First of all, for a finite number of subchannels, the probability distribution will be a discrete one, described by a probability mass function (PMF). Second, unlike the Gaussian, the supports of the involved distributions are bounded and there exists a maximum value that occurs with finite probability. So, what could be more interesting than to figure out the exact PMFs involved? Ultimately, we are interested in the statistics of the sample powers. Often, these are given as a peak-to-average-power ratio (PAPR) [1]. We will discuss why this has almost no meaning in just a little while.
We’ll start with OFDM in this article since it allows us to use some results from combinatorics to save some work. OTDM, on the other hand, will require a brute force – or comprehensive – approach, and we’ll save that for later.
Caveat emptor: As with almost all articles on this blog, the content presented is not just reproduced from some book, but derived with the help of the cited sources only. I cannot guarantee that it is correct or that it is not treated elsewhere in a more complete way. If you find it discussed somewhere not cited here, or if you find an error, please let me know in the comments.
OFDM Recap
In the article on OFDM Basics an OFDM symbol $C$ was described as
$$C(t) = \sum_{k=0}^{K-1} C_k(t) = \sum_{k=0}^{K-1} c_k \cdot \exp\bigl(i\omega_k t\bigr)\tag{1}$$
where the $c_k$ are the coefficients containing the data. Often, these are taken from a complex M-QAM alphabet which resembles a square in phase space when $M = 2^{2m}$ with integer $m$, e.g. $M=4$, $M=16$ or $M=64$. If we (initially) restrict our analysis to the case $t=0$, i.e. a particular sampling point in the duration of the symbol, then the exponentials vanish for all $k$ and all subchannel symbols $C_k$ would use the same alphabet – without this restriction, the alphabet for each symbol would be rotated by a $t$- and $k$-dependent amount.
We start by looking at the signal statistics for a single quadrature (I or Q), since both are equal with square alphabets / constellation diagrams, and then combine these statistics into a PMF for the symbol power.
Single Quadrature
The signal in a single quadrature of an M-QAM symbol can take $N = \sqrt{M}$ different values. The exact possible values depend on our normalization. If we normalize to average power $\bigl\langle P \bigr\rangle = 1$ then we have the following alphabets
$$\begin{gather}
A_\mathrm{4-QAM} = \frac{1}{\sqrt{K} \sqrt{2}} \left\lbrace -1, 1 \right\rbrace \\
A_\mathrm{16-QAM} = \frac{1}{\sqrt{K} \sqrt{10}} \left\lbrace -3, -1, 1, 3 \right\rbrace \tag{2}\\
A_\mathrm{64-QAM} = \frac{1}{\sqrt{K} \sqrt{42}} \left\lbrace -7, -5, -3, -1, 1, 3, 5, 7 \right\rbrace
\end{gather}$$
Since all coefficients $C_k(t=0)$ are chosen from the same alphabet, the order in which they appear in the sum $C(t=0)$ does not matter, so we are reduced to the problem of figuring out the possible combinations of symbols in $A$ in a sequence of length $K$ and the probability of each possibility occurring. The problem is equivalent to the combinatorical problem of choosing $K$ elements from the alphabet $A$ with repetitions, but without order (since the order of the occurrence of the various symbols is not relevant for their sum), and the number of combinations is given by the binomial coefficient [2]
$$\begin{pmatrix} N + K - 1 \\ K \end{pmatrix} \tag{3}$$
The probability of each such combination is then determined by the multinomial distribution [3].
$$p = \begin{cases}\frac{K!}{\prod x_n!}\left(\frac{1}{N}\right)^K & \text{when} \; \sum_{n=1}^N x_n = K \\ 0\vphantom{\frac{1}{1}} & \text{otherwise}\end{cases} \tag{4}$$
where $x_n$ is the number of occurrences of symbol $n$ from alphabet $A$. In (4) we assumed that all symbols appear with equal probability $N^{-1}$. Figure 1 compares the number of combinations for various $N$ as described by (3) and with a comprehensive approach, which has $N^K$ combinations. The comprehensive approach produces combination counts which are orders of magnitude larger for even small values of the sequence length $K$, and can only be analyzed using Monte Carlo techniques except for small $N$ and $K$.
Figure 1: Number of possible combinations of choosing K times from a set of size N (with repetitions); regarding order (solid lines, comprehensive) and disregarding order (dashed lines).
