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Topic: Signal loss, Hi/Lo over a distance? (Read 2526 times) previous topic - next topic

jonisonvespa

Sep 04, 2012, 01:00 pm Last Edit: Sep 04, 2012, 01:02 pm by jonisonvespa Reason: 1
hi,
just wondering whats best to use a hi or a low, when transmitting a signal over a distance of say 30 metres
does a high or low transmit further/better?
i did a test using a high over 50m of 2.5mm twin and earth and got a loss of 0.1v seemed to work ok, can under stand that loss, but is there a loss in signal wen using a 0v over a distance?
thanks  

majenko

You can't just use a high OR a low to transmit a signal.  A signal is made up of a series of high AND low states.

I mean, what would this signal mean?

Code: [Select]
Low, low low low, low, low, low low low low low, low, low.

It makes no sense.

I think you probably mean, is it better to use "Active High" or "Active Low" switching.

With "Active High", the line is normally LOW, and when you "switch on" it goes high.  The opposite for "Active Low".

Whichever you choose, over long distances at low switching frequencies you can think of your wire as a resistor. 

For a "switched" signal (connected to a switch or button) that resistor forms one half of a resistive divider to the input that is reading it.

You need to ensure that your pull-up or pull-down resistor, coupled with that wire resistance, keeps the voltage levels presented to the input within the static discipline of the chip (see the Vil and Vih in the datasheet).

For an "active" signal, one being driven by an electrical device, then the losses caused by the resistance of the wire have to be low enough that the HIGH doesn't drop below the Vih level of the static discipline - as there is no resistive divider there is no possibility of offsetting Vil (LOW).

michael_x

Calculating the resistance of a 30m wire can be done, of course,  and getting the voltage divider maths done should be rather easy.
However, that would result in a high pullup / pulldown resistor, the higher the better.

On the other side, high pullups are also called "weak" pullup resistors, and usually you are better off running a higher current, aren't you ?

Isn't the more interesting part some non-static influences? Isn't that about the capacity of the wire, rather than the resistance?
How do you calculate maximum frequency of your signal over a long wire?
If you deal with very seldom changing signals, wouldn't it be better to also add an extra capacitor, to short down electric noise?

How are the maths done when it's about considering wire length and signals?
I know Ohm's law, now I want to learn something beyond ...

majenko

High speed cabled communications is a rather involved subject...

At higher frequencies you have lots of things to consider:

1. The resistance of the cable
2. The capacitance of the cable
3. The inductance of the cable

All these add up to basically form a low-pass filter, which limits the maximum frequency that can be transmitted down the line with minimum attenuation.

Note that this frequency isn't the data speed, but the signal frequencies.  Your perfect digital signal is a perfect square wave.  However, any square wave has many high frequency harmonics above the base frequency, and the perfect square wave has infinite harmonics (not actually achievable, but good for illustration purposes).

The low-pass nature of the cable causes these high frequency harmonics to be attenuated, so the nice clean, sharp edges of the square wave become more and more rounded, until at the worst case, you end up with a sine wave at the base frequency.  Not ideal.

It's this rounding of the signal that, to a large extent, dictates the maximum data rate that can be transmitted down a line.  You have to have enough time for the signal to rise to the HIGH level, be sampled at the far end, and fall to the LOW level, to be sampled again.

Then on top of all that there is, for very long communications lines, the actual time taken for the signal to travel down the line.  Normally you don't think about that kind of thing at low speeds and short distances, because the times are so unmeasurably small as to be ignored.  But, with extremely long lines there may be a considerable lag between the sending of a bit at one end and that bit appearing at the other end.

It all gets a bit messy.

And to top it all off, there's something called "The Skin Effect".  This only affects the sending of an AC signal down wire - which any communication signal effectively is - whereby the signal only travels at the surface of the conductor.  The higher the frequency, the thinner the layer that the signal is actually in.  The thinner the layer, the greater the losses due to resistance.

Like I said, it's a very involved subject.

fm

You will also need to consider induced noise on a long wire. Noise could, and will, cause false level detection at the other end. Shielding the wire will surely be necessary if you run the wires through your house wiring.
   

majenko


You will also need to consider induced noise on a long wire. Noise could, and will, cause false level detection at the other end. Shielding the wire will surely be necessary if you run the wires through your house wiring.

Oh, don't get me started on noise ...

Balanced differential communication lines, modulation systems ... ;)

fm

   

jonisonvespa

ok maby i asked the wrong question , whats the best to use a high or a low over a distance? but thanks for the infomation

MarkT


High speed cabled communications is a rather involved subject...

At higher frequencies you have lots of things to consider:

1. The resistance of the cable
2. The capacitance of the cable
3. The inductance of the cable

All these add up to basically form a low-pass filter, which limits the maximum frequency that can be transmitted down the line with minimum attenuation.


Well this was the state of understanding before Oliver Heaviside solved the telegraph equations.  So long as a cable has a constant cross-sectional geometry it can be modelled with a single characteristic impedance.  Capacitance and inductance effectively cancel across all frequencies - provided you drive and terminate the cable at the correct impedance. 

There is attenuation due to series resistance (which increases at very high frequencies due to the skin-effect) and dielectric losses in the insulation (air or foam insulation is often used to reduce the latter).   CAT5 ethernet cable is designed as a transmission line (well 4 transmission lines actually).  100Mbit signalling over 100m works nicely over CAT5 because its driven at its characteristic impedance of 100 ohms.  Try driving it at 1k and it simply will not work at all at that speed or anything like it.

At 100Mbit there are 50 or so bits in transit along the 100m length at any one time(!) - a transmission line carries a wave, it cannot be modelled as a simple RC filter.  There is still complexity, but its to do with skin-effect, radiation losses, dielectric losses, cross-talk, impedance mismatching and bending radius.
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