Respuesta :
The solution of the inverse Laplace transforms is mathematically given as
- [tex]f_{1}(t)=e^{-t}\sin (2 t)[/tex]
- [tex]f_{2}(t)=\frac{7}{9} e^{-t}+\frac{2}{3} e^{-t}+\frac{2}{9} e^{-4 t}[/tex]
- [tex]f_{3}(t)=2 e^{-t}-2 e^{-2 t} \cos (2 t)-e^{-2 t} \sin (2 t)[/tex]
What is the inverse Laplace transform?
1)
Generally, the equation for the function is mathematically given as
[tex]$F_{1}(s)=\frac{6 s^{2}+8 s+3}{s\left(s^{2}+2 s+5\right)}$[/tex]
By Applying the Partial fractions method
[tex]\frac{6 s^{2}+8 s+3}{s\left(s^{2}+2 s+5\right)}=\frac{A}{s}+\frac{B s+C}{s^{2}+2 s+5}[/tex]
[tex]$6 s^{2}+8 s+3=A\left(s^{2}+2 s+5\right)+(B s+C) s$[/tex]
[tex]\begin{aligned}&3=5 A \\&A=\frac{3}{5}\end{aligned}[/tex]
Considers s^2 coefficient
[tex]\begin{aligned}&6=A+B \\&B=6 \cdot A \\&B=\frac{27}{5}\end{aligned}[/tex]
Consider s coeffici ent
[tex]\begin{aligned}&8=2 A+C \\&C=8-2 A \\&C=\frac{34}{5}\end{aligned}[/tex]
Putting these values into the previous equation
[tex]&F_{1}(s)=\frac{3}{5 s}+\frac{27 s+34}{5\left(s^{2}+2 s+5\right)} \\\\&F_{1}(s)=\frac{3}{5 s}+\frac{27(s+1)}{5\left((s+1)^{2}+4\right)}+\frac{7 \times 2}{10\left((s+1)^{2}+4\right)}[/tex]
By taking Inverse Laplace Transforms
[tex]f_{1}(t)=\frac{3}{5}+\frac{27}{5} e^{-t} \cos (2t) + \frac{7}{10}\\\\[/tex]
[tex]f_{1}(t)=e^{-t}\sin (2 t)[/tex]
For B
[tex]$F_{2}(s)=\frac{s^{2}+5 s+6}{(s+4)(s+1)^{2}}$[/tex]
By Applying Partial fractions method
[tex]\begin{aligned}&\frac{s^{2}+5 s+6}{(s+4)(s+1)^{2}}=\frac{A}{s+1}+\frac{B}{(s+1)^{2}}+\frac{C}{s+4} \\\\&s^{2}+5 s+6=A(s+1)(s+4)+B(s+4)+C(s+1)^{2}\end{aligned}[/tex]
at s=-1
[tex]1-5+6=3 B \\\\B=\frac{2}{3}[/tex]
at s=-4
[tex]&16-20+6=9 C \\\\&9 C=2 \\\\&C=\frac{2}{9}[/tex]
at s^2 coefficient
1=A+C
A=1-C
A=7/9
inputting Variables into the Previous Equation
[tex]\begin{aligned}&F_{2}(s)=\frac{A}{s+1}+\frac{B}{(s+1)^{2}}+\frac{C}{s+4} \\&F_{2}(s)=\frac{7}{9(s+1)}+\frac{2}{3(s+1)^{2}}+\frac{2}{9(s+4)}\end{aligned}[/tex]
By taking Inverse Laplace Transforms
[tex]f_{2}(t)=\frac{7}{9} e^{-t}+\frac{2}{3} e^{-t}+\frac{2}{9} e^{-4 t}[/tex]
For C
[tex]$F_{3}(s)=\frac{10}{(s+1)\left(s^{2}+4 s+8\right)}$[/tex]
Using the strategy of Partial Fractions
[tex]\frac{10}{(s+1)\left(s^{2}+4 s+8\right)}=\frac{A}{s+1}+\frac{B s+C}{s^{2}+4 s+8}[/tex]
[tex]10=A\left(s^{2}+4 s+8\right)+(B s+C)(s+1)[/tex]
S=-1
10=(1-4+8) A
A=10/5
A=2
Consider constants
10=8 A+C
C=10-8 A
C=10-16
C=-6
Considers s^2 coefficient
0=A+B
B=-A
B=-2
inputting Variables into the Previous Equation
[tex]&F_{3}(s)=\frac{2}{s+1}+\frac{-2 s-6}{\left((s+2)^{2}+4\right)} \\\\&F_{3}(s)=\frac{2}{s+1}-\frac{2(s+2)}{\left((s+2)^{2}+4\right)}-\frac{2}{\left((s+2)^{2}+4\right)}[/tex]
Inverse Laplace Transforms
[tex]f_{3}(t)=2 e^{-t}-2 e^{-2 t} \cos (2 t)-e^{-2 t} \sin (2 t)[/tex]
Read more about Laplace Transforms
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