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Is the identification of each point on the constellation diagram arbitrary as long as the sender and receiver know the mapping?

I have seen some diagrams which order in an outward spiral, while others go left to right. Does it matter?

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2 Answers 2

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Yes and no.

The mapping is arbitrary as long as the receiver correctly determines which constellation point a symbol is. If the receiver makes a mistake, though, it is most likely going to pick a "neighbor" constellation point (i.e. a constellation point that is only one spot away). It is highly unlikely that a correctly implemented receiver will pick a constellation point far away from the transmitted point. Communication systems try to mitigate the impact of the "neighbor" mistakes by minimizing the Hamming distance between neighbors. A map where all of the nearest neighbors have a Hamming distance of 1 is called a Gray map. If you use a Gray map almost all of the symbol errors will only result in one bit error.

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It all depends on what you mean by identification. The four bits associated with each constellation point are arranged so that along any row (or any column) two of the four bits are fixed and the other two change in Gray code order. Different people (and standards) might use different choices for which two bits change along rows and which two change along columns. For example, if the first two bits change on the Q axis and the last two bits on the I axis, the assignment is as shown. $$\begin{array}{l|l|l|l} 1111 & 1110 & 1100 & 1101\\ \hline 1011 & 1010 & 1000 & 1001\\ \hline 0011 & 0010 & 0000 & 0001\\ \hline 0111 & 0110 & 0100 & 0101 \end{array} $$ but different Gray code orderings could also be used (e.g. $00, 01, 11, 10$ instead of the $11,10,00,01$ from top to bottom and left to right shown above. If you want to label the constellation points as $s_0, s_1, \ldots, s_{15}$, you can do so in any way you wish including the spiraling out from the origin, but if you want the subscripts to match any straightforward and commonly used mapping from integers $[0, 15]$ to $4$-bit nybbles, that is more complicated.

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