Step 1
Formula Used:
\(L\left\{e^{at}\sin bt\right\}=\frac{b}{(s-a)^{2}+b^{2}}\)
\(L\left\{e^{at}\cos bt\right\}=\frac{s-a}{(s-a)^{2}+b^{2}}\)
Step 2
\(\text{Consider }F(s)=\frac{s+8}{s^2+5s+5}\)
\(=\frac{s+8}{(s+2)^{2}+1}\)
\(=\frac{s+2+6}{(s+2)^{2}+1}\)
\(=\frac{(s+2)}{(s+2)^{2}+1}+\frac{6}{(s+2)^2+1}\)
Step 3
\(\text{Consider }F(s)=\frac{(s+2)}{(s+2)^{2}+1}+\frac{6}{(s+2)^2+1}\)
We now apply Inverse Laplace Transform, we get
\(L^{-1}\left\{F(s)\right\}=L^{-1}\left\{\frac{(s+2)}{(s+2)^2+1}\right\}+6L^{-1}\left\{\frac{1}{(s+2)^2+1}\right\}\)
\(\Rightarrow\ f(t)=e^{-2t}\cos t+6e^{-2t}\sin t\)
\(=e^{-2t}(\cos t+6\sin t)\)
Step 4
ANSWER
: The inverse Laplace Transform of F(s) is
\(f(t)=e^{-2t}(\cos t+6\sin t)\)
Find Laplace transform for frac{s+8}{s^2+4s+5}
Find Laplace transform for frac{s+8}{s^2+4s+5}
Question
Answers (1)
2021-02-01
Relevant Questions
asked 2020-11-09
Use Laplace transform to evaluate the integral
\(\int_0^\infty te^{-2t} \sin(2t)dt\)
a) \(\frac{1}{B}\)
b) not defined
c) \(\frac{4s}{(s^2+4)^2}\)
d) \(\frac{4(s+2)}{(s^2+4s+8)^2}\)
\(\int_0^\infty te^{-2t} \sin(2t)dt\)
a) \(\frac{1}{B}\)
b) not defined
c) \(\frac{4s}{(s^2+4)^2}\)
d) \(\frac{4(s+2)}{(s^2+4s+8)^2}\)
asked 2020-12-17
determine a function f(t) that has the given Laplace transform
\(F(s)=\frac{4s+5}{s^{2}+9}\)
\(F(s)=\frac{4s+5}{s^{2}+9}\)
asked 2020-10-28
find the inverse Laplace transform of the given function
\(F(s)=\frac{e^{-2}+e^{-2s}-e^{-3s}-e^{-4s}}{s}\)
\(F(s)=\frac{e^{-2}+e^{-2s}-e^{-3s}-e^{-4s}}{s}\)
asked 2020-11-07
Write down the qualitative form of the inverse Laplace transform of the following function. For each question first write down the poles of the function , X(s)
a) \(X(s)=\frac{s+1}{(s+2)(s^2+2s+2)(s^2+4)}\)
b) \(X(s)=\frac{1}{(2s^2+8s+20)(s^2+2s+2)(s+8)}\)
c) \(X(s)=\frac{1}{s^2(s^2+2s+5)(s+3)}\)
a) \(X(s)=\frac{s+1}{(s+2)(s^2+2s+2)(s^2+4)}\)
b) \(X(s)=\frac{1}{(2s^2+8s+20)(s^2+2s+2)(s+8)}\)
c) \(X(s)=\frac{1}{s^2(s^2+2s+5)(s+3)}\)
asked 2020-12-27
Find inverse Laplace transform \(L^{-1}\left\{\frac{s-5}{s^2+5s-24}\right\}\)
Please provide supporting details for your answer
asked 2020-10-25
Please provide steps
The inverse Laplace transform for
\(\displaystyle{F}{\left({s}\right)}=\frac{8}{{{s}+{9}}}-\frac{6}{{{s}^{2}-\sqrt{{3}}}}\) is
a) \(\displaystyle{8}{e}^{{-{9}{t}}}-{6} \sin{{h}}{{\left({3}{t}\right)}}\)
b) \(\displaystyle{8}{e}^{{-{9}{t}}}-{6} \cos{{h}}{\left({3}{t}\right)}\)
c) \(\displaystyle{8}{e}^{{{9}{t}}}-{6} \sin{{h}}{\left({3}{t}\right)}\)
d) \(\displaystyle{8}{e}^{{{9}{t}}}-{6} \cos{{h}}{\left({3}{t}\right)}\)
The inverse Laplace transform for
\(\displaystyle{F}{\left({s}\right)}=\frac{8}{{{s}+{9}}}-\frac{6}{{{s}^{2}-\sqrt{{3}}}}\) is
a) \(\displaystyle{8}{e}^{{-{9}{t}}}-{6} \sin{{h}}{{\left({3}{t}\right)}}\)
b) \(\displaystyle{8}{e}^{{-{9}{t}}}-{6} \cos{{h}}{\left({3}{t}\right)}\)
c) \(\displaystyle{8}{e}^{{{9}{t}}}-{6} \sin{{h}}{\left({3}{t}\right)}\)
d) \(\displaystyle{8}{e}^{{{9}{t}}}-{6} \cos{{h}}{\left({3}{t}\right)}\)
asked 2020-11-26
Solve for the Inverse Laplace Transformations. Show the solution.
