# Circuit current from series resistor and inductor + pulse voltage using Laplace method

A series $RL$ circuit experiences a pulse of voltage, $V$, occurring during the interval $t_0 \lt t \lt t_1$. Determine the circuit current $i(t)$.

First I've broken up the pulse into $\pm V u(t - t_j), \ j=0,1$ and will add the two solutions $i_j(t), \ j=0,1$ when done, via superposition.

So the differential equation modeling the system is:

$\pm V u(t - t_j) = R i_j(t) + L \dfrac{d i_j(t)}{dt}$.

Taking the Laplace transform and rearranging I get:

$I_j(s) = \dfrac{e^{-t_j s}}{s}\dfrac{ \pm \dfrac{V}{L}}{\dfrac{R}{L} + s}$

and taking the inverse transform I get:

$\pm \dfrac{V}{L} \int\limits_0^t u(\tau - t_j) e^{-\frac{R}{L}(t - \tau)} d\tau$.

My question is how do I evaluate this integral, or am I doing it wrong?

Hint: For $\tau < t_j$,$$u(\tau-t_j) = 0,$$ and $$u(\tau-t_j) = 1$$ otherwise. Consequently, $$\int_{0} ^{t} u(\tau-t_j) f(\tau)d\tau = 0,$$ if $t \le t_j$, and $$\int_{0} ^{t} u(\tau-t_j) f(\tau)d\tau = \int_{t_j} ^{t} f(\tau)d\tau$$ for $t > t_j$.

• That's brilliant! So to clarify, I just want to compute $\int_{t_j}^t f(\tau) d\tau$? – Shine On You Crazy Diamond Feb 21 '18 at 18:22
• Oh I see, I write the answer as a multiple of $u(t)$. Cool – Shine On You Crazy Diamond Feb 21 '18 at 18:24

Notice that in general:

$$\text{V}_{\space\text{in}}\left(t\right)=\text{V}_{\space\text{R}}\left(t\right)+\text{V}_{\space\text{C}}\left(t\right)\space\implies\space\text{V}_{\space\text{in}}'\left(t\right)=\text{I}_{\space\text{in}}'\left(t\right)\cdot\text{R}+\text{I}_{\space\text{in}}\left(t\right)\cdot\frac{1}{\text{C}}\tag1$$

Let:

$$\mu\left(t\right)=\exp\left\{\int\frac{1}{\text{C}\cdot\text{R}}\space\text{d}t\right\}=\frac{t}{\text{C}\cdot\text{R}}\tag2$$

Apply the reverse product rule:

$$\int\frac{\partial}{\partial t}\left(\text{I}_{\space\text{in}}\left(t\right)\cdot\exp\left(\frac{t}{\text{C}\cdot\text{R}}\right)\right)\space\text{d}t=\int\frac{1}{\text{R}}\cdot\text{V}_{\space\text{in}}'\left(t\right)\cdot\exp\left(\frac{t}{\text{C}\cdot\text{R}}\right)\space\text{d}t\tag3$$

Simplyfing a bit:

$$\text{I}_{\space\text{in}}\left(t\right)\cdot\exp\left(\frac{t}{\text{C}\cdot\text{R}}\right)=\text{k}_1+\frac{1}{\text{R}}\cdot\int\text{V}_{\space\text{in}}'\left(t\right)\cdot\exp\left(\frac{t}{\text{C}\cdot\text{R}}\right)\space\text{d}t\space\Longleftrightarrow\space$$ $$\text{I}_{\space\text{in}}\left(t\right)=\exp\left(-\frac{t}{\text{C}\cdot\text{R}}\right)\cdot\left\{\text{k}_1+\frac{1}{\text{R}}\cdot\int\text{V}_{\space\text{in}}'\left(t\right)\cdot\exp\left(\frac{t}{\text{C}\cdot\text{R}}\right)\space\text{d}t\right\}\tag4$$