1. Gains provided by multichannel transmissions
Month Year
doc.: IEEE yy/xxxxr0
Gains provided by multichannel transmissions
Date:
Authors:
John Doe, Some Company

2. Interests toward multi-channel
Currently, the increase of bandwidth seems to be the only solution to increase the single user throughput.
However, the probability of being allowed to transmit at 80MHz could be quite low in dense environment
Interests of multi-channel
Multi-channel is an efficient solution to increase that probability and ensure an increase of single-user throughput compared to 11n.
In [1], we proposed 3 different methods for multichannel transmissions

3. Most favoured solution for multi-channel Non contiguous synchronous
Aggregation of non contiguous channel wherever in the band
Only one CSMA-CA on the primary channel
Secondary, tertiary and quaternary can be on any channel of the band
Primary Channel
AIFS + backoff
Ack
time
Secondary Channel
Ack
time
PIFS
Tertiary Channel
Ack
time
Quaternary Channel
PIFS
Ack
time
PIFS
SIFS 30 23 17
Primary
Secondary
Tertiary
Quaternary

4. Most favoured solution for multi-channel Non contiguous synchronous
Without channel bonding within 80MHz
With channel bonding within 80MHz
Primary Channel
AIFS + backoff
Ack
time
Secondary Channel
Adj. BSS trans.
SIFS
time
Tertiary Channel
time
Quaternary Channel
PIFS
time
PIFS
Primary Channel
AIFS + backoff
Ack
Secondary Channel
Adj. BSS trans.
time
time
PIFS
Tertiary Channel
Ack
time
Quaternary Channel
PIFS
Ack
time
PIFS
SIFS

5. Most favoured solution for multi-channel Non contiguous synchronous
Synchronous non contiguous implementation allows easy implementation of channel bonding
Seg 1
Seg 2
Transmitter
Segment 1
Digital processing
DAC I
DAC
5GHz GHz
Seg 1
Seg 2
DAC
DAC
Segment 2 IF RF
Seg 1
Seg 2
Busy channel

6. Goal of this presentation
Demonstrate the gains provided by multichannel to increase the probability to reach the desired bandwidth
Clarify the advantages of multichannel transmissions
 We ran some simulations for that purpose

7. Simulation scenarios
Simulate different load configurations on the whole 5GHz band
Select a number of busy channels among the 19 available in Europe: called density
Restrict the set of scenarios by setting a unique load for each busy channel (fixed to 0.3, 0.6 and 0.9)
For each density from 0 to 19, we simulate all possible configurations 30 23 36 40 44 48 52 56 60 64
100
104
108
112
116
120
124
128
132
136
140
Example: density=8, load =0.3, one configuration among all possible

8. Simulation scenarios
Simulate different load configurations on the whole 5GHz band
For each configuration, for each multichannel solution (non contiguous, bonding, nb of aggregated channels) and for each channel allocation, we calculate the equivalent bandwidth (bandwidth that you would be able to use 100% of the time)
And select the best channel allocation and its corresponding optimal equivalent bandwidth
Average the equivalent bandwidth for each density

9. Calculation of the equivalent bandwidth
Assume we have 4 aggregated channels, each with a specific probability Pi of being idle.
We can calculate the probability to transmit at 20, 40, 60 or 80MHz
P1, P2, P3, P4
Primary
P60
P60
Without channel bonding
P80 = P1*P2*P3*P4.
P60 = P1*P2*P3*(1-P4)
P40 = P1*P2*(1-P3)
P20 = P1*(1-P2)
With channel bonding
P80 = P1*P2*P3*P4.
P60 = P1*P2*P3*(1-P4)
+P1*P3*P4*(1-P2)
+P1*P2*P4*(1-P3) …

10. Calculation of the equivalent bandwidth
Assume we have 4 aggregated channels, each with a specific probability Pi of being idle.
We can calculate the probability to transmit at 20, 40, 60 or 80MHz
And calculate the equivalent bandwidth (bandwidth that you would be able to use 100% of the time)
P1, P2, P3, P4
Primary

11. Simulations Compare the equivalent bandwidth vs density for
Contiguous mode (with or without channel bonding)
Ideal synchronous non contiguous mode (with or without channel bonding)
Restricted* synchronous non contiguous mode (with or without channel bonding)
*Restricted: only two non contiguous sets of contiguous channels
Tested for different multichannel bandwidth (40, 60, 80MHz)
Tested for different loads for busy channels: 0.3, 0.9
- 40MHz = 2*20MHz channels
- 60MHz = 3*20MHz channels
- 80MHz = 4*20MHz channels

12. Simulation results: 80MHz, load=0.9
Significant improvements with ideal non contiguous
Still very good gains with restricted non contiguous
Importance of the channel bonding

13. Simulation results: 80MHz, load=0.3
More compact results
Still significative gains with restricted non contiguous
Importance of the channel bonding

14. Simulation results: Restricted non contiguous gains, load=0
Simulation results: Restricted non contiguous gains, load=0.9, 40MHz up to 80MHz
Leads to a potential reduction of the bandwidth used by a BSS
Example: Density 12,
Eq bandwidth required 50MHz:
- contiguous required used bandwidth: 80 MHz
- non contiguous required used bandwidth: 60 MHz
Example: Density 13,
- contiguous: saturated

15. Advantages of synchronous multi-channel (Conclusions)
Clear increase of the probability to reach the desired bandwidth
Increase of the equivalent bandwidth
Or reduce the bandwidth occupied by a BSS for the same equivalent bandwidth
Importance of the channel bonding
The implementation with two front-end segments enables easy channel bonding, even in contiguous mode
Better use of the whole band
Assuming the knowledge of the 5GHz band occupancy, it offers a good flexibility in the use of the band: possible switch for the best channel allocation
The two front-end segments implementation enables the possibility to sound the band while maintaining the link

16. References
[1] Cariou, L. and Benko, J., Multichannel transmissions, IEEE /1022r0, Sep. 2010
Slide 16
Youhan Kim, Atheros