how does inverse polarity works ?

[gryfon]

i've seen several uses for inverse polarity. if i understand correctly, there's at least two channel. The second channel contains the first channel with an inverted polarity. Because it's inverted, therefore when you combine it, it eliminates noise. Right ? (if it's wrong i'm happily accept the corrections) Some common applications are utp cable for ethernet, and audio / mixer connector with 3 cable / xlr. The thing that i don't understand is, how does it inverted ? how can it eliminates noise ? Does it really combines both (sory if i'm wrong on this) isn't the contained channel will be gone too ? Can someone lighten up ?

thanks
gryfon

 

[klaviernista]

see below for much better (and certainly more correct) anwer then I gave. Hey! I'm a chemist qith a background in applied magnetic materials..I said I wasn't an EE. Thanks for correcting me guys.

 

[FrankSchwab]

Sheesh.

"inverse polarity" in the sense that you're describing is known as "Differential transmission" and simply means that you have two wires with opposite voltages on them - i.e. when one wire has a high voltage, the second has a low volatage, and vice-versa. The technical answer is phrased a bit differently - rather than referring to a high and a low voltage, what you're really measuring is the voltage difference between the two wires. Imagine hooking a voltmeter between the two wires - a "1" may be transmitted by having a positive voltage appear on the voltmeter; a "0" may be transmitted by having a negative voltage appear on the voltmeter. It's really no harder than this.

The reason is relatively simple - every wire in the universe is also an antenna, and picks up stray signals. You can imagine that this can be a huge problem when you imagine running a 100 meter length of twisted-pair ethernet cable. That's one big antenna. So, how do you eliminate the stray signals?

You start off by twisting the wires together. If you open up any modern communications cable (USB, Ethernet, Firewire, etc) you'll notice that the data wires are twisted around each other. Because of this twisting and the consequent extremely close coupling between the wires, almost all of the stray signals picked up by one wire are also picked up by the other one. So, now you have two noisy signals rather than one - doesn't seem like we've gotten anywhere, ehh?

Well, now we do some mathematics. Imagine the signal in one wire is (S+N), where S is the original signal, and N is the noise it's picked up. The signal in the second wire will then be (-S + N), where the signal is exactly the opposite of the first wire, but the noise is the same (it didn't get inverted). What happens when you subtract the second signal from the first?

Result = (S+N) - (-S + N)
Result = S + N + S -N
Result = 2S

What happened? We ended up with twice the amplitude of the first signal, and all of the noise got cancelled out!

As to how you generate a signal, it's pretty easy. In order to transmit a signal down a cable, you normally run it through an amplifier inside whatever electronics it starts from. Note that in this sense, the "amplifier" is a tiny circuit, not a big black box with a dial that goes from 0 to 11, though the big black box probably has a bunch of the tiny amplifiers.

Anyway, in order to generate this differential signal, you run it through two seperate amplifiers - one an Inverting amplifier (google it, or simply accept that this exists), one a non-inverting amplifier. Presto, the two outputs are attached to a wire, and you have a differential signal.

/frank

 

[gryfon]

thanks to both frank and klaviernista. It think i understand the largest part now. But there's still things that i don't understand, pardon me if it sounds noobie, or part of a very basic physics .

Imagine the signal in one wire is (S+N), where S is the original signal,

am i right that the s is the electrical potential energy measured in volts ? How does it related to wave ? Does a high amplitude wave also have a high voltage ? Oh, and how's the form of the noise that you said got picked up by the cable ? is it spikes in the voltage ? or does it only visible using osciloscope ?

what you're really measuring is the voltage difference between the two wires. Imagine hooking a voltmeter between the two wires - a "1" may be transmitted by having a positive voltage appear on the voltmeter; a "0" may be transmitted by having a negative voltage appear on the voltmeter. It's really no harder than this.

i don't think i understand this, the high / low voltage seems easier to understand for me. What do you mean by voltage difference ? If a two wires, A and B, is using Differential transmission A is high, B is low, let's just say A is +5v , then B is -5v ? then the difference (i assume it's a simple math) is 10V. is that what you mean ?

You can imagine that this can be a huge problem when you imagine running a 100 meter length of twisted-pair ethernet cable

i've heard that ethernet cables are only reliable at < 100m max without active repeaters. If it's using differential transmission then i assume noise is not a problem, is it because the original signal strength is already too weak ?

