Electronics/RCL time domain1

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. This is a step function given by $$V\cdot u(t)$$ where V is the magnitude of the step and $$u(t)=1$$ for $$t\geq0$$ and zero otherwise.

To analyse the circuit response using transient analysis, a differential equation which describes the system is formulated. The voltage around the loop is given by:

$$ Vu(t)=v_c(t)+\frac{di(t)}{dt}L+Ri(t) \mbox{ (1)} $$

where $$v_c(t)$$ is the voltage across the capacitor, $$\frac{di(t)}{dt}L$$ is the voltage across the inductor and $$Ri(t)$$ the voltage across the resistor.

Substituting $$i(t)=\frac{dv_c(t)}{dt}C$$ into equation 1:

$$ Vu(t)=v_c(t)+\frac{d^2v_c(t)}{dt^2}LC+R\frac{dv_c(t)}{dt}C $$ $$ \frac{d^2v_c(t)}{dt^2}+\frac{R}{L}\frac{dv_c(t)}{dt}+\frac{1}{LC}v_c(t)=\frac{Vu(t)}{LC} \mbox{ (2)} $$

The voltage $$v_c(t)$$ has two components, a natural response $$v_n(t)$$ and a forced response $$v_f(t)$$ such that:

$$ v_c(t)=v_f(t)+v_n(t)\mbox{ (3)} $$

substituting equation 3 into equation 2.

$$ \bigg[\frac{d^2v_n(t)}{dt^2}+\frac{R}{L}\frac{dv_n(t)}{dt}+\frac{1}{LC}v_n(t)\bigg]+\bigg[\frac{d^2v_f(t)}{dt^2}+\frac{R}{L}\frac{dv_f(t)}{dt}+\frac{1}{LC}v_f(t)\bigg]=0+\frac{Vu(t)}{LC} $$

when $$t>0s$$ then $$u(t)=1$$:

$$ \bigg[\frac{d^2v_n(t)}{dt^2}+\frac{R}{L}\frac{dv_n(t)}{dt}+\frac{1}{LC}v_n(t)\bigg]=0 \mbox{ (4)} $$ $$ \bigg[\frac{d^2v_f(t)}{dt^2}+\frac{R}{L}\frac{dv_f(t)}{dt}+\frac{1}{LC}v_f(t)\bigg]=\frac{V}{LC} \mbox{ (5)} $$

The natural response and forced solution are solved separately.

Solve for $$v_f(t):$$

Since $$\frac{V}{LC}$$ is a polynomial of degree 0, the solution $$v_f(t)$$ must be a constant such that:

$$ v_f(t)=K $$ $$ \frac{dv_f(t)}{dt}=0 $$ $$ \frac{d^2v_f(t)}{dt}=0 $$

Substituting into equation 5:

$$ \frac{1}{LC}K=\frac{V}{LC} $$ $$ K=V $$ $$ v_f=V \mbox{ (6)} $$

Solve for $$v_n(t)$$:

Let: $$ \frac{R}{L}=2\alpha $$ $$ \frac{1}{LC}=\omega_n^2 $$ $$ v_n(t)=Ae^{st} $$

Substituting into equation 4 gives:

$$ \frac{d^2Ae^{st}}{dt^2}+2\alpha\frac{dAe^{st}}{dt}+\omega_n^2Ae^{st}=0 $$ $$ s^2Ae^{st}+2\alpha Ae^{st}+\omega_2^2Ae^{st}=0 $$ $$ s^2+2\alpha s+\omega_n^2=0 $$

$$ s=\frac{-2\alpha\pm\sqrt{4\alpha^2-4\omega_n^2}}{2}=-\alpha\pm\sqrt{\alpha^2-\omega_n^2} $$

Therefore $$v_n(t)$$ has two solutions $$Ae^{s_1t}$$ and $$Ae^{s_2t}$$

where $$s_1$$ and $$s_2$$ are given by:

$$ s_1=-\alpha+\sqrt{\alpha^2-\omega_n^2} $$ $$ s_2=-\alpha-\sqrt{\alpha^2-\omega_n^2} $$

The general solution is then given by:

$$ v_n(t)=A_1e^{s_1t}+A_2e^{s_2t} $$

Depending on the values of the Resistor, inductor or capacitor the solution has three posibilies.

1. If $$\alpha > \omega_n$$ the system is said to be overdamped. The system has two distinct real solutions: $$ v_n(t)=A_1e^{\big(-\alpha+\sqrt{\alpha^2-\omega_n^2}\big)t}+A_2e^{\big(-\alpha-\sqrt{\alpha^2-\omega_n^2}\big)t} $$ 2. If $$\alpha = \omega_n$$ the system is said to be critically damped. The system has one real solution: $$ v_n(t)=\big(A_1+A_2\big)\big(e^{-\alpha t}\big) $$
 * Let $$B_1=A_1+A_2$$:

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

3. If $$\alpha < \omega_n$$ the system is said to be underdamped. The system has two complex solutions: $$ s_1=-\alpha+j\sqrt{\omega_n^2-\alpha^2} $$ $$ s_2=-\alpha-j\sqrt{\omega_n^2-\alpha^2} $$

$$ v_n(t)=e^{-\alpha t}\Big[A_1e^{j\sqrt{\omega_n^2-\alpha^2}t}+A_2e^{-j\sqrt{\omega_n^2-\alpha^2}t}\Big] $$
 * By Euler's formula ($$e^{j\phi}=\cos{\phi}+j\sin{\phi}$$):

$$ v_n(t)=e^{-\alpha t}\big[(A_1+A_2)\cos(\sqrt{\omega_n^2-\alpha^2} t)+j(-A_1+A_2)\sin(\sqrt{\omega_n^2-\alpha^2} t)\big] $$


 * Let $$B=A_1+A_2$$ and $$B_2=j(-A_1+A_2)$$

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