\(F(s)=\frac{6}{(s^2+4s+20)^2}\)
\(F(s)=\frac{6}{(s^2+4s+20)^2}\)
asked 2021-02-25
Use properties of the Laplace transform to answer the following
(a) If \(f(t)=(t+5)^2+t^2e^{5t}\), find the Laplace transform,\(L[f(t)] = F(s)\).
(b) If \(f(t) = 2e^{-t}\cos(3t+\frac{\pi}{4})\), find the Laplace transform, \(L[f(t)] = F(s)\). HINT:
\(\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha) \sin(\beta)\)
(c) If \(F(s) = \frac{7s^2-37s+64}{s(s^2-8s+16)}\) find the inverse Laplace transform, \(L^{-1}|F(s)| = f(t)\)
(d) If \(F(s) = e^{-7s}(\frac{1}{s}+\frac{s}{s^2+1})\) , find the inverse Laplace transform, \(L^{-1}[F(s)] = f(t)\)
(a) If \(f(t)=(t+5)^2+t^2e^{5t}\), find the Laplace transform,\(L[f(t)] = F(s)\).
(b) If \(f(t) = 2e^{-t}\cos(3t+\frac{\pi}{4})\), find the Laplace transform, \(L[f(t)] = F(s)\). HINT:
\(\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha) \sin(\beta)\)
(c) If \(F(s) = \frac{7s^2-37s+64}{s(s^2-8s+16)}\) find the inverse Laplace transform, \(L^{-1}|F(s)| = f(t)\)
(d) If \(F(s) = e^{-7s}(\frac{1}{s}+\frac{s}{s^2+1})\) , find the inverse Laplace transform, \(L^{-1}[F(s)] = f(t)\)
asked 2020-11-02
Find the Laplace transform \(L\left\{u_3(t)(t^2-5t+6)\right\}\)
\(a) F(s)=e^{-3s}\left(\frac{2}{s^4}-\frac{5}{s^3}+\frac{6}{s^2}\right)\)
\(b) F(s)=e^{-3s}\left(\frac{2}{s^3}-\frac{5}{s^2}+\frac{6}{s}\right)\)
\(c) F(s)=e^{-3s}\frac{2+s}{s^4}\)
\(d) F(s)=e^{-3s}\frac{2+s}{s^3}\)
\(e) F(s)=e^{-3s}\frac{2-11s+30s^2}{s^3}\)
\(a) F(s)=e^{-3s}\left(\frac{2}{s^4}-\frac{5}{s^3}+\frac{6}{s^2}\right)\)
\(b) F(s)=e^{-3s}\left(\frac{2}{s^3}-\frac{5}{s^2}+\frac{6}{s}\right)\)
\(c) F(s)=e^{-3s}\frac{2+s}{s^4}\)
\(d) F(s)=e^{-3s}\frac{2+s}{s^3}\)
\(e) F(s)=e^{-3s}\frac{2-11s+30s^2}{s^3}\)
asked 2020-12-25
Let x(t) be the solution of the initial-value problem
(a) Find the Laplace transform F(s) of the forcing f(t).
(b) Find the Laplace transform X(s) of the solution x(t).
\(x"+8x'+20x=f(t)\)
\(x(0)=-3\)
\(x'(0)=5\)
\(\text{where the forcing } f(t) \text{ is given by }\)
\(f(t) = \begin{cases} t^2 & \quad \text{for } 0\leq t<2 ,\\ 4e^{2-t} & \quad \text{for } 2\leq t < \infty . \end{cases}\)
(a) Find the Laplace transform F(s) of the forcing f(t).
(b) Find the Laplace transform X(s) of the solution x(t).
\(x"+8x'+20x=f(t)\)
\(x(0)=-3\)
\(x'(0)=5\)
\(\text{where the forcing } f(t) \text{ is given by }\)
\(f(t) = \begin{cases} t^2 & \quad \text{for } 0\leq t<2 ,\\ 4e^{2-t} & \quad \text{for } 2\leq t < \infty . \end{cases}\)
