Electronic – How to be sure that a multi-bit-per-symbol encoding schema exists

baudratebit rateencoding

(I came up with this question when I try to understand bit rate and baud rate.)

Suppose I have some data to transfer. And the data is binary encoded as a data amount of N bits.

If I use 2 symbols to represent the binary data, which means one symbol for 0 and another symbol for 1, then I can only transfer 1 bit of data a time. And the effective bit rate is the same as baud rate.

If I use more than 2 symbols to represent the binary data, and each symbol represents multiple bits, then I can transfer the same N bits of effective data much faster. And the symbol change rate on the line (baud rate) is lower than the effective bit rate.

But how can I know such a multiple-bits-per-symbol encoding schema exists for a given data trunk?

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I once had some difficulty in understanding the baud rate and bit rate. I think the difficulty comes from the first impression I got from pictures similar to below one:

The pic gives me an impression that what get physically transferred over the wire is the digit 0 and 1. And for each digit, a different voltage level is assigned. So there are always only 2 different signal/symbol/voltage types on the wire. And thus the bit rate is always the same as baud rate.

Now I think this pic just shows some effective result. It is symbol that is get transferred over the wire. The amount of possible symbols is determined by the physical nature of the channel/medium. When a symbol is transferred, the one or more bits it carries get transferred effectively. And how symbols represent bits is a mathematical agreement among the communication parties.

Best Answer

There are three elements of a carrier that you can modulate: the amplitude, the phase and the frequency.

A very popular digital modulation scheme uses one of four possible phases (QPSK). So it can convey two bits on each symbol.

Other often used digital modulation schemes use combinations of several amplitude and phases. For example, 16QAM can send one between 16 possible combinations of phase and amplitude. So each 16QAM symbol can convey four bits.

There are other digital modulation variations similar to those mentioned, like 8PSK or 64QAM, 256QAM, etc.

To be able to decode a multi-bit symbol you need rather complex receivers. So those multi-bit per symbol protocols need mechanisms for data synchronization, you have to analyze the path to know if the SNR is enough to differentiate each symbol, etc.

This is it really in a nutshell, I hope that it is clear as an introduction.

Related Solutions

Electronic – How do two UARTS know which baud rate to use

Ordinary UARTs have to be pre-configured with the desired baud rate (as well as word length, stop bits, parity, etc) traditionally by a human.

For several decades now though there have been implementations of "auto baud" detection found in some settings, which typically works by timing key features of the waveform to deduce the baud rate. Early versions needed a known character to be transmitted, but more sophisticated versions might be able to find the rate from more arbitrary data.

A receiving UART typically has a local clock that runs at a faster rate - typically 8 or 16 times the baud rate. This is used to sample the incoming signal and detect the bits within a word in a way that can tolerate a few percent of error. Even two crystal oscillators wouldn't match rates perfectly, but the error tolerance can permit use of some less precise sources, sometimes including trimmed on-chip oscillators, etc. It can also help accommodate the fact that dividing down popular oscillator frequencies may only produce an inaccurate approximation to certain baud rates - in the old days, UART master clocks sometimes needed particular frequencies to access popular baud rates, for example 11.0592 MHz on the 8051 family.

Electronic – Understanding the relationship between LoRa chips, chirps, symbols and bits

LoRa is a chirp-based spread-spectrum modulation. A symbol is a chirp.

To generate symbols/chirps, the modem modulates the phase of an oscillator. The number of times per second that the modem adjusts the phase is called the chip rate and defines the modulation bandwidth. Chip rate is a direct subdivision of the quartz frequency (32 MHz).

Example for 125 kHz LoRa:

125 kHz modulation bandwidth
    = 125000 chips per second
    = 8 µs per chip

modulation bandwidth < occupied spectral bandwidth < channel spacing (typ 200 kHz)

Basic chirps are simply a ramp from fmin to fmax (up-chirp) or fmax to fmin (down-chirp). Data-carrying chirps are chirps that are cyclically-shifted, and this cyclical shift carries the information.

The spreading factor defines two fundamental values:

the number of chips contained in each symbol is \$ 2^{SF} \$
the number of raw bits that can be encoded by that symbol is SF

The reason is that a symbol, with a length of N chips, can be cyclically shifted from 0 to N-1 positions. The "reference" position is given by the un-shifted symbols at the beginning of the LoRa frame. So this cyclical shift can carry log2(N) bits of information. If N is a power of two, the math works nicely.

Example for SF 7

A SF 7 symbol is 128 chips long
    = 1.024 ms @125kHz modulation bandwidth
    = 512 µs @250kHz modulation bandwidth
    = 256 µs @500kHz modulation bandwidth

A 128-chip long symbol can by cyclically shifted from 0 to 127 positions, and that shift
carries 7 bits of raw information:
    ~ 6.8 kbps raw @125kHz modulation bandwidth
    ~ 13.7 kbps raw @250kHz modulation bandwidth
    ~ 27.3 kbps raw @500kHz modulation bandwidth

Due to noise, this modulation/demodulation process introduces errors, and that's why an error correction code is added. For a typical payload, 25% (CR1) or 50% (CR2) of redundancy is added before modulating chirps. In practice, the data sent by the user is also mixed to get better error correction properties.

Raw data-rate and error correction define the nominal data-rate. To get the effective maximum data-rate a device can transmit at, you have to take into account:

legal duty-cycle limit, if applicable, of the band you emit in
overhead of the LoRa preamble, header and CRC for each frame sent (significant influence when short frames are sent)
overhead of your protocol for each frame (also very important for short frames)

Edit:

I have added (in red) the boundaries of chirps so that the effect of cyclical-shifts is easier to understand. Except for a few special symbols at the end of the preamble signaling a start of frame, all chirps in a LoRa frame are the exact same length. Frequency seems to "jump around" quite a bit, but there is no discontinuity in phase that would lead to copious amounts of unwanted harmonics all around the band.

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Related Solutions

Electronic – How do two UARTS know which baud rate to use

Electronic – Understanding the relationship between LoRa chips, chirps, symbols and bits

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