CN104283597A - Method and device for shaping wave beams - Google Patents

Method and device for shaping wave beams Download PDF

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Publication number
CN104283597A
CN104283597A CN201310607975.6A CN201310607975A CN104283597A CN 104283597 A CN104283597 A CN 104283597A CN 201310607975 A CN201310607975 A CN 201310607975A CN 104283597 A CN104283597 A CN 104283597A
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mrow
msub
msup
mfrac
munderover
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CN104283597B (en
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崔高峰
王卫东
吕程程
赵丽娜
唐明环
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Aerospace Xingyun Technology Co ltd
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Beijing University of Posts and Telecommunications
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    • HELECTRICITY
    • H04ELECTRIC COMMUNICATION TECHNIQUE
    • H04BTRANSMISSION
    • H04B7/00Radio transmission systems, i.e. using radiation field
    • H04B7/02Diversity systems; Multi-antenna system, i.e. transmission or reception using multiple antennas
    • H04B7/04Diversity systems; Multi-antenna system, i.e. transmission or reception using multiple antennas using two or more spaced independent antennas
    • H04B7/06Diversity systems; Multi-antenna system, i.e. transmission or reception using multiple antennas using two or more spaced independent antennas at the transmitting station
    • H04B7/0613Diversity systems; Multi-antenna system, i.e. transmission or reception using multiple antennas using two or more spaced independent antennas at the transmitting station using simultaneous transmission
    • H04B7/0615Diversity systems; Multi-antenna system, i.e. transmission or reception using multiple antennas using two or more spaced independent antennas at the transmitting station using simultaneous transmission of weighted versions of same signal
    • H04B7/0617Diversity systems; Multi-antenna system, i.e. transmission or reception using multiple antennas using two or more spaced independent antennas at the transmitting station using simultaneous transmission of weighted versions of same signal for beam forming

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  • Engineering & Computer Science (AREA)
  • Computer Networks & Wireless Communication (AREA)
  • Signal Processing (AREA)
  • Mobile Radio Communication Systems (AREA)

Abstract

The invention provides a method and device for shaping wave beams. The method includes the steps of obtaining position information of users to be served by a current time slot, determining a handling capacity function of the total handling capacity of the users relative to downward inclination angle combinations according to the position information of the users, estimating the downward inclination angle combination by which the maximum value of the handling capacity function is obtained with the DIRECT algorithm, and then adjusting the downward inclination angles of the two wave beams according to the downward inclination angle combination obtained through calculation. By means of the technical scheme, the system hanging capacity can be improved.

Description

Beam forming method and device
Technical Field
The present invention relates to the field of communications technologies, and in particular, to a beamforming method and apparatus.
Background
Dual beam downtilt refers to having the same base station transmit two beams with different downtilt angles to provide service to the user. The prior art schemes for setting the downtilt angle of a dual beam generally include two types: (1) setting a fixed down tilt angle for the two beams; (2) the two beams are respectively directed to the served users, for the first scheme, real-time adjustment cannot be performed according to the positions of the users, the maximum throughput of the users cannot be guaranteed, and for the second scheme, the two users which need to be served at present are close to each other, so that greater interference is generated, and the overall throughput of the system is reduced.
Disclosure of Invention
The invention provides a beam forming method and device, which can improve the throughput of a system.
The invention provides a beam forming method, which is applied to a dual-beam communication system and comprises the following steps:
acquiring the position information of each user to be served by the next time slot service;
selecting a combination (beta) of the total throughput of each user with respect to the downtilt angle according to the acquired location information1,β2) Total throughput function ofβ 1 and β 2 denote downtilt angles of the 1 st and 2 nd beams, respectively, Ri12) Throughput versus downtilt combination (β) for the ith user1,β2) Is composed ofNumber, I is the number of users served by two beams;
computing using a DIRECT algorithmMaximum down dip angle combination (beta)1M,β2M);
Correspondingly adjusting the downtilt angles of the two beams to beta1M,β2M
Preferably, the selecting of the combination (β) of the total throughput of each user with respect to the downward inclination angle according to the acquired location information1,β2) Total throughput function ofThe method specifically comprises the following steps:
for each user, the received power of each beam is calculated, and the receiving gain of each beam is combined with the declination angle (beta)1,β2) A function of (a);
getAs a combination of total throughput versus downtilt (β) for each user1,β2) Total throughput function ofWherein,
