WO2010149049A1

WO2010149049A1 - Method and apparatus for improving downlink coverage and throughput in ofdma system

Info

Publication number: WO2010149049A1
Application number: PCT/CN2010/074352
Authority: WO
Inventors: 唐枫
Original assignee: 中兴通讯股份有限公司
Priority date: 2009-12-04
Filing date: 2010-06-23
Publication date: 2010-12-29
Also published as: CN102088310B; CN102088310A

Abstract

A method for improving downlink coverage and throughput in an OFDMA system is provided in the present invention, which includes: selecting a Zone for a terminal, and traversing all Zones in turn (401); judging whether Zone Boosting of the downlink Zone is predetermined (402), if yes, then performing the step 403, otherwise, performing the step 405; calculating the maximum power value of data sub-carrier of all terminals in a Zone (403); allocating Data Boosting for the terminal based on the maximum power value and the downlink Carrier Interfere Noise Ratio (CINR) of the terminal (404), then end; allocating Data Boosting for the terminal based on the downlink CINR of the terminal (405); calculating the maximum value of Zone Boosting based on the number of the sub-channel occupied by the terminal and the allocated Data Boosting value (406); enabling the maximum value of Zone Boosting for said Zone (407). By using the method for combining Zone Boosting and Data Boosting, the present invention resolves the problem of power overflow caused by allocating downlink power unreasonably when enabling Zone Boosting and Data Boosting at the same time.

Description

Method and device for improving downlink coverage and throughput of O A system

This paper relates to the field of communication wood, especially to the method and device for improving the downward coverage and throughput of OFD A Othogo a Feq e cyDvso peAccess system.

In the communication system, the terminal provides each and every communication between the upper and lower terminals. Downstream refers to the direction of the terminal, and uplink refers to the direction of the terminal. The terminal can be connected to the station and can also receive the downlink.

OFD A wood has become the mainstream wood of communication physics wood due to its effectiveness and efficiency. F A+ O, pe- p fpe-O -p ) wood is more than 3 generations of A, Co vso peAcces ) In the communication system. Global microwave compatibility (W A , Wo dwde e opeab y o c owave Acce ) OF AA Muli Physics and W 802.16e and 802.16 under development are strong contenders for next generation communication.

In the OF A system, the upper component OF A is in the upper component, and the multi-member meson in the upper component is the set of group subcarriers. Usually the meson channel and the inter- or cross-o) So are the smallest assigned sheep of the OF A system. It is possible to represent the rectangular table of the physical resource sub of OF A. The same matrix that can be divided into multiple units in the downward direction of the OF A system, you can force different functional requirements of Zoe, than the phase carrier arrangement (Ba dA, ad ce S bca e pe ao ) O and AA, etc. Zo e force, as shown. Zoe does not have to occupy all the sub- 5s in the downlink, nor does it have to occupy all of the downlink but its size must be a matrix.

Currently, 802 6e Boo g is a commonly used method in OF A system that does not boost the power of certain subcarriers (including a) subcarriers and P o) subcarriers. Boos g components aaBoos g and Zo e oos g aa oos g are mainly All the subcarriers of the terminal pass through all the subcarriers of the terminal 1 aaBoo g 12 north to 9 ), which can improve the downlink of the terminal, and the downlink throughput of the uplink terminal is the same. Zo eBoo g is all subcarriers and subcarriers of the downlink Zo e g, pass some Zo e, especially the Zoe of the limiter, do not P CZo e, Pa a eageo bcha e Zo e) Zo e Boo g can effectively improve the accuracy of the downlink, and can improve the downlink cover. Where aa oos g Chinese gain, representing the power gain of the subcarrier, Zo e boo g Chinese gain, yes

Power gain of subcarriers and subcarriers.

The power required by the RF sheep in communication

The purpose is to control the power of the RF sheep according to the power instead of 2 power, and the power, then cover the deviation, the effect of the group.

However, the existing wood has the following defects: the existing wood in the same ZoeBoost g and aaBoo g power consumption overflows due to the inability to reasonably allocate the downlink power. The problem to be solved in the content is to provide a method for improving the downlink coverage and throughput of the OF A system, and reasonablely allocate the downlink power without increasing the power and increase the coverage and throughput of the OF A system.

