# Examples of The Inverse Laplace Transform

Example 1: Find the inverse Laplace transform of$$F(s)=\frac{3}{s}-\frac{5}{s+1}+\frac{6}{s^{2}+4}$$ Solution: The inverse transform is given by
\begin{aligned}f(t)=\mathcal{L}^{-1}[F(s)] &=\mathcal{L}^{-1}\left(\frac{3}{s}\right)-\mathcal{L}^{-1}\left(\frac{5}{s+1}\right)+\mathcal{L}^{-1}\left(\frac{6}{s^{2}+4}\right) \\&=3 u(t)-5 e^{-t}+3 \sin 2 t, \quad t \geq 0\end{aligned}
where Table $2$ has been consulted for the inverse of each term.
Example 2: Find $f(t)$ given that
$$F(s)=\frac{s^{2}+12}{s(s+2)(s+3)}$$
Solution: Unlike in the previous example where the partial fractions have been provided, we first need to determine the partial fractions. Since there are three poles, we let
$$\frac{s^{2}+12}{s(s+2)(s+3)}=\frac{A}{s}+\frac{B}{s+2}+\frac{C}{s+3} \tag{1}$$
where $A, B$, and $C$ are the constants to be determined. We can find the constants using two approaches.
METHOD I Residue method:
$$\begin{array}{c}A=\left.s F(s)\right|_{s=0}=\left.\frac{s^{2}+12}{(s+2)(s+3)}\right|_{s=0}=\frac{12}{(2)(3)}=2 \\B=\left.(s+2) F(s)\right|_{s=-2}=\left.\frac{s^{2}+12}{s(s+3)}\right|_{s=-2}=\frac{4+12}{(-2)(1)}=-8 \\C=\left.(s+3) F(s)\right|_{s=-3}=\left.\frac{s^{2}+12}{s(s+2)}\right|_{s=-3}=\frac{9+12}{(-3)(-1)}=7\end{array}$$
METHOD 2 Algebraic method: Multiplying both sides of Eq. (1) by $s(s+2)(s+3)$ gives
$$s^{2}+12=A(s+2)(s+3)+B s(s+3)+C s(s+2)$$
or
$$s^{2}+12=A\left(s^{2}+5 s+6\right)+B\left(s^{2}+3 s\right)+C\left(s^{2}+2 s\right)$$
Equating the coefficients of like powers of $s$ gives
$$\begin{array}{lll}\text { Constant: } & 12=6 A \quad \Longrightarrow \quad A=2 \\s: & 0=5 A+3 B+2 C \quad \Longrightarrow \quad 3 B+2 C=-10 \\s^{2}: & 1=A+B+C \quad \Longrightarrow \quad B+C=-1\end{array}$$
Thus $A=2, B=-8, C=7$, and Eq. (1) becomes
$$F(s)=\frac{2}{s}-\frac{8}{s+2}+\frac{7}{s+3}$$
By finding the inverse transform of each term, we obtain
$$f(t)=2 u(t)-8 e^{-2 t}+7 e^{-3 t}, \quad t \geq 0 .$$
Example 3: Calculate $v(t)$ given that
$$V(s)=\frac{10 s^{2}+4}{s(s+1)(s+2)^{2}}$$
Solution: While the previous example is on simple roots, this example is on repeated roots. Let
\begin{aligned}V(s) &=\frac{10 s^{2}+4}{s(s+1)(s+2)^{2}} \\&=\frac{A}{s}+\frac{B}{s+1}+\frac{C}{(s+2)^{2}}+\frac{D}{s+2}\end{aligned} \tag{2}
METHOD I Residue method:
$$\begin{array}{c}A=\left.s V(s)\right|_{s=0}=\left.\frac{10 s^{2}+4}{(s+1)(s+2)^{2}}\right|_{s=0}=\frac{4}{(1)(2)^{2}}=1 \\B=\left.(s+1) V(s)\right|_{s=-1}=\left.\frac{10 s^{2}+4}{s(s+2)^{2}}\right|_{s=-1}=\frac{14}{(-1)(1)^{2}}=-14 \\C=\left.(s+2)^{2} V(s)\right|_{s=-2}=\left.\frac{10 s^{2}+4}{s(s+1)}\right|_{s=-2}=\frac{44}{(-2)(-1)}=22 \\D=\left.\frac{d}{d s}\left[(s+2)^{2} V(s)\right]\right|_{s=-2}=\left.\frac{d}{d s}\left(\frac{10 s^{2}+4}{s^{2}+s}\right)\right|_{s=-2} \\=\left.\frac{\left(s^{2}+s\right)(20 s)-\left(10 s^{2}+4\right)(2 s+1)}{\left(s^{2}+s\right)^{2}}\right|_{s=-2}=\frac{52}{4}=13\end{array}$$
METHOD 2 Algebraic method: Multiplying Eq. (2) by $s(s+1)(s+2)^{2}$, we obtain
\begin{aligned}10 s^{2}+4=& A(s+1)(s+2)^{2}+B s(s+2)^{2} \\&+C s(s+1)+D s(s+1)(s+2)\end{aligned}
or
\begin{aligned}10 s^{2}+4=& A\left(s^{3}+5 s^{2}+8 s+4\right)+B\left(s^{3}+4 s^{2}+4 s\right) \\&+C\left(s^{2}+s\right)+D\left(s^{3}+3 s^{2}+2 s\right)\end{aligned}
