Discover a wealth of information and get your questions answered on IDNLearn.com. Discover comprehensive answers to your questions from our community of experienced professionals.
It looks like we're given the Laplace transform of f(t),
[tex]F(p) = L_p\left\{f(t)\right\} = \dfrac6{p(2p^2+4p+10)} = \dfrac3{p(p^2+2p+5)}[/tex]
Start by splitting up F(p) into partial fractions:
[tex]\dfrac3{p(p^2+2p+5)} = \dfrac ap + \dfrac{bp+c}{p^2+2p+5} \\\\ 3 = a(p^2+2p+5) + (bp+c)p \\\\ 3 = (a+b)p^2 + (2a+c)p + 5a \\\\ \implies \begin{cases}a+b=0 \\ 2a+c=0 \\ 5a=3\end{cases} \implies a=\dfrac35,b=-\dfrac35, c=-\dfrac65[/tex]
[tex]F(p) = \dfrac3{5p} - \dfrac{3p+6}{5(p^2+2p+5)}[/tex]
Complete the square in the denominator,
[tex]p^2+2p+5 = p^2+2p+1+4 = (p+1)^2+4[/tex]
and rewrite the numerator in terms of p + 1,
[tex]3p+6 = 3(p+1) + 3[/tex]
Then splitting up the second term gives
[tex]F(p) = \dfrac3{5p} - \dfrac{3(p+1)}{5((p+1)^2+4)} - \dfrac3{5((p+1)^2+4)}[/tex]
Now take the inverse transform:
[tex]L^{-1}_t\left\{F(p)\right\} = \dfrac35 L^{-1}_t\left\{\dfrac1p\right\} - \dfrac35 L^{-1}_t\left\{\dfrac{p+1}{(p+1)^2+2^2}\right\} - \dfrac3{5\times2} L^{-1}_t\left\{\dfrac2{(p+1)^2+2^2}\right\} \\\\ L^{-1}_t\left\{F(p)\right\} = \dfrac35 - \dfrac35 e^{-t} L^{-1}_t\left\{\dfrac p{p^2+2^2}\right\} - \dfrac3{10} e^{-t} L^{-1}_t\left\{\dfrac2{p^2+2^2}\right\} \\\\ \implies \boxed{f(t) = \dfrac35 - \dfrac35 e^{-t} \cos(2t) - \dfrac3{10} e^{-t} \sin(2t)}[/tex]