RF, reflections, impedance matching - Page 1
Forum section: electronics / beginners

author: npelov
author_rank: Frequent Contributor
timestamp: February 01, 2014, 12:25:22 pm
content: Hi,I have very limited knowledge about how higher frequency signals behave. I'm trying to build experience to investigate any problems with parasitic inductance, capacitance, data line length. Currently I'm trying to understand reflections off the other end of a very long cable. I know that's not the proper way to do it but that's what I have now. I'm using (about) 5 meters long UTP cable. I'm using 2 wires of different pair so they are not twisted together. What I expect to see is the reflected signal off the other side. 5 meters both directions is 10m which should be 10/300 000 000 ~ 33ns. Well that's in vacuum, so the speed in the cable should be less than that.I attached 3 screens from the scope - first the signal without the cable measured with the spring attachment of the scope. It's a 10 MHz square wave from an ebay C-MAC STP-2390C OCXO. Second is when the cable is attached to the signal and nothing on the other end. Third is taken with second probe attached to the other end of the cable. Can someone explain why the signal has this form and which part of it is the reflection, which part is the signal itself, how do they sum ... etc. I think the time between the signal from signal to the other end is 40ns, but I might be wrong. Then signal comes back in probably 40ns more. And ... does it reflect third time? Well I guess it depends on the impedance of the source. If it's high impedance then it could reflect multiple times. If it's low impedance it won't. I don't have the datasheet of the oscillator and I don't know how much load I can put on it to lower the impedance. Maybe I can use MCU with clock out as a buffer.One of the problems could be the limited bandwidth of the scope - 70Mhz. I guess that's the reason the edges are not that sharp. That could also explain why the form of the reflection is distorted.

author: valentinc
author_rank: Regular Contributor
timestamp: February 01, 2014, 03:19:14 pm
content: If you are measuring 2 wires that are not twisted together, the result is irrelevant, since the total impedance of the cable can vary depending of how the cable is twisted (the space between the 2 wires may vary across the length of the cable, which in turn varies the local capacitance...). Twisted pairs in UTP cables are meant to be used with differential signals... That's the reason they are twisted together... You should do this measurements with a coaxial cable instead...Second, consider the probe of your scope... You are probably using a high impedance probe... Which also affects the measurement... Ideally, if your oscilloscope has a 50 ohm input impedance mode, you should connect the end of the cable directly to the BNC of the scope... If not, you could use a 50 ohm terminator, but note that the input capacitance of the scope (~16pF) will have a reactance that is in parallel with the 50 ohm terminator, and at high frequencies it will have an impedance lower than 50 ohm, which will make the total impedance lower than 50 ohm...About the bandwidth of your scope, it is indeed a limitation, that is why the edges of the signals are curved...The signal seems to reflect 2 times: once, at the connection between the signal source and the cable and second at the end of the cable, where it "sees" the high impedance of the probe...

author: Rory
author_rank: Frequent Contributor
timestamp: February 01, 2014, 03:53:52 pm
content: As valentinc said, you are seeing multiple reflections. Try terminating the scope end of the line with a pad to isolate the oscillator from the line. Select values for the pad that will match the oscillator's output impedance to the line impedance. Can you explain why you would need to do this?Do you know how to simply measure the output impedance of a signal generator?As an experiment, you can terminate the far end with different resistance values and observe the scope traces. Find the value that gives the least amount of distortion of the square wave. This is the characteristic impedance of the line. Don't forget to use identical length scope probe cables, or terminate directly at the scope inputs.Try this on the twisted pairs too. What do you find?One thing I would do is move the scope trigger point over to the left to remind me that the reflections occur after the transmitted pulse. Did you notice the rising edge of channel 2 was about 33nS after channel 1? What is it with the end terminated? Use the cursor functions to find the delays.

author: sacherjj
author_rank: Frequent Contributor
timestamp: February 01, 2014, 04:24:34 pm
content: Here is a good video that might help you wrap your mind around things.

author: npelov
author_rank: Frequent Contributor
timestamp: February 01, 2014, 05:30:25 pm
content: Thanks to everyone.@sacherjj that's the video that triggered my interest, but often it's harder to get it from the first time and that's why I decided to do some experiments.The problem is that I don't know how much load can my oscillator take. I don't want to blow it so I'll try to get/make one that can handle 50 ohms. Well I still learned something. If the impedance of the generator is high it's possible to have multiple reflections. Maybe I'll use some 74XX series IC to buffer the signal.Also my idea is to have a feeling for strange looking signals and to know what's causing the distorsion. Because the higher the frequency the more reflections will become a thing.So next thing I'll try to buffer the signal through something and put a termination of some kind. Also if the buffer has too low resistance I'll put a series resistance at the start of the cable.One of the reasons is that I'm going to work with RS485 and other protocols for long line transmissions. I'll post the results if I have questions.Thanks again.