<math> <mrow> <mi>R</mi> <mrow> <mo>(</mo> <msub> <mi>&gamma;</mi> <mn>1</mn> </msub> <mo>,</mo> <msub> <mi>&gamma;</mi> <mn>2</mn> </msub> <mo>,</mo> <mi>M</mi> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mi>E</mi> <mi>X</mi> </msub> <mo>[</mo> <msub> <mi>log</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <mi>X</mi> <mo>)</mo> </mrow> <mo>]</mo> <mo>=</mo> <msub> <mi>log</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>e</mi> <mo>)</mo> </mrow> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> <mrow> <mi>M</mi> <mo>-</mo> <mn>1</mn> </mrow> </munderover> <munderover> <mi>&Sigma;</mi> <mrow> <mi>l</mi> <mo>=</mo> <mn>0</mn> </mrow> <mi>i</mi> </munderover> <mfrac> <msubsup> <mi>&gamma;</mi> <mn>1</mn> <mrow> <mi>l</mi> <mo>+</mo> <mn>1</mn> <mo>-</mo> <mi>i</mi> </mrow> </msubsup> <mrow> <msub> <mi>&gamma;</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>i</mi> <mo>-</mo> <mi>l</mi> <mo>)</mo> </mrow> <mo>!</mo> </mrow> </mfrac> <mo>&CenterDot;</mo> <msub> <mi>I</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mfrac> <mn>1</mn> <msub> <mi>&gamma;</mi> <mn>1</mn> </msub> </mfrac> <mo>,</mo> <mfrac> <msub> <mi>&gamma;</mi> <mn>1</mn> </msub> <msub> <mi>&gamma;</mi> <mn>2</mn> </msub> </mfrac> <mo>,</mo> <mi>i</mi> <mo>,</mo> <mi>l</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> </mrow> </math>
wherein I1(a, b, m, n) is an integral expression, and
<math> <mrow> <msub> <mi>I</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mi>m</mi> <mo>,</mo> <mi>n</mi> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>n</mi> </munderover> <mfrac> <msup> <mrow> <mo>(</mo> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mrow> <mi>i</mi> <mo>-</mo> <mn>1</mn> </mrow> </msup> <msup> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <mi>b</mi> <mo>)</mo> </mrow> <mi>i</mi> </msup> </mfrac> <mo>&CenterDot;</mo> <msub> <mi>I</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mi>m</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mi>i</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> <mo>+</mo> <mfrac> <mrow> <msub> <mi>I</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mn>1</mn> <mo>,</mo> <mi>m</mi> <mo>,</mo> <mn>1</mn> <mo>)</mo> </mrow> </mrow> <msup> <mrow> <mo>(</mo> <mi>b</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mi>n</mi> </msup> </mfrac> <mo>;</mo> </mrow> </math>
<math> <mrow> <msub> <mi>I</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mi>m</mi> <mo>,</mo> <mi>n</mi> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mo>&Integral;</mo> <mn>0</mn> <mo>&infin;</mo> </munderover> <mfrac> <mrow> <msup> <mi>x</mi> <mi>m</mi> </msup> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mi>ax</mi> </mrow> </msup> </mrow> <msup> <mrow> <mo>(</mo> <mi>x</mi> <mo>+</mo> <mi>b</mi> <mo>)</mo> </mrow> <mi>i</mi> </msup> </mfrac> <mi>dx</mi> <mo>=</mo> <msup> <mi>e</mi> <mi>ab</mi> </msup> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> <mi>m</mi> </munderover> <mfenced open='(' close=')'> <mtable> <mtr> <mtd> <mi>m</mi> </mtd> </mtr> <mtr> <mtd> <mi>i</mi> </mtd> </mtr> </mtable> </mfenced> <msup> <mrow> <mo>(</mo> <mo>-</mo> <mi>b</mi> <mo>)</mo> </mrow> <mrow> <mi>m</mi> <mo>-</mo> <mi>i</mi> </mrow> </msup> <msub> <mi>I</mi> <mn>3</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mi>i</mi> <mo>-</mo> <mi>n</mi> <mo>)</mo> </mrow> <mo>;</mo> </mrow> </math>
<math> <mrow> <msub> <mi>I</mi> <mn>3</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mi>m</mi> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mo>&Integral;</mo> <mi>b</mi> <mo>&infin;</mo> </munderover> <msup> <mi>x</mi> <mi>m</mi> </msup> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mi>ax</mi> </mrow> </msup> <mi>dx</mi> <mo>=</mo> <mfenced open='{' close=''> <mtable> <mtr> <mtd> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mi>ab</mi> </mrow> </msup> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> <mi>m</mi> </munderover> <mfrac> <mrow> <mi>m</mi> <mo>!</mo> </mrow> <mrow> <mi>i</mi> <mo>!</mo> </mrow> </mfrac> <mfrac> <msup> <mi>b</mi> <mi>i</mi> </msup> <msup> <mi>a</mi> <mrow> <mi>m</mi> <mo>-</mo> <mi>i</mi> <mo>+</mo> <mn>1</mn> </mrow> </msup> </mfrac> </mtd> <mtd> <mi>m</mi> <mo>&GreaterEqual;</mo> <mn>0</mn> </mtd> </mtr> <mtr> <mtd> <msub> <mi>E</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>ab</mi> <mo>)</mo> </mrow> </mtd> <mtd> <mi>m</mi> <mo>=</mo> <mo>-</mo> <mn>1</mn> </mtd> </mtr> <mtr> <mtd> <mfrac> <mrow> <msub> <mi>E</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>ab</mi> <mo>)</mo> </mrow> </mrow> <mrow> <msup> <mrow> <mo>(</mo> <mo>-</mo> <mi>a</mi> <mo>)</mo> </mrow> <mrow> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mn>1</mn> </mrow> </msup> <mrow> <mo>(</mo> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mo>!</mo> </mrow> </mfrac> <mo>+</mo> <mfrac> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mi>ab</mi> </mrow> </msup> <msup> <mi>b</mi> <mrow> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mn>1</mn> </mrow> </msup> </mfrac> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> <mrow> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mn>2</mn> </mrow> </munderover> <mfrac> <mrow> <msup> <mrow> <mo>(</mo> <mo>-</mo> <mi>ab</mi> <mo>)</mo> </mrow> <mi>i</mi> </msup> <mrow> <mo>(</mo> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mi>i</mi> <mo>-</mo> <mn>2</mn> <mo>)</mo> </mrow> <mo>!</mo> </mrow> <mrow> <mrow> <mo>(</mo> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mo>!</mo> </mrow> </mfrac> </mtd> <mtd> <mi>m</mi> <mo>&le;</mo> <mo>-</mo> <mn>2</mn> </mtd> </mtr> </mtable> </mfenced> </mrow> </math>
wherein,received power of s beam for i user, Gsis) Represents the downtilt angle beta of the reception gain of the ith user to the s beam relative to the s beamsAs a function of (c).