5 To the above purpose, the wood scheme provides a way to improve the coverage and throughput of the OF A system.

A, terminal Zo e, in turn all Zo e

Whether the gain Zo eBoo g of the downlink Zo e is in advance, if yes, then special step C otherwise special step E

C, the maximum power of the number of subcarriers of all terminals in Zo e

The maximum power and the downlink carrier ratio of the terminal, the terminal allocation gain aaBoo g ends

E. The downlink carrier ratio of the terminal is allocated by the terminal ataBoos g F, the child occupied by the terminal, and the assigned aa oo g , the maximum value of Zo e 15 oo g

G, Zo e maximum Zo eBoos g

Step C includes

- According to the formula below

8

" b a e e mb * ) / GaW Where, "Where. "s c e" h

The maximum power of the number of subcarriers of all terminals in Zo e, where ax w maximum power "" Ga of the front subcarrier Zoe Boos g Zo e

The sub-carriers contained in the "e" meson.

Steps include

According to the order in which the downlink carrier ratio of the terminal is small to large, Zo e All terminal line sorting

2. According to the order of sorting, if the maximum power value is not greater, the terminal allocates aBoo

Step 2 Hugh includes:

终端 Terminal with downlink carrier ratio less than the previous value gdb aa Boo g The terminal with a downlink carrier ratio greater than or equal to the previous value is not aaBoo g

Formula a

" , . W

" , .

『 G

"The power of several subcarriers of which , ,

""", the child occupied by the terminal's aB s,

The subcarriers contained in the meson,

a oos g subcarrier non-gain of the terminal

Compare the power of several subcarriers, the maximum value ax . ,

Ax is less than or equal to the maximum power, and if so, end No "". Yes

Reassign aaBoos

Step E includes

终端 The downlink carrier ratio is less than the pre-valued terminal g aa Boos g The downlink carrier ratio is greater than or equal to the previous value of the terminal is not aaBoos g.

The step F Hugh includes: Formula Gan Fan 8 / ((

, ," , n e oW oh r ) s bcha e, , 1

" P s b )

"Maximum value of ZO e OOS g

Ga, where, "pre-subcarrier, S

" "", the subcarrier occupied by a "S" of the terminal,

"" "n" contains the number contained in the meson

G aa terminal

" BOOS g number subcarrier ungained gain Where the Zo e is occupied by,

" " " Subcarriers contained in the meson.

Subcarrier C R

",", on CZ R, "directly on the self-encoding judgment.

The subcarriers C R , C R on the terminal are increased by aaBoost g

" "

Ga

" , and then from the judgment.

The terminal subcarrier CZ R

"amhe,

The gain of the aaBoo g Gan ",""e and the gain of the Zo eBoo g Ganz, then the self-encoding decision.

The device for improving the coverage and throughput of the OF A system includes at least the overlay, the second, and the second.

, in the terminal Zo e, in turn all Zo e in the Zoe oos of the downstream Zo e is in advance

To the first pass, the Zo e oo g on the down Zo e is not in advance, to the second second pass

The maximum power obtained by receiving the maximum power of the subcarriers of all terminals in the first Zo e and the downlink carrier of the terminal

Ratio, terminal allocation aaBoo g

Secondly, the downlink carrier ratio of the second terminal is received, the terminal allocates the aa oos g terminal and the allocated aa Boos g , the maximum value of Zo eBoos g Zo e Boo g of the maximum value

Existing wood compared to this beneficial effect

Ben Mingtong solves the problem that Zo e oos g and aaBoos g are combined in the same Zoe Boo g and aaBoo g because the downlink power cannot be allocated reasonably. The problem of power overflow is that under the premise that the power does not overflow, the reasonable allocation of downlink power will use Zo eBoos g to mediate P Czo e and concentrate aBoos g for users, which improves the coverage and throughput of OF A system. 1 is the downward Zoe of the existing wood

2 is the downward coverage of the OFD A system.

Is the downlink allocation of this

4 is the flow of the method for improving the downlink coverage and throughput of the OF A system.

5 Forces The composition of the device that raises the coverage and throughput of the OF A system. Hugh way

The following combines and , the rest of the way. The following is a description, but there is no need to limit this.

The downward coverage of the OF AA system is shown in Figure 2. It can be seen that there is no Zoe oo g and aaBoos g. The downlink coverage is 21 in Zo eBoo g and a oos g , because the power coverage of the subcarriers of Zo e is improved. When the power does not overflow, the maximum value of a Boos g is not improved, and the throughput of the D Boos g is also improved.

The required subcarriers of the strategy are not fixed in the OF AA system because the pre-eamb subcarriers are not fixed, so the power of the P ea be subcarriers must satisfy the formula 1: e mbc + P , (1) In Equation 1), the power of the e P ea be subcarrier "power of a subcarrier, power of P o subcarrier.

