# Examples of The Laplace Transform

Example 1: Obtain the Laplace transform of $f (t) = δ(t) + 2u(t) − 3e^{−2t}$, $t ≥ 0$
Solution: By the linearity property,
\begin{aligned}F(s) &=\mathcal{L}[\delta(t)]+2 \mathcal{L}[u(t)]-3 \mathcal{L}\left[e^{-2 t}\right] \\&=1+2 \frac{1}{s}-3 \frac{1}{s+2}=\frac{s^{2}+s+4}{s(s+2)}\end{aligned}
Example 2: Determine the Laplace transform of $f(t)=t^{2} \sin 2 t u(t)$.
Solution: We know that
$$\mathcal{L}[\sin 2 t]=\frac{2}{s^{2}+2^{2}}$$
Using frequency differentiation,
\begin{aligned}F(s)=\mathcal{L}\left[t^{2} \sin 2 t\right] &=(-1)^{2} \frac{d^{2}}{d s^{2}}\left(\frac{2}{s^{2}+4}\right) \\&=\frac{d}{d s}\left(\frac{-4 s}{\left(s^{2}+4\right)^{2}}\right)=\frac{12 s^{2}-16}{\left(s^{2}+4\right)^{3}}\end{aligned}
Example 3: Find the Laplace transform of the gate function in Fig. 1.
Fig. 1: The gate function
Solution: We can express the gate function in Fig. $1$ as
$$g(t)=10[u(t-2)-u(t-3)]$$
Since we know the Laplace transform of $u(t)$, we apply the time-shift property and obtain
$$G(s)=10\left(\frac{e^{-2 s}}{s}-\frac{e^{-3 s}}{s}\right)=\frac{10}{s}\left(e^{-2 s}-e^{-3 s}\right)$$
Example 4: Calculate the Laplace transform of the periodic function in Fig. 2.
Fig. 2: The gate function
Solution: The period of the function is $T=2$.
$$F(s)=\frac{F_{1}(s)}{1-e^{-T s}} \tag{1}$$
To apply Eq. (1), we first obtain the transform of the first period of the function.
\begin{aligned}f_{1}(t) &=2 t[u(t)-u(t-1)]=2 t u(t)-2 t u(t-1) \\&=2 t u(t)-2(t-1+1) u(t-1) \\&=2 t u(t)-2(t-1) u(t-1)-2 u(t-1)\end{aligned}
Using the time-shift property,
$$F_{1}(s)=\frac{2}{s^{2}}-2 \frac{e^{-s}}{s^{2}}-\frac{2}{s} e^{-s}=\frac{2}{s^{2}}\left(1-e^{-s}-s e^{-s}\right)$$
Thus, the transform of the periodic function in Fig. $1$ is
$$F(s)=\frac{F_{1}(s)}{1-e^{-T s}}=\frac{2}{s^{2}\left(1-e^{-2 s}\right)}\left(1-e^{-s}-s e^{-s}\right)$$
Example 5: Find the initial and final values of the function whose Laplace transform is
$$H(s)=\frac{20}{(s+3)\left(s^{2}+8 s+25\right)}$$
Solution: Applying the initial-value theorem,
\begin{aligned}h(0) &=\lim _{s \rightarrow \infty} s H(s)=\lim _{s \rightarrow \infty} \frac{20 s}{(s+3)\left(s^{2}+8 s+25\right)} \\&=\lim _{s \rightarrow \infty} \frac{20 / s^{2}}{(1+3 / s)\left(1+8 / s+25 / s^{2}\right)}=\frac{0}{(1+0)(1+0+0)}=0\end{aligned}
To be sure that the final-value theorem is applicable, we check where the poles of $H(s)$ are located. The poles of $H(s)$ are $s=-3,-4 \pm j 3$, which all have negative real parts: they are all located on the left half of the $s$ plane . Hence the final-value theorem applies and
\begin{aligned}h(\infty) &=\lim _{s \rightarrow 0} s H(s)=\lim _{s \rightarrow 0} \frac{20 s}{(s+3)\left(s^{2}+8 s+25\right)} \\&=\frac{0}{(0+3)(0+0+25)}=0\end{aligned}
Both the initial and final values could be determined from $h(t)$ if we knew it.