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I'm taking an intro communications module and I'm quite confused on what the term "symbol" really means?

Consider this setup:

enter image description here

My understanding is as folllows

  1. After the information source, we have a stream of bits coming in 0 1 0 1 0 0 0 1 1.....
  2. The source encoder takes these bits and groups them into sets of N bits.
  3. The channel encoder takes these sets of N bits and based on what coding scheme we use (gray coding) for example, it will multiply it by a certain coefficient of a orthogonal basis function (carrier signal) depending on what exact gray code it is.
  4. The digital modulator now receives a coefficient of it's carrier/basis function and just mulitplies it by the fixed carrier/basis function (either sin or cos of something) and sends it onto the channel.

My question is - what point in this transmitter do we call "the symbol"? Is it the group of N bits after the source encoder? Or is it the coefficient of the orhtogonal basis function after the channel encoder?

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In digital communications, information is transmitted in quantized form. We are constrained to transmit items that belong to a finite, discrete set. The items are called "symbols", and the set is called a "constellation". The symbols may be transmitted by changing a signal's frequency, phase and/or amplitude.

As an example, pulse amplitude modulation (PAM) is a technique where the symbols are pulse amplitudes, and the transmitted signal is a sequence of pulses. Let's say that you are using the 4-PAM constellation $\mathcal{C} = \lbrace -3, -1, 1, 3 \rbrace$, and the pulse shape is $h(t)$. Then, you group your bits 2 by 2 and assign them symbols as follows:

bits | symbol | pulse
-----------------------
 00      -3     -3h(t)
 01      -1     - h(t)
 11       1       h(t)
 10       3      3h(t)

Recall that a PAM signal can be written as $$s(t) = \sum_k c_k h(t-kT_p),$$ where $T_p$ is the pulse interval (the pulse rate is $R_p = 1/T_P$) and $c_k \in \mathcal{C}$ are the symbols. Then, transmission of the bit sequence $00111001$ would be done with the signal $$h(t) = -3h(t) + h(t-T_p) + 3h(t-2T_p) - h(t-3T_p).$$ Note that the information (bits) is conveyed by the "symbols" or pulse amplitudes (hence, PAM = pulse amplitude modulation). The task of the receiver is to recover the transmitted symbols from the received signal.

(Note that there's no unique or "official" way to map bits to symbols in 4-PAM or any other constellation; the mapping above is just an example. The only rule is to use Gray encoding, in order to minimize the bit error rate. And, or course, the mapping must be known at the receiver!)

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  • $\begingroup$ For pulse amplitude modulation, the symbols just correspond to a real valued amplitude but the symbols could be complex and have magnitude and phase components. Just to clarify that you saying "amplitude" means a possibly complex valued amplitude $\endgroup$ – Engineer Mar 17 at 18:59
  • $\begingroup$ @Engineer That's a good point! Indeed, in quadrature modulation the symbols are complex. $\endgroup$ – MBaz Mar 17 at 19:29
  • $\begingroup$ I think of "symbol" as any generic item from a collection of items that we can choose to transmit over a single time interval- the symbol duration. The more items in the collection the higher the achieved bit error rate over that bandwidth that is defined by the symbol duration. So this allows for collections of amplitude (as in ASK), quadrature-amplitude (as in QAM), phase (as in M-ary PSK), frequency (as in M-ary FSK) or codes when not used for multiple access (so didn't want to say CMDA specifically but we can create an orthogonal symbol space as well with orthogonal codes). $\endgroup$ – Dan Boschen Mar 18 at 1:28
  • $\begingroup$ So generically any collection from which we can discriminate in the receiver--each item of that collection is a symbol! So probably "Most often..." is not the right intro here for use of amplitude of symbols, but rather one example of many (IMHO) $\endgroup$ – Dan Boschen Mar 18 at 1:28
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    $\begingroup$ @DanBoschen Thanks a lot for the feedback. You are of course correct; I've edited my answer and hopefully it's better now. $\endgroup$ – MBaz Mar 18 at 16:07

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