The MATLAB code to compute the PMF of a single quadrature signal is quite straightforward$^2$:
clear variables
NoS = 8; % number of subchannels
NoV = 4; % number of possible values in each quadrature for 16-QAM - only powers of 2 allowed
normalization = 1 / sqrt(NoS) / sqrt(2/3*(NoV^2 - 1)); % alphabet normalization according to (2)
% see http://www.dsplog.com/2007/09/23/scaling-factor-in-qam/
% for the origin of the factor (2/3*(NoV^2 - 1)
probability = 1 / NoV; % (uniform) probability of each value to occur
values = 0:2:2*(NoV-1); values = values - NoV + 1; % generate integer part of the alphabet
support = min(values)*NoS:max(values)*NoS; % range of integers that the signal can take
[count, tuples] = nsumk(NoV, NoS);
% listing of (ordered) NoV-tuples of non-negative integers adding up to NoS
% see http://www.mathworks.com/matlabcentral/fileexchange/28340-nsumk for details on nsumk()
p = factorial(NoS) ./ prod(factorial(tuples), 2) * probability^NoS; % multinomial PDF of each tuple row
sums = sum(tuples .* repmat(values, [size(tuples, 1), 1]), 2); % sum of all symbols
% add PMFs of equal sum values but different addends (if any)
PMF = zeros(size(support));
for ii = 1:size(p)
index = find(support == sums(ii));
PMF(index) = PMF(index) + p(ii);
end
support = normalization * support;
and a typical PMF plot will look like this:
Figure 2: Probability mass functions for two exemplary combinations of signal levels (N) and number of subchannels (K). More signal levels will lead to more combination possibilities and thus a finer-grained PMF.
The PMF is unfortunately not a very clear tool due to the colored dots all over the place – often it is easier to compare distributions with discrete values by their cumulative density function which is – like for continuous probability densities – the integral of the $\mathrm{PDF}_C(C)$.
$$\mathrm{CDF}_C(C) = \intop_{-\infty}^C \mathrm{PDF}_C(C’) \, dC’ \tag{5}$$
which in MATLAB can be done via
CDF = (sum(tril(repmat(PMF, [size(PMF, 2), 1])),2);
The CDFs are step functions:
Figure 3: Cumulative distribution functions of single-quadrature signal levels for various QAM formats and K = 8 (top), K = 16 (middle) and K = 32 (bottom) subchannels, together with the CDF for a Gaussian distribution with equal power.
We can see that each QAM format and subchannel count will result in a slightly different distribution of values. As $N$ and $K$ increase, the CDFs (and thus PDFs) approach the Gaussian distribution. However, each $(N,K)$ pair has a distinctive minimum value $C_\mathrm{min} = K \cdot \min(A)$ and associated probability $\mathrm{PDF}_C(C_\mathrm{min})$ for which we will give a closed expression later. The CDF can be used to determine clipping probabilities and the CDF of the clipped portion of the signal (where we also have to pay attention to the positive side which does, however, have the same statistics due to symmetry).
Signal Power
For each in-phase amplitude $C_i$ with associated probability $\mathrm{PMF}_C(C_i)$ and quadrature amplitude $C_q$ with associated probability $\mathrm{PMF}_C(C_q)$ we can assign a sample power
$$P = C_i^2 + C_q^2\tag{6a}$$
with joint probability
$$\mathrm{PMF}_P(P) = \mathrm{PMF}_C(C_i) \cdot \mathrm{PMF}_C(C_q) \tag{6b}$$
since the data in both quadratures is independent. We simply calculate all power values and associated probabilities, clean the list by combining the probabilities of equal power values and sort the values$^2$:
ll = length(support);
power = repmat(support,[ll, 1]).^2 + repmat(support.',[1, ll]).^2; % all possible power values
p = repmat(PMF_C,[ll, 1]) .* repmat(PMF_C.',[1, ll]); % joint probability
support_P = unique(power(:)); % find unique power values and sort them
PMF_P = zeros(size(support_P));
for ii = 1:length(p(:))
index = find(support_P == power(ii));
PMF_P(index) = PMF_P(index) + p(ii);
end
The result looks like this:
Figure 4: Probability mass function of the sample power for N = 8 (64-QAM), K = 8 subchannels and unit mean power.
Again, it’s more useful to plot the CDF. In particular, the complimentary CDF can be used to determine the probability of the sample power being larger than some value of interest, which can be related to the PAPR.
Figure 5: Cumulative distribution functions of sample power for various QAM formats and K = 8 (top), K = 16 (middle) and K = 32 (bottom) subchannels, together with the CDF for a χ-squared distribution with two degrees of freedom and equal power.
While there are significant deviations from the $\chi^2$ distribution$^3$ visible in Figure 5, these occur mostly at low probabilities.
Other Sample Times
Generally, a single OFDM symbol comprising $K$ subchannels is made up of at least $K$ samples, more if there is oversampling and/or cyclic prefix. Since sampling times other than multiples of $\Delta\omega \cdot t = \pi/2$ result in different coefficient alphabets$^1$ for different subchannels, we cannot apply our previous simplifications and have to resort to comprehensive methods. We’ll discuss these in more detail in a separate post on Nyquist WDM signal statistics.
