Rectangular Pulse on a Transmission Line

[ask_LXXXVI]

Homework Statement

In Fig a line is used to produce a short rectangular pulse of width 12ns and peak value 800V. With S-2 open. S-1 is closed to charge the line to V_dc; after charging, S-1 is opened. Then, at t = 0, S-2 is closed to discharge the line through R_R and form the pulse. Find the length of line and V_dc
[PLAIN]http://img51.imageshack.us/img51/4968/capturezc.jpg

Answer :- 1.2 m , 1600 V.

Homework Equations

v = 1/\sqrt{\mu\xi}

The Attempt at a Solution

Velocity in the Tx line would be 2 x 10^8 m/s by making use of the formula.

Also as R_R = R_o , equivalent ckt while discharging would be a voltage divider circuit , so Vout will be 1/2(Vdc) .So we need a Vdc of 1600 V.


But I am unable to work out how the pulse would be generated.

I mean how is the Tx line charged and discharged in general.After switch S2 is closed do we treat the lines as a voltage source and for how much time will the discharge take place?

Confused , help please.

 

[berkeman]

Homework Statement

In Fig a line is used to produce a short rectangular pulse of width 12ns and peak value 800V. With S-2 open. S-1 is closed to charge the line to V_dc; after charging, S-1 is opened. Then, at t = 0, S-2 is closed to discharge the line through R_R and form the pulse. Find the length of line and V_dc
[PLAIN]http://img51.imageshack.us/img51/4968/capturezc.jpg

Answer :- 1.2 m , 1600 V.

Homework Equations

v = 1/\sqrt{\mu\xi}

The Attempt at a Solution

Velocity in the Tx line would be 2 x 10^8 m/s by making use of the formula.

Also as R_R = R_o , equivalent ckt while discharging would be a voltage divider circuit , so Vout will be 1/2(Vdc) .So we need a Vdc of 1600 V.


But I am unable to work out how the pulse would be generated.

I mean how is the Tx line charged and discharged in general.After switch S2 is closed do we treat the lines as a voltage source and for how much time will the discharge take place?

Confused , help please.

I think you have it right. The problem statement is a little confusing, but the switch S1 is closed for however long it takes to charge up the TL, and then opened. Switch S2 is closed at t=0 and stays closed. Since the TL Zo equals the termination resistance, you are correct that it acts as a 2:1 voltage divider. So that gives you the Vdc that the TL is charged up to, and the propagation time gives you the length of the cable.

It seems that things would get a lot weirder if the termination did not match the Zo -- you'd have reflections and re-reflections to deal with then...

Of course, it would take a bounceless switch to make this work in the real world. Kind of a clever way to make short pulses, though.

 

[ask_LXXXVI]

can you please explain how exactly the discharge takes place.


see before t=0 entire Tx line is charged to a voltage V. I am unable to visualise what happens after switch S2 is closed.

What is voltage across the line at different points of time ? it would be great if someone plots the V(x) plots for different values of t , i.e V(x,t1) V(x,t2) and so on. That would help the visualisation process.