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# Standing waves in Long cables and at 100K-200Khz frequencies.

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#### sh-eda

##### Member level 1
Hi

I wonder if someone could help with a question I have.
I am investigating standing waves on signal (100KHz-200Khz) sent over on long cables, over 1km -3Km.
Ultimately with a known load and source impedance and the details of the cable, I need to be able to predict what the signal level will be at various points along a fixed length cable, and especially where along the cable we are going to see lowest signals levels, dues to standing waves (It is very unlikely that the cable will ever be properly matched.)

I have done some testing on shorter lengths of cables and done some simulations (using Multisim), with mixed results.

I am partly using the following formulas (from https://en.wikipedia.org/wiki/Standing_wave_ratio)
and I have plotted this in Excel to give me an idea.

Reflective Coefficient = (Z-Zo)/(Z+Zo)
VSWR = (1+p)/(1-p)=Vr/Vf

Vmot = A*sqrt(4*p*(coskx)^2+(1-p)^2)

Where
x=is the distance along the cable from the source
p= (rho) is magnitude of the voltage standing wave ratio.
A=voltage amplitude
K is the wave number (ie. = radial freqeuncy / speed of the wave)

1. I am surprised that the distance from the source to the terminating load is not taken in to account.
Surely more signal will be reflected back If it is stituated in a peak, less signal will be reflected back if it is in a trough

2. In my case resistance of the cable at 2-3km could more than the source impedance or the load impedance.
Surely the loss in such a cable has to be taken into account when considering the amount of signal reflected back.
Maybe its's when calculating the reflective coefficient Vr/Vf

Hope someone can help me clear this up for me.

Thanks

1. there are no peaks and troughs until you get the missmatch.
2. its called a self terminating line. The classic formula you used does not take into account the line loss. Suppose you are putting a certain voltage on to the line and only getting one half of it at the far end, the the reflected wave would be only one half of what you would expect, so the VSWR at that end of the line is given by your formula, but at the sending end the reflection would be one quarter of that you would expect due to the poor load (1/2 of the voltage = reflection,X 1/2 due to the line attenuation back to the source).
Frank

At 200kHz, a wavelength is 1499 in air, and maybe 0.66*1499 ~= 989m of cable.
If the cable is mismatched, the first null is about 247m (quarter wave).

This is a classic for some work with a Smith Chart (quite a fast way to get the right answers).
Multisim is perfectly good for getting the right answers provided you use the right lossy transmission line model.
You need accurate data on your cable, and you can do this with even simple kit.

If you can open-circuit the far end, and feed the cable with signal generator via a SWR bridge, you can find out everything about it. Probably the fastest way is to borrow a antenna analyser from a ham-radio operator. Where the nulls are will depend on the frequency, but for a given single frequency, they will always be in the same place. Only the depth of the nulls varies with the amount of mismatch.

Yet another way to go at it is a time domain reflectometer cable tester - or use a pulse generator and hang an oscilloscope on the connection. Make the rise time of the pulse no shorter than about 20% of the rise time of the highest frequency, or you will see reflections from frequencies you are not interested in. Sine-squared pulses are best is available.

Terminations impedances higher than the characteristic impedance of the cable return pulses the same way up as they were sent.

Open circuit returns all of it. The height of the return pulse compared to the height of the sent pulse gives you the loss for the 2-way trip.

Terminations lower than the characteristic impedance return pulses upside down. Short circuit returns all of it (less the loss), all upside down. I expect you already know why.

If the cable resistance gets up to the terminating impedance, it could look rather well matched at the input, even with a poor match at the other end.

Thanks for the responses.

Unfortunately I don’t have access the actual cable. I have tried to work out some of the properties of some similar cables by running some tests.

Basically one of the things I want to be able to check the results that Multisim gave me, which didn't seem always seems to be correct. I am using the lossy model. Obviously I need to understand how to do this manually in the future so I am looking at the smith chart which look interesting. I haven't quite figured how to use it yet. I'm trying to find some similar examples that I can work through.

My cable is;

Capacitance 82pF/m

Resist 0.00935ohms/m

Inductance 360nH/m

Signal is 1v rms at 170Khz.

Output impedance from source is around 80ohms.

So I was going to start by to calculating signal level at 1.8km.

With a resistive load impedance of 10ohms at 1.8km.I've calculated the cable impedance Zo to be 66.3ohms.

Resistance of the line is 17.251ohms per line at 1.8km. (so total of 34ohms)

Unfortunately I don't have G.?

I assume when normalizing the smith chart you would use the cable resistance and series and parallel impedances over the whole cable length, or is it just per metre? It doesn't seem clear to me.

Am I correct in that series inductors are added to the smith chart /calculated as reactance, whereas the cable (i.e parallel) capitance are calculated / added to the chart as conductance?

The source output impedance would be added to the cable series resistance?

Thanks

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