Demonstrative Video
Inverse Laplace Transform
- \[F(s) = \dfrac{N(s)}{D(s)}\]has the general form Suppose
Roots of \(N(s) = 0\) are known as Zeros of \(F(s)\)
Roots of \(D(s) = 0\) are called Poles of \(F(s)\)
Use partial fraction expansion to break \(F(s)\) down into simple terms
Steps to Find the Inverse Laplace Transform:
Decompose \(F(s)\) into simple terms using partial fraction expansion
Find inverse of each term by matching entries from Table
Three possible forms \(F(s)\) may take and two step approach to each form.
- \[\begin{aligned} F(s) &=\frac{N(s)}{\left(s+p_{1}\right)\left(s+p_{2}\right) \cdots\left(s+p_{n}\right)}\qquad p_i \neq p_j\\ F(s) &=\frac{k_{1}}{s+p_{1}}+\frac{k_{2}}{s+p_{2}}+\cdots+\frac{k_{n}}{s+p_{n}} \\ & \text{(Assuming degree of }~ N(s)<D(s))\\ \color{magenta}{\textbf{Residues}} \quad &\boxed{k_{i} =\left.\left(s+p_{i}\right) F(s)\right|_{s=-p_{i}}} \\ &\boxed{f(t) =\left(k_{1} e^{-p_{1} t}+k_{2} e^{-p_{2} t}+\cdots+k_{n} e^{-p_{n} t}\right) u(t)} \end{aligned}\]Simple Poles :
- \[\begin{aligned} & F(s) =\frac{k_{n}}{(s+p)^{n}}+\frac{k_{n-1}}{(s+p)^{n-1}}+\cdots+\frac{k_{2}}{(s+p)^{2}}+\frac{k_{1}}{s+p} + \underset{\text{remaining part}}{F_1(s)} \\ & k_{n} =\left.(s+p)^{n} F(s)\right|_{s=-p} \\ & k_{n-1} =\left.\frac{d}{d s}\left[(s+p)^{n} F(s)\right]\right|_{s=-p} \\ & k_{n-2} =\left.\frac{1}{2 !} \frac{d^{2}}{d s^{2}}\left[(s+p)^{n} F(s)\right]\right|_{s=-p} \\ & k_{n-m} =\left.\frac{1}{m !} \frac{d^{m}}{d s^{m}}\left[(s+p)^{n} F(s)\right]\right|_{s=-p} \\ & \mathcal{L}^{-1}{\left[\frac{1}{(s+a)^{n}}\right] } =\frac{t^{n-1} e^{-a t}}{(n-1) !} u(t) \end{aligned}\]Repeated Poles :
Complex Poles
A pair of complex poles is simple if it is not repeated
it is a double or multiple pole if repeated.
Simple complex poles may be handled the same way as simple real poles, but because complex algebra is involved the result is always cumbersome.
An easier approach is a method known as completing the square
\[\begin{aligned}
F(s) &=\frac{A_{1} s+A_{2}}{s^{2}+a s+b}+F_{1}(s) \\
F(s)
&=\frac{A_{1}(s+\alpha)}{(s+\alpha)^{2}+\beta^{2}}+\frac{B_{1}
\beta}{(s+\alpha)^{2}+\beta^{2}}+F_{1}(s) \\
&\boxed{f(t) =\left(A_{1} e^{-\alpha t} \cos \beta t+B_{1}
e^{-\alpha t} \sin \beta t\right) u(t)+f_{1}(t) }
\end{aligned}\]
\[\begin{aligned}
s^{2}+a s+b &=s^{2}+2 \alpha s+\alpha^{2}+\beta^{2}
=(s+\alpha)^{2}+\beta^{2} \\
A_{1} s+A_{2} &=A_{1}(s+\alpha)+B_{1} \beta
\end{aligned}\]
Problem-1
\[F(s)=\frac{3}{s}-\frac{5}{s+1}+\frac{6}{s^{2}+4}\]
\[\begin{gathered}
f(t)=\mathcal{L}^{-1}[F(s)]=\mathcal{L}^{-1}\left(\frac{3}{s}\right)-\mathcal{L}^{-1}\left(\frac{5}{s+1}\right)+\mathcal{L}^{-1}\left(\frac{6}{s^{2}+4}\right)
