# Homework Help: Transmission line Secondary Coefficients

1. Jan 31, 2017

### David J

1. The problem statement, all variables and given/known data
A transmission line has the primary coefficients as given below.

$R=2\Omega/m$
$L=8 nH/m$
$G=0.5 mS/m$
$C=0.23 pF/m$

Determine the lines secondary coefficients $Z0$. $\alpha$ and $\beta$ at a frequency of $1 GHz$

2. Relevant equations

In my notes I am given

$\alpha=\frac{R}{2}\sqrt\frac{C}{L} +\frac{G}{2}\sqrt\frac{L}{C}$ and $\beta=\omega\sqrt{LC}$

3. The attempt at a solution

$\alpha=\frac{2}{2}\sqrt\frac{0.23 X 10^-12}{8 X 10^-9} +\frac{0.5 X 10^-3}{2}\sqrt\frac{8 X 10^-9}{0.23 X10^-12}$

$\alpha=1 X \sqrt{2.875 X 10^-5} + (2.4 X 10^-4)\sqrt{34,782.6}$

$\alpha=\sqrt{2.875 X 10^-5} + (2.4 X 10^-4)\sqrt{34,782.6}$

$\alpha=(5.362 X 10^-3) +0.044760 = 0.050121902$ nepers per meter

I think this is correct. I am unsure how to input the single multiplication sign X in LaTeX form. I think my " to the power of`s" are correct for $R, L, G$ and $C$ but I am unsure about the final result in nepers per meter

For the second part I got the following:-

$\beta=\omega\sqrt{LC}$

$(2\pi)(1 X 10^9) \sqrt{(8 X 10^-9)(0.23 X 10^-21)}$

So I have $6,283,185,307\sqrt{1.84 X 10^-21}$

So $6,283,185,307(4.289522 X 10^-11)=0.269518623$ radians

So $\beta= 0.269518623$ radians

This second answer I am not so sure as I have some very large numbers but I have followed the examples in my notes.

Any comments on the two attempts above would be appreciated.

Thanks

2. Jan 31, 2017

### Staff: Mentor

The values look okay. $\beta$ should be radians per meter. You'll want to round to the appropriate number of significant figures to match your "givens".

For multiplication in LaTeX you can use \times or \cdot : $a \cdot b = a \times b$.

3. Jan 31, 2017

### David J

Thanks a lot for your help with this

4. Apr 6, 2018

### Connorm1

l Following onto this question i got all the same workings however we needed to find Zo aswell. Using with the values above I got Zo=179.427+j26.5060 Ω or in polar Zo=181.375 /_+8.403° Ω (sorry don't currently have software to do the polar expression so used /_ to signify the angle).Does this sound correct for this answer?

5. Apr 6, 2018

### Connorm1

https://www.wolframalpha.com/input/?i=√((2+(16π)i)/(0.0005+i(0.00046π))) Here's my workings using wolframalpha

6. Apr 6, 2018

Looks good.

7. Apr 6, 2018

### Connorm1

Thanks @gneill! Only bit i was struggling on! But I hoped it'll be as simple as use the equation and plug in values. Helpful as always :)

8. Apr 11, 2018