Sometimes I see bandwidth of e.g. ethernet cable mentioned in hertz instead of bits/sec. What does this tells about transmission rate? Can it be converted to bits/sec?

For example:

If bandwidth of some cable in hertz is 20 Mhz then can it assumed to be 20 Mbit/s?

3 Answers 3


A cable does not have any inherent bit transmission rate, whereas bandwidth is a measurable property. The cable can be thought of as a low pass filter. All cables can pass DC (f = 0Hz), but some cables have signal loss in the megaHertz range, while coaxial cable can go up to the gigaHertz range.

A cable can carry signal voltage, which is a continuous (i.e. analog) form of electrical energy. Voltage cannot be quantized, so the cable cannot carry digital data unless that digital information is modulated (or converted) to a waveform for transmission. The simplest modulation used for digital data is amplitude modulation, and is commonly known as digital or logic levels. Logic levels are okay for on-board circuitry and cables of very short length. Logic level is not used for longer transmission cables because, although trival and cheap to implement, it uses bandwidth inefficiently and is not energy balanced (consider the worst case of a long string of logical ones or zeroes).

The choice of modulation is a compromise based on the cable's bandwidth and attenuation properties and the complexity of the transmit/receiver electronics. Modulation for digital information can employ phase as well as amplitude. The modulation will determine the number of bits transmitted per symbol. For instance QAM256 can transmit 8 data bits per one symbol, but requires sophisticated transceiver electronics.


what exactly are these cycles if not the rate of bits?

Cycles of a pure sine wave (i.e. a single frequency, just the fundamental, no harmonics).

A digital data signal (using high/low logic levels) would have frequency components 10x to 100x the basic clock rate. If you're acquainted with Fourier series, then you're aware that a square wave is equivalent to an infinite summation of sine waves of related frequencies.

Bandwidth is measured in Hertz or cycles per second. For typical cables the bottom frequency of the band is DC or 0HZ, and the upper frequency would be the frequency of a pure sine wave of a given voltage that does not significantly attenuate (e.g. -3dB) over a given length of cable. Signal attenuation is an intrinsic property of cable; it is usually tabulated in the cable's spec sheet as a function of frequency (pure sine wave) and length.

The transfer rate (e.g. bits per sec) is established by design specification, and is a function of (at least):

  • cable type and length,
  • the capability & cost of transceiver electronics,
  • connector design,
  • the chosen modulation, and
  • the protocol (e.g. Ethernet chooses powers of 10 (& then doubles it for full duplex), SATA chooses multiples of 1500).

For operational reliability the transfer rate actually chosen may be below the theoretical maximum rate.

A cable does not have any inherent bit transmission rate, whereas bandwidth is a measurable property. Therefore there is no conversion formula between the two numbers.


They are not the same. Bits per second refers to the rate that bits of data pass through a transmission media (cable, radio or optical channel). The ethernet cable's hertz refers to its ability to carry high enough frequencies to faithfully reproduce a signal over the cable's length. The better the cable, the higher its frequency rating and the more bits per second of coded signal it can handle. How much will also depend on other factors such as the type of signal used to carry the data and length of cable run.

  • 5
    "hertz per second" -- um no, frequency cannot accelerate. Hertz is "cycles per second".
    – sawdust
    Jan 21, 2012 at 20:39
  • Can you explain what exactly are these cycles if not the rate of bits? electrical pulses or something?
    – Ryan
    Jan 22, 2012 at 10:36

As Wikipedia says:

the Shannon capacity of a ... line is bandwidth \times \log_2(1 + \frac{P_u}{P_n}), with \frac{P_u}{P_n} the (linear) signal-to-noise ratio.

So, there need the second line characteristic: SNR for the line for the bits per second bandwidth to be able to computation.

  • @DavidSchwartz Confused by the link which is anchored at a section titled, “V.34/28.8k and 33.6k.” Cleaned up and upvoted. Apr 10, 2015 at 5:28
  • Sorry, I don't get what you mean with that. Can you please explain? (I mean, the author of the answer)
    – José Leal
    Aug 18, 2015 at 18:43
  • @JoséLeal If you want to convert bandwith of the line measured in Hz to "bit per seconds" you should know SNR (signal to noise ration) for the line in addition to "Hz bandwidth" value. The value you can obtain using formula from the answer is theoretical maximum, called Shannon capacity. Apr 9, 2016 at 11:45

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