author: Rory
author_rank: Frequent Contributor
timestamp: February 01, 2014, 05:56:39 pm
content: Alan's video is really good. (Alan's videos are always good!)The technique works for conductors of almost any impedance - open wire balanced lines, twisted pairs, zip wire, even mismatched wires in a bundle of pairs as npelov is trying. It even works for a single wire against ground (as in Beverage antenna...) One thing Alan mentions, in an offhand way, is that the 50 ohm source impedance of the pulse generator absorbs the reflected pulse. This is only true if the impedance of the generator is close to the line impedance. There is an exception though, if terminating the end of the line to find its impedance, the generator's impedance is not critical. Once the line is terminated by its characteristic impedance there are no reflections for the generator to reflect back into the line.npelov, you didn't mention if you measured the output Z of your oscillator. It's easy. Measure its output voltage with no load, then place a load across it of such a value that you measure 1/2 of the unloaded voltage. You can use this value to design a voltage divider/series-parallel circuit (pad) that presents the proper terminating resistance to the line. If your generator isstiffor has a high fanout it will have a low impedance. If it has high impedance you can always select the pad element values that provide a matching impedance to the generator, but a different/lower impedance of the line. If it's a low impedance it may be as simple as a series resistor as you mentioned.

author: npelov
author_rank: Frequent Contributor
timestamp: February 01, 2014, 06:48:37 pm
content: I monitored something interesting. I was playing around while waiting for the monday to come and get some BNCs, 74HCxxx ICs etc, not expecting any meaningful results and I got the result shown on the screen capture. The first picture shows the connection schematic. I've put series resistance at the start of the cable where the oscillator is. I connected channel one before the resistor the reflections were not back to the channel 1 anymore. What else happend is they cancel on the other end of the cable in some way that I have nearly 2.5V DC. I've set channel 2 to 500mV in order to see that there is some kind of pulse there, but it's very little. Can someone tell me why this is happening. The pulse is (50% duty) 50ns, the signal travels to the end of the cable for about 35 ns - I can clearly see the little pulse with about 3.5 divisions length. So it's not like reflections sum to make that voltage.I thought about impedance of the cable if it's 10k, but then I would see just 2.5V pulses.Maybe the cable works like low pass LC filter so I get (almost) DC at the end (got that idea for the video that sacherjj suggested) and the resistor together with the cable form a voltage divider and it happens to be exactly half of the initial amplitude. So that's why you can't use high impedance generators to send over long line - the cable will just filter it right down to DC.@Rory I tried to measure the impedance of the oscillator, but the amplitude drops and I stopped to avoid damaging it. I can't continue without having a proper buffer. Can you draw a schematic of what you mean by series-parallel circuit. Is it a series resistor after the oscillator and load at the end of the cable?

author: npelov
author_rank: Frequent Contributor
timestamp: February 01, 2014, 07:18:28 pm
content: Well I tried it anyway. I think the signal is almost half amplitude at 100 ohm. Then I tried to put series and load resistance like this:http://fourier.eng.hmc.edu/e84/lectures/TransmissionLine1.gifI used resistors for Z1 and Z2. I tried 100 Ohm and I attached the results. I don't have 75Ohm resistors to try.Is it normal that the amplitudeis half? Probably. The resistors form a divider. I'll have to buy some low value potentiometers in order to do that right.Edit. Amplitude is not half. The p-p value is 1/3 probably because the impedance of the oscillator is about 100 ohm and the series 100 ohm resistor adds to 200 ohm. Is that what I should expect as an amplitude? I didn't notice the amplitude go down in Alan's video. Maybe I didn't pay attention.

author: Rory
author_rank: Frequent Contributor
timestamp: February 01, 2014, 07:55:30 pm
content: Quote from: npelov on February 01, 2014, 07:18:28 pmIf you watch beginning at 4:29 in Alan's video, he shows the effect when you remove the termination from the generator. It is 2x the amplitude. His explanation is that the signal reflects back to the generator and adds, perfectly valid. I was taught the voltage divider method, which works all the way down to DC.The impedance the generator sees is not really important, as long as it is something it can safely handle. Here is an example of an L-pad series-parallel network used to match the generator impedance to the line. The voltage it presents to the line is up to you, and there will always be attenuation.The point of all this is to determine the Z of the line, and with that you can then present the line with a matched source, which doesn't reflect back into the line, giving clean waveforms to measure.TDR 101...