# Please, ive been stuck on this inverse laplace for awhile

1. Oct 14, 2005

### mathrocks

I have to find the laplace inverse of a function y(s) which has repeated complex roots.
Y(s)=s^2 / (s^2+4)^2
so s=2i, s=2i, s=-2i, s=-2i.

My partial fraction is as follows:
A/(s-2i) + B/(s-2i)^2 + C/(s+2j) + D/(s+2j)^2

I use the standard method for finding regular repeated roots but I get stuck trying to calculating C and D. My values are undefined. My work is below...

A= d/ds[(s-2i)^2*Y(s)]=8s(3s^2-4)/(s^2+4)^3 + 4s^2(s^2-12)*i/(s^2+4)^3-->then you set s=2i which then results in A=-6i.

And B=1

But now for C, when I use the same process as A but instead of 2i, I use -2i, my answer is a number over 0 which results in undefined.

Am I even doing this problem correctly? Any help would be appreciated...

Thanks!

2. Oct 14, 2005

### Tide

You didn't do your partial fraction decomposition correctly. Because the polynomials are quadratic you need terms like (As + B) in the numerators.

BTW - if you are familiar with the Cauchy integral formula you could save yourself a lot of work and just evaluate the residues directly to evaluate the integral.

3. Oct 14, 2005

### Disar

I gave it a go.

The values I have are:
A=undefined
B=1
C=undefined
D=1/4

Checked with maple 10
A=0
B=1
C=0
D=-4

However this was not done in complex numbers (for maple)

For my results I followed the method of residues. When evaulating for the Unknowns A and C (where you use d/ds) both resulted in undefined which I chose to interpret as just leave it alone

Obviously, you could evaluate this another way which would be advantageous right now but on a test where class lasts fifty minutes you need to know which method works most accurately. Since I am a novice at this I cannot be of much more help.

4. Oct 15, 2005

### mathrocks

Actually I just realized my Y(s)=s^2 / [(s^2+4)^2*(s^2+2)^2]
so s=2i, s=-2i, s=sqrt(2)i, s=-sqrt(2)i
Thus, resulting in no repeating roots.

But I still have problems coming up with the coefficients.

I think you setup the partial fraction like this:

As+2B/(s^2+4)^2 + Cs+sqrt(2)D/(s^2+2)^2

But I don't know where to go from there....

5. Oct 15, 2005

### Disar

Here is what I got

Ok, given:

Y(s) = s^2
[ (s-j2)^2*(s+j2)^2*(s-jsqrt(2))^2*(s+jsqrt(2))^2]

which equates to:

=K1/(s-j2)^2 + K2/(s-j2) + {plus the conjugates of K1 and K2}

Q1/(s-jsqrt(2))^2 + Q2/(s-jsqrt2) + {plus the conjugates of Q1 and Q2}

Solving for K1, K2, Q1 ,Q2 using the method of residues.

K1=6.1
K2=0
Q1=1/16
Q2=0

I do not know if this is correct but this is what I got. I really can't stand complex partial fraction decomposition.

6. Oct 15, 2005

### mathrocks

See, I haven't learned the method of residues...This problem is for my signals and systems class and I'm only allowed to use inverse laplace methods.

7. Oct 15, 2005

### Disar

If you you've taken circuit analysis 2 then you should have seen the method of residues. Maybe you should refer to it. If not google it. If you are interested I can explain it to you quickly.