Preferably, for each user, the combination (β) of the reception gain of each beam with respect to the downtilt angle is calculated separately1,β2) Comprises:
calculating the receiving gain G of the ith user to the s beam by the following formulasis):
Wherein,and thetaiAzimuth angle and pitch angle, SLL, corresponding to the ith user respectivelyazAnd SLLelAre horizontal and vertical, respectivelySide lobe level of directional diagram, SLLtotIs the total side-lobe level of the signal,for the received power of the ith user on the s-th beam,is the angle between the aiming axis of the antenna array of the base station where the ith user is located and the x axis.
The invention provides a beam forming device, which is used as a base station in a wireless communication system, and comprises:
the system comprises a position information acquisition module, a position information acquisition module and a time slot processing module, wherein the position information acquisition module is used for acquiring the position information of each user to be served by the next time slot service;
a function calling module for selecting the total throughput of each user relative to the downward inclination angle combination (beta) according to the position information acquired by the position information acquisition module1,β2) Total throughput function ofβ 1 and β 2 denote downtilt angles of the 1 st and 2 nd beams, respectively, Ri12) Throughput versus downtilt combination (β) for the ith user1,β2) I is the number of users served by the two beams;
a calculation module for calculating the equation using the DIRECT algorithmMaximum down dip angle combination (beta)1M,β2M);
A shaping module for correspondingly adjusting the downtilt angles of the two beams to beta1M,β2M
Preferably, the function call module is specifically configured to determine that each user receives each item according to the acquired location informationSignal strength of individual beams combined with respect to downtilt angle (beta)1,β2) Is taken as a function ofAs a combination of total throughput versus downtilt (β) for each user1,β2) Total throughput function ofWherein,
<math> <mrow> <mi>R</mi> <mrow> <mo>(</mo> <msub> <mi>&gamma;</mi> <mn>1</mn> </msub> <mo>,</mo> <msub> <mi>&gamma;</mi> <mn>2</mn> </msub> <mo>,</mo> <mi>M</mi> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mi>E</mi> <mi>X</mi> </msub> <mo>[</mo> <msub> <mi>log</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <mi>X</mi> <mo>)</mo> </mrow> <mo>]</mo> <mo>=</mo> <msub> <mi>log</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>e</mi> <mo>)</mo> </mrow> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> <mrow> <mi>M</mi> <mo>-</mo> <mn>1</mn> </mrow> </munderover> <munderover> <mi>&Sigma;</mi> <mrow> <mi>l</mi> <mo>=</mo> <mn>0</mn> </mrow> <mi>i</mi> </munderover> <mfrac> <msubsup> <mi>&gamma;</mi> <mn>1</mn> <mrow> <mi>l</mi> <mo>+</mo> <mn>1</mn> <mo>-</mo> <mi>i</mi> </mrow> </msubsup> <mrow> <msub> <mi>&gamma;</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>i</mi> <mo>-</mo> <mi>l</mi> <mo>)</mo> </mrow> <mo>!</mo> </mrow> </mfrac> <mo>&CenterDot;</mo> <msub> <mi>I</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mfrac> <mn>1</mn> <msub> <mi>&gamma;</mi> <mn>1</mn> </msub> </mfrac> <mo>,</mo> <mfrac> <msub> <mi>&gamma;</mi> <mn>1</mn> </msub> <msub> <mi>&gamma;</mi> <mn>2</mn> </msub> </mfrac> <mo>,</mo> <mi>i</mi> <mo>,</mo> <mi>l</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> </mrow> </math>
wherein I1(a, b, m, n) is an integral expression, and
<math> <mrow> <msub> <mi>I</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mi>m</mi> <mo>,</mo> <mi>n</mi> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>n</mi> </munderover> <mfrac> <msup> <mrow> <mo>(</mo> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mrow> <mi>i</mi> <mo>-</mo> <mn>1</mn> </mrow> </msup> <msup> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <mi>b</mi> <mo>)</mo> </mrow> <mi>i</mi> </msup> </mfrac> <mo>&CenterDot;</mo> <msub> <mi>I</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mi>m</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mi>i</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> <mo>+</mo> <mfrac> <mrow> <msub> <mi>I</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mn>1</mn> <mo>,</mo> <mi>m</mi> <mo>,</mo> <mn>1</mn> <mo>)</mo> </mrow> </mrow> <msup> <mrow> <mo>(</mo> <mi>b</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mi>n</mi> </msup> </mfrac> <mo>;</mo> </mrow> </math>
<math> <mrow> <msub> <mi>I</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mi>m</mi> <mo>,</mo> <mi>n</mi> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mo>&Integral;</mo> <mn>0</mn> <mo>&infin;</mo> </munderover> <mfrac> <mrow> <msup> <mi>x</mi> <mi>m</mi> </msup> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mi>ax</mi> </mrow> </msup> </mrow> <msup> <mrow> <mo>(</mo> <mi>x</mi> <mo>+</mo> <mi>b</mi> <mo>)</mo> </mrow> <mi>i</mi> </msup> </mfrac> <mi>dx</mi> <mo>=</mo> <msup> <mi>e</mi> <mi>ab</mi> </msup> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> <mi>m</mi> </munderover> <mfenced open='(' close=')'> <mtable> <mtr> <mtd> <mi>m</mi> </mtd> </mtr> <mtr> <mtd> <mi>i</mi> </mtd> </mtr> </mtable> </mfenced> <msup> <mrow> <mo>(</mo> <mo>-</mo> <mi>b</mi> <mo>)</mo> </mrow> <mrow> <mi>m</mi> <mo>-</mo> <mi>i</mi> </mrow> </msup> <msub> <mi>I</mi> <mn>3</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mi>i</mi> <mo>-</mo> <mi>n</mi> <mo>)</mo> </mrow> <mo>;</mo> </mrow> </math>
