short-circuited termination?

Hi everybody, I try to simulate some inductors which are required to have short-circuit terminations, what does it mean? can I just put a port with very small impedance, say 0.00001, at the output? It seems that 0 is not acceptable. Please help. Thanks in advance.

Hi,

You need only to use a short at the end of a transmission line to create inductance. As long as it is between 0 and 90 degrees long.

Look at a Smith Chart. Image a short transmission line with 0 degress of length. Obviuosly this is a dead short and will appear at the leftmost point on the Smith.

As the line length inceases from 0 degrees, this point will begin to rotate clock-wise around the Smith Chart into the Inductive region. When the line reaches 90 degrees inductance is infinite and becomes an open circuit. This why a quarter-wave short = an open circuit.

This all has to with the signal reflecting back to the input of the transmission and how it interact with the income signal

0 degree short = short
Between 0-90 degree short = inductive
90 degree short = open
Between 90-180 degree short = capacitive
1800 degree short = short

0 degree open = open
Between 0-90 degree open = cacapitive
90 degree open = short
Between 90-180 degree open = inductive
180 degree open = open

Rod