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The length of a transmission line is limiting the highest possible data rate on that line.

Why are faster signals more likely to become corrupted on long transmission lines than on shorter ones, with focus on frequency-BW of the signal, raise time, etc... ?

Would appreciate any explanations.

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    \$\begingroup\$ AFAIK your primary statement "The length of a transmission line is limiting the highest possible data rate" is not (or at least not always) true. \$\endgroup\$ – Wouter van Ooijen Sep 6 '14 at 12:56
  • \$\begingroup\$ There may be dispersion in a transmission line. This would cause different frequencies to travel at different velocities. That corrupts the sharp edges of any pulsed data. \$\endgroup\$ – George Herold Sep 6 '14 at 15:02
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SPI uses clock and data. From the sending (master) end, clock and data fly down their respective cables in sync (but delayed by the cable) and they reach the slave end and hey presto, the slave clocks in the data and does what it has to do but, what if it has to send back some data such as a value of something.

OK, it transmits its data synchronized to the local clock it is receiving and doesn't worry any more BUT, that local clock it receives is delayed by the cable and, the data the slave sends back to the master is further delayed by the cable and what happens at the master (when receiving data) is a mess unless the data rate is slow or the cable is short.

The main problem is that data sent from a slave is "timed" to the clock edges at the slave. The data received by the master is clocked into the master by the local clock at the master - the slave clock and master clock are not aligned due to the cable delay.


The OP has changed the question so here's some extra things about cable: -

Longer cables attenuate more - think about powering a motor from a battery - it works fine up close on short leads but if you make the leads longer the terminal voltage seen on the motor gets smaller and smaller as cable length increases. Copper is not zero-ohms.

It gets worse as frequency rises due to a phenomena called skin effect. skin effect reduces the conductivity of a copper wire by forcing currents to only be present in the skin of the conductor. This means smaller cross sectional area for the current hence higher resistance hence greater losses.

Dielectric loss in cable is proportional to frequency - basically energy is stolen from the signal to heat-up the insulating material between the two wires that form the transmission line or cable. This is what wiki says: -

Attenuation (loss) per unit length, in decibels per meter. This is dependent on the loss in the dielectric material filling the cable, and resistive losses in the center conductor and outer shield. These losses are frequency dependent, the losses becoming higher as the frequency increases. Skin effect losses in the conductors can be reduced by increasing the diameter of the cable. A cable with twice the diameter will have half the skin effect resistance. Ignoring dielectric and other losses, the larger cable would halve the dB/meter loss. In designing a system, engineers consider not only the loss in the cable but also the loss in the connectors.

If losses are proportional to frequency then the likelihood of data corruption is also proportional to faster signals. Above a certain point there is another mechanism when cable (such as coax) starts to acts like a waveguide. Again wiki has the word: -

In radio-frequency applications up to a few gigahertz, the wave propagates primarily in the transverse electric magnetic (TEM) mode, which means that the electric and magnetic fields are both perpendicular to the direction of propagation. However, above a certain cutoff frequency, transverse electric (TE) or transverse magnetic (TM) modes can also propagate, as they do in a waveguide. It is usually undesirable to transmit signals above the cutoff frequency, since it may cause multiple modes with different phase velocities to propagate, interfering with each other. The outer diameter is roughly inversely proportional to the cutoff frequency. A propagating surface-wave mode that does not involve or require the outer shield but only a single central conductor also exists in coax but this mode is effectively suppressed in coax of conventional geometry and common impedance. Electric field lines for this [TM] mode have a longitudinal component and require line lengths of a half-wavelength or longer.

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  • \$\begingroup\$ Thanks for the answer, that makes sense. but the question seemed to be unclear. im not worried about spi-signals getting out of sync. Im looking for a more physical explanation about signal deterioration with respect to frequency-BW of the signal or raise time or so. i edited the question therefore. \$\endgroup\$ – Fry Sep 6 '14 at 13:39
  • \$\begingroup\$ @Fry thanks for making my answer largely redundant! Can I ask you to restore your question to it's previous state and add an informal "edited" section so that folk reading my answer won't downvote it thinking it misappropriate. \$\endgroup\$ – Andy aka Sep 6 '14 at 13:56
  • \$\begingroup\$ After a valid answer has be provided, I'd now like to add one more consideration which affects half-duplex bus-systems (like CAN). At high frequency and long transmission lines bus participants have to wait until the last bit of the last message has arrived at the end of the line before sending something. the higher the frequency, the shorter the bit time. Due to time of propagation, the maximum length of the signal line is limited by the bit time (and frequency). If lines get to long, then participants may think the bus if free, even if it isn't (since bits are still on the bus but far..) \$\endgroup\$ – Fry Apr 27 '15 at 21:41
  • \$\begingroup\$ Data will travel down a cable at about 8" per nano second so if the cable is a mile long from some central hub the closest will receive the data transmission 7.9 us before the furthest. If the data transmission (packet) is less than 7.9 us then conceivably the closest could entirely read the message before the furthest begins to see the message but so what. It's up to the protocol to handle stuff like this. \$\endgroup\$ – Andy aka Apr 27 '15 at 21:54
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Imagine you're a pizza delivery guy, and you're told that you have 10min to deliver one pizza right around the corner. Let's say that's 50 meters away, so that's really easy. Imagine now that you have 10min to deliver one pizza at the other side of the city. The throughput (like a datarate, since that's 0.0017 pizza/second) is the same, but the increased distance made it very difficult to achieve it.

Every circuit has a speed capability due to signals transport delays, capacitances etc., and signals beyond that capability will struggle to come through. The bandwidth is expressed in units of frequency (the point where signals are attenuated by a certain factor) and is very closely linked to datarate though they are not the same (other factors come into play).

I'll leave it to physicists to explain the bandwidth in details, but it must be something in the lines of: if you change the input too fast when the output is too far away the signals won't have the time to reach the other end before they have to change again, and this will result in more and more attenuation.

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  • \$\begingroup\$ In your analogy bandwidth would be the the number off pizza guys that can pass through a certain point per unit of time. I don't see how that would be affected by the delivery distance. \$\endgroup\$ – Wouter van Ooijen Sep 6 '14 at 13:01
  • \$\begingroup\$ This analogy is not meant to be a match to bandwidth (I didn't find anything good enough for this), but to give an intuitive example on why distance is a factor in the achievable rate of the transmission of information. I edited the post to make it clearer, though I don't think it deserved a downvote. \$\endgroup\$ – Mister Mystère Sep 6 '14 at 13:11
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    \$\begingroup\$ Your main argument seems to be attenuation, but that can be solved by higher transmission power and/or a more sensitive receiver. \$\endgroup\$ – Wouter van Ooijen Sep 6 '14 at 13:29
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    \$\begingroup\$ Except that the attenuation is not linear with frequency so more and more power would be needed. However you are right that I had analog bandwidth in mind and that topic seems to be about SPI, which is well explained by Andy. Still. \$\endgroup\$ – Mister Mystère Sep 6 '14 at 13:36
  • \$\begingroup\$ attenuation might be the thing im looking for. so the attenuation could be a reason why one would use a linedriver in order to still being able to sent fast signals over longer lines ? \$\endgroup\$ – Fry Sep 6 '14 at 13:45

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