As hatsunearu said, the radio modulation described is grossly incorrect. WiFi never uses 8-PSK (encoding 3 bits per symbol). 802.11n and 11ac encode 1/2/4/6/8 bits using a BPSK/QPSK/16-QAM/64-QAM/256-QAM symbol (256-QAM is for 11ac only). The modulation scheme is negotiated based on signal quality. Here is a quick reference: http://mcsindex.com/ (MCS = modulation coding scheme) On Linux you can find the MCS negotiated with "iw dev wlan0 link | grep -i mcs"
14 channels are defined in the 2.4GHz band. For example channel 6 is centered on 2437 MHz. Each channel is 20MHz wide and divided in 52 "data" subcarriers, each occupying a different frequency and spaced out by 312.5 kHz (52 × 312.5 kHz is less than 20 MHz because there are "control" subcarriers and additional spacing.) So 52 different symbols can be sent in parallel at the same time, which is what we call OFDM https://en.wikipedia.org/wiki/Orthogonal_frequency-division_... (basically, I'm simplifying!)
Remember this is for just 1 channel. So with 14 channels each composed of 52 subcarriers, we could have 728 symbols transmitted at the same time. If they are 256-QAM symbols that's basically 728 × 8 = 5824 bits being transmitted at the same time in the air. And they will all be received and demodulated independently. This high level of parallelism of OFDM is how WiFi can achieve very high throughput.
Then, with wide channels of 40 MHz, which basically aggregate two 20 MHz channels, we get a few more data subcarriers because we don't need as many control subcarriers so a few of them become used as data subcarriers. Hence a 40 MHz channel will have not 52 × 2 = 104 but actually 108 data subcarriers. And 802.11ac defines 80 MHz and 160 MHz channels with respectively 234 and 468 data subcarriers.
Let's calculate the maximum usable throughput of a single 802.11ac 160 MHz channel using 256-QAM modulation... It sends 468 symbols at the same time on 468 data subcarriers. Each symbol encodes 8 bits and takes in the best case 3.6us to be transmitted: 3.2us for the actual symbol + a short guard interval of 0.4us (the GI is normally 0.8us but can be a short GI of 0.4us if negotiated). The raw physical bitrate is:
1/3.6e-6 × 468 × 8 = 1.04 Gbit/s
However there is a mandatory error correction which is 5/6 in the best case so the actual usable bandwidth is:
Note that, although the 2.4GHz spectrum is formally divided into 12-14 channels (depending on local regulations), these are very narrow channels; in practice there are only 3 non-overlapping 20MHz-wide channels. This is a small fraction of the width of the 5GHz band.
There are 3 non-overlapping 2.4Ghz channels (1,6,11) for 802.11b, because of the channel spectrum shape. Not only "b" uses 22Mhz channels with is just a bit too wide, but also, due to the way how single-carrier PSK modulation works, especially on older hardware, "b" has considerable amount of spurious emissions adjacent to the main carrier that widen it even more - could be seen here as smaller "hills" to the left and right [1],[2].
802.1g/n/ac can easily have 4 non-overlapping channels (1,5,9,13) because (thanks to OFDM) channel spectrum is much neater with rather square 20 (40) Mhz channels with practically no energy outside [3].
Yet, everybody's stuck with 1,6,11 channel scheme which is wasting precious bandwidth. [4] (middle graph) Notice gaps between the channels that could be eliminated by moving 6->5 and 11->9, and gap on the right where channel 13 can fit after that.
14 channels are defined in the 2.4GHz band. For example channel 6 is centered on 2437 MHz. Each channel is 20MHz wide and divided in 52 "data" subcarriers, each occupying a different frequency and spaced out by 312.5 kHz (52 × 312.5 kHz is less than 20 MHz because there are "control" subcarriers and additional spacing.) So 52 different symbols can be sent in parallel at the same time, which is what we call OFDM https://en.wikipedia.org/wiki/Orthogonal_frequency-division_... (basically, I'm simplifying!)
Remember this is for just 1 channel. So with 14 channels each composed of 52 subcarriers, we could have 728 symbols transmitted at the same time. If they are 256-QAM symbols that's basically 728 × 8 = 5824 bits being transmitted at the same time in the air. And they will all be received and demodulated independently. This high level of parallelism of OFDM is how WiFi can achieve very high throughput.
Then, with wide channels of 40 MHz, which basically aggregate two 20 MHz channels, we get a few more data subcarriers because we don't need as many control subcarriers so a few of them become used as data subcarriers. Hence a 40 MHz channel will have not 52 × 2 = 104 but actually 108 data subcarriers. And 802.11ac defines 80 MHz and 160 MHz channels with respectively 234 and 468 data subcarriers.
Let's calculate the maximum usable throughput of a single 802.11ac 160 MHz channel using 256-QAM modulation... It sends 468 symbols at the same time on 468 data subcarriers. Each symbol encodes 8 bits and takes in the best case 3.6us to be transmitted: 3.2us for the actual symbol + a short guard interval of 0.4us (the GI is normally 0.8us but can be a short GI of 0.4us if negotiated). The raw physical bitrate is:
1/3.6e-6 × 468 × 8 = 1.04 Gbit/s
However there is a mandatory error correction which is 5/6 in the best case so the actual usable bandwidth is:
1.04 × 5/6 = 866.67 Mbit/s