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  1. From http://en.kioskea.net/contents/pc/bus.php3

    A computer bus is characterised by the amount of information that can be transmitted at once. This amount, expressed in bits, corresponds to the width of the bus, i.e. the number of physical lines over which data is sent simultaneously. A 32-wire ribbon cable can transmit 32 bits in parallel. The term "width" is used to refer to the number of bits that a bus can transmit at once.

    Additionally, the bus speed is also defined by its frequency (expressed in Hertz), the number of data packets sent or received per second. Each time that data is sent or received is called a cycle.

    This way, it is possible to find the maximum transfer speed of the bus, the amount of data which it can transport per unit of time, by multiplying its width by its frequency.

    A bus with a width of 16 bits and a frequency of 133 MHz, therefore, has a transfer speed equal to: 16 * 13310^6 = 212810^6 bit/s.

  2. If I understand correctly, a signal/data transmitted in a computer bus is actually an electrical current with its electric frequency. The power supply unit on the power cord for a computer is a AC->DC converter, so is the electricity frequency zero in computer buses?

I wonder if the bus frequency and the electric current frequency are the same concept. If not, how are they related?

Also if I understand correctly, cables such as USB cables and Ethernet cables are also computer buses, si the above questions are asked for them too.

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  • \$\begingroup\$ possible duplicate of Generation of clock signals \$\endgroup\$
    – stevenvh
    Aug 28, 2012 at 16:35
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    \$\begingroup\$ How is it duplicate? \$\endgroup\$
    – Tim
    Aug 28, 2012 at 16:37

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I think what might be missing in your understanding is that when we send information on a computer bus, we are sending digital information. Bits are represented with two states, '1' or '0' - or simply "on" and "off". A string of bits represents a number in base 2. We represent these two states (1 and 0) with discrete voltage levels, for the sake of simplicity, something like +5V and 0V respectively (different devices may use different levels to represent their bits but the principle is the same).

So when we talk about a computer bus having a frequency, we're talking about how quickly it can toggle a bit - or a single wire in this case - from a logic '1' to a logic '0' or vice versa as this will determine the maximum bandwidth the bus is capable of (at the bit rate level). To further divide computer buses into categories, there are parallel buses and serial buses. A parallel bus breaks out the individual bits of a base 2 number (typically an address or data value) and gives each one a wire. Thus, the calculation that was quoted holds true for parallel buses. Then there are serial buses for something like USB like you mentioned. On a serial bus, all bits are sent on a single wire (two wires in USB's case, but they are a differential pair, so you can think of it as one wire). The bit rate calculation for that would simply be its maximum "toggling" frequency.

You should not try to relate the above with a sinusoidal AC signal frequency found in AC power. Yes, you are correct in that a DC voltage should have a theoretical frequency of 0 - but in the context of digital communication, we are not really interested in that.

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  • \$\begingroup\$ Thanks! In digital electrical circuits, does electromagnetic field/wave affect the transmission of the electronic signal, as it does to electrical transmission of radio frequency? \$\endgroup\$
    – Tim
    Aug 28, 2012 at 17:58
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    \$\begingroup\$ @Tim, it certainly can! When operating at high speeds or in a very noisy environment, it can become a problem. It's only really a problem though if it is strong enough to flip a bit or make a bit ambiguous to interpret on the other end. Some related topics that you may find interesting: cross-talk, differential signalling, PCB noise reducing techniques in general. A lot of protocols tend to have checksums with their transmissions for this reason. \$\endgroup\$
    – Jon L
    Aug 28, 2012 at 18:22
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The quote you provided regarding the flow of information on a computer bus is basically correct, but I think you are confusing the electrical signals in a computer that convey information with those that convey only power. The wires that carry power generally do not also carry information, so they are d.c. signals. The voltage on wires that carry information must change with time. Your example of the USB bus actually includes both types of wires. Two of the wires in a USB cable carry power and two other wires carry information.