As an example, we’ll look at the highest sample power at various sampling instances of OFDM symbols with eight 16-QAM subchannels obtained by exhaustively searching all combinations (which takes about half an hour PER POINT to compute on my machine and which could probably be done much faster with a little more thinking). We would expect the maximum to occur at $\Delta\omega \cdot t = m \cdot \pi/2$ (quarter rotations of the constellation alphabet between neighboring subchannels) because only then do the constellation points with the highest powers in each subchannel potentially add in phase for all subchannels. Figure 6 confirms our assumption:
Figure 6: Maximum possible single-sample power depending on the position of the sample within the OFDM symbol for N = 4 and K = 8. To reduce calculation times, only the non-repeating part (solid) was actually calculated and then repeated (dashed). The symbols correspond to the colors in Figure 7.
The CDFs will of course also differ with sample timing, but not by much as Figure 7 shows.
Figure 7: Cumulative distribution functions of sample power for 16-QAM format and K = 8 for different sample times within the OFDM symbol, together with the CDF for a χ-squared distribution with two degrees of freedom and equal power.
Limit Probabilities and PAPR
Usually, it’s computationally not feasible to calculate the CDFs when there are either a lot of QAM levels $M$ or subchannels $K$. We can, however, quite easily determine the probability of the highest possible sample power to occur. We know that it appears for $t=0$ and we can thus use the derivations made above. We also know that it is just the square of $K$ times the largest-amplitude signal in a single subchannel, or with (6),
$$\max(P) = \max\left(C_i^2 + C_q^2\right) = 2 K^2 \cdot \bigl[\max(A)\bigr]^2 \tag{7}$$
with the alphabet $A$ as in (2). Since or normalization is such that $\left\langle P \right\rangle=1$, $\mathrm{max}(P)$ is also the “true” PAPR of an OFDM signal.
Let us check how probable an occurrence of $\max(P)$ actually is. We start with the probability for all coefficients $C_k$ being equal in a single quadrature, which corresponds to all $x_n$ in (4) being zero except for the one denoting that particular coefficient which is equal to $K$ (it does not matter which one that is when all coefficients are equally likely).
$$p_\text{equal quadrature} = \left(\frac{1}{N}\right)^K\tag{8}$$
The probability that the other quadrature also has the same coefficient in each subchannel (but possibly a different one than the first quadrature) is
$$p_\text{equal symbol} = p^2_\text{equal quadrature} = \left(\frac{1}{N}\right)^{2K} = \left(\frac{1}{M}\right)^{K}\tag{9}$$
where $M$ is the number QAM constellation points, since we still assume the data in both quadratures to be independent. This is the probability that any single constellation point appears simultaneously in all subchannels. There are four constellation points which cause the symbol to have the maximum possible power – the corner points. Hence the probability for the maximum power to occur is
$$p_{\max(P)} = 4 \left(\frac{1}{M}\right)^{K} \tag{10}$$
This can be very, very small, as we can also see in Figure 5, easily down to the $10^{-50}$s. Furthermore, (10) is valid only for the sample times within the OFDM symbol for which $\Delta\omega \cdot t = m \cdot \pi/2$, as discussed above. For all other sample times, the probability for $\max(P)$ to occur is zero.
Such probabilities are utterly irrelevant for communication even at Terabit/s. If just observing an OFDM signal for a while and giving the PAPR as the ratio involving the highest observed sample power, the result would be somewhat random and should only be given with proper confidence bounds. It would make more sense to define a power value (or rather, ratio) which is only exceeded with a probability lower than some value like $10^{-3}$ or $10^{-6}$ maybe. At these probabilities, the differences between the different sample timing (cf. Figure 7) seem to be negligible. As far as I know, such a definition does not exist (yet).
Anyway, if PAPR is a serious issue in a system, one should maybe look at Nyquist WDM which has nicer overall sample statistics (non-Gaussian).
1 Here, $\Delta\omega$ is the subchannel frequency separation. When $\Delta\omega \cdot t = \pi/2$, then all subchannel alphabets are rotated by multiples of a quarter circle (neighboring subchannels are rotated a quarter circle against each other, the next subchannels a half circle, and so forth) which results in identical alphabets for square QAM. The length of the “core” OFDM symbol is given by $\Delta\omega \cdot t = 2\pi$.
2 This is the fastest code I could hack together in a short time. Take it as a starting point.
3 The (central) $\chi^2$ distribution with 2 degrees of freedom describes the distribution of power in a signal with two-dimensional Gaussian amplitude distribution [4].
[1] PAPR [Wikipedia]
[2] Combinations [Wikipedia]
[3] Multinomial Distribution [Wikipedia]
[4] $\chi^2$ distribution [Wikipedia]
