# Modeling a transmission line

I want to simulate a transmission line in PSPICE, however I ran into a problem. I got floating nodes error (I looked it up and I was suggested to put a ground on the far side of the transmission line which sounds wrong, it means that the voltage is zero when I short the load). After simulating I got weird results that I did not expect. I'm attaching a photo of my circuit, Help would be appreciated!

Edit solved my problem with changint the LEN parameter of the line,

However that fact that I need to place a ground on the right side still seems wrong to me.

• What if you placed a 1gigaohm resistor to ground on the far side instead of the short? May 16 at 16:26

I'm not a PSpice user, but I believe I correctly duplicated your circuit in LTspice. The fact that you required the len parameter specified tells me that the RG58/U coax symbol you're using infers a lossy transmission line model (i.e. the LTRA model). I plugged in some per-unit RLC values for 20AWG RG58/U coax I found on the web and came up with what's below for a 1 meter length of it. I disabled LTspice's topology checker via the .options statement since it tries to "fix" certain problems on its own and I want it to behave more like PSpice right now. It of course gives me an error because of this.
Node N002 is the node at the bottom of R2 and is considered "floating" to the underlying SPICE engine. Knowing roughly how the SPICE engine calculates things, every node needs a DC path to ground (i.e. the "zero" node) to be able to formulate and solve the circuit matrix. This far side of the transmission line needs a DC reference in order for the matrix math to work. Like you mentioned in the question, you can put a ground directly at this node to solve the problem. However, if that bothers you then you can also opt to connect it to ground via a very low conductance (high resistance) such as 1t or 1g. This still satisfies the requirement of the SPICE engine, and will allow the simulation to execute properly. No current ends up flowing through this resistor, except maybe some noise due to numerical artifacts of the calculations (as shown below in the sub-femtoamp range).
• @tobalt Great question. You can definitely do that as well to solve the circuit matrix problem. Your guess is correct. However, now the out node will be fixed at zero volts. So to get the correct waveform you must do a differential probe across the resistor from bottom to top, i.e. V(n002,out). May 18 at 15:41