There are many modulation types found in the IEEE 802.11-2020 standard, and regardless of the size of their symbol sets, they all have one thing in common—amplitude spacings of two intervals. The amplitude levels in 16‑QAM, for example, are ‑3, ‑1, +1, and +3—two intervals between each value. Although the actual amplitudes—or voltages—may differ in practice, this spacing remains constant, and as higher-order modulation schemes are introduced with ever increasing symbol sets, their respective constellations also increase in size. Naturally, this results in more and more constellation points positioned farther from the origin, and the farther a constellation point is from the origin—the more power is needed to generate that symbol.
By scaling—or normalizing—the constellations, we can keep these escalating power requirements under control without distorting the relative spacing of points within each constellation.
Review
In Part 1, we learned that a modulated symbol—or constellation point—can be mathematically represented by
$$ \large {\color{Green} A}\cos\left ( 2\pi f_{c}t+\theta \right )\:+\:{\color{Red} B}\cos\left ( 2\pi f_{c}t+\theta \right ) $$where ${\color{Green} A}$ and ${\color{Red} B}$ are the respective amplitudes of the cosine and sine components, $2\pi f_{c}$ is the frequency of the carrier waves in Hertz (Hz), $t$ is the time period, and $θ$ is the phase. We also discussed how we can represent this symbol as the following complex number $z$—where the amplitude of the cosine component ($A$) becomes the real part ($a$), and the amplitude of the sine component ($B$) becomes the imaginary part ($b$).
$$ \large z={\color{Green} a}+j{\color{Red} b} $$Since the above complex number will be representing an actual signal composed of an I and a Q component, let’s rewrite this equation to reflect that.
$$ \large z={\color{Green} I}+j{\color{Red} Q} $$Accordingly, the absolute value—or magnitude (when referencing vectors)—of the above can be expressed as the following:
$$ \large \left | {\color{Red} z} \right |= \sqrt{{\color{Green} I}^{2}\:+\:{\color{Red} Q}^{2}} $$
Normalization Factor
When data needs to be transmitted wirelessly, the incoming bits are broken into groups—the size of which is determined by the modulation scheme in use (4 bits for 16-QAM, 6 bits for 64-QAM, etc.). Each group is mapped to a constellation point and converted into the complex number that corresponds to that point. This number is then multiplied by a normalization factor—$K_{MOD}$—forming the output value. This is defined in Equation (17-20) in the IEEE 802.11-2020 standard as the following, where the output value is denoted by $d$:
$$ \large d= ({\color{Green} I} +j{\color{Red} Q} ) \times K_{MOD} $$A single transmission may be comprised of more than one modulation type. For example, if a client decides to transmit data using 256-QAM (8 bits per symbol), only the data payload of the ensuing 802.11 frame will use this modulation; the preamble and the signal field—as well as the pilot subcarriers—will still be sent using BPSK (1 bit per symbol). The normalization factor—which is dependent upon modulation type—is used so that the average power will be the same for all constellation mappings.
Table 1 below lists the various modulations used in 802.11*, along with their respective $K_{MOD}$ values.
*4096-QAM will be introduced in the upcoming 802.11be amendment.
So, how do we determine these $K_{MOD}$ values?
Well, we first need to find the Root Mean Square (RMS) value of a given constellation.
Root Mean Square (RMS)
To calculate the Root Mean Square—or RMS value—of a constellation, we take the square root of the mean of the square of each of the constellation points.
What?
Don’t be confused. It may sound complicated, but it’s really pretty straightforward. Here are the steps:
- Determine the maximum—or peak—amplitude of each constellation point, and square each value.
- Calculate the mean—or average—by adding all values and dividing by the number of symbols in the given constellation.
- Take the square root of the mean.
To make this a bit clearer, let’s do a few examples—starting with 16-QAM.
RMS of 16-QAM
Let’s calculate the Root Mean Square of 16-QAM.
To begin, we will first need to find the peak amplitude of each constellation point. To do this, we’ll represent each symbol as a complex number and determine the absolute value. Then, we’ll square it.
As an example, let’s look at the constellation point in the upper right-hand corner of Figure 1:
$$ \large z={\color{Green} 3}+j{\color{Red} 3} $$Now, we stated above that the absolute value of a complex number is:
$$ \large \left | {\color{Red} z} \right |= \sqrt{{\color{Green} I}^{2}\:+\:{\color{Red} Q}^{2}} $$However, the number we are really after is the square of the absolute value. Squaring both sides of the equation, gives us:
$$ \large \left | {\color{Red} z} \right |= {\color{Green} I}^{2}\:+\:{\color{Red} Q}^{2} $$So, returning to our example, the square of the constellation point $3+j3$ is:
$$ \large \left | {\color{Red} z} \right |= {\color{Green} 3}^{2}\:+\:{\color{Red} 3}^{2} = {\color{Red} 18} $$We will need to do this for the rest of the symbols, after which we will add up all 16 values and find the average. See Table 2 below.
