Hi all,

I need some help analyzing the circuit below. I understand that at t=0- the steady state Load Voltage is 9 V, and that when the switch is thrown it needs to bounce to 10.285 V, but I can't figure out how to get to the first voltage propagation (V+).



My first try is to start at t=0+, R_eq = 25 Ohms, then V = 12 V * (Z0/(Z0+R_eq)) to get 8 V, but bouncing from this value puts me below 9V when i know that the value at infinity should be 10.285V.

My professor gave us solutions to the homework, but i don't know where he got the equation for V+. I've included his equation below:



I don't need help with bouncing the voltage, just getting the conditions at t=0+ (i.e. i need to find V+ after 0). Any help would be appreciated.

Thank you for your time,
-Barry

 

Hi,

A few questions...
So your transmission line is lossless?
Is that a 150 ohm load?
What position along the line are you solving for?

 

Hi all,

I need some help analyzing the circuit below. I understand that at t=0- the steady state Load Voltage is 9 V, and that when the switch is thrown it needs to bounce to 10.285 V, but I can't figure out how to get to the first voltage propagation (V+).

View attachment 136221

My first try is to start at t=0+, R_eq = 25 Ohms, then V = 12 V * (Z0/(Z0+R_eq)) to get 8 V, but bouncing from this value puts me below 9V when i know that the value at infinity should be 10.285V.

My professor gave us solutions to the homework, but i don't know where he got the equation for V+. I've included his equation below:

View attachment 136222

I don't need help with bouncing the voltage, just getting the conditions at t=0+ (i.e. i need to find V+ after 0). Any help would be appreciated.

Thank you for your time,
-Barry

I would do it a little differently to your prof, in this sort of problem I prefer to work simply with the traveling waves, even for the dc condition. At t=0-, you have two waves on the line, V+=6V, V-=3V. At t=0+ at the left end of the line, you can only change V+, and you can write two equations for the tx line relating V and I at that point:
V = V+ + V-
I = (V+ - V-) / Z0

along with V = 12- 25*I for the source.

Using V- = 3V from before, you can find the new V+ which I get as 7V. As this is 1V above the dc value, I guess this corresponds to your prof's solution.