Why is it that usually lower digitisation rates are used for long distance communications? e.g. sound is highly distorted in undersea communications, is this because of the lower digitisation rate used?

why lower digitisation rate is preferable?

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    $\begingroup$ Hi! You have your nice answers, but in your question the term digitisation rate can be confusing. Do you mean (possibly) a-) digital data transmission rate or b-) analog signal sampling rate? (which are related to some extend but different things otherwise)... Also make sure what you mean by sound is highly distorted in undersea communications? Do you mean a baseband underwater acoustic transmission distortion (?) or the speech quality deterioration at low bit-rates (usual) $\endgroup$
    – Fat32
    Aug 9, 2017 at 15:36
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    $\begingroup$ Hi, thanks for your questions. I meant: a) Digital transmission rate & b) speech quality is distorted but still usable for such application. My question is mainly related to data compression. If that helps? $\endgroup$
    – S Fateri
    Aug 10, 2017 at 8:22

3 Answers 3


At long distances, your transmit signal loses a lot of power. As a result of that, the SNR at the receiver is possibly relatively low.

A low SNR means that you cannot transport many bits per second over that channel (Shannon limit).

Not having many bit/s means you can't push much data across, which means you must reduce the account of data in the digitized speech. It's that simple!


(As a complement to Marcus' fine answer)

Usually, a transmitter has a fixed amount of available power. You either have a regulatory or economic constraint, or your RF front-end has a power amp of a certain wattage and that's what you're stuck with.

The transmitted power is attenuated over the distance between transmitter and receiver. The actual attenuation depends on the propagation environment, but in general the power decreases exponentially with the distance. So, as a rule of thumb, $$\frac{P_r}{P_t} \propto d ^ {-\gamma},$$ where $d$ is the distance and the path loss exponent $\gamma$ is typically between 2 and 4.

The error probability at the receiver depends on the energy per bit. The energy per bit $E_b$ is equal to $$E_b=\frac{P_r}{R},$$ where $P_r$ is the signal power at the receiver and $R$ is the bit rate. You're spreading your available watts over a number of bits per second.

Let's say that you require a certain error probability in the receiver, which implies a certain amount of energy per bit. If you can't increase $P_t$ (see my first paragraph) nor decrease $d$, your only option is to decrease the rate: clearly, if you transmit at rate $R_{slow} < R_{fast}$, you have a larger $E_b$ and a consequenty a smaller error rate.


Since you used ocean acoustics as an example, for a nominally deep water situation, dispersion of a signal because of multipath tends to stretch a pulse out. Stretching includes the case of discontinuous arrivals. A simple receiver would process a single carrier symbol at a time before having the next transmitted symbol arrives. This is true even is the SNR is high, and often convergence zones exhibit situations where received power is higher at some further distances and lower closer to the transmitter. There isn't just the relative motion of the source and receiver(often unknown) the medium itself has random motions so frequency tones have spreads and shifts. Matched filters are no longer matched. The preponderance of uncertainties limits gains that can be achieved by channel equalization. If the receiver and transmitter have knowledge of all the random propagation effects between them, including knowledge of relative location and speeds, gains can be made but such knowledge is more typically unknown. One symbol at a time for a pessimistically assumed channel, and channels is a reliable strategy.

One might think that this corresponds to a low digitization rate, or alternatively there is a lot of redundancy for each distinct symbol.

  • $\begingroup$ This. Regardless of power level, dispersion will limit the rise time (thus pulse width, thus modulation frequency) - the further the signal travels through a dispersive medium, the worse it gets. Thus the need for mono-mode fiber (no dispersion due to multipath, only due to refractive index), and repeaters with signal recovery... get the high frequencies back before they disappear. $\endgroup$
    – Floris
    Aug 9, 2017 at 18:57

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