Now, we need to determine the square root of the average that we just calculated. The result is the Root Mean Square value—which represents the average power of the constellation.
$$ \large RMS_{16-QAM}^{} = \sqrt{{\color{Red} 10}} $$To visualize the difference between peak and RMS values, let’s take a look at Figure 2 below. The radius of the red circle depicts the RMS value, which we’ve just discovered to be $\sqrt{10}$—or about 3.162 (after rounding). The radius of the yellow circle shows us the peak value of the constellation, which—for a square constellation like 16-QAM—is the absolute value of one of the corner constellation points. For 16-QAM, this is $\sqrt{18}$—or about 4.243.
Let’s solidify this concept with a few more examples.
RMS of BPSK
The BPSK modulation scheme consists of only two symbols, making it relatively easy to determine the RMS value of the constellation.
Let’s first convert the constellation points to complex numbers. If you examine Figure 3 above, you’ll notice that both constellation points lie on the real, or I axis, meaning neither symbol has an imaginary component—or more appropriately—they both have an imaginary component of zero.
For example, the symbol on the left can be represented as the following complex number:
$$ \large z= {\color{Green} -1} + j{\color{Red} 0} $$The square of the absolute value is:
$$ \large \left | {\color{Red} z} \right |= {\color{Green} -1}^{2}\:+\:{\color{Red} 0}^{2} = {\color{Red} 1} $$We will then do the same for the other symbol, add the two values, and find the average. See Table 3 below.
Next, we need to determine the square root of the average—and the square root of 1 is 1.
$$ \large RMS_{BPSK}^{} = \sqrt{1} = {\color{Red} 1} $$You will find that BPSK is unique in that its peak and RMS values are both one. This is illustrated in Figure 4 below.
RMS of 64-QAM – First Shortcut
We are now going to find the RMS value of 64-QAM—but this time, we’re going to do something a bit different. There are two shortcuts we can use to determine the RMS value that can save a lot of time—especially for the larger modulation schemes, such as 4096-QAM.
The first shortcut is to simply calculate the RMS value using only a single quadrant of the constellation. Since, in 802.11, we are mainly dealing with square QAM constellations, the number of constellation points in each quadrant—and their absolute values—are identical. This means that finding the RMS value over only a single quadrant will still give us the same value as if we used all four quadrants for our calculation. Table 4 below lists the squared values—and their average—for the first quadrant of 64-QAM.
Then, just like before, we need to take the square root of the average.
$$ \large RMS_{64-QAM}^{} = \sqrt{{\color{Red} 42}} $$
RMS of 64-QAM – Second Shortcut
The second shortcut is probably the fastest and easiest way to determine the RMS value of the 802.11 QAM constellations. Simply use the following equation, where $M$ is the number of symbols in the constellation:
$$ \large RMS_{{\color{goldenrod} M}-QAM}^{} = \sqrt{\frac{2}{3}({\color{goldenrod} M}-1)} $$If we use the above equation to find the RMS value of 64-QAM, we get:
$$ \large RMS_{{\color{goldenrod} M}-QAM}^{} = \sqrt{\frac{2}{3}({\color{goldenrod} M}-1)} = \sqrt{\frac{2}{3}({\color{Cyan} 63})} = \sqrt{{\color{Red} 42}} $$
Normalizing to One
We now know how to calculate the RMS value of a given constellation, but we’re still not done. We’ve already mentioned that the RMS value represents the average power of a given constellation. Well, now we want to normalize the average power to one.
So, how do we do that?
It’s pretty simple, really. We divide 1 by the RMS value.
This is how we derived those normalization factors—or $K_{MOD}$ values—listed in Table 1. For example, to find the normalization factor for 64-QAM, we would divide 1 by $\sqrt{42}$.
$$ \large K_{MOD_{64-QAM}} = \frac{{\color{Cyan} 1}}{\sqrt{{\color{Red} 42}}} $$If you recall, a single 802.11 transmission may contain more than one modulation type, so we want to minimize the power disparity between constellations. To visualize this, look at Figure 6 below for a side-by-side comparison of the BPSK and 64-QAM constellations before normalization.
Now, compare that with Figure 7 below, which shows the BPSK and 64-QAM constellations after normalization, where the normalized RMS value—$RMS_{norm}$—is 1.
The BPSK constellation is unchanged, since the RMS and Peak values were already 1. However, you can see that the peak power of the 64-QAM constellation has been reduced a great deal—from approximately 9.899 to about 1.528.
The impact of normalization would be even greater for higher-order modulations, such as 4096-QAM, where the normalized peak power— $Peak_{norm}$ —would be approximately 1.705, which is a massive reduction from about 89.095. The RMS and Peak values of the constellations found in 802.11—before and after normalization—are listed in Table 5 below.
While normalization helps constrain the rising power requirements of higher-order modulations, in doing so, it forces the constellation points in each constellation closer together. Illustrating this point, Figure 8 below displays the constellations for 16-QAM and 1024-QAM after normalization.
You can see how the constellation points in 1024-QAM are much more densely packed as compared to 16-QAM. As the number of symbols increases, the distance between constellation points decreases—which greatly increases the accuracy requirements at the transmitter.
This leads us into Part 3, where we discuss the Error Vector Magnitude, or EVM.