In studying for the CWNA (107) exam, I found one topic to be glossed over repeatedly. Going from the digital data stream to the RF stream. In particular, page 361 state the following:

“When 802.11a/g radios transmit at 54Mpbs, each OFDM symbol contains 288 bits; 216 of these bits are data, and 72 of the bits are error-correction bits. All of the data bits of an OFDM symbol are transmitted across the 48 data subcarriers of a 20 MHz non-HT channel.”

CWNA Certified Wireless Network Administrator Study Guide, 5th Edition, Exam CWNA-107, Coleman and Westcott, 2018.

I didn’t find the preceeding text to have prepared me well enough to easily unpack the unwritten leaps in those two sentences.

- 802.11a/g (see pg 44-46 of the above-cited book)
- 802.11a – Operates in the 5GHz range, uses OFDM, and can handle data rates up to 54Mbps
- 802.11g – Operates in 2.4GHz, uses (ERP-)OFDM, and can handle data rates up to 54Mbps
- ERP-OFDM: Extended Rate Physical OFDM
- Data rate: IEEE only mandates 802.11g supporting up to 24Mbps

- Take-away: OFDM modulation and data rate of 54Mbps, 20MHz wide channels

- OFDM (see Chapter 10)
- Orthogonal Frequency-Division Multiplexing is an RF technique that allows multiple signals to be transmitted and received without causing interference to one another. I found the picture in [1] to be very helpful in understanding this concept. For a given waveform (signal), say the red line, the peak of its waveform happens at the same time that the other signals (green, blue, purple) have 0 amplitude. That means the receiver will be able to uniquely pick out the red signal from the others.
- The above is a very simplistic, non-RF engineering explanation. Feel free to help me make it better for general consumption 🙂

- Take-away: OFDM uses multiple signals transmitted at the same time for data transmission. For 802.11a and g, each Wi-Fi channel is 20MHz wide. Each channel consists of 64 subcarriers (pg. 356). So, each Wi-Fi channel will have 64 different colored lines in it – some are for data transfer and some for other things. In this situation, 48 subcarriers are used for data. These signals are simultanesouly transmitted in a 3.2us symbol time (p 201)

- Data Rate: 54Mbps
- The data rate of 54Mbps is the useful, meaningful data, say coming from an software application.
- For this data rate, looking back to Chapter 6 pg 203, in order to transmit 54Mbps of data from the source to the destination, we will need to encode it.
- Encoding data is a way of using a “thing” to stand in for another “thing”.
- I like to think of pictographic languages where a single picture is used to convey a longer meaning. In this case, we are going to use waveforms to as a stand in for a group of data bits.

- If we have 54Mbps, and we have 48 subchannels to use for our data, then we will need each waveform to stand in for 6 data bit OFDM symbol (signal, colored line from above) will need to be able to represent 64 different states.

Information we have now to understand the sentence on pg. 361.

Protocol: 802.11a/g

Data rate: 54,000bps

OFDM technology: 20 MHz channels, each channel comprised of 64 subcarriers of different types. 48 subcarriers are used for data transmission.

Symbol time: while p. 201 indicates that the symbol rate is 3.2us, use 4us to account for the guard interval (p.362)

I think that working through the problem with the above values will lead us to the need for encoding.

An application is generating data at a data rate of 54Mbps, represented by the Data block above. The binary data will be encoded into one of 64 values, shown coming out of the Encoding block and going into the Modulation block. The Modulation block will map each one of these encoded values into a waveform which will then be combined and sent to the antenna. For OFDM, each of the waveforms selected is orthogonal to one another and can be combined without interference to one another.

Let’s walk through the math, starting with calculating the OFDM symbol rate.

1 OFDM symbol/4us = 250,000 OFDM symbols/second. Recall that we are using 4us to include the guard interval which helps against *Intersymbol Interference* (pg. 361).

In order to have an effective data rate of 54Mbps, how much data needs to be carried by each OFDM symbol?

(54Mbits/1 second) * (1 second/250,000 OFDM symbols) = 216 data bits per OFDM symbol

At this point, we’ve really not considered anything specific about OFDM other than the symbol time. I think this is where some other discipline comes in to evaluate how we might balance data transmission, encoding and modulation techniques to come up with the “right” option for a given application. For example, if we were to use a different (non-OFDM) technology with the same symbol time, our math would be the same so far.

However, given that 802.11 a and g specify OFDM for this data rate, we know that 216 data bits out per symbol time will be sent across some number of the 48 subcarriers capable of transmitting information. This means that the data bits will need to be encoded and the protocol specifies 64-QAM for the modulation which means that 6 bits of data will be represented by one symbol (colored line, not OFDM symbol which is the collection of waves in a 4us time period).

(216 data bits) * (1 sub-carrier/6 data bits) = 36 sub-carriers needed to transmit data

This leaves 48-36 = 12 sub-carriers available for use in each OFDM symbol. In this specific example, the 12 sub-carriers carry error-correcting bits, calculated using Convolutional Code Keying (CCK).

(12 sub-carriers) * (6 data bits/sub-carrier) = 72 data bits for error-correction

## Conclusion

I hope that this helps others understand what is going on in one sentence in the CWNA study guide. It took me awhile to pull this all together and by sharing, I hope we can learn together. Please leave a comment or correction.