Partial Fractions.

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April 12th 2010, 01:49 PM
TheBerkeleyBoss

Partial Fractions.

I need to use partial fractions on this term:

$ \dfrac {1} {(s + 1)[(s-1)^2 +1]} $

this term (note: it said $\omega \neq 4)$ :

$ \dfrac {s} {(s^2 + 4)(s^2 + \omega ^2)} + \dfrac {s} {(s^2 + \omega ^2)} $

and lastly this term

$ \dfrac {s-3} {s^2 - 4s + 4} = \dfrac {a} {(s-2)^2} + \dfrac {b} {s-2} $

I am stumped because I have tried all the methods that worked on the prior problems (which were easier no doubt) but I keep running into an impasse. Can anyone please help? I'm in differential equations (doing the chapter on LaPlacian's) but this is definitely from calculus. Thank you!

Kind Regards,
Andy
April 12th 2010, 02:51 PM
Deadstar

Quote:

Originally Posted by TheBerkeleyBoss

$ \dfrac {s-3} {s^2 - 4s + 4} = \dfrac {a} {(s-2)^2} + \dfrac {b} {s-2} $

Plenty of folk looking at this thread so I'll do the last one.

$ \dfrac {s-3} {s^2 - 4s + 4} = \dfrac {a} {(s-2)^2} + \dfrac {b} {s-2} $

You want to multiply the right fraction by $\frac{s-2}{s-2}$ .

This gives,
$ \dfrac {s-3} {s^2 - 4s + 4} = \dfrac {a} {(s-2)^2} + \dfrac {bs - 2b} {(s-2)^2} $

Hence you get,

$a + bs - 2b = s-3$ .

=> $b = 1$ ,
$a - 2 = -3$

Hence $b = 1$ , $a = -1$ .
April 12th 2010, 02:56 PM
drumist

Quote:

Originally Posted by TheBerkeleyBoss

I need to use partial fractions on this term:

$ \dfrac {1} {(s + 1)[(s-1)^2 +1]} $

First of all, it's helpful to know the rule that the number of constants will always equal the total degree of the original denominator. In this case, the degree is 3, so there should be exactly 3 constants used.

Since the second term can't be factored and doesn't have degree 1, we need to use multiple constants. (The number of constants is determined by the degree of that term.) So:

$\frac{1}{(s+1)[(s-1)^2+1]} = \frac{A}{s+1} + \frac{Bs+C}{(s-1)^2+1}$

That's it. You can now try to solve for the constants.

Quote:

this term (note: it said $\omega \neq 4)$ :

$ \dfrac {s} {(s^2 + 4)(s^2 + \omega ^2)} + \dfrac {s} {(s^2 + \omega ^2)} $

For this one, we should probably combine it into a single fraction first. (Alternatively, you can use partial fractions separately for each term or whatever, but this simplifies somewhat.)

$\frac{s}{(s^2+4)(s^2+\omega^2)}+\frac{s}{s^2+\omeg a^2} = \frac{s+s(s^2+4)}{(s^2+4)(s^2+\omega^2)} = \frac{s^3+5s}{(s^2+4)(s^2+\omega^2)}$

Now, our denominator is comprised of two terms, both with multiplicity of 2, and neither can be reduced any further, so:

$\frac{s^3+5s}{(s^2+4)(s^2+\omega^2)} = \frac{As+B}{s^2+4} + \frac{Cs+D}{s^2+\omega^2}$

Quote:

and lastly this term

$ \dfrac {s-3} {s^2 - 4s + 4} = \dfrac {a} {(s-2)^2} + \dfrac {b} {s-2} $

You already broke this one apart correctly. Is there a problem solving for the constants?
April 12th 2010, 02:57 PM
zzzoak

$\frac{1}{(s+1)(s^2-2s+2)}=\frac{A}{s+1}+\frac{Bs+C}{s^2-2s+2}$
A=1/5
B=-1/5
C=3/5
April 12th 2010, 03:25 PM
TheBerkeleyBoss

wow!

Wow thank you everyone! I love you all in a very platonic way. Thanks around the table.