What happens when you subtract the second signal from the first?

How does this subtraction happened ? I don't think it's even close to binary subtraction using logical gates, am i right ?

once again, sory if this thing sounds too noobie.
thanks,
gryfon

 

[harrkev]

For the record, I *AM* an electrical engineer...

gryfon, your assumptions on the post above yours are mostly correct. The receiving is done by special parts, which do NOT have any logic in them. So forget all of that digital stuff about true/false/and/or/etc. Differential pair receivers and transmitters are mainly analog devices that turn one voltage into two voltages (or two into one for a receiver).

And yes, the energy is transmitted as voltage. Assume that you have two wires, A and B. A transmitter sends A=+5, B=-5V. The total difference is 10V. If you get noise, A=10+1=11. B=-10+1=9. The difference is STILL 10V. So differential pairs are fairly immune to noise.

Another factor is that this type of setup RADIATES less noise. If wire A induces an electromagnetic field, wire B would induce almost the exact opposite field, so they cancel out. This is also a big plus.

Differential signaling is used in such standards as RS-485 and LVDS, along with MANY others.

The reason for the ethernet size limit is a result of the early "shard bus" topology of ethernet (back when it used coax). This length limit is related to the physical size of one minimum-sized packet when traveling down a wire at the speed of light. I do not want to go into it too much, but it is possible for two eternet ports to talk at the same time, and the collisions have to be detected reliably. And if the physical length of the packet (in feet, at the speed of light) is shorter than the wire, then not all collisions can be detected.

 

[TuxDave]

I'm kinda curious about your background (to the OP). You know about digital circuits but have no knowledge of analog circuits? What did you study?

 

[FrankSchwab]

Thanks, harrkev, and an informed engineer you are...

gryfon - the addition and subtraction of two signals is, as harrkev noted, done by purely analog circuitry. Try googling for "Op amp" . For example,
This (at least the first page or so) appears to be a reasonable overview of an opamp.

This page, about the second or third page down, shows the connections you might make for an op-amp based receiver (see the diagram labeled "Differential Mode = Acceptance").

There are probably better pages out there

/frank

 

[gryfon]

thanks to all who replied. All is clear now. However there's a part of my question that you guys haven't answered, maybe you don't notice, so i'll ask again

How does the electrical signal related to wave (i.e the one that you see on a oscilloscope) ? Does a high amplitude wave also have a high voltage ?

Oh yeah, i don't know if it's still related or not, but i've seen a guy measured output/transmit powers for a wireless router linkysys wrt54g on something (is it osciloscope ?) . How can he measures the output when it's wireless ? link

To TuxDave :

Currently i'm on computer science major, so i'm not used to analog circuit at all. For the digital circuit, i got some lesson on college about a little logic algaebra and logic circuit.

thanks,
gryfon

 

[f95toli]

Originally posted by: gryfon

Oh yeah, i don't know if it's still related or not, but i've seen a guy measured output/transmit powers for a wireless router linkysys wrt54g on something (is it osciloscope ?) . How can he measures the output when it's wireless ? link

gryfon

He probably just connected the oscilloscope to where the antenna of the router would normaly be connected.

 

[gryfon]

He probably just connected the oscilloscope to where the antenna of the router would normaly be connected.

actually that does make sense

 

[harrkev]

Originally posted by: gryfon
thanks to all who replied. All is clear now. However there's a part of my question that you guys haven't answered, maybe you don't notice, so i'll ask again

How does the electrical signal related to wave (i.e the one that you see on a oscilloscope) ? Does a high amplitude wave also have a high voltage ?

Oh yeah, i don't know if it's still related or not, but i've seen a guy measured output/transmit powers for a wireless router linkysys wrt54g on something (is it osciloscope ?) . How can he measures the output when it's wireless ? link

To TuxDave :

Currently i'm on computer science major, so i'm not used to analog circuit at all. For the digital circuit, i got some lesson on college about a little logic algaebra and logic circuit.

thanks,
gryfon

The typical oscilloscope measures voltage vs. time. So if the signal has more voltage, the wiggles get higher. Just plugging an oscope in place of an antenna will not let you measure absolute power (you have to match the impedances, do some math, etc.), but should be close enough to get you a relative power reading. Larger wiggles = more power.

 

[gryfon]

ok, thanks to all who replied, i get the whole point now.