# propagation coefficient - Complex Numbers

Discussion in 'Homework Help' started by Kayne, Feb 23, 2011.

1. ### Kayne Thread Starter Active Member

Mar 19, 2009
105
0
Hi There,

I am a little confused to how to work out a complex number that needs to be squared. For example

$y =a+jB = sqrt[(R+jwL) (G+jwC)]$

where
R=0.2Ω/km
L=1.1mH/km
C=0.01uF/km
G= Assume ≈ Negligible
Freq = 50Hz

$= sqrt[(0.2+j2*Pi*50*1.1mH) (0+j2*Pi*50*0.01uF)]$
$= sqrt[(0.2+j0.3455) (0+j3.141*10^-6)]$

I know that with complex numbers that to multiply the process is

$(ac-bd)+j(ad+bc)$
so

$= sqrt[(0.2*0)- (j0.3455*j3.141*10^-6) + (0.2*j3.141*10^-6)+(j0.3455*0)]$

$= sqrt[(1.085*10^-6) + (628*10^-9)]$

Then If i square roots this the answer that I get is not correct. The answer that is given is

$= 0.29*10^-3 +j1.082*10^-3$

I am not sure what I am doing incorrect. So any guidance would be appricated.

Thanks

2. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
You have an error here. The expression $(0.2+j0.3455) (0+j3.141*10^-6)$ written in the form (a+jb)(c+jd), you have a=0.2, b=0.3455 (not j0.3455), c=0 and d=3.141E-6.

Therefore the expression equals $-0.3455 \cdot 3.141 \cdot 10^{-6}+j \cdot (0.2 \cdot 3.141 \cdot 10^{-6})$

Can you take it from here?

3. ### Kayne Thread Starter Active Member

Mar 19, 2009
105
0
So

$= sqrt[(-1.085*10^{-6})+j(0.628*10^{-6})]$

then I take out the $10^{-6}$

$= 10^{-6}sqrt[-1.085+j(0.628)]$

when i sovle this I still get the incorrect answer..... Still unclear sorry

Is there any tutorials on this type of questions, that anyone knows of?

4. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
Could you post the original question? I noticed that the quantities R, L and C are expressed per kilometer, so the size may differ.

However, in the exercise's defence, for 1km, the correct answer is the one suggested and you mentioned. I, from my part, suggest you take more care of your number crunching.

As an example, you forgot to change $10^{-6}\ to\ 10^{-3}$ when you took it out of the root.

Last edited: Feb 24, 2011
5. ### Kayne Thread Starter Active Member

Mar 19, 2009
105
0
Hi Georacer,

The question I have attached,

I will have another look at the question and work though it again slowly.

I guess what I am after is the steps that need to be taken so I can work this out on paper without using a computer program.

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6. ### Kayne Thread Starter Active Member

Mar 19, 2009
105
0
Georacer,

I have tried to solve this with Matlab but the code I am using is incorrect.

I have tried to input

$complex sqrt{(R+jwL)(G+jwC)}$ with the corresponding values above and there are errors..

If this what you used to solve the problem? If so how did you enter it?

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
perhaps more like

$\gamma=sqrt (complex(R,\omega L)*complex(0,\omega C))$

but using the function 'sqrt(operand)'

The '*' just denotes multiplication

Last edited: Feb 25, 2011
8. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Here's an approximately equivalent sequence in Scilab ...

• ###### Scilab_1.png
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9. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
Here is an exact Matlab code snippet:
Code ( (Unknown Language)):
1.
2. R=0.2;
3. L=1.1*10^(-3);
4. C=0.01*10^(-6);
5. G=0;
6. f=50;
7.
8. omega=2*pi*f;
9. y=sqrt((R+omega*L*i)*(G+omega*C*i))
10.

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10. ### Kayne Thread Starter Active Member

Mar 19, 2009
105
0
Thanks for that, I will try that code now and see how I go. This seems an easier way to solve these types of questions.

11. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
It won't help you in an exam, though. Better practice by hand.

12. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
It might also help if you remembered there is a polar form of complex numbers.

For instance it can be helpful to use the polar form in the square root operation you have been considering. Convert the rectangular form to polar form and then use this realtionship

$\sqrt{(X\angle \theta)(Y\angle \phi )}=\sqrt{XY}\angle( \frac{\theta + \phi}{2})$