# 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 ${\displaystyle V\cdot u(t)}$ where V is the magnitude of the step and ${\displaystyle u(t)=1}$ for ${\displaystyle t\geq 0}$ 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:

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

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

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

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

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

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

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

substituting equation 3 into equation 2.

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

when ${\displaystyle t>0s}$ then ${\displaystyle u(t)=1}$:

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

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

The natural response and forced solution are solved separately.

Solve for ${\displaystyle v_{f}(t):}$

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

${\displaystyle v_{f}(t)=K}$

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

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

Substituting into equation 5:

${\displaystyle {\frac {1}{LC}}K={\frac {V}{LC}}}$

${\displaystyle K=V}$

${\displaystyle v_{f}=V{\mbox{ (6)}}}$

Solve for ${\displaystyle v_{n}(t)}$:

Let:

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

${\displaystyle {\frac {1}{LC}}=\omega _{n}^{2}}$

${\displaystyle v_{n}(t)=Ae^{st}}$

Substituting into equation 4 gives:

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

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

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

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

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

where ${\displaystyle s_{1}}$ and ${\displaystyle s_{2}}$ are given by:

${\displaystyle s_{1}=-\alpha +{\sqrt {\alpha ^{2}-\omega _{n}^{2}}}}$

${\displaystyle s_{2}=-\alpha -{\sqrt {\alpha ^{2}-\omega _{n}^{2}}}}$

The general solution is then given by:

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

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

1. If ${\displaystyle \alpha >\omega _{n}}$ the system is said to be overdamped. The system has two distinct real solutions:

${\displaystyle v_{n}(t)=A_{1}e^{{\big (}-\alpha +{\sqrt {\alpha ^{2}-\omega _{n}^{2}}}{\big )}t}+A_{2}e^{{\big (}-\alpha -{\sqrt {\alpha ^{2}-\omega _{n}^{2}}}{\big )}t}}$

2. If ${\displaystyle \alpha =\omega _{n}}$ the system is said to be critically damped. The system has one real solution:

${\displaystyle v_{n}(t)={\big (}A_{1}+A_{2}{\big )}{\big (}e^{-\alpha t}{\big )}}$

Let ${\displaystyle B_{1}=A_{1}+A_{2}}$:

${\displaystyle v_{n}(t)=Be^{-\alpha t}}$

3. If ${\displaystyle \alpha <\omega _{n}}$ the system is said to be underdamped. The system has two complex solutions:

${\displaystyle s_{1}=-\alpha +j{\sqrt {\omega _{n}^{2}-\alpha ^{2}}}}$

${\displaystyle s_{2}=-\alpha -j{\sqrt {\omega _{n}^{2}-\alpha ^{2}}}}$

${\displaystyle v_{n}(t)=e^{-\alpha t}{\Big [}A_{1}e^{j{\sqrt {\omega _{n}^{2}-\alpha ^{2}}}t}+A_{2}e^{-j{\sqrt {\omega _{n}^{2}-\alpha ^{2}}}t}{\Big ]}}$

By Euler's formula (${\displaystyle e^{j\phi }=\cos {\phi }+j\sin {\phi }}$):

${\displaystyle v_{n}(t)=e^{-\alpha t}{\big [}(A_{1}+A_{2})\cos({\sqrt {\omega _{n}^{2}-\alpha ^{2}}}t)+j(-A_{1}+A_{2})\sin({\sqrt {\omega _{n}^{2}-\alpha ^{2}}}t){\big ]}}$

Let ${\displaystyle B=A_{1}+A_{2}}$ and ${\displaystyle B_{2}=j(-A_{1}+A_{2})}$

${\displaystyle v_{n}(t)=e^{-\alpha t}{\big [}B_{1}\cos({\sqrt {\omega _{n}^{2}-\alpha ^{2}}}t)+B_{2}\sin({\sqrt {\omega _{n}^{2}-\alpha ^{2}}}t){\big ]}}$