<math> <mrow> <msub> <mi>I</mi> <mn>3</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mi>m</mi> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mo>&Integral;</mo> <mi>b</mi> <mo>&infin;</mo> </munderover> <msup> <mi>x</mi> <mi>m</mi> </msup> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mi>ax</mi> </mrow> </msup> <mi>dx</mi> <mo>=</mo> <mfenced open='{' close=''> <mtable> <mtr> <mtd> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mi>ab</mi> </mrow> </msup> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> <mi>m</mi> </munderover> <mfrac> <mrow> <mi>m</mi> <mo>!</mo> </mrow> <mrow> <mi>i</mi> <mo>!</mo> </mrow> </mfrac> <mfrac> <msup> <mi>b</mi> <mi>i</mi> </msup> <msup> <mi>a</mi> <mrow> <mi>m</mi> <mo>-</mo> <mi>i</mi> <mo>+</mo> <mn>1</mn> </mrow> </msup> </mfrac> </mtd> <mtd> <mi>m</mi> <mo>&GreaterEqual;</mo> <mn>0</mn> </mtd> </mtr> <mtr> <mtd> <msub> <mi>E</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>ab</mi> <mo>)</mo> </mrow> </mtd> <mtd> <mi>m</mi> <mo>=</mo> <mo>-</mo> <mn>1</mn> </mtd> </mtr> <mtr> <mtd> <mfrac> <mrow> <msub> <mi>E</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>ab</mi> <mo>)</mo> </mrow> </mrow> <mrow> <msup> <mrow> <mo>(</mo> <mo>-</mo> <mi>a</mi> <mo>)</mo> </mrow> <mrow> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mn>1</mn> </mrow> </msup> <mrow> <mo>(</mo> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mo>!</mo> </mrow> </mfrac> <mo>+</mo> <mfrac> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mi>ab</mi> </mrow> </msup> <msup> <mi>b</mi> <mrow> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mn>1</mn> </mrow> </msup> </mfrac> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> <mrow> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mn>2</mn> </mrow> </munderover> <mfrac> <mrow> <msup> <mrow> <mo>(</mo> <mo>-</mo> <mi>ab</mi> <mo>)</mo> </mrow> <mi>i</mi> </msup> <mrow> <mo>(</mo> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mi>i</mi> <mo>-</mo> <mn>2</mn> <mo>)</mo> </mrow> <mo>!</mo> </mrow> <mrow> <mrow> <mo>(</mo> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mo>!</mo> </mrow> </mfrac> </mtd> <mtd> <mi>m</mi> <mo>&le;</mo> <mo>-</mo> <mn>2</mn> </mtd> </mtr> </mtable> </mfenced> </mrow> </math>
wherein,received power of s beam for i user, Gsis) Is shown asThe reception gain of the i users for the s-th beam is declined by an angle beta relative to the s-th beamsAs a function of (c).
Preferably, the function call module calculates the receiving gain G of the ith user to the ith beam by the following formulasis):
Side lobe levels, SLL, of the horizontal and vertical patterns, respectivelytotIs the total side-lobe level of the signal,for the received power of the ith user on the s-th beam,is the angle between the aiming axis of the antenna array of the base station where the ith user is located and the x axis.
In the invention, the position information of the user is obtained, the throughput function of the total throughput of each user relative to the downtilt combination is determined according to the position information of the user, the downtilt combination which enables the throughput function to take the maximum value is estimated by adopting a DIRECT algorithm, and then the downtilt angles of two beams are adjusted according to the downtilt combination obtained by calculation. The technical scheme provided by the invention can improve the throughput of the system.
Drawings
Fig. 1 is a schematic flowchart of a beamforming method according to an embodiment of the present invention;
fig. 2 is a schematic diagram of an application scenario of a beamforming method according to an embodiment of the present invention;
fig. 3 is a schematic diagram of an application scenario of a beamforming method according to an embodiment of the present invention;
fig. 4 is a schematic structural diagram of a beamforming apparatus according to an embodiment of the present invention.
Detailed Description
The following description of the embodiments of the present invention will be made with reference to the accompanying drawings. The following examples are only for illustrating the technical solutions of the present invention more clearly, and the protection scope of the present invention is not limited thereby.
An embodiment of the present invention provides a beamforming method, which is applied to a dual-beam communication system, where a base station in the communication system uses two beams to serve users in a cell, as shown in fig. 1, the method includes:
step 101, obtaining the location information of each user to be served by the next time slot service.
Step 102, selecting the combination (beta) of the total throughput of each user relative to the declination angle according to the acquired position information1,β2) Total throughput function ofβ 1 and β 2 denote downtilt angles of the 1 st and 2 nd beams, respectively, Ri12) Throughput versus downtilt combination (β) for the ith user1,β2) I is the number of users served by the two beams.
103, calculating by using a DIRECT algorithmMaximum down dip angle combination (beta)1M,β2M)。
Step 104, correspondingly adjusting the downtilt angles of the two beams to beta1M,β2M
In the embodiment of the invention, the position information of the user is obtained, the throughput function of the total throughput of each user relative to the downtilt combination is determined according to the position information of the user, the downtilt combination which enables the throughput function to take the maximum value is estimated by adopting a DIRECT algorithm, and then the downtilt angles of two beams are adjusted according to the downtilt combination obtained by calculation. The technical scheme provided by the invention can improve the throughput of the system.
