Electronics/RCL time domain

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} 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

2. If \alpha = \omega_n the system is said to be critically damped

3. If \alpha < \omega_n the system is said to be underdamped


Example:Edit

Given the general solution

R L C V
0.5H 1kΩ 100nF 1V


\alpha=\frac{R}{2L}=1000


\omega_n=\frac{1}{\sqrt{LC}}\approx 4472


s_1=-1000-4359j


s_2=-1000+4359j


v_n(t)=A_1e^{(-1000-4359j)t}+A_2e^{(-1000+4359j)t}

Thus by Euler's formula (e^{j\phi}=\cos{\phi}+j\sin{\phi}):


v_n(t)=e^{-1000}\big[A_1\big(\cos(-4359t)+j\sin(-4359t)\big)+A_2\big(\cos(4359t)+j\sin(4359t)\big)\big]


v_n(t)=e^{-1000t}\big[(A_1+A_2)\cos(4359t)+j(-A_1+A_2)\sin(4359t)\big]

Let B_1=A_1+A_2 and B_2=j(-A_1+A_2)


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

Solve for B_1 and B_2:

From equation \ref{eq:vf}, v_f=1 for a unit step of magnitude 1V. Therefore substitution of v_f and v_n(t) into equation \ref{eq:nonhomogeneous} gives:


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

for t=0 the voltage across the capacitor is zero, v_c(t)=0


0=1+B_1\cos(0)+B_2\sin(0)


B_1=-1\mbox{ (7)}

for t=0, the current in the inductor must be zero, i(0)=0


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


i(0)=100\cdot10^{-9}\big[e^{-1000t}\big(-4359B_1\sin(4359t)+4359B_2\cos(4359t)\big)-1000e^{-1000t}\big(B_1\cos(4359t)+B_2\sin(4359t)\big)\big]


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

substituting B_1 from equation \ref{eq:B1} gives


B_2\approx-0.229

For t>0, v_c(t) is given by:


v_c(t)=1-e^{-1000t}\big[\cos(4359t)+0.229\sin(4359t)\big]

v_{out} is given by:


v_{out}=V_{in}-v_c(t)


v_{out}=Vu(t)-v_c(t)

For t>0, v_{out} is given by:


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

Last modified on 29 November 2011, at 23:03