The power of the 802.16e subcarrier is higher than the power of the subcarrier. The pre ( ea b ) subcarrier power is higher than the number of subcarriers 9 so that

.

The power P of the subcarrier is the subcarrier power 16/9P, and the pea be subcarrier power P. Get the formula (2)

Ga 2) The power of all the subcarriers of the terminal in equation (2) and the power P of the subcarrier without a Boo g

R, G * Where, . * * 3 ne mb e b "8 P 4 where Ga means Zo eBoo g Zo e No gain Sheep

Where the P ea be subcarrier is represented, the FFTSze is different.

"Every" e is different, for example, the FFTSze 024 system does not say that a be 284 represents the child occupied by Zo e

Indicates the power and frequency of the subcarriers of all terminals

Where os represents the subcarriers contained in the meson. Because of the fixed Zo e, e, where, G,

n e is already so the formulas ( 1) (2), (3), (4) can be found in equation (5): g where " * /G - where, .

Os b e 1 5) , ,"

Since the number of sub-allocations of different terminals is different from that of the sheep terminal, Equation 6) is not mentioned: .

h e W, Ga

" "w * r 6 where Gan. . ,,

" " indicates the ata Boost g terminal's respectful carrier carrier gain, sheep, due to a oo g at the terminal a

( aB s you indicate that the allocation to the terminal is determined by

) So it’s said at the sheep terminal.

Indicates that the number of subcarriers contained in the meson refers to the subcarrier occupied by the terminal's D aB .

Different aaBoos can be enabled on different terminals but must meet the formula (5 limits, in OF A system, terminal aa B size and is determined by base, so S

" "", is the downlink allocation shown in 3, which can occupy 1 meson 2 and occupy 2 mesons.

Others, Fan, Ga Z. n e and where ",.

" "", only "AaBoo g to be assigned to the terminal satisfies Equation 5", no downlink power overflow.

4 The process of increasing the OF A system downlink coverage and throughput method is shown in the case of Zo e oos g in the downlink Zo e, the maximum value of the subcarrier power in Zo e, the maximum The value of the downstream RF power is not different. The different terminals can have different aa oos g than the user's highest g north aaBoos g. Compared with the better terminal, the north ao oo g or aaBoo g is determined in the case of Zo e oo g. Terminal a oo g, can be considered first

(

"The pre-requisite is the terminal that has the downlink and reversed." a Boo g max to 9 and then Zo e Zo eBoo g¥ maximum value. Refer to 4, this includes the following steps

Step 401 Terminal Zo e, in turn all Zo e. The media terminal can only be switched between different Zo e in the Zoe. Control (AC e ce co oae , control can be used for different functional requirements terminal Zoe, than STCzo) intelligent system (zo), from this Czo e), p Czo e and other Zo e oo g configurable traffic It can be fixed to Zoe oos g -gdb to g north. It can also be used as the maximum, Zo e maximum of aaBoo g Zo eBoo g. This requires all Zo e in the downlink

Step 402: If the Zo e oos g of the downlink Zo e is in advance, then step 403 is performed, otherwise step 405 is performed.

Step 403 Maximum power of subcarriers of all terminals in Zo e. In this, the formula "e e be* 8) /G - chan.

" "" " * 1 / g The maximum power of the subcarriers of all terminals in Zo e , where ax ,

"Max power" Ga power of the front subcarrier Zoe Boo g Zo e The gain of no, "", the sub-occupant of Zo e. ", "Subcarriers contained in the meson.

Step 404: The maximum power and the downlink carrier ratio (C) of the terminal, and the terminal allocation a is also Boos g, step 408. First, all the terminal lines in the Zo e are sorted according to the descending order of the terminal. In the order described, the terminal allocates aBoos g when the value is not greater than the maximum power value. Which assigns aaBoos g

First, the downlink C R is smaller than the previous ratio in QP /2 ] mode) ¥J terminal g aa oo g, the terminal C R is greater than or equal to the previous value of the terminal is not aa oo g

Pick up the formula

. W, s bc , . Ga

" ""

The power of the number of subcarriers where the "a" occupied by the terminal's aaB Where "," the number of subcarriers contained in the meson

" The power of the subcarrier of the aBoos g terminal is not the gain of the subcarrier.