Preferably, the step 102 specifically includes:
for each user, the received power of each beam is calculated, and the receiving gain of each beam is combined with the declination angle (beta)1,β2) A function of (a);
getAs a combination of total throughput versus downtilt (β) for each user1,β2) Total throughput function ofWherein,
<math> <mrow> <mi>R</mi> <mrow> <mo>(</mo> <msub> <mi>&gamma;</mi> <mn>1</mn> </msub> <mo>,</mo> <msub> <mi>&gamma;</mi> <mn>2</mn> </msub> <mo>,</mo> <mi>M</mi> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mi>E</mi> <mi>X</mi> </msub> <mo>[</mo> <msub> <mi>log</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <mi>X</mi> <mo>)</mo> </mrow> <mo>]</mo> <mo>=</mo> <msub> <mi>log</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>e</mi> <mo>)</mo> </mrow> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> <mrow> <mi>M</mi> <mo>-</mo> <mn>1</mn> </mrow> </munderover> <munderover> <mi>&Sigma;</mi> <mrow> <mi>l</mi> <mo>=</mo> <mn>0</mn> </mrow> <mi>i</mi> </munderover> <mfrac> <msubsup> <mi>&gamma;</mi> <mn>1</mn> <mrow> <mi>l</mi> <mo>+</mo> <mn>1</mn> <mo>-</mo> <mi>i</mi> </mrow> </msubsup> <mrow> <msub> <mi>&gamma;</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>i</mi> <mo>-</mo> <mi>l</mi> <mo>)</mo> </mrow> <mo>!</mo> </mrow> </mfrac> <mo>&CenterDot;</mo> <msub> <mi>I</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mfrac> <mn>1</mn> <msub> <mi>&gamma;</mi> <mn>1</mn> </msub> </mfrac> <mo>,</mo> <mfrac> <msub> <mi>&gamma;</mi> <mn>1</mn> </msub> <msub> <mi>&gamma;</mi> <mn>2</mn> </msub> </mfrac> <mo>,</mo> <mi>i</mi> <mo>,</mo> <mi>l</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> </mrow> </math>
wherein I1(a, b, m, n) is an integral expression
<math> <mrow> <msub> <mi>I</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mi>m</mi> <mo>,</mo> <mi>n</mi> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>n</mi> </munderover> <mfrac> <msup> <mrow> <mo>(</mo> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mrow> <mi>i</mi> <mo>-</mo> <mn>1</mn> </mrow> </msup> <msup> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <mi>b</mi> <mo>)</mo> </mrow> <mi>i</mi> </msup> </mfrac> <mo>&CenterDot;</mo> <msub> <mi>I</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mi>m</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mi>i</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> <mo>+</mo> <mfrac> <mrow> <msub> <mi>I</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mn>1</mn> <mo>,</mo> <mi>m</mi> <mo>,</mo> <mn>1</mn> <mo>)</mo> </mrow> </mrow> <msup> <mrow> <mo>(</mo> <mi>b</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mi>n</mi> </msup> </mfrac> <mo>;</mo> </mrow> </math>
<math> <mrow> <msub> <mi>I</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mi>m</mi> <mo>,</mo> <mi>n</mi> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mo>&Integral;</mo> <mn>0</mn> <mo>&infin;</mo> </munderover> <mfrac> <mrow> <msup> <mi>x</mi> <mi>m</mi> </msup> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mi>ax</mi> </mrow> </msup> </mrow> <msup> <mrow> <mo>(</mo> <mi>x</mi> <mo>+</mo> <mi>b</mi> <mo>)</mo> </mrow> <mi>i</mi> </msup> </mfrac> <mi>dx</mi> <mo>=</mo> <msup> <mi>e</mi> <mi>ab</mi> </msup> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> <mi>m</mi> </munderover> <mfenced open='(' close=')'> <mtable> <mtr> <mtd> <mi>m</mi> </mtd> </mtr> <mtr> <mtd> <mi>i</mi> </mtd> </mtr> </mtable> </mfenced> <msup> <mrow> <mo>(</mo> <mo>-</mo> <mi>b</mi> <mo>)</mo> </mrow> <mrow> <mi>m</mi> <mo>-</mo> <mi>i</mi> </mrow> </msup> <msub> <mi>I</mi> <mn>3</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mi>i</mi> <mo>-</mo> <mi>n</mi> <mo>)</mo> </mrow> <mo>;</mo> </mrow> </math>
<math> <mrow> <msub> <mi>I</mi> <mn>3</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mi>m</mi> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mo>&Integral;</mo> <mi>b</mi> <mo>&infin;</mo> </munderover> <msup> <mi>x</mi> <mi>m</mi> </msup> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mi>ax</mi> </mrow> </msup> <mi>dx</mi> <mo>=</mo> <mfenced open='{' close=''> <mtable> <mtr> <mtd> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mi>ab</mi> </mrow> </msup> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> <mi>m</mi> </munderover> <mfrac> <mrow> <mi>m</mi> <mo>!</mo> </mrow> <mrow> <mi>i</mi> <mo>!</mo> </mrow> </mfrac> <mfrac> <msup> <mi>b</mi> <mi>i</mi> </msup> <msup> <mi>a</mi> <mrow> <mi>m</mi> <mo>-</mo> <mi>i</mi> <mo>+</mo> <mn>1</mn> </mrow> </msup> </mfrac> </mtd> <mtd> <mi>m</mi> <mo>&GreaterEqual;</mo> <mn>0</mn> </mtd> </mtr> <mtr> <mtd> <msub> <mi>E</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>ab</mi> <mo>)</mo> </mrow> </mtd> <mtd> <mi>m</mi> <mo>=</mo> <mo>-</mo> <mn>1</mn> </mtd> </mtr> <mtr> <mtd> <mfrac> <mrow> <msub> <mi>E</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>ab</mi> <mo>)</mo> </mrow> </mrow> <mrow> <msup> <mrow> <mo>(</mo> <mo>-</mo> <mi>a</mi> <mo>)</mo> </mrow> <mrow> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mn>1</mn> </mrow> </msup> <mrow> <mo>(</mo> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mo>!</mo> </mrow> </mfrac> <mo>+</mo> <mfrac> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mi>ab</mi> </mrow> </msup> <msup> <mi>b</mi> <mrow> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mn>1</mn> </mrow> </msup> </mfrac> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> <mrow> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mn>2</mn> </mrow> </munderover> <mfrac> <mrow> <msup> <mrow> <mo>(</mo> <mo>-</mo> <mi>ab</mi> <mo>)</mo> </mrow> <mi>i</mi> </msup> <mrow> <mo>(</mo> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mi>i</mi> <mo>-</mo> <mn>2</mn> <mo>)</mo> </mrow> <mo>!</mo> </mrow> <mrow> <mrow> <mo>(</mo> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mo>!</mo> </mrow> </mfrac> </mtd> <mtd> <mi>m</mi> <mo>&le;</mo> <mo>-</mo> <mn>2</mn> </mtd> </mtr> </mtable> </mfenced> </mrow> </math>
Wherein,received power of s beam for i user, Gsis) Represents the downtilt angle beta of the reception gain of the ith user to the s beam relative to the s beamsAs a function of (c). In this way, the calculation of the down dip angle combination (β) can be reduced1,β2) Of the system.
