pulse duration in transmission line

screen1988

Joined Mar 7, 2013
310
I got stucked in the question 5 from worksheet:http://www.allaboutcircuits.com/worksheets/charz.html

I got the right answer with the value of pulse approximately 480μV but the pulse duration is wrong.
Here is what I calculate the pulse duration:
1) With velocity factor 0.63,
t1 = 10 miles/(0.63*186 000 miles/s) = 85.34 μs
2) With velocity factor 0.66,
t1 = 10 miles/(0.66*186 000 miles/s) = 81.46 μs

Attachments

• 34.3 KB Views: 61

crutschow

Joined Mar 14, 2008
25,686
The pulse current will be constant until the reflected pulse returns from the far end of the line (assuming the far end is unterminated).

screen1988

Joined Mar 7, 2013
310
Yes, it is 480μV but I want to know the time that pulse last. Can you check my result and tell me what I am wrong?

crutschow

Joined Mar 14, 2008
25,686
You calculated the time for the pulse to get to the end of the line. But you won't see the effect of that until the reflection from that is returned to the source. Thus the pulse time is double of what you calculated.

Ron H

Joined Apr 14, 2005
7,014
You will get more than one pulse.
Are you assuming that RG-58 is lossless?

screen1988

Joined Mar 7, 2013
310
You will get more than one pulse.
Are you assuming that RG-58 is lossless?
Yes, it is ideal.
Do you mean that two pulses are incident pulse and reflected pulse?
I am thinking about it but they have the same amplitude and time duration.
The second pulse will continue the first pulse right after the first pulse end.
I am having a hard time in imagining reflected wave in terms of electrons.
When electrons go to the end and there is no path for it to continue, it will come back to source?

Tesla23

Joined May 10, 2009
411
I think the answer to this question in the workbook is wrong. Ignoring cable losses and the 1mΩ resistor, the voltage on the far end of the cable is a square wave that switches between 0V and 48V and the current from the source is a ±480mA square wave (in quadrature with the voltage on the far end).

screen1988

Joined Mar 7, 2013
310
Tesla23,
How can you get 48V at the far end?

Tesla23

Joined May 10, 2009
411
When you close the switch, a +24V amplitude wave is sent down the cable, at the far end the open circuit reflects a +24V wave back to the source, these two waves add to 48V at the open circuit.

You should draw a reflection diagram to understand what is happening here.

To get a single pulse on the line you need to source terminate - make the 1mΩ resistor 50Ω, then the far end goes to 24V (after the line delay) and stays there.

screen1988

Joined Mar 7, 2013
310
Tesla23,
Can you help me explain how 24V amplitude wave is sent down the cable in terms of electrons movement? I have read it here but still don't understand much:http://www.allaboutcircuits.com/vol_2/chpt_14/3.html.
I want to understand how wave is propagated and reflexed at the far end in terms of electrons movement.
The transmission lines include many capacitor and inductors. In the picture in the site, it seems that these capacitors are charged in turn, one after another.
At first the first capacitor is charged fully to maximum value and then the capacitor will be a source and charge the second capacitor. The progess goes on to the far end.
My question is the understaning above right?

Ron H

Joined Apr 14, 2005
7,014
When you close the switch, a +24V amplitude wave is sent down the cable, at the far end the open circuit reflects a +24V wave back to the source, these two waves add to 48V at the open circuit.

You should draw a reflection diagram to understand what is happening here.

To get a single pulse on the line you need to source terminate - make the 1mΩ resistor 50Ω, then the far end goes to 24V (after the line delay) and stays there.
That is the answer I hinted at in my previous post:
You will get more than one pulse.
The square wave theoretically would go on forever, in the absence of any resistor.
I failed to realize that this was not homework, (I didn't see the link), so I didn't want to give away the answer.