havin trouble solving this partial fraction:

$\displaystyle 1/s^2-2s+2$

i think i can break it up into:

$\displaystyle 1/(s-1)^2+1$

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- Nov 16th 2006, 05:49 PMChadly724Partial Fraction
havin trouble solving this partial fraction:

$\displaystyle 1/s^2-2s+2$

i think i can break it up into:

$\displaystyle 1/(s-1)^2+1$ - Nov 16th 2006, 05:51 PMQuick
- Nov 16th 2006, 06:13 PMChadly724correction
sorry there should be partheses around the entire bottom function

- Nov 16th 2006, 07:00 PMThePerfectHacker
You have,

$\displaystyle \frac{1}{s^2-2s+2}$

Express as,

$\displaystyle \frac{1}{s^2-2s+1+1}$

Thus,

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

Now, I presume you have an inverse Laplace transform for $\displaystyle s>1$.

In that case if you have,

$\displaystyle \frac{1}{s^2+1}$ $\displaystyle s>0$

Then, that is the function,

$\displaystyle f(x)=\sin x$

Since you have a substitution you use the shift rule.

Thus,

$\displaystyle f(x)=e^x\sin x$

If the unique inverse LaPlace Transform. - Nov 16th 2006, 10:37 PMCaptainBlack
Do you want to break this up into partial fractions or just transform into a

standard form of LT that can be inverted by table look-up?

My table of LT's has:

$\displaystyle

\frac{1}{(s-b)^2+a^2}=\mathcal{L}\ \frac{e^{bt}\sin(at)}{a}

$

which should take case of your last form.

However we can do the same thing with partial fractions if you will:

$\displaystyle

\frac{1}{s^2-2s+2}=\frac{1}{ (s-(1+\bold{i})) (s-(1-\bold{i}))}=

\frac{\bold{i}}{2(s-(1-\bold{i})) } - \frac{\bold{i}}{2(s-(1+\bold{i})) }

$

(please check the arithmetic)

RonL - Nov 17th 2006, 08:37 AMSoroban
Hello, Chadly724!

Quote:

Havin trouble solving this partial fraction: $\displaystyle \frac{1}{s^2-2s+2}$

i think i can break it up into: $\displaystyle \frac{1}{(s-1)^2+1}$. . Yes, you can . . . but why?

Partial Fractions are used when the denominator can be**factored**.

. . Like: .$\displaystyle \frac{1}{x^2+2x-15}\:=\:\frac{1}{(x-3)(x+5)}$ or: .$\displaystyle \frac{1}{x^3+4x} \:=\:\frac{1}{x(x^2 +4)}$

As Captain Black pointed out, it can be factored: .$\displaystyle (s - 1 - i)(s - 1 + i)$

. . but I doubt that you want to introduce complex numbers.

If this is an integration problem:

. . $\displaystyle \int\frac{ds}{s^2 - 2x + 2} \;=\;\int\frac{ds}{(s-1)^2+1} \;=\;\tan^{-1}(s-1) + C $

- Nov 17th 2006, 08:43 AMCaptainBlack