Partial Fractions

1. Mar 25, 2007

FrogPad

I have: $$\frac{(1+j\omega)(3-j\omega)}{(3+j\omega)(3-j\omega)}$$

When I perform the partial fraction expansion I get:

$$\frac{-2}{3+j\omega}$$

Where my calculator gets:
$$1 - \frac{-2}{3+j\omega}$$.

Why am I wrong?

I am performing the expansion as follows:

$$\bar F(s) = \frac{(1+s)(3-s)}{(3+s)(3-s)}$$

and,
$$K_i = (s+p_i)\bar F (s)$$ where: $s = - p_i [/tex] note: $$p_i$$ corresponds to 3 and -3 respectively. I am getting: $$K_1 = -2$$ and $$K_2 = 0$$ (this does not match my calculator. I am assuming simple poles. Is this not proper? thanks in advance! 2. Mar 25, 2007 jpr0 You don't even need to partial fraction this expression $$\frac{(1+j\omega)(3-j\omega)}{(3+j\omega)(3-j\omega)}$$ The term [itex](3-j\omega)/(3-j\omega)=1$, so you're left with

$$\frac{(1+j\omega)}{(3+j\omega)}$$

Then rewrite this as

$$\frac{(1+j\omega)}{(3+j\omega)} = \frac{(3+j\omega - 2)}{(3+j\omega)} = 1 - \frac{2}{3+j\omega}$$

3. Mar 25, 2007

FrogPad

Well I like what you did, that is nice way of doing it.

My original term was:
$$\frac{1+jw}{3+jw}$$

I then multiplied by:
$$\frac{3-jw}{3-jw} = 1$$

and was trying to expand it as such.

Is there a reason why this method does not work (in this case)?

I want to know how to generalize it. The original expression, written as: $$\frac{1+jw}{3+jw}$$ does not match my transform table, so I wanted to express it differently. I'll definitely remember the way you showed me for similar expressions, but in general I should be applying partial fractions... so why didn't it work here?

thanks man, I do appreciate it

4. Mar 26, 2007

HallsofIvy

Staff Emeritus
??jpr0 did use "partial fractions", he just didn't make it overly complicated by introducing new factors into the numerator and denominator. The technique of "partial fractions" assumes you have already reduced the numerator and denominator as much as possible. For example, if you applied partial fractions to
$$\frac{x^2- 5x+3}{x-2}$$
you would get similarly incorrect results- first do the division to to get a "proper" fraction.

5. Mar 26, 2007

FrogPad

Gotchya'. I wasn't trying to be overly complicated, I was just applying the technique (blindly in this case) that I remembered. I didn't know that it had to be a "proper" fraction first. Thanks halls! That clears it up.

6. Mar 27, 2007

HallsofIvy

Staff Emeritus
I know you weren't trying to be- but sometimes it just happens!

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