Register to reply

Laplace transform into partial fractions, then I'm stuck!

by _Greg_
Tags: fractions, laplace, partial, stuck, transform
Share this thread:
_Greg_
#1
Aug8-10, 12:54 PM
P: 38
1. The problem statement, all variables and given/known data

It's hard to explain, I can do everything exept get the answer to what I'v pointed out below.
I just don't know what order to solve it in to get the correct answer, I must have tried every method except the right one!


2. Relevant equations

Click image for larger version

Name:	problem.JPG
Views:	33
Size:	49.1 KB
ID:	27424

3. The attempt at a solution

I'v tried lots of things but I'm obviously doing it completely wrong.

Thanks for looking
Phys.Org News Partner Science news on Phys.org
New type of solar concentrator desn't block the view
Researchers demonstrate ultra low-field nuclear magnetic resonance using Earth's magnetic field
Asian inventions dominate energy storage systems
╔(σ_σ)╝
#2
Aug8-10, 01:04 PM
╔(σ_σ)╝'s Avatar
P: 849
[tex]\frac{8}{9(-0.5 + 0.441i)(2*0.441i)}[/tex] = [tex]\frac{8}{-3.50066 -3.969 i}[/tex]

Now you can multiply the top and bottom by -3.50066 + 3.969i ( which is the complex conjugate of the denominator) . When you simpify you will arrive at the answer.
_Greg_
#3
Aug8-10, 01:24 PM
P: 38
How did you simplify the bottom line down to -3.50066 - 3.969i ?
What do you do with the 9? I find it confusing because its (...+...)(...x...) not ++ or xx
My basic maths has left me lol.
Thanks

╔(σ_σ)╝
#4
Aug8-10, 01:46 PM
╔(σ_σ)╝'s Avatar
P: 849
Laplace transform into partial fractions, then I'm stuck!

Quote Quote by _Greg_ View Post
How did you simplify the bottom line down to -3.50066 - 3.969i ?
What do you do with the 9? I find it confusing because its (...+...)(...x...) not ++ or xx
My basic maths has left me lol.
Thanks
I multiplied the complex numbers out .

9(a +bi) = 9a + 9bi

(a+bi)(c+di) = a*c + (a*d)i + (b*c)i - b*d
(2*ai) = (2*a)i

(*) denotes multiplication.
_Greg_
#5
Aug8-10, 04:09 PM
P: 38
I kinda get you, see if this is right:

8
-----------------------
9(-0.5+0.441i)(2*0.441i)

8
-----------------------
(-4.5+3.969i)(2*0.441i)

8
-------------------------
-8 -1.9845i + 7.938i +1.75i

8
---------------
-8 + 7.7035i

which equals!! -1 + 1.038

is that more or less correct do you think?
vela
#6
Aug8-10, 04:29 PM
Emeritus
Sci Advisor
HW Helper
Thanks
PF Gold
P: 11,722
No, it's not correct. How did you go from
Quote Quote by _Greg_ View Post
(-4.5+3.969i)(2*0.441i)
to
-8 -1.9845i + 7.938i +1.75i
?
_Greg_
#7
Aug8-10, 04:31 PM
P: 38
-4.5 * 2 = -8
-4.5 * 0.441i = 1.9845i
3.969i * 2 = 7.938i
3.969i * 0.441i = 1.75i

just used that 'foil' method
vela
#8
Aug8-10, 04:48 PM
Emeritus
Sci Advisor
HW Helper
Thanks
PF Gold
P: 11,722
You can't FOIL because the second factor isn't a sum. Also, 4.5*2 isn't 8; it's 9.

Just out of curiosity, what's the goal of this problem? To find y(t)? I just ask because I find the method used kind of questionable.
_Greg_
#9
Aug8-10, 05:06 PM
P: 38
oh yea, woops

The question is:

obtain the process response Y(t) as deviation from its initial steady state condition y(0). Use Laplace transforms and expansion by partial fractions. Assume a unit step forcing function x(t) = u(t)

The initial equation is 9y''(t) + 9y'(t) +4y(t) = 8x(t) - 4
vela
#10
Aug8-10, 05:26 PM
Emeritus
Sci Advisor
HW Helper
Thanks
PF Gold
P: 11,722
OK, the solutions you have took a less-than-ideal approach. It'll work, but it relies on working with complex numbers more than is necessary. When you work with factors of i, it greatly increases your chance of making sign mistakes, so if you can avoid working with them, it's usually for the best. Also, the solutions evaluated the roots numerically too early. it's usually not a good idea to evaluate numerical expressions midway through the problem because you end up with decimals, like 3.969, that you have to copy from line to line and sometimes you drop a digit or transpose two of them, etc. It's just asking to make a mistake. Plus, like any time you plug numbers in too early, it obscures possible cancellations down the road.

Anyway, back to your problem... You made one other mistake. You did something like

[tex]\frac{a}{x+iy} \rightarrow \frac{a}{x}+i\frac{a}{y}[/tex]

which isn't correct. What you need to do is multiply the top and bottom by the complex conjugate of the denominator, so you get

[tex]\frac{a}{x+iy} = \frac{a}{x+iy}\times\frac{x-iy}{x-iy} = \frac{a(x-iy)}{x^2+y^2} = \frac{ax}{x^2+y^2} - i\frac{ay}{x^2+y^2}[/tex]


Register to reply

Related Discussions
Partial fractions (for laplace) Calculus & Beyond Homework 3
Partial fractions/Laplace Calculus & Beyond Homework 1
Partial fractions and Laplace inverse.. Calculus 0
Partial Fractions/Laplace Transforms Calculus & Beyond Homework 2
Laplace Transform - (stinking partial fractions) Calculus & Beyond Homework 4