I read a calculation about determining the bandwidth of USB https://www.edn.com/electronics-blogs/all-aboard-/4425691/Rule-of-Thumb--2--Signal-bandwidth-from-clock-frequency

Roughly what was said is:

USB 2.0 = 480 Mbps 
Thus the clock = 240 MHz
Multiply by 5 to get Bandwidth and get ~ 1GHz

So trying this for USB 3.0 I think I'd end up with this:

USB3.0 = 640MBps
clock = 2560 Mhz
Bandwidth = ~12Ghz

As far as I know however, USB 3 noise runs somewhere in the 2.5Ghz range. Is the formula given in the article incorrect? Or is there a different way I should be going about calculating the frequency of different signals?


The "multiply by 5" suggestion would be a way to estimate the oscilloscope bandwidth required to get a good-fidelity waveform display. That's way overkill in terms of the minimum bandwidth required to transfer the data reliably.

USB uses NRZI line coding, which basically means that the data lines can switch states at the clock period, or some multiple of the clock period, with an upper limit that is determined by the specific speed of operation.

This results in an energy distribution that has the overall shape of the sinc function (sin(x)/x). The main peak occurs at half the clock frequency, and there are "nulls" (frequencies of little to no energy) on either side of this peak. With NRZI, the first nulls occur at DC and at the clock frequency.

If you can transmit all of this frequency-domain "main lobe" to a receiver, you will have little trouble recovering the data from the signal, even though there will be some visible distortion of the actual time-domain waveform. If you try to reduce the bandwidth below this value, you will start to see significant amounts of ISI (intersymbol interference). Therefore, most people would say that the bandwidth of this type of signal is equal to the clock frequency.

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    \$\begingroup\$ USB 3+ no longer uses NRZI - which significantly reduces the required bandwidth. \$\endgroup\$ – Turbo J Sep 6 '18 at 7:37
  • \$\begingroup\$ @TurboJ: Well, no, it doesn't reduce the bandwidth -- it just prevents the bandwidth from growing beyond 2.56 GHz. Once you move beyond binary encoding, it's the symbol rate that determines the analog bandwidth requirement. \$\endgroup\$ – Dave Tweed Sep 6 '18 at 11:18

There is no 2.56 GHz. USB 3.0 Gen.1 (5 Gbps) operates exactly at 5000 Mbps bitrate, 8b/10b encoding, (=500 MBytes/s minus overhead), 200 ps unit interval, fundamental frequency 2.5GHz.

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The best signal waveform is nearly sinusoidal, so the factor "5X" is severely overblown. Theoretically sufficient channel bandwidth is 5 GHz for Gen1 speed, and cable specification don't include anything higher than 7.5 GHz.


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