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Algebraic Expansion with complex numbers

  • Thread starter h_ngm_n
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  • #1
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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.
 

Answers and Replies

  • #2
lewando
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4i+8i2 = 4i-8 = [-8+4i, edit]
 
Last edited:
  • #3
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Ah yes, simple error... my bad. But it still doesn't match the answer in my text :(
 
  • #4
eumyang
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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:
[itex]K_2 = \frac{3}{-8 + 4i}[/itex]
Multiply the top and bottom by the conjugate of the denominator, and eventually, you will get:
[itex]K_2 = -\frac{3}{20}\left( 2 + i \right)[/itex]
 
  • #5
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Ok, thank you for your time, I needed to simplify my fraction to get the complex term out of the denominator didn't I?
 
  • #6
eumyang
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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.
 
  • #7
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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.
 
  • #8
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Hi Millennial,

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

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+e-jθ)/2

and: sinθ=(e-e-jθ)/2j

I get a bit stuck using this to convert from exponential form to trig form...
 
  • #9
LCKurtz
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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.
 

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