# 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 ${\displaystyle V\cdot u(t)}$ 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 ${\displaystyle i(t)={\frac {dv_{c}(t)}{dt}}}$ 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

2. If ${\displaystyle \alpha =\omega _{n}}$ the system is said to be critically damped

3. If ${\displaystyle \alpha <\omega _{n}}$ the system is said to be underdamped

## Example:

Given the general solution

 R L C V 0.5H 1kΩ 100nF 1V

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

${\displaystyle \omega _{n}={\frac {1}{\sqrt {LC}}}\approx 4472}$

${\displaystyle s_{1}=-1000-4359j}$

${\displaystyle s_{2}=-1000+4359j}$

${\displaystyle v_{n}(t)=A_{1}e^{(-1000-4359j)t}+A_{2}e^{(-1000+4359j)t}}$

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

${\displaystyle v_{n}(t)=e^{-1000}{\big [}A_{1}{\big (}\cos(-4359t)+j\sin(-4359t){\big )}+A_{2}{\big (}\cos(4359t)+j\sin(4359t){\big )}{\big ]}}$

${\displaystyle v_{n}(t)=e^{-1000t}{\big [}(A_{1}+A_{2})\cos(4359t)+j(-A_{1}+A_{2})\sin(4359t){\big ]}}$

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

${\displaystyle v_{n}(t)=e^{-1000t}{\big [}B_{1}\cos(4359t)+B_{2}\sin(4359t){\big ]}}$

Solve for ${\displaystyle B_{1}}$ and ${\displaystyle B_{2}}$:

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

${\displaystyle v_{c}(t)=1+e^{-1000t}{\big [}B_{1}\cos(4359t)+B_{2}\sin(4359t){\big ]}}$

for ${\displaystyle t=0}$ the voltage across the capacitor is zero, ${\displaystyle v_{c}(t)=0}$

${\displaystyle 0=1+B_{1}\cos(0)+B_{2}\sin(0)}$

${\displaystyle B_{1}=-1{\mbox{ (7)}}}$

for ${\displaystyle t=0}$, the current in the inductor must be zero, ${\displaystyle i(0)=0}$

${\displaystyle i(t)={\frac {dv_{c}(t)}{dt}}C}$

${\displaystyle i(0)=100\cdot 10^{-9}{\big [}e^{-1000t}{\big (}-4359B_{1}\sin(4359t)+4359B_{2}\cos(4359t){\big )}-1000e^{-1000t}{\big (}B_{1}\cos(4359t)+B_{2}\sin(4359t){\big )}{\big ]}}$

${\displaystyle 0=100\cdot 10^{-9}{\big [}4359B_{2}-1000B_{1}{\big ]}}$

substituting ${\displaystyle B_{1}}$ from equation \ref{eq:B1} gives

${\displaystyle B_{2}\approx -0.229}$

For ${\displaystyle t>0}$, ${\displaystyle v_{c}(t)}$ is given by:

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

${\displaystyle v_{out}}$ is given by:

${\displaystyle v_{out}=V_{in}-v_{c}(t)}$

${\displaystyle v_{out}=Vu(t)-v_{c}(t)}$

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

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