*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, 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:

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

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

The voltage $v_{c}(t)$ has two components, a natural 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.

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

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:

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

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 |

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

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

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

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

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

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, $i(0)=0$

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

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

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

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:

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