\\
=\left(3-5 e^{-t}+3 \sin 2 t\right) u(t), \quad t \geq 0
\end{gathered}\]
? Find the inverse Laplace Transform of
Problem-2
\[\begin{gathered}
F(s) = \frac{s^{2}+12}{s(s+2)(s+3)}
=\frac{A}{s}+\frac{B}{s+2}+\frac{C}{s+3} \\
A=\left.s
F(s)\right|_{s=0}=\left.\frac{s^{2}+12}{(s+2)(s+3)}\right|_{s=0}=\frac{12}{(2)(3)}=2
\\
B=\left.(s+2)
F(s)\right|_{s=-2}=\left.\frac{s^{2}+12}{s(s+3)}\right|_{s=-2}=\frac{4+12}{(-2)(1)}=-8
\\
C=\left.(s+3)
F(s)\right|_{s=-3}=\left.\frac{s^{2}+12}{s(s+2)}\right|_{s=-3}=\frac{9+12}{(-3)(-1)}=7
\\
F(s)=\frac{2}{s}-\frac{8}{s+2}+\frac{7}{s+3} \\
f(t)=\left(2-8 e^{-2 t}+7 e^{-3 t}\right) u(t)
\end{gathered}\]
? Find the inverse Laplace Transform of
Problem-3
\[\small
\begin{gathered}
V(s)=\frac{10 s^{2}+4}{s(s+1)(s+2)^{2}}
=\frac{A}{s}+\frac{B}{s+1}+\frac{C}{(s+2)^{2}}+\frac{D}{s+2} \\
A=\left.s V(s)\right|_{s=0}=\left.\frac{10
s^{2}+4}{(s+1)(s+2)^{2}}\right|_{s=0}=\frac{4}{(1)(2)^{2}}=1 \\
B=\left.(s+1) V(s)\right|_{s=-1}=\left.\frac{10
s^{2}+4}{s(s+2)^{2}}\right|_{s=-1}=\frac{14}{(-1)(1)^{2}}=-14 \\
C=\left.(s+2)^{2} V(s)\right|_{s=-2}=\left.\frac{10
s^{2}+4}{s(s+1)}\right|_{s=-2}=\frac{44}{(-2)(-1)}=22 \\
D=\left.\frac{d}{d s}\left[(s+2)^{2}
V(s)\right]\right|_{s=-2}=\left.\frac{d}{d s}\left(\frac{10
s^{2}+4}{s^{2}+s}\right)\right|_{s=-2} \\
=\left.\frac{\left(s^{2}+s\right)(20 s)-\left(10
s^{2}+4\right)(2
s+1)}{\left(s^{2}+s\right)^{2}}\right|_{s=-2}=\frac{52}{4}=13 \\
V(s)=\frac{1}{s}-\frac{14}{s+1}+\frac{13}{s+2}+\frac{22}{(s+2)^{2}}
\\
v(t)=\left(1-14 e^{-t}+13 e^{-2 t}+22 t e^{-2 t}\right) u(t)
\end{gathered}\]
Problem-4
\[\begin{aligned}
H(s)=& \frac{20}{(s+3)\left(s^{2}+8 s+25\right)}
=\frac{A}{s+3}+\frac{B s+C}{\left(s^{2}+8 s+25\right)} \\
A=&\left.(s+3) H(s)\right|_{s=-3}=\left.\frac{20}{s^{2}+8
s+25}\right|_{s=-3}=\frac{20}{10}=2 \\
s^{2}+& 8 s+25=0 \Rightarrow \quad s=-4 \pm j 3 . \\
s = 0 ~~\Rightarrow & \frac{20}{75}=\frac{A}{3}+\frac{C}{25}
\quad \Rightarrow 20=25 A+3 C \quad \Rightarrow C=-10 \\
s = 1~~ \Rightarrow &
\frac{20}{(4)(34)}=\frac{A}{4}+\frac{B+C}{34} \Rightarrow ~ 20 =
34A+4B+4C ~ \Rightarrow B = -2
\end{aligned}\]
\[\begin{aligned}
H(s)=& \frac{2}{s+3}-\frac{2 s+10}{\left(s^{2}+8
s+25\right)}=\frac{2}{s+3}-\frac{2(s+4)+2}{(s+4)^{2}+9} \\
=& \frac{2}{s+3}-\frac{2(s+4)}{(s+4)^{2}+9}-\frac{2}{3}
\frac{3}{(s+4)^{2}+9} \\
h(t)=&\left(2 e^{-3 t}-2 e^{-4 t} \cos 3 t-\frac{2}{3} e^{-4
t} \sin 3 t\right) u(t) \\
\Rightarrow h(t)=&\left(2 e^{-3 t}-2.108 e^{-4 t} \cos
\left(3 t-18.43^{\circ}\right)\right) u(t)
\end{aligned}\]