Wherein, the receiving gain of the s beam by any ith user
Wherein,and thetaiAzimuth angle and pitch angle, SLL, corresponding to the ith user respectivelyazAnd SLLelSide lobe levels, SLL, of the horizontal and vertical patterns, respectivelytotIs the total side-lobe level of the signal,for the received power of the ith user on the s-th beam,is the angle between the aiming axis of the antenna array of the base station where the ith user is located and the x axis.
The following describes in detail the steps and principles of the beam forming method provided by the embodiment of the present invention with reference to a specific application scenario, and in practical applications, two ways of beam downtilt are provided, one of which is based on user downtilt; the other is vertical domain based downtilt, and the following respectively describes the steps and principles of the beamforming method in the two downtilt modes.
User-based downtilt
As shown in fig. 2, for an application scenario of the beamforming method provided in the embodiment of the present invention, it is assumed that the base station divides the transmitting antennas into two groups, each antenna group has 4 transmitting antennas, the two groups of antennas simultaneously transmit two beams, and each beam serves only one user in each time slot.
As shown in FIG. 3, assume thatAn included angle between a connecting line of a user and a base station antenna and an x axis (an antenna array aiming axis) on an xoy plane, namely an azimuth angle of the user; theta is an included angle between a connecting line of the user and the base station antenna and a horizontal line, namely a pitch angle of the user. The pitch angle theta of the user i can be obtained by geometric calculationiAnd azimuth angleComprises the following steps:
θi=tan-1((hbs-hi)/di)
wherein h isbsIs the base station antenna height, hiFor the user antenna height, (x)i,yi) As coordinates of the user, diFor the distance of the user from the base station, i.e.
d i = x i 2 + y i 2 .
Let the service user of the ith beam be i. Antenna gain from base station to user i is
Wherein,
<math> <mrow> <msub> <mi>G</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>V</mi> </mrow> </msub> <mrow> <mo>(</mo> <msub> <mi>&theta;</mi> <mi>i</mi> </msub> <mo>)</mo> </mrow> <mo>=</mo> <mo>-</mo> <mi>min</mi> <mo>[</mo> <mn>12</mn> <msup> <mrow> <mo>(</mo> <mfrac> <mrow> <msub> <mi>&theta;</mi> <mi>i</mi> </msub> <mo>-</mo> <msub> <mi>&beta;</mi> <mi>i</mi> </msub> </mrow> <msub> <mi>&theta;</mi> <mrow> <mn>3</mn> <mi>dB</mi> </mrow> </msub> </mfrac> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>,</mo> <msub> <mi>SLL</mi> <mi>el</mi> </msub> <mo>]</mo> <mo>;</mo> </mrow> </math>
that is to say that the first and second electrodes,
then the received signals of the users in the two vertical domains are:
wherein:
ri,jrepresents the signal received by user i from its serving beam j;
(□)*represents a conjugate transpose of a vector;
xirepresents the transmission signal from the beam i to the user i and satisfies the power constraint condition E [ | x [ ]i|2]=1;
N represents complex white Gaussian noise at the user i, obeying N to N (0, 1);
hirepresenting n between user i and base stationtX 1 dimensional channel vector, and h is given to rayleigh fading channeli,jThe elements of (1) are independent and equally distributed random variables, and obey to N (0, 1);
fi,iis the precoding vector of the base station beam i to the user i served by it, which is normalized to satisfy fi,i 2=1;
Pi rRepresenting the received power of the user, the invention adopts a path loss model of
Pi r=P0(D0/di)α
Wherein, P0Is the distance D of the user from the base station0Received power of time, also called reference power, diIs the distance between the user and the base station, and D is set in the invention0R, then P0Indicating the received power of the user at the cell edge. The present invention assumes that the power of the two beams emanating from the base station is equal;
the first term in the formula represents the expected received signal of the user i, the second term represents the interference of other vertical area beams to the user, and the third term represents the normalized additive white gaussian noise at the user i.
The SINR of the user is expressed as
The throughput of the user is
R1=E[log2(1+SINR(β12))]
The total throughput of two users is
R=R1+R2
The aim of the invention is to find the optimal combination that maximizes the throughput of both users.