"" Most,

Whether it is less than or equal to the maximum power, if it is, then 5, otherwise reassign aaBoo g

Step 405: Downstream C of the terminal, the terminal allocates aaBoos g. The middle and lower downlink C is smaller than the preamble of the phase keyed (P / ) mode terminal g aBoo g, and the terminal user performance downlink C R is greater than or equal to the previous value of the terminal is not aaBoos g

Step 406 The maximum value of the sub-and the assigned aa Boo g and Zo eBoo g. In this, the formula Ga is 8)/

,

16/9)) Zo e Boo g's maximum value Ga, where, where, "pre-subcarrier, S, end occupied,

" " End aa Son Taoism, W Jie

" " child "

Subcarriers included in the media, Ga. aaBoo g final

"

The subcarriers of the end are not gaining, where, Zo,

"" e is occupied by the child " " " subcarriers contained in the meson.

Step 407: Zo eBoo of the maximum value of Zo e

Step 408 consider Boos g downlink C. Because the downlink C R of the terminal is reversed, it is necessary to consider the impact of the uplink and downlink C of the Zo eBoos g and the DaaBoos g terminal, depending on the support capability of the terminal (the capability is reported at the terminal end capability end). In this case, the terminal subcarrier C R C R

"" Since Zo eBoo g and aaBoo g both act on subcarriers So on the CR

"Without the gain of J Boos g, you can directly go to A

The decision of the downlink C decision downlink coding mode is the ideal state of God, and depends on the external capability of the terminal, the subcarrier C C R, . ,

" Since Zo eBoos g acts on subcarriers and subcarriers, aa oos g only acts on subcarriers, so CZWR

"" needs the gain Ga of aa Boos g

"" can be used for A C decision on the terminal pea be subcarrier C R, wave force C 『,

",", because Zo eBoo g acts on subcarriers and subcarriers a oo g only acts on subcarriers, so C R ,

"" "Need the gain Ga of J ataBoos g

"" " , "and the gain of Zo e oo g GanZ " " can be used for A C's decision.

In step 403, step 404 shown in FIG. 4, that is, in the case of Zo e Boo g of the downlink Zo e, since OF is a system, o is the smallest allocation sheep, and the downlink shown by 3 is composed of a multi-o matrix, terminal allocation The physical resource is the resource of o sheep, which occupies 5 mediat than terminal 1 ( ), and 2 occupies 4 mediations. The number of sub-elements occupied by the terminal (including all sub-) terminals is different than that of the meson at o4, 2 does not occupy the sub-portion, 3 occupies 2 mesons, and 4 occupies 2 mesons.

o The child occupied by the terminal, calculates the power of the unused subcarriers at o and satisfies the formula (7): =ws b ha e W Ga

"" ( e be 8) / nz , ,,.

" 1

Under the premise, the excess power can be allocated to the terminal that needs a oos g to boost the throughput. The OF AA of FFTSze 024 does not say """284, where "h, 30, ,,. ,

" "" "" 4, W

"," " " twenty four

Ganz is equal to 9 north, formula (7), gets s bcha ab snn 1.62

Because the downside is low, aa OOS g9 2 goes down. 5 North BOOS g, 3 and 4 down are better. -3 BOOS g

Resource allocation as shown in 3

O , S O 2 S O 3 , 2, 3 occupy 1 meson respectively 4 occupy 2 meson Z

Bch e n ab s"

O4, occupy 1 meson 2 no occupyer 3 occupy 2, 4 occupy 2 meson St e * h r -3

O 5 , occupy 1 meson 2 no occupyer 3 occupy 2 15 , 4 occupy 1 meson S , Ga rw 0. O6, no occupant, 2 occupies 1 child, 3 occupies 1 and 4 occupies 2 meson bc, a n mn n -8

0 The maximum power of all O is less than 1.629. The scheme is feasible.

In step 405, step 407 shown in 4, that is, in the case where the downlink ZO e zO e BOOS g is not, the OF A of the FFT Sze 024 is not said. 3

.

Pe mbe 284 w 30, W. s h e 4, W 24 z ( 8 / "ch e ' W

Ten (" n S

16/9)) " ) / ch " . (( m kb "" s

Gan b sr „ bc "n e 16/9 8 Get GanZ 9. 36, so the maximum of Zo e oo g 9 .

Ben Mingtong's method of combining Zo e oo g and aaBoo g solves the problem that in the same ZoeBoos g and aaBoos g, the downlink power can be allocated reasonably, and the downlink power is allocated reasonably. Zo e oo g used the P Czo e and the aaBoo g to improve the coverage and throughput of the OF A system.

