Electronics/RCL time domain

Figure 1: RCL circuit

Template:- 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:

$V u (t) = v_{c} (t) + \frac{d i (t)}{d t} L + R i (t) (1)$

where $v_{c} (t)$ is the voltage across the capacitor, $\frac{d i (t)}{d t} L$ is the voltage across the inductor and $R i (t)$ the voltage across the resistor.

Substituting $i (t) = \frac{d v_{c} (t)}{d t}$ into equation 1:

$V u (t) = v_{c} (t) + \frac{d^{2} v_{c} (t)}{d t^{2}} L C + R \frac{d v_{c} (t)}{d t} C$

$\frac{d^{2} v_{c} (t)}{d t^{2}} + \frac{R}{L} \frac{d v_{c} (t)}{d t} + \frac{1}{L C} v_{c} (t) = \frac{V u (t)}{L C} (2)$

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

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

substituting equation 3 into equation 2.

$[\frac{d^{2} v_{n} (t)}{d t^{2}} + \frac{R}{L} \frac{d v_{n} (t)}{d t} + \frac{1}{L C} v_{n} (t)] + [\frac{d^{2} v_{f} (t)}{d t^{2}} + \frac{R}{L} \frac{d v_{f} (t)}{d t} + \frac{1}{L C} v_{f} (t)] = 0 + \frac{V u (t)}{L C}$

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

$[\frac{d^{2} v_{n} (t)}{d t^{2}} + \frac{R}{L} \frac{d v_{n} (t)}{d t} + \frac{1}{L C} v_{n} (t)] = 0 (4)$

$[\frac{d^{2} v_{f} (t)}{d t^{2}} + \frac{R}{L} \frac{d v_{f} (t)}{d t} + \frac{1}{L C} v_{f} (t)] = \frac{V}{L C} (5)$

The natural response and forced solution are solved separately.

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

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

$v_{f} (t) = K$

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

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

Substituting into equation 5:

$\frac{1}{L C} K = \frac{V}{L C}$

$K = V$

$v_{f} = V (6)$

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

Let:

$\frac{R}{L} = 2 α$

$\frac{1}{L C} = ω_{n}^{2}$

$v_{n} (t) = A e^{s t}$

Substituting into equation 4 gives:

$\frac{d^{2} A e^{s t}}{d t^{2}} + 2 α \frac{d A e^{s t}}{d t} + ω_{n}^{2} A e^{s t} = 0$

$s^{2} A e^{s t} + 2 α A e^{s t} + ω_{2}^{2} A e^{s t} = 0$

$s^{2} + 2 α s + ω_{n}^{2} = 0$

$s = \frac{- 2 α \pm \sqrt{4 α^{2} - 4 ω_{n}^{2}}}{2} = - α \pm \sqrt{α^{2} - ω_{n}^{2}}$

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

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

$s_{1} = - α + \sqrt{α^{2} - ω_{n}^{2}}$

$s_{2} = - α - \sqrt{α^{2} - ω_{n}^{2}}$

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 $α > ω_{n}$ the system is said to be overdamped

2. If $α = ω_{n}$ the system is said to be critically damped

3. If $α < ω_{n}$ the system is said to be underdamped

Example

Given the general solution

R	L	C	V
0.5H	1kΩ	100nF	1V

$α = \frac{R}{2 L} = 1000$

$ω_{n} = \frac{1}{\sqrt{L C}} \approx 4472$

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

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

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

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

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

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

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

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

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:

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

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

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

$B_{1} = - 1 (7)$

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

$i (t) = \frac{d v_{c} (t)}{d t} C$

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

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

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

$B_{2} \approx - 0.229$

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

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

$v_{o u t}$ is given by:

$v_{o u t} = V_{i n} - v_{c} (t)$

$v_{o u t} = V u (t) - v_{c} (t)$

For $t > 0$ , $v_{o u t}$ is given by:

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

Template:BookCat

Electronics/RCL time domain

Example

Navigation menu

Search