# Partial Fractions

1. Apr 21, 2007

### snowJT

if you have

$$\frac{3s + 1}{(s+2)^2 + 4^2}$$

does it become...

$$3s + 1 = \frac{A}{(s+2)} + \frac{B}{(s+2)^2} + \frac{C}{4} + \frac{D}{4^2}$$

or...

$$3s + 1 = \frac{A}{(s+2)} + \frac{B}{(s+2)^2} + \frac{C}{4^2}$$

2. Apr 21, 2007

### arildno

What is the difference between a FACTOR, and a TERM?

3. Apr 21, 2007

### snowJT

no idea? I take it your saying its just $$\frac{C}{4^2}$$

Edit: oh nvm, you just ignore the part after B.. then add it in later...

Last edited: Apr 21, 2007
4. Apr 21, 2007

### Dan 1st

$$3s + 1 = \frac{A}{(s+2)} + \frac{B}{(s+2)^2} + \frac{C}{4} + \frac{D}{4^2}$$

seems the more logic answer to me ; but I personally have not yet studies partial fractions . If , underneath you have :
(s+2)² + 4²
, then why would it not be A divided by (s+2) + B divided by (s+2) instead of B being divided by (s+2)² ?

thought Like I said, I have not yet learnt these. If anybody can correct me , as I'm probably wrong , it would be appreciated of course.

5. Apr 21, 2007

### Gib Z

Partial Fractions Only work like that on FACTORS of the denominator. Is 4 a factor or just a term?

6. Apr 21, 2007

Looks to me like a Lapace Transform. No partial fractions required, split them and do some algebra work and you shell get your transforms. Yeah, definately a Lapace Transform, hey If you need any help, you are welcome to pm me, I can help you out. I saw your recent threads, that give me alittle clue. Here is a big hint, there will be a cosine and a sine answer.

Last edited: Apr 22, 2007
7. Apr 22, 2007

### robert Ihnot

This has an interesting solution which comes from splitting (s+2)^2+4^2 into imaginary parts. However we then have a new term in i, so that we need three equations to solve (for the numerical value, for the value in s and for the value in i):

$$\frac{3s+1}{(s+2)^2+4^2}=\frac{A}{s+2+4i}+\frac{B}{s+2-4i}+\frac{C}{(s+2)^2+4^2}$$

Then cross multiplying on the first two terms and adding the third gives:
A(s+2-4i)+B(s+2+4i)+C=3s+1. Giving A=B=3/2, C=-5.

Now should this be done in order to facilate integration, since A and B are the same, by cross multiplying the first two terms, we can eliminate the i term and are left with:

$$\frac{3(s+2)}{(s+2)^2+4^2}-\frac{5}{(s+2)^2+4^2}$$ So that integration is easy to perform, the first resulting in a log form and the second ,with the substitution 4u=s+2, in an arctangent form:

(3/2)In(s+2)^2+4^2) -(5/4)arctangent((s+2)/4)

Of course for integration it is now clear the only form needed is: $$\frac{3(s+2)}{(s+2)^2+4^2}-\frac{5}{(s+2)^2+4^2}$$

Last edited: Apr 22, 2007
8. Apr 22, 2007

### Gib Z

Umm Tiny error robert, (s+2)^2 + 4^2 factors into (s+2+2i)(s+2-2i). And Why did you put the 3rd fraction?

9. Apr 22, 2007

My bad, I didn't mean separate them into two equations. I mean split them and use the Lapace Transform identity of sine and cosine functions.

10. Apr 22, 2007

### Gib Z

Ahh actually I see Why We need the C, because I tried it without that and It turned out 3 had to equal 1/2, So i needed a constant term..Let me do it one second

11. Apr 22, 2007

why go through all that robert? you can do it so much quicker by separation?

12. Apr 22, 2007

### Gib Z

$$\frac{3s+1}{(s+2)^2+4^2}=\frac{A}{s+2+4i}+\frac{B} {s+2-4i}+\frac{C}{(s+2)^2+4^2}$$

Cross multiply, equate co efficents.

$$3s+1 = (A+B)s + (2A+2B + C) + (4B-4A)i$$

A+B=3
2A+2B + C =1
Since there are no imaginary terms in 3s+1, 4B-4A = 0.

Since 4B-4A=0, B=A. But A+B=3, A=B=1.5
Putting those values is to 2A+2B+C=1 gives C= -5.

$$\frac{3s+1}{(s+2)^2+4^2}=\frac{3}{2(s+2+4i)}+\frac{3} {2(s+2-4i)}-\frac{5}{(s+2)^2+4^2}$$

13. Apr 22, 2007

### Gib Z

Ok thanks for that, I just thought i could be treated like any other constant therefore didnt need any special treatment, but I see why im wrong thanks.

14. Apr 22, 2007

### robert Ihnot

mathPimpDaddy:why go through all that robert? you can do it so much quicker by separation?

I don't know, except the original question ask to find a partial fraction form. I am adding the assumption that this is for integration, since that is frequently the case.

15. Apr 22, 2007

I've checked his previous threads, they are all forms of laplace transforms. Look at his original form, doesn't it look familiar?

Last edited: Apr 22, 2007
16. Apr 26, 2007

### euler_fan

Personally, I avoid imaginary terms whenever possible, so factor the 3 out of the numerator and proceed with the common denominator.

$$3$$\frac{s+\frac{1}{3}} {(s+2)^2+4^2}$$$$

Last edited: Apr 26, 2007
17. Apr 27, 2007

### trickae

simple partial fractions help (warning complex analysis :P )

1. The problem statement, all variables and given/known data
the question can be ignored - it involves laplace and Z transforms of RLC ckts.

Code (Text):

Vc(s) =          0.2
-----------------
s^2 + 0.2s + 1

find the partial fraction equivalent such that it is :
Code (Text):

-j(0.1005)     +    j (0.1005)
--------------    ------------------
s + 0.1-(0.995)    s + 0.1 + j(0.995)

2. Relevant equations

none

3. The attempt at a solution
Code (Text):

0.2                      A                     B
---------------  =  ---------------------  +  -------------------
s^2 + 0.25 + 1      s + (0.1 - j(0.995)))     s + (0.1 + j(0.995))

0.2 = A(s + 0.1 + j(0.995)) + B(s + (0.1 - j0.995))

0.2 = As + A(0.1 + j(0.995)) + Bs + B(0.1 - j0.995)

so As + Bs = 0
or (A + B) = 0
or A = -B
so
0.2 = j(0.995A) - j(0.995B)

somethings not right - if i evaluate this I don't get anywhere near the answer

18. Jul 22, 2007

### EugP

The first one seems right, but I think you should simplify the denominator:

$$\frac{3s + 1}{s^2 + 2s + 4 + 16} = \frac{3s + 1}{s^2 + 2s + 20} = \frac{3s + 1}{(s + 1 + \sqrt{19}j)(s + 1 - \sqrt{19}j)}$$

So now you can do partial fractions:

$$\frac{3s + 1}{(s + 1 + \sqrt{19}j)(s + 1 - \sqrt{19}j)} = \frac{A}{(s + 1 + \sqrt{19}j)} + \frac{A^*}{(s + 1 - \sqrt{19}j)}$$

I'm going to assume you know how to solve for A and A* so i'll just post the answer that I got:

$$3.078e^{-t}cos(\sqrt{19}t - 0.225)$$