Inverse Laplace of 1st Order RC Parallel Circuit Differential Equation

Question

The differential equation for the first order parallel RC circuit is:

$$ v' + \frac{v}{RC} = \frac{i}{C} $$

The laplace transform is then:

$$ V(s) = \frac{I(s)}{sC + \frac{1}{R}} $$

where $ I(s) = \frac{I_ow}{s^2+w^2} $ and $I_o$, $w$, $R$ and $C$ are constants.

What is the inverse laplace of $V(s)$ ?

Answer 1

Well, in a general can a parallel RC-circuit be described by the following equation:

$$ \begin{cases} \text{I}_{\space\text{in}}\left(t\right)=\text{I}_{\space\text{R}}\left(t\right)+\text{I}_{\space\text{C}}\left(t\right)\\ \\ \text{V}_{\space\text{R}}\left(t\right)=\text{I}_{\space\text{R}}\left(t\right)\cdot\text{R}\\ \\ \text{I}_{\space\text{C}}\left(t\right)=\text{V}_{\space\text{C}}'\left(t\right)\cdot\text{C}\\ \\ \text{V}_{\space\text{in}}\left(t\right)=\text{V}_{\space\text{R}}\left(t\right)=\text{V}_{\space\text{C}}\left(t\right) \end{cases}\tag1 $$

So, we can write:

$$\text{I}_{\space\text{in}}\left(t\right)=\text{V}_{\space\text{in}}\left(t\right)\cdot\frac{1}{\text{R}}+\text{V}_{\space\text{in}}'\left(t\right)\cdot\text{C}\tag2$$

Using Laplace transform, we can write:

$$\text{i}_{\space\text{in}}\left(\text{s}\right)=\text{v}_{\space\text{in}}\left(\text{s}\right)\cdot\frac{1}{\text{R}}+\left(\text{s}\cdot\text{v}_{\space\text{in}}\left(\text{s}\right)-\text{V}_{\space\text{in}}\left(0\right)\right)\cdot\text{C}\space\Longleftrightarrow\space$$ $$\text{v}_{\space\text{in}}\left(\text{s}\right)=\frac{\text{i}_{\space\text{in}}\left(\text{s}\right)+\text{V}_{\space\text{in}}\left(0\right)\cdot\text{C}}{\frac{1}{\text{R}}+\text{s}\cdot\text{C}}=\frac{\text{i}_{\space\text{in}}\left(\text{s}\right)}{\frac{1}{\text{R}}+\text{s}\cdot\text{C}}+\frac{\text{V}_{\space\text{in}}\left(0\right)\cdot\text{C}}{\frac{1}{\text{R}}+\text{s}\cdot\text{C}}\tag3$$

Now, applying the inverse Laplace transform and the convolution theorem:

$$\text{V}_{\space\text{in}}\left(t\right)=\frac{1}{\text{C}}\cdot\int_0^t\text{I}_{\space\text{in}}\left(t\right)\cdot\exp\left(\frac{\tau-t}{\text{R}\cdot\text{C}}\right)\space\text{d}\tau+\text{V}_{\space\text{in}}\left(0\right)\cdot\exp\left(-\frac{1}{\text{R}\cdot\text{C}}\cdot t\right)\tag4$$