I would view the pulse source of 12V amplitude as being imposed on a series combination comprising R1 and Z01 - much like a voltage divider.

Equal values of R1 and Z01 would give equal voltage drops - 6V each.

Doesn't that apply only for circuit with R1 and Z01? In this the pulse would die before it even reaches the second TL...

Edit: Looks like you're right. I found a similiar problem and the pulse was evenly distributed between R1 and Z01. I still don't understand why. But there was no voltage drop in the transmission lines.

Remember energy is proportional to the square of the voltage. I realize it is confusing if you just add the two voltages algebraically.

You ask what appears to be a logical question - "Isn't 0.5*E+0.75*E greater than the original E value - that means more energy! ...?"

You need to think it through carefully.

You're right. I was going the easy way.

On more question. Since the point P is in the middle of the TL, can I assume that the voltage has dropped half the value of the total voltage drop in the line.

 

You are still thinking of the line impedance as a resistance which absorbs energy. The line in this problem is regarded as a lossless wave guide - it "transports" energy without absorbing it. The 50Ω is the characteristic impedance of the line. Even "free space" has a characteristic impedance to radio waves - 377Ω - and not a resistor or piece of wire in sight! [OK there is lot's of junk in space thanks to various space programs - North Korea claims to have added more recently.]

As I stated in an earlier post, since the pulse is traveling along the line with time, the position of the observer will determine what is observed at that point. Imagine what you would see as the observer at the mid-point. Say you've just seen (you've got to be quick ) the pulse pass by at some time - when will you see it next? - given it has to wander down to one end or the other and then make its way back to you. And so on ......

The lossless condition assumption means that the pulse amplitude will remain constant during any single pass between the ends of the (second) line. The amplitude reductions occur at the reflections as some energy either passes back to the source or reaches the load which, being a resistor, does absorb the energy.

Hopefully that is the end of the story.

Good luck with the rest of the work.

 

Got it now, hopefully its right. I still can't really understand the meaning of the characteristic impedance, though. So in ideal conditions, like this, the impedance only affects reflection and transmission. But I was going through this site's volumes, and it said that a 50Ω line acts as 50Ω resistor for certain amount of time (obviously depending on the lenght of the line and speed of the signal), and from there on acts according to the rest of the configuration. Maybe our teacher hasn't gone in-depth enough with resistor, 'cause all I know resistor do is consume energy, and hence says where the current'll go.

Thanks for the help, t_n_k. This cleared alot of things.

 

Hi again,

No problem

I have just found a link that might be of interest - it includes a discussion of pulse propagation on lines - also discusses the transmission factor.

http://books.google.com.au/books?id...X&oi=book_result&ct=result&resnum=3#PPA999,M1

 

But I was going through this site's volumes, and it said that a 50Ω line acts as 50Ω resistor for certain amount of time (obviously depending on the length of the line and speed of the signal), and from there on acts according to the rest of the configuration.

You also made an important point I had missed here - the characteristic impedance would be seen at the source because the pulse has not yet propagated down the line. The conditions further along the line at time zero are unobservable from the perspective of the source.

For continuous signals the outcome is different and at steady state will be dependent on remote line conditions.

 

can i get any note on transmission lines

 

can i get any note on transmission lines

Whilst I don't condone hijacking other members threads, the answer to this will be of help to the OP: Volume II Chapter 14

If you have further questions, please start a new thread. Thank you.

Dave

 

t_n_k, I've got say one more thing. You had said everything perfectly clear multiple times, but for some reason I was bangin my (and your) head to the wall. I guess my small brains need some time to digest all the information... Your posts all clicked today as we started on steady-state transmission lines in AC. Oh, and I got full points for the problem! Thanks again!

Edit: Dave, your avatar is too hypnotic! I stared at it for a while and couldn't stop...

 

Hi whalum,

You are very welcome - congratulations on a good score.

It's a relief when things click into place - I remember being in your situation as a student - many moons ago!

All the best with your studies.