)}{dt}C

0=100\cdot10^{-9}\big[4359B_2-1000B_1 \big]

Therefore B_2\approx-0.229. Once v_c(t) is known, the voltage across the inductor and resistor (V_{out}) is given by:


v_{out}=e^{-1000t}\big[\cos(4359t)+0.229\sin(4359t)\big]

You have missed a lot of steps, where are they?

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

Image:Example1 underdamped.png