Equating coefficients,
$$\begin{array}{ll}\text { Constant: } & 4=4 A \quad \Longrightarrow \quad A=1 \\s: & 0=8 A+4 B+C+2 D \quad \Longrightarrow \quad 4 B+C+2 D=-8 \\s^{2}: & 10=5 A+4 B+C+3 D \quad \Longrightarrow \quad 4 B+C+3 D=5 \\s^{3}: & 0=A+B+D \quad \Longrightarrow \quad B+D=-1\end{array}$$
Solving these simultaneous equations gives $A=1, B=-14, C=22$, $D=13$, so that
$$V(s)=\frac{1}{s}-\frac{14}{s+1}+\frac{13}{s+2}+\frac{22}{(s+2)^{2}}$$
Taking the inverse transform of each term, we get
$$v(t)=u(t)-14 e^{-t}+13 e^{-2 t}+22 t e^{-2 t}, \quad t \geq 0$$
Example 4: Find the inverse transform of the frequency-domain function:
$$H(s)=\frac{20}{(s+3)\left(s^{2}+8 s+25\right)}$$
Solution: In this example, $H(s)$ has a pair of complex poles at $s^{2}+8 s+25=0$ or $s=-4 \pm j 3$. We let
$$H(s)=\frac{20}{(s+3)\left(s^{2}+8 s+25\right)}=\frac{A}{s+3}+\frac{B s+C}{\left(s^{2}+8 s+25\right)} \tag{3}$$
We now determine the expansion coefficients in two ways.
METHOD I Combination of methods: We can obtain $A$ using the method of residue,
$$A=\left.(s+3) H(s)\right|_{s=-3}=\left.\frac{20}{s^{2}+8 s+25}\right|_{s=-3}=\frac{20}{10}=2$$
Although $B$ and $C$ can be obtained using the method of residue, we will not do so, to avoid complex algebra. Rather, we can substitute two specific values of $s$ [say $s=0,1$, which are not poles of $F(s)$ ] into Eq. (3). This will give us two simultaneous equations from which to find $B$ and $C$. If we let $s=0$ in Eq. (3), we obtain
$$\frac{20}{75}=\frac{A}{3}+\frac{C}{25}$$
or
$$20=25 A+3 C \tag{4}$$
Since $A=2$, Eq. (4) gives $C=-10$. Substituting $s=1$ into Eq. (3) gives
$$\frac{20}{(4)(34)}=\frac{A}{4}+\frac{B+C}{34}$$
or
$$20=34 A+4 B+4 C$$
But $A=2, C=-10$, so that Eq. (15.11.3) gives $B=-2$
METHOD 2 Algebraic method: Multiplying both sides of Eq. $(3)$ by $(s+3)\left(s^{2}+8 s+25\right)$ yields
\begin{aligned}20 &=A\left(s^{2}+8 s+25\right)+(B s+C)(s+3) \\&=A\left(s^{2}+8 s+25\right)+B\left(s^{2}+3 s\right)+C(s+3)\end{aligned}
Equating coefficients,
$$\begin{array}{ll}s^{2}: & 0=A+B \quad \Longrightarrow \quad A=-B \\s: & 0=8 A+3 B+C=5 A+C \quad \Longrightarrow \quad C=-5 A \\\text { Constant: } & 20=25 A+3 C=25 A-15 A \quad \Longrightarrow \quad A=2\end{array}$$
That is, $B=-2, C=-10$. Thus
\begin{aligned}H(s) &=\frac{2}{s+3}-\frac{2 s+10}{\left(s^{2}+8 s+25\right)}=\frac{2}{s+3}-\frac{2(s+4)+2}{(s+4)^{2}+9} \\&=\frac{2}{s+3}-\frac{2(s+4)}{(s+4)^{2}+9}-\frac{2}{3} \frac{3}{(s+4)^{2}+9}\end{aligned}
Taking the inverse of each term, we obtain
$$h(t)=2 e^{-3 t}-2 e^{-4 t} \cos 3 t-\frac{2}{3} e^{-4 t} \sin 3 t \tag{5}$$
It is alright to leave the result this way. However, we can combine the cosine and sine terms as
$$h(t)=2 e^{-3 t}-A e^{-4 t} \cos (3 t-\theta) \tag{6}$$
To obtain Eq. (6) from Eq. (5), we apply Eq. (A).
$$C = \sqrt{A^2 + B^2}, \, θ = \tan^{−1} {B \over A} \tag{A}$$
Next, we determine the coefficient $A$ and the phase angle $\theta$ :
$$A=\sqrt{2^{2}+\left(\frac{2}{3}\right)^{2}}=2.108, \quad \theta=\tan ^{-1} \frac{\frac{2}{3}}{2}=18.43^{\circ}$$
Thus,
$$h(t)=2 e^{-3 t}-2.108 e^{-4 t} \cos \left(3 t-18.43^{\circ}\right)$$