Find the inverse Laplace transform of 0.4 times e to the minus s, plus 0.2 times e to the minus 2s, plus 0.4 times e to the minus 3s.

Step-by-step solution

1. 1

   Check the sign of the exponent first. Laplace-domain functions carry $e^{-as}$ with $a > 0$; a factor $e^{+as}$ describes a shift backwards in time and is not the transform of any causal function. So an expression written with $e^{s}$, $e^{2s}$, $e^{3s}$ has to be read as $e^{-s}$, $e^{-2s}$, $e^{-3s}$ before it can be inverted at all.

2. 2

   Note that there is no F(s) attached to the exponentials. The second shifting theorem says

   $\mathcal{L}^{-1}\left\{e^{-as}F(s)\right\} = u(t-a)\,f(t-a)$

   Here each exponential stands alone, i.e. $F(s) = 1$. The inverse transform of $1$ is not an ordinary function — it is the Dirac delta $\delta(t)$, since $\mathcal{L}\{\delta(t)\} = 1$.

3. 3

   Apply the delta shift rule to each term. Combining the two facts:

   $\mathcal{L}^{-1}\left\{e^{-as}\right\} = \delta(t - a)$

   so the three terms invert to impulses at $t = 1$, $t = 2$ and $t = 3$ respectively.

4. 4

   Use linearity to assemble the answer. The inverse transform is linear, so the coefficients pass straight through:

   $\mathcal{L}^{-1}\{F(s)\} = \frac{2}{5}\delta(t-1) + \frac{1}{5}\delta(t-2) + \frac{2}{5}\delta(t-3)$

   This is a train of three impulses of strengths $0.4$, $0.2$ and $0.4$.

5. 5

   Verify by transforming back. Using $\int_0^\infty \delta(t-a)e^{-st}\,dt = e^{-as}$ on each impulse recovers $\tfrac25 e^{-s} + \tfrac15 e^{-2s} + \tfrac25 e^{-3s}$ exactly. Note the coefficients sum to $\tfrac25 + \tfrac15 + \tfrac25 = 1$, so the total impulse strength is $1$ — the same total area as a single $\delta(t)$, just split across three instants.