# Transmission Lines

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1. Apr 7, 2016

### Sum Guy

1. The problem statement, all variables and given/known data

I am having problems with the second part of the question - proving that the relationship given is true.

2. Relevant equations
See question.

3. The attempt at a solution
Firstly, consider a single pair of transmission lines with characteristic impedances $Z_{1}$ and $Z_{2}$. My interpretation is that each of these segments have no load impedances on their own. We can say that this pair of transmission lines is equivalent to one transmission line whereby the load impedance is the input impedance as seen by the transmission line with characteristic impedance $Z_{2}$. So the overall input impedance of this pair is:
$$Z_{in} = Z_1 \times \frac{Zcos(kl) + iZ_{1}sin(kl)}{Z_{1}cos(kl) + iZsin(kl)}$$ where $Z = Z_{2}itan(kl)$.

Following this through we end up with:
$$Z_{in} = \frac{isin(kl)cos(kl)[Z_{1} + Z_{2}]}{cos^{2}(kl) - \frac{Z_2}{Z_1}sin^{2}(kl)}$$
I then thought about adding another pair of these transmission lines and enforcing the rule that the input impedance shouldn't change, but by my flawed reasoning would reduce this to adding a resistor in parallel whilst ensuring that the overall resistance did not change (which cannot be the case for something non-trivial).

2. Apr 8, 2016

### rude man

This part is incorrect I think (always "I think" heh heh).
This problem is just very messy. Once you have Zin for the two-section concatenated line, add two more identical concatenated ones to get Zin for the 4-section line in the same manner, then force the ensuing Zin to be the same as for the 2-section line.

3. Apr 8, 2016

### Sum Guy

My reasoning was as follows:
$$Z_{in 2} = Z_{2} \times \frac{Zcos(kl) + iZ_{2}sin(kl)}{Z_{2}cos(kl) + iZsin(kl)}$$ where $Z = 0$ (?)
Giving $$Z_{in 2} = Z_{2} \times \frac{iZ_{2}sin(kl)}{Z_{2}cos(kl)} = Z_{2}itan(kl)$$
What is wrong here?

4. Apr 8, 2016