What is the difference between an inductor and transmission line?

suppose we have a transmission line and inductor, both have one end shorted to ground, if the TLINE increase its length and the inductor increase its value. What do they look like in the Smith Chart?

I was asked this question in an interview, my answer is the TLINE will go around the Smith Chart from the short (left side) while the inductor will never get to the negative part of the Smith Chart, it will get to the open (right side) if the L is infinite.

Then I was asked: why?

I said because L=sL or jwL, whose imaginary part is always positive. So it will stay on the upper side of Smith Chart.

Then I was asked: why?

I said because the way Smith Chart is setup determine that the positive part is on the top and the negative part is on the bottom.

Then I was asked: why is that? what is the different between an inductor and TLINE?

---------- Post added at 22:43 ---------- Previous post was at 22:18 ----------



Hi BigBoss:

At that time I was stuck in the question that why the inductor won't go beyond the open point to the negative part of Smith Chart.

Thanks for your answer. The inductor starts from the short (left) point btw.

In fact they behave the same

A shorted transmission line acts as an inductor when the length is less than lamda/4 and as a capacitor when the length is greater than lamda/4. Using the following expression of input impedance for shorted transmission line: Zin = jZo*tan(beta*l).

At a single frequency they can look identical. However, if you sweep the frequency, and inductors reactance follows X=jωL, while the shorted transmission line acts as a complex tangent function.