Partial fraction method (in s-space)

• Apr 18th 2011, 10:15 AM
courteous
Partial fraction method (in s-space)
The following is a part of a current I(s) (in $s$-space):
http://img191.imageshack.us/img191/4021/giflatex2.gif

What am I doing wrong in using this formula for finding $A$ and $B$:
http://img546.imageshack.us/img546/4265/giflatex3.gif ; where $a$ is a pole.

PROBLEM
I easily find A, but not B:
http://img269.imageshack.us/img269/7302/giflatex4.gif
end of PROBLEM

The solution (which is correct) is http://img251.imageshack.us/img251/315/giflatex5q.gif

What am I doing wrong inside the PROBLEM section? Or is the formula for $A$ (or $B$) wrong? (Crying)
• Apr 18th 2011, 08:59 PM
ojones
The mistake is at the beginning:

$B/L=\displaystyle{\lim_{s\rightarrow b}(s-b)F(s)}}$.
• Apr 19th 2011, 02:37 AM
courteous
http://img228.imageshack.us/img228/9864/giflatex6g.gif
You wrote this, right?

So, is the formula I used for $A$ wrong? Why do you divide $B$ by $L$ ( B/L )?

Seeing the correct solution, shouldn't you multiply by $L$?
• Apr 19th 2011, 12:37 PM
ojones
Yes, I wrote that. For some reason Math Help Forum wasn't compiling Latex at the time I wrote it. I'm still having trouble with their Latex and so I won't use it.

No, the formula for A is correct. You divide B by L because sL is in the denominator of F(s).

You don't multiply by L; B/L gives the correct answer.
• Apr 20th 2011, 07:25 AM
courteous
Quote:

Originally Posted by ojones
No, the formula for A is correct. You divide B by L because sL is in the denominator of F(s).

Aren't s and ( $R_2$ + sL) the two denominators of $F(s)$? Or is $F(s)$ something entirely different (than the one I used inside PROBLEM tags)?

In short, what am I doing wrong inside the PROBLEM section?

Thank you for the help.
• Apr 20th 2011, 08:11 AM
TaylorM0192
Why are you bothering with the formalism of limits? This is just a (very trivial) partial fraction decomposition you should have learned in precalculus --

You have the following after clearing fractions:

(1). A * (R + sL) + B * (s) = U

Let s = 0 => A = U/R
Let s = -R/L => B = -UL/R

Therefore, after substitution for A and B into your original decomposition, we obtain:

(A/s) + (B/(R + sL)) = (U/(Rs)) - (UL/(R(R+sL))).

The answer is easily verifiable by recombining fractions.
• Apr 20th 2011, 08:45 AM
courteous
I'm not questioning whether the solution you (and ojones) gave is correct (and that the problem is (very) trivial).

But what the heck is wrong with "the limit formula" (or my way of (blindly) using it)?
• Apr 20th 2011, 08:56 AM
TaylorM0192
Quote:

Originally Posted by courteous
I'm not questioning whether the solution you (and ojones) gave is correct (and that the problem is (very) trivial).

But what the heck is wrong with "the limit formula" (or my way of (blindly) using it)?

Because you're not setting up the correct expression for the limit. It should be the limit as s => -R/L of the expression [U/s] because the factors (R + sL) will cancel. Where you obtain the factor (s + R/L) I have no idea; this is why you don't get a perfect cancellation, have to combine denominators and introduce an extra L.

By the way, this is known as the Heaviside "cover up" method. Here is a helpful resource you should consider: http://www.math-cs.gordon.edu/course.../heavyside.pdf. SO now, you can forget about this limit formalism and just "cover up" the factors (only applies when all factors are linear).
• Apr 20th 2011, 01:27 PM
ojones
I am looking at what's in the PROBLEM section. The statement B=lim (s-b)F(s) is not correct. Why do you think it should be? (Latex still not working. Fix it MHF!)