Electronics/RCL time domain

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 $t\geq 0$ and zero otherwise.

To analyse the circuit response using transient analysis, an 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:

$Vu(t)=v_{c}(t)+{\frac {d^{2}v_{c}(t)}{dt^{2}}}LC+R{\frac {dv_{c}(t)}{dt}}C$

${\frac {d^{2}v_{c}(t)}{dt^{2}}}+{\frac {R}{L}}{\frac {dv_{c}(t)}{dt}}+{\frac {1}{LC}}v_{c}(t)={\frac {Vu(t)}{LC}}{\mbox{ (2)}}$

The voltage $v_{c}(t)$ has two components, a natural responce $v_{n}(t)$ and a forced reponse $v_{f}(t)$ such that:

$v_{c}(t)=v_{f}(t)+v_{n}(t){\mbox{ (3)}}$

substituting equation 3 into equation 2.

${\bigg [}{\frac {d^{2}v_{n}(t)}{dt^{2}}}+{\frac {R}{L}}{\frac {dv_{n}(t)}{dt}}+{\frac {1}{LC}}v_{n}(t){\bigg ]}+{\bigg [}{\frac {d^{2}v_{f}(t)}{dt^{2}}}+{\frac {R}{L}}{\frac {dv_{f}(t)}{dt}}+{\frac {1}{LC}}v_{f}(t){\bigg ]}=0+{\frac {Vu(t)}{LC}}$

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

${\bigg [}{\frac {d^{2}v_{n}(t)}{dt^{2}}}+{\frac {R}{L}}{\frac {dv_{n}(t)}{dt}}+{\frac {1}{LC}}v_{n}(t){\bigg ]}=0{\mbox{ (4)}}$

${\bigg [}{\frac {d^{2}v_{f}(t)}{dt^{2}}}+{\frac {R}{L}}{\frac {dv_{f}(t)}{dt}}+{\frac {1}{LC}}v_{f}(t){\bigg ]}={\frac {V}{LC}}{\mbox{ (5)}}$

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:

$v_{f}(t)=K$

${\frac {dv_{f}(t)}{dt}}=0$

${\frac {d^{2}v_{f}(t)}{dt}}=0$

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:

${\frac {d^{2}Ae^{st}}{dt^{2}}}+2\alpha {\frac {dAe^{st}}{dt}}+\omega _{n}^{2}Ae^{st}=0$

$s^{2}Ae^{st}+2\alpha Ae^{st}+\omega _{2}^{2}Ae^{st}=0$

$s^{2}+2\alpha s+\omega _{n}^{2}=0$

$s={\frac {-2\alpha \pm {\sqrt {4\alpha ^{2}-4\omega _{n}^{2}}}}{2}}=-\alpha \pm {\sqrt {\alpha ^{2}-\omega _{n}^{2}}}$

Therefore $v_{n}(t)$ has two solutions $Ae^{s_{1}t}$ and $Ae^{s_{2}t}$

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:

$v_{n}(t)=A_{1}e^{s_{1}t}+A_{2}e^{s_{2}t}$