Let fi,iAnd channel matrix hiNot related, then there areWhereinRepresenting a chi-square distribution with a degree of freedom n.
Order to
<math> <mrow> <mi>X</mi> <mo>=</mo> <mfrac> <mrow> <msub> <mi>&gamma;</mi> <mn>1</mn> </msub> <mi>Z</mi> </mrow> <mrow> <mn>1</mn> <mo>+</mo> <msub> <mi>&gamma;</mi> <mn>2</mn> </msub> <mi>Y</mi> </mrow> </mfrac> </mrow> </math>
Wherein the random variable isAnd are independent of each other. Then there is
<math> <mrow> <mi>R</mi> <mrow> <mo>(</mo> <msub> <mi>&gamma;</mi> <mn>1</mn> </msub> <mo>,</mo> <msub> <mi>&gamma;</mi> <mn>2</mn> </msub> <mo>,</mo> <mi>M</mi> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mi>E</mi> <mi>X</mi> </msub> <mo>[</mo> <msub> <mi>log</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <mi>X</mi> <mo>)</mo> </mrow> <mo>]</mo> <mo>=</mo> <msub> <mi>log</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>e</mi> <mo>)</mo> </mrow> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> <mrow> <mi>M</mi> <mo>-</mo> <mn>1</mn> </mrow> </munderover> <munderover> <mi>&Sigma;</mi> <mrow> <mi>l</mi> <mo>=</mo> <mn>0</mn> </mrow> <mi>i</mi> </munderover> <mfrac> <msubsup> <mi>&gamma;</mi> <mn>1</mn> <mrow> <mi>l</mi> <mo>+</mo> <mn>1</mn> <mo>-</mo> <mi>i</mi> </mrow> </msubsup> <mrow> <msub> <mi>&gamma;</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>i</mi> <mo>-</mo> <mi>l</mi> <mo>)</mo> </mrow> <mo>!</mo> </mrow> </mfrac> <mo>&CenterDot;</mo> <msub> <mi>I</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mfrac> <mn>1</mn> <msub> <mi>&gamma;</mi> <mn>1</mn> </msub> </mfrac> <mo>,</mo> <mfrac> <msub> <mi>&gamma;</mi> <mn>1</mn> </msub> <msub> <mi>&gamma;</mi> <mn>2</mn> </msub> </mfrac> <mo>,</mo> <mi>i</mi> <mo>,</mo> <mi>l</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> </mrow> </math>
Wherein I1(a, b, m, n) is an integral expression
<math> <mrow> <msub> <mi>I</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mi>m</mi> <mo>,</mo> <mi>n</mi> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>n</mi> </munderover> <mfrac> <msup> <mrow> <mo>(</mo> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mrow> <mi>i</mi> <mo>-</mo> <mn>1</mn> </mrow> </msup> <msup> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <mi>b</mi> <mo>)</mo> </mrow> <mi>i</mi> </msup> </mfrac> <mo>&CenterDot;</mo> <msub> <mi>I</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mi>m</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mi>i</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> <mo>+</mo> <mfrac> <mrow> <msub> <mi>I</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mn>1</mn> <mo>,</mo> <mi>m</mi> <mo>,</mo> <mn>1</mn> <mo>)</mo> </mrow> </mrow> <msup> <mrow> <mo>(</mo> <mi>b</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mi>n</mi> </msup> </mfrac> </mrow> </math>
<math> <mrow> <msub> <mi>I</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mi>m</mi> <mo>,</mo> <mi>n</mi> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mo>&Integral;</mo> <mn>0</mn> <mo>&infin;</mo> </munderover> <mfrac> <mrow> <msup> <mi>x</mi> <mi>m</mi> </msup> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mi>ax</mi> </mrow> </msup> </mrow> <msup> <mrow> <mo>(</mo> <mi>x</mi> <mo>+</mo> <mi>b</mi> <mo>)</mo> </mrow> <mi>i</mi> </msup> </mfrac> <mi>dx</mi> <mo>=</mo> <msup> <mi>e</mi> <mi>ab</mi> </msup> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> <mi>m</mi> </munderover> <mfenced open='(' close=')'> <mtable> <mtr> <mtd> <mi>m</mi> </mtd> </mtr> <mtr> <mtd> <mi>i</mi> </mtd> </mtr> </mtable> </mfenced> <msup> <mrow> <mo>(</mo> <mo>-</mo> <mi>b</mi> <mo>)</mo> </mrow> <mrow> <mi>m</mi> <mo>-</mo> <mi>i</mi> </mrow> </msup> <msub> <mi>I</mi> <mn>3</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mi>i</mi> <mo>-</mo> <mi>n</mi> <mo>)</mo> </mrow> </mrow> </math>
<math> <mrow> <msub> <mi>I</mi> <mn>3</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mi>m</mi> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mo>&Integral;</mo> <mi>b</mi> <mo>&infin;</mo> </munderover> <msup> <mi>x</mi> <mi>m</mi> </msup> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mi>ax</mi> </mrow> </msup> <mi>dx</mi> <mo>=</mo> <mfenced open='{' close=''> <mtable> <mtr> <mtd> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mi>ab</mi> </mrow> </msup> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> <mi>m</mi> </munderover> <mfrac> <mrow> <mi>m</mi> <mo>!</mo> </mrow> <mrow> <mi>i</mi> <mo>!</mo> </mrow> </mfrac> <mfrac> <msup> <mi>b</mi> <mi>i</mi> </msup> <msup> <mi>a</mi> <mrow> <mi>m</mi> <mo>-</mo> <mi>i</mi> <mo>+</mo> <mn>1</mn> </mrow> </msup> </mfrac> </mtd> <mtd> <mi>m</mi> <mo>&GreaterEqual;</mo> <mn>0</mn> </mtd> </mtr> <mtr> <mtd> <msub> <mi>E</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>ab</mi> <mo>)</mo> </mrow> </mtd> <mtd> <mi>m</mi> <mo>=</mo> <mo>-</mo> <mn>1</mn> </mtd> </mtr> <mtr> <mtd> <mfrac> <mrow> <msub> <mi>E</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>ab</mi> <mo>)</mo> </mrow> </mrow> <mrow> <msup> <mrow> <mo>(</mo> <mo>-</mo> <mi>a</mi> <mo>)</mo> </mrow> <mrow> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mn>1</mn> </mrow> </msup> <mrow> <mo>(</mo> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mo>!