The method provides a schematic diagram showing the composition of the device for improving the downward coverage and throughput of the OF A system, and includes at least, at least, and at least the terminal Zo e in sequence all Zo e

, in judging that the Zo e oo g of the downlink Zo e is in advance, to the first pass, Zo e Boos g on the downlink Zo e is not in advance to the second second pass

The maximum power obtained by the maximum power of the digital carriers of all the terminals in the Zoe and the Zo e, and the downlink carrier of the terminal

Ratio, terminal allocation aBoos g

Secondly, the downlink carrier ratio of the second communication terminal is received, the terminal allocates the sub-dao and the assigned aa Boo g occupied by the aa Boo g terminal, and the Zoe Boos g of the maximum ZoE of the Zo eBoo g The above is the way of this, and it is pointed out that the ordinary wood in the field of the wood has not been said to be able to produce a number of sums, and also the protection of the basics.

Claims

Claim

1. Method for improving the downward coverage and throughput of the OF A system, which is characterized in that it includes the following steps

A, terminal Zo e, in turn all Zo e

And determine whether the gain of the downstream Zo e Zoe oos g is in advance, then the special step C, otherwise the special step E

C, the maximum power of the number of subcarriers of all terminals in Zo e

The maximum power and the downlink carrier ratio of the terminal are allocated to the terminal aaBoos g

E, the downlink carrier ratio of the terminal, the terminal allocation aaBoo g F, the sub-dao and the assigned aa oo g, the maximum value of the Zo e Boost g

G, Zo eBoo g of the maximum value of Zo e

2. The method of claim 1 wherein the step C includes

Formula

" s bc e

Z

The maximum power of the number of subcarriers in all terminals in oe, where the maximum power, "the former subcarrier, Ga Zo e oos g Zo e the gain of the """" er Zo e occupied by the Taoist, "e The subcarriers contained in meson 0.

3. The method according to claim 1, wherein the step is performed, and all terminal lines in the Zo e are sorted according to a sequence in which the downlink carrier ratio of the terminal is small to large.

2. In the order of sorting, if the maximum power value is not greater than 5, the terminal allocates aBoo g

4. The method according to claim 3, wherein the step 2 is: the terminal with a downlink carrier ratio smaller than a previous value g aa Boo g, and the terminal with a downlink carrier ratio greater than or equal to the previous value is not aaBoo g

formula

Subcarrier power where S

"Sub-channels occupied by D aB of the terminal, where "," c sub-carriers are included in the sub-carrier, G

".. aBoo g terminal number of subcarriers without gain comparison of the number of subcarriers, maximum

Ax. Whether it is less than or equal to the maximum power

" , if yes, then reassign DaaBoos g

5. The method of claim 1, wherein the step E includes: a terminal with a downlink carrier ratio less than a previous value, a g aa Boost, a downlink carrier ratio greater than or equal to a previous value, not aaBoo g

6. The method of claim 1 wherein the step F is comprised of: a formula Ganz 凡 " , 8 / ((

" , " , "

'

Gaf ab rn ) where n"e,r s b , 1 / ) , Zo e oo g

Ga, where, ,

" " Front subcarrier , S

" "", the sub-Tao of the terminal a, the subcarriers contained in the meson,

"" "

Ga. . , aa

"" Boo g

"" The number of subcarriers of the terminal does not gain the subcarriers contained in the sub-memes occupied by Zo e .

7. The method of claim 1 to 6 is characterized in that it is on the terminal Subcarrier CR

The CW is directly determined by the self. 8. The method of any one of claims 1 to 6 is characterized in that the subcarrier C R on the terminal is

",. C R,

",., aa oo g gain,

"" ",," and then the judgment.

The method of any one of claims 1 to 6 is characterized in that the terminal is on the subcarrier C R CZ

"" , WR

" , , " aa oos g gain noPe and Zo eBoo g gain Ga, "," and then from this judgment.

10. A device for improving the downlink and throughput of the OF A system, wherein at least, the second, and the second are in the terminal Zo e in sequence all Zo e

Zo e oo g in the down Zo e is the first

To the first pass, the Zo e Boo g on the down Zo e is not in advance to the second second pass

Ratio, terminal allocation aaBoo g

Secondly, the downlink carrier ratio of the second terminal is received, and the terminal allocates a sub Boo g terminal sub- and assigned a Boo g , Zo eBoo g maximum value Zo e maximum Zoe Boos g