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  • \$\begingroup\$ Thanks! For USB cable, do the two wires that carry power carry analogue current, and the two wires that carry information carry digital signal? \$\endgroup\$
    – Tim
    Aug 29, 2012 at 3:40
  • \$\begingroup\$ To me, an analog voltage/current conveys information by its magnitude...the magnitude of the voltage/current is somehow related to some physical parameter of interest (temperature, pressure, etc.). The voltage and current in power wires convey no information at all, they just provide power to the connected device. You might say that the voltage/current is analog in the sense that its magnitude is typically measured in volts/amps instead of 1 or 0, but the intent is different. \$\endgroup\$
    – Joe Hass
    Aug 29, 2012 at 14:37
  • \$\begingroup\$ @JoeHass Came across this years down the line, but I wanted to note that information can be carried by power wires; consider the 4-20 mA current loop, for instance, or the 1-wire protocol for a digital example. \$\endgroup\$
    – Hearth
    Oct 16, 2021 at 22:46
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  1. The quote from the link is only superficially accurate. It is good enough for this discussion, but that's about it. Reality imposes some limitations. For example, a 32-wire ribbon cable will require some grounds and other control signals so the actual number of data bits will be less-- sometimes much less. Also, that link assumes that you get one word per clock. That is OK for a theoretical maximum bandwidth, but the actual bandwidth will usually be lower than that.

  2. Bus speed and the frequency of the AC power are completely independent. There are devices inside that generate the proper frequency (whatever that is). These devices are usually quartz crystals that resonate at the require frequency.

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Everything inside a computer is electric. Electrons are needed to go through devices in order to perform their duties. Starting from the PCB, where the connections between devices are made, then going up to basic resistors, capacitors, then diodes, transistors, ICs, everything powered by electricity and needs electricity to work.

When you say electric, don't think large currents, power lines, and such, think electrons, and when you say bus, think a number of wires in parallel which have (small) electrical current running through. When you say frequency, think switching frequency, going from zero to full and back ("full" being different limits for each device) -- like Indian smoke signals.

As for the way the electricity is used by each, the power supply deals large currents, converts them into high-frequency, high-voltage pulses, and then smoothens them out to supply the power ( http://en.wikipedia.org/wiki/Switched-mode_power_supply ). The DC voltage is then used to supply all the ICs inside, which are working in a similar switching manner, only at a higher frequency and lower voltage. Also, they are inter-connected by buses, which are nothing more than, as the description you gave says, parallel wires.

Therefore, what you name " bus frequency" and "electricity current frequency" are one and the same, only performed differently. Think two teachers: one at junior high and the other at university, they're both teachers but they both do it differently.

So, the computer buses are electrical, as are power supplies, they work and communicate with switching frequency, but each does it differently. While sticking your hand inside a power supply will grant you the chance of electrocution, doing it inside a bus will not. In fact, there are other phenomenons that may even make you burn the ICs on the bus (static electricity). Hopefully, I managed to clear up the fog a bit.

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A computer bus is comprised of on/off signals. The computer clock is simply square-wave alternating current. Or to be more accurate, it is usually pulsed DC which can be considered a type of AC since it is alternating between the bus voltage and zero.

What comes out of an electrical outlet is also alternating current, but it is a sine wave, and the current truly flows in both directions as opposed to turning it on and off.

Now, on a bus, you have many on/off signals that occur at different rates. So the frequency would be erratic. So that is partly what the system clock is for, to give context for these signals. Otherwise, you couldn't tell identical data apart. So is it the same information or two separate pieces of the same thing? So the system clock allows you to know that 2 ones or zeroes in a row are intended to be two different things. If a separate device has to read the signals, then it would need to read them at the same rate, even if it uses the data at a different rate. So in that context, the clock would be the baud clock as it sets the baud rate of the data.

Yes, external cables would be a part of the bus. They connect the buses of 2 different devices.

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