# Transmission lines

Discussion in 'Homework Help' started by Whalum, Apr 3, 2009.

1. ### Whalum Thread Starter Member

Mar 10, 2009
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0

Graph the voltage as a function of time in point P. The voltage E is given in the picture.

I'm so lost I don't even know where to start. All I know is, that there's no reflection between R1 and Z01.

Any tips on how to start are appreciated!

2. ### beenthere Retired Moderator

Apr 20, 2004
15,815
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Transmission lines tend to be sensitive to frequency in terms of length and so on. But think of the duration of the pulse and the propagation time through the TL - recalling the speed of light. That may clarify your problem.

3. ### Whalum Thread Starter Member

Mar 10, 2009
11
0
Nope, not getting it. This whole concept of transmission line is somehow so different from anything I've done before. Let's try baby steps. So it takes 2.5 microseconds for the pulse to reach point P. But how do I determine the voltage drop in R1? Can I use normal voltage division? So it would be $U_{R1}= \frac{R_1}{R_1+Z_{01}+Z_{02}+R_L}E=2V$

4. ### KL7AJ AAC Fanatic!

Nov 4, 2008
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Alas...our ultraparanoid server at work has excised your drawings....or I'd be able to help. Could you attach the files to an email to me?

eric.nichols@eielson.af.mil

Eric

5. ### KL7AJ AAC Fanatic!

Nov 4, 2008
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I'm currently teaching a comprehensive section on the Smith Chart in my A.C. theory class.

Mar 10, 2009
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7. ### wr8y Active Member

Sep 16, 2008
232
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I wish I was there.

8. ### KL7AJ AAC Fanatic!

Nov 4, 2008
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You can enroll in the Fall semester.

eric

Nov 25, 2008
394
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Lefty

10. ### t_n_k AAC Fanatic!

Mar 6, 2009
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There are two points where reflections will occur. Where the two transmission lines meet (with unequal characteristic impedances) and at the load - due to load mismatch with the second line.

Work out the reflection coefficients at each point and draw a time graph with a superposition of the forward and backward traveling pulses, having regard to the reflection coefficient signs.

11. ### wr8y Active Member

Sep 16, 2008
232
1
Alaska? I moved to Georiga from Michigan 8 years ago 'cause 40 years of winter was enough! No thanks, I'll stay here and enjoy the warm weather and year around hiking. Besides, "Georgia HOPE" pays for me to go to school here.

But I WOULD like to sit in your class...

12. ### Whalum Thread Starter Member

Mar 10, 2009
11
0
Will the reflected signals turn to heat completely at R1? And will the signal that reaches the load turn to heat as well?

Last edited: Apr 4, 2009
13. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Yes.

You may have deduced by now that owing to the match discontinuity at the junction of the two lines and the load mismatch at the end of the second line you will end up with a pulse of ever decreasing magnitude and alternating sign "bouncing" back and forth between the two reflection points in the second line at light speed. Depending on where you "stand" on the second line as an observer, determines what you record as the time plot of the pulses as they come and go. In your original question I'm assuming the diagram intends to indicate the observer's position as mid-point on the second line.

14. ### Whalum Thread Starter Member

Mar 10, 2009
11
0
So what I get now, is: voltage drop in the first resistor and TL is 4V, so the voltage goin into the second TL is 8V. Reflection coefficient between first and second TL is .5, so the magnitude of signal that gets transmitted, is 1.5 times greater, 12V. And the pulse that gets "trapped" in the second TL gets half the magnitude and opposite sign every reflection.

There are two things bothering me. My logics are against my course material, which says that the transmission coefficient can be calculated using $\tau=1+ \rho$, where rho is the reflection coefficient. Are we getting free energy? And the second thing: is the trapped pulse really going to be there for ever?

Last edited: Apr 5, 2009
15. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Not sure where the 4V came from.
Since the first line has Z0=100 and R1=100 I would expect the pulse amplitude at the line input would be 6V.

Also not sure about $\tau=1+\rho$ .... never used it myself.

If I wanted to find the pulse amplitude when it first enters the second line at the line junction, I would do an energy balance for the incident, reflected & forward values.

I think it would go something like ...

W_incident = W_reflected +W_forward [W being the energy]

Why?: The sum of reflected and forward energies must equate to the incident energy.

So..
$Wi=(E^2/Z01)*\delta t$

where $E=6V \ and\ \delta t = 1\mu sec$

I'd need to keep in mind that there is a different impedance on the second line - Z02=300Ω.

As to the pulse trapped on the second line, I think it would gradually loose energy each time it hits the mismatch at each end. Some energy leaks back to the source and some goes into the load. Does it go on forever? I guess it stops when all the original energy is exhausted - which would be at time=∞. This is an ideal situation in which no line losses are included.

I'm happy to be corrected on any of these assumptions.

Do you actually have access to a solution for the problem? Say from the source text ...

16. ### Whalum Thread Starter Member

Mar 10, 2009
11
0
Unfortunately, no. This is an assignment for extra points in the next mid-term.

I got the 4V from voltage division: $U=\frac{R_1+Z_{01}}{R_1+Z_{01}+Z_{02}+R_L}E=4V$. I'm with you, though, that the transmission coefficient, $\tau$, can't be right. It would make more sense to subtract the reflection coefficient from 1. Maybe a typo in the material?

I'll try the energy method and see what I'll get.

17. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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I'm thinking the transmission coefficient should be something like ....

TC=1-ρ^2

So if ρ=0.5

TC=0.75

Hence a pulse of E volts magnitude traversing the line intersection would have a reflected component of 0.5*E and a transmitted component of 0.75*E.

18. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Doubt if that's the way to go ....

Just consider the first transmission line as having 100Ω impedance as "viewed" from the source feed point.

I know this all sounds crazy - especially when you first start on transmission lines. We get so used to solving problems with Kirchoff's laws we often balk at these new concepts at the first (second ...third ...) encounter.

The problem is we are dealing with guided waves traveling along these lines with distributed parameters - not discrete circuits. I remember an RF engineer saying once - "50 ohm coax has 50 ohms impedance whether its 100 miles long or 1 inch long". Knock me over with a feather .....!

19. ### Whalum Thread Starter Member

Mar 10, 2009
11
0
This sounds a bit more logical to me, but aren't we still getting free energy?

Does this really change something? The pulse goes through the whole line, so isn't it same, if we replace the line with resistor with the same resistance. If not, how do you get the 6V?

20. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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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.

Free energy - not really.

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.