Question 4.4.8: The differential equation for the current i(t) in a single-l......

Using the result in (15)

of the preceding example, the Laplace transform of the \mathrm{DE} is

L s I(s)+R I(s)=\frac{1}{s\left(1+e^{-s}\right)} \quad \text { or } \quad I(s)=\frac{1 / L}{s(s+R / L)} \cdot \frac{1}{1+e^{-s}}. (17)

To find the inverse Laplace transform of the last function, we first make use of geometric series. With the identification x=e^{-s}, s>0, the geomeric series

\frac{1}{1+x}=1-x+x^{2}-x^{3}+\cdots \quad becomes \quad \frac{1}{1+e^{-s}}=1-e^{-s}+e^{-2 s}-e^{-3 s}+\cdots.

From     \frac{1}{s(s+R / L)}=\frac{L / R}{s}-\frac{L / R}{s+R / L}

we can then rewrite (17) as

By applying the form of the second translation theorem to each term of both series we obtain

or, equivalently,

i(t)=\frac{1}{R}\left(1-e^{-R t / L}\right)+\frac{1}{R} \sum_{n=1}^{\infty}(-1)^{n}\left(1-e^{-R(t-n) L}\right) \mathscr{U}(t-n) .

To interpret the solution, let us assume for the sake of illustration that R=1, L=1, and 0 \leq t<4. In this case

i(t)=1-e^{-t}-\left(1-e^{t-1}\right) \mathscr{U}(t-1)+\left(1-e^{-(t-2)}\right) \mathscr{U}(t-2)-\left(1-e^{-(t-3)}\right) \mathscr{U}(t-3) ;

The graph of i(t) for 0 \leq t<4, given in FIGURE 4.4.5, was obtained with the help of a CAS.