Inverse Laplace transform of $\frac{2s+1}{s^2 - 4s + 5}$

Question

I approached this problem as follow:

1. Complete the square in the denominator and obtain $$(s-2)^2 + 1$$

2. Now break the function into 2 parts.

$$ \frac{2s}{(s-2)^2 + 1}+ \frac{1}{(s-2)^2 + 1}$$

Now the inverse laplace transform is straight forward.

$$e^{2t}(2cos(t) + sin(t))$$

However! I am wrong. The solution to this problem is actually

$$e^{2t}(2cos(t) + 5*sin(t))$$

What is wrong with my approach?

Thanks.

Answer 1

Remember, the Laplace transform tables have the following kinds of terms

$$ \frac{s-a}{(s-a)^2+b^2},\frac{b}{(s-a)^2+b^2} $$ Your term with an $s$ in the numerator has to look like $s-2$ in order to use that form. So your problem is at this step:

$$ \frac{2s}{(s-2)^2 + 1}+ \frac{1}{(s-2)^2 + 1} = \frac{2(s-2)+4}{(s-2)^2 + 1}+ \frac{1}{(s-2)^2 + 1} = 2\frac{(s-2)}{(s-2)^2 + 1}+ 5\frac{1}{(s-2)^2 + 1} $$

Basically, you have to have the same form for $s$ in the top and bottom, so you add and subtract $4$ from the numerator of that first term so that you can put it in terms of $(s-2)$. Then the extra $4$ you added goes over to the other term to make a total of $5$, the coefficient of the $\sin$

Answer 2

$s^2-4 s+5$ has zeroes at $s_{\pm} = 2 \pm i$. The ILT $f(t)$ is simply the sum of the residues of

$$\frac{2 s+1}{s^2-4 s+5} e^{s t}$$

at these poles. Then

$$f(t) = \frac{2 s_+ + 1}{2 s_+-4} e^{s_+ t} + \frac{2 s_- + 1}{2 s_--4} e^{s_- t}$$

Expanding this a bit:

$$f(t) = e^{2 t} \left [ \left (1-i \frac52 \right ) (\cos{t}+i \sin{t}) + \left(1+i \frac52 \right ) (\cos{t}-i \sin{t}) \right ]$$

Simplifying, I get

$$f(t) = e^{2 t} (2 \cos{t} + 5 \sin{t})$$

Answer 3

I think if you do the fraction as follows:

$$\frac{2s+1}{(s-2)^2 + 1}= \frac{2(s-2)+5}{(s-2)^2 + 1}=\frac{2(s-2)}{(s-2)^2 + 1}+\frac{5}{(s-2)^2 + 1}$$ and considering very useful fact:

$\mathcal{L}^{-1}(F(s-a))=e^{at}f(t)$

then we get:

$$\mathcal{L}^{-1}\left(\frac{2s+1}{(s-2)^2 + 1}\right)=2\mathcal{L}^{-1}\left(\frac{(s-2)}{(s-2)^2 + 1}\right)+5\mathcal{L}^{-1}\left(\frac{1}{(s-2)^2 + 1}\right)=2e^{2t}\cos t+5e^{2t}\sin t$$