• Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

Explain this Transmission Line

Status
Not open for further replies.

Mikebits

Well-Known Member
Greetings; I have a delay line from an old Tektronix scope. It is a 120nS delay line. I connected an ARB (with ~8Ns pulse at 1 KHz) on one end (50Ω) and clipped a 1Meg scope probe to other end, yes I know improper match (I wanted to see response of open line). What has me befuddled is the initial pulse amplitude. As seen in image, Yellow trace is input to cable, purple trace is output end. The amplitude for the initial pulse should be same as the purple trace or greater as the cable has attenuation.
Yet, as you can see the input signal is less than the output in fact if you had the second yellow pulse to the first, then you would equal the purple trace.
So we have reflections, but the reflection takes 240 nS round trip to effect the input, so how is it the initial amplitude is as shown below.
Does this make sense? I got myself confused here :banghead:
Mucho Gracias
------------------------------------edit----------------------------------------
I should clarify. The delay line is actually about 80 feet of twinax cable, I was driving a single wire with shield as ground.I guesstimated 80 feet based on typical numbers of 1.5nS delay/ft for twinax.
------------------------------------edit----------------------------------------
ARB - Arbitrary waveform generator
---------------------------------
transmissionLine.png
 
Last edited:

spec

Well-Known Member
Most Helpful Member
Hi Mike,

I am not sure of what is going on with the delay line, but may I suggest that you terminate the scope end with 50 Ohms. Many scopes have a 50 Ohm input which helps.

The other thing, are you sure the coax characteristic impedance is 50 Ohms: could be 75 Ohms or even higher.

On an old wide-band scope I got for £5UK, the whole top deck (the scope was the size of a ship) was taken up with a huge delay-line; it was a real work of art.:)

spec
 

JimB

Super Moderator
Most Helpful Member
Looking at the far end of the line:

If the line was terminated in its characteristic impedance, you would see the incident pulse minus the line loss. There is no return pulse.
If the line was terminated in a short circuit, you would see no pulse, the incident pulse is cancelled out by the return pulse which is of equal amplitude and opposite phase.
When the line is open circuit, you see the incident pulse added to the return pulse, they are in phase and so add together as we see here.

Now consider the frequency domain:
At some frequency, the line will be a quarter wavelength long.
With the far end open circuit, the line will appear as a short circuit at the generator end.
If we inject a frequency equivalent to the quarter wave length of the line, and measure the voltage at the generator and at the open end, we will find that there is a high voltage at the open end and zero volts at the generator end.

This RF stuff is black magic I tell you!

JimB
 

Mikebits

Well-Known Member
I am not sure of what is going on with the delay line, but may I suggest that you terminate the scope end with 50 Ohms. Many scopes have a 50 Ohm input which helps.
Well I was using a 1Meg scope probe, so the little prong on the probe forces the scope impedance to 1Meg by default. Your right though, I should have used better termination techniques like solder a SMA connector on the end of the cable then I could have terminated in a more proper impedance, but I was just playing around to see the delay.

The other thing, are you sure the coax characteristic impedance is 50 Ohms: could be 75 Ohms or even higher.
No, I was not sure of the impedance at all. The only thing I was sure of is that it was a 120nS delay line as I found that info on the web somewhere, but I did not have any other info. I suppose I could try and determine the real impedance, I could possibly try this method I found looking on Ye Ol' Web; http://www.irb.hr/users/stipcevi/ele/coaxweb/index.html
I just might try it later, after I do a few other things I have on my to do list :) Thanks for the advice.
 

Mikebits

Well-Known Member
Just when I think I got some semblance of understanding of this stuff, I am constantly reminded just how much I still yet have to learn or have forgotten and must relearn, the latter becoming more true than not :)

Now consider the frequency domain:
At some frequency, the line will be a quarter wavelength long.
With the far end open circuit, the line will appear as a short circuit at the generator end.
If we inject a frequency equivalent to the quarter wave length of the line, and measure the voltage at the generator and at the open end, we will find that there is a high voltage at the open end and zero volts at the generator end.
Time for me to break out my ARRL handbook :) I need to ponder your input and try to compute.

This RF stuff is black magic I tell you!
Dr. Howard Johnson would disagree with you, but I think your right :)
Thanks Jim
 

Mikebits

Well-Known Member
Just for giggles, I wanted to see what I could gleen from LTSpice. I used the 10:1 scope probe file in the education sub-directory example files. I tweeked a few things and results pretty much match the screenshot I showed in my original email. For your viewing pleasure, here are a couple screenshots.



120nSdelayLineASC.PNG

120nSdelayLine.PNGtransmissionLine.png

I think that is cool:)
I can post the ASC file if anyone wants it...
Thanks
 

JimB

Super Moderator
Most Helpful Member
It would be entertaining to get a small variable resistor and adjust it until the reflected pulse was zero.

Then the scope should see a pulse which is the same as the input to the line, give or take the losses.

JimB


Edited to add:
When you have adjusted the variable resistor for no reflected pulse, you can measure the resistance of the variable resistor using a simple DMM and the DC resistance will give a good indication of the characteristic impedance of the line.
 
Last edited:

JimB

Super Moderator
Most Helpful Member

Mikebits

Well-Known Member
Maybe a bit late to ask, but what is an "ARB" ?

JimB
Oh sorry. I should of not used abbreviations
ARB is short for arbitrary waveform generator.
 

JimB

Super Moderator
Most Helpful Member
OK, understood.
JimB
 

Mikebits

Well-Known Member
It would be entertaining to get a small variable resistor and adjust it until the reflected pulse was zero.

Then the scope should see a pulse which is the same as the input to the line, give or take the losses.

JimB


Edited to add:
When you have adjusted the variable resistor for no reflected pulse, you can measure the resistance of the variable resistor using a simple DMM and the DC resistance will give a good indication of the characteristic impedance of the line.
Well you made me curious, so I went ahead and ran some test. Just so happens I had a connectorized 2k pot, and I soldered connectors on the delay line.
So I ran 3 test:

1. Tuned pot for best match on load end (turns out pot was set at 42 ohm, when I connected an actual 50 ohm load I got a mismatch, so I assume the pot is adding reactance)
2. Shorted the load
3. Open load

The test results are shown below in respective order as mentioned above, and I should say I am glad I ran these test because it all makes sense to me now as to what I was seeing earlier, and started this post.

translinematch42.png

translineshort.png
translineopen.png
 
Last edited:

Mikebits

Well-Known Member
For me, the coolest part of this exercise was how the simulations came so close to real world, and this sim tool is free no less. Simply amazing :woot:
 

Mikebits

Well-Known Member
Just so happens I had a connectorized 2k pot,
Somebody PM'd me asking what a connectorized pot is. Well, I had this really ultra stable awesome multi-turn pot that I got from an old Tektronix scope. It seemed such a waste to not use it somehow, so I soldered a BNC connector to its terminals and mounted it into a old pill bottle. Pictures do a much better explanation than I can, so here is a couple shots.

IMG_1824.jpg IMG_1825.jpg
 

Mikebits

Well-Known Member
Oops, I plugged it in upside down so, the dial does not match...
 
Status
Not open for further replies.

Latest threads

EE World Online Articles

Loading
Top