Algebraic Expansion with complex numbers

[h_ngm_n]

Hi all,

I am a bit rusty and have hit a snag with decomposition of partial fractions, I am taking an Engineering course dealing with Laplace transforms. The example is:

F(s)= 3 / s(s²+2s+5)

Now I get that there are complex roots in the denominator and that there are conjugate complex roots (s+1±2i) giving:

F(s)= 3 / s(s+1-2i)(s+1-2i)

So partial fraction decomposition would give:

F(s) = K₁/s + K₂/(s+1+2i) + K₃/(s+1-2i)

Ok, so solving for K₁= 3/5 is easy and I get that, but when it comes to K₂ which involves substituting (-1-2i) in for 's' and then expanding, I can't seem to get the answer. What I did was:

K₂=3[STRIKE](s+1+2i)[/STRIKE]/s(s+1-2i)[STRIKE](s+1+2i)[/STRIKE]
K₂=3/s(s+1-2i)
K₂=3/(-1-2i)[(-1-2i)+1-2i)]

It is the algebraic expansion on the denominator that is getting me. When I do it I get:

(-1-2i)[(-1-2i)+1-2i)]
= (-1-2i)(0-4i)
=-1(-4i)-2i(-4i)
=4i+8i²
=8+4i
=4(2+i)

However my textbook gets -3/20 (2+i) for the final partial fraction... I'm lost any help would be greatly appreciated.

 

[lewando]

4i+8i² = 4i-8 = [-8+4i, edit]

 

[h_ngm_n]

Ah yes, simple error... my bad. But it still doesn't match the answer in my text :(

 

[eumyang]

Ah yes, simple error... my bad. But it still doesn't match the answer in my text :(

That's because you haven't simplified K₂ completely. With lewando's correction the denominator simplifies to -8+4i, so K₂ is:
$K_2 = \frac{3}{-8 + 4i}$
Multiply the top and bottom by the conjugate of the denominator, and eventually, you will get:
$K_2 = -\frac{3}{20}\left( 2 + i \right)$

 

[h_ngm_n]

Ok, thank you for your time, I needed to simplify my fraction to get the complex term out of the denominator didn't I?

 

[eumyang]

Ok, thank you for your time, I needed to simplify my fraction to get the complex term out of the denominator didn't I?

Yes, the fraction wouldn't be considered simplified otherwise.

 

[Millennial]

It seems simpler to me to directly find the residues and then sum them to obtain the inverse Laplace, given that all poles here are simple. I do not know if you have any knowledge of the inverse formula, but it is certainly easier to apply here.

 

[h_ngm_n]

Hi Millennial,

So say that the residues are (took me a while to get this :)

K₁=-0.4615
K₂=0.2308-j2.0110
K₃=0.2308+j2.0110

I can make this

g(t)= -0.4615e^(-1.4t)+[(0.2308-j2.0110) e^(-(1-j0.7)t)+(0.2308+j2.0110) e^(-(1+j0.7)t) ] just by 'inspecting' the residues ? As opposed to breaking the partial fractions down and then doing inverse Laplace transform?

And I am little confused with the method to express it as trig identities:

The formulae: cosθ=(e^jθ+e^-jθ)/2

and: sinθ=(e^jθ-e^-jθ)/2j

I get a bit stuck using this to convert from exponential form to trig form...

 

[LCKurtz]

Of course, you can avoid all the complex number stuff by using the partial fraction expansion$$
\frac 3 {s(s^2+2s+5)}= \frac A s + \frac{Bs+C}{s^2+2s+5}$$and writing $s^2+2s+5=(s+1)^2+4$ when taking the inverse Laplace Transform.