Electronics/RCL time domain1

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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. 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]