How can I derive the Capacitance from the open circuit line and the
inductance of the short circuit line? How is phase shift related to
these values.
A short open circuit line can be used to measure the capacitance per metre but you need to do it at a low enough frequency so that the inductance isn't starting to affect readings. Because the line is open circuit and you are using low frequencies, small volt drops due to inductance are minimized. I'm also assuming the the parallel conductance of the line is negligible.
You can measure line capacitance by measuring the impedance the short open-line presents to a voltage source - this is largely Xc if you ensure the frequency is relatively low i.e. 1 kHz for example.
You could then do a short circuit test and calculate inductance but, if you already know the characteristic impedance of the t-line then you can use the following formula: -
\$Z_0 = \sqrt{\frac{L}{C}}\$
You have measured C and you know Z0 so L is calculated.
How is phase shift related to these values
I assume you mean the phase difference between input and output. This is determined by the velocity of propagation of the cable. As a percentage of the speed of light it is: -
Velocity factor = \$\dfrac{1}{c_0\sqrt{LC}}\$
So, if your short cable is 1 metre long and VF = 50% then it will take 6.667 ns for a signal to reach the end. This means that if you stimulus frequency is a sinewave of 75 MHz (period = 13.333 ns), it will undergo a phase shift of 180 degrees on that 1 metre length of cable.