</mo> </mrow> </mfrac> <mo>+</mo> <mfrac> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mi>ab</mi> </mrow> </msup> <msup> <mi>b</mi> <mrow> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mn>1</mn> </mrow> </msup> </mfrac> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> <mrow> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mn>2</mn> </mrow> </munderover> <mfrac> <mrow> <msup> <mrow> <mo>(</mo> <mo>-</mo> <mi>ab</mi> <mo>)</mo> </mrow> <mi>i</mi> </msup> <mrow> <mo>(</mo> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mi>i</mi> <mo>-</mo> <mn>2</mn> <mo>)</mo> </mrow> <mo>!</mo> </mrow> <mrow> <mrow> <mo>(</mo> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mo>!</mo> </mrow> </mfrac> </mtd> <mtd> <mi>m</mi> <mo>&le;</mo> <mo>-</mo> <mn>2</mn> </mtd> </mtr> </mtable> </mfenced> </mrow> </math>
The total system throughput can be expressed as
R=R1+R2=R1(P1 rG111),P1 rG211),1)+R1(P2 rG222),P2 rG122),1)
When beta is1、β2Satisfies the constraint condition of beta being less than or equal to 0 degree2≤β1When the angle is less than or equal to 90 degrees, the optimal (beta) is obtained for each pair of users with known positions12) The combination maximizes the throughput of both users.
The problems to be solved by the present invention are: according to the position information of the current served user, selecting the optimal downward inclination angle for each beam sent by the base station to enable the sum rate of all users to be maximum, namely solving the following problems:
<math> <mrow> <mrow> <mo>(</mo> <msup> <mi>&beta;</mi> <mo>*</mo> </msup> <mo>)</mo> </mrow> <mo>=</mo> <mi>arg</mi> <munder> <mi>max</mi> <mrow> <msub> <mi>&beta;</mi> <mi>min</mi> </msub> <mo>&lt;</mo> <mi>&beta;</mi> <mo>&lt;</mo> <msub> <mi>&beta;</mi> <mi>max</mi> </msub> </mrow> </munder> <munderover> <mi>&Sigma;</mi> <mrow> <mi>s</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>S</mi> </munderover> <msub> <mi>R</mi> <mi>s</mi> </msub> <mrow> <mo>(</mo> <msub> <mi>&beta;</mi> <mi>s</mi> </msub> <mo>)</mo> </mrow> </mrow> </math>
wherein beta isminAnd betamaxRespectively minimum and maximum downtilt combinations. The present invention solves this problem using the scaling recovery (DIRECT) algorithm, which is considered to be an effective method to solve the continuity optimization problem with simple constraints. The DIRECT algorithm does not need to know the prior information of the target function, and is very suitable for solving the complex target function under the condition similar to the invention. The algorithm treats the function variable domain as a hyper-rectangle, and performs space segmentation on the hyper-rectangle to obtain a smaller hyper-rectangle. By representing each hyper-rectangle by its center, the objective function will be computed only at these center points, and the computational complexity is reduced. In addition, the DIRECT only selects a potentially optimal hyper-rectangle in each iteration to carry out the next iteration, and the convergence speed is increased. The algorithm continuously divides a function variable domain, and finally, the value of a target function, namely the cell throughput, is maximum at a certain point.
Two, vertical domain based downtilt
Vertical domain antenna downtilt differs from user antenna downtilt in that the latter downtilt angle is constantly changing, varying with the served users at each time slot, while the former downtilt angle is substantially constant over a period of time, each serving a plurality of users within its area. The vertical domain based antenna downtilt study is based on the following assumptions: the base station divides the transmitting antennas into two groups, each antenna group has 4 transmitting antennas, and the two groups of antennas simultaneously send out two wave beams to serve a plurality of users in the sector. The optimum downtilt angle for each beam is selected based on the user location. After the downtilt angles of the two groups of antennas are determined, we refer to the user area served by one group of antennas as a vertical domain.
With the 3GPP antenna model and cell configuration, the parameter configuration used in the present invention can be as shown in table 1 below:
TABLE 1
The sector is divided into 2 vertical domains, and users are uniformly distributed in the sector. As shown in table 1, assuming a cell radius is 500m, the base station adopts a three-sector model, and the sector is divided into two circular effective Vertical domains (Vertical regions) VS1 and VS 2.
The downward inclination angle of the antenna is too large, the propagation characteristic diagram in the horizontal direction becomes flat, and particularly when the downward inclination angle of the antenna exceeds 20 degrees, gaps and side lobes can occur, so that the interference to cells with the same frequency is increased. As known from the literature, the maximum downtilt angle of the antenna cannot exceed 24 °.
100 users are scattered uniformly in a sector, and the beam to which the user is served is determined according to the SINR of the received signals from the users to two beams. Taking a certain user as an example, as can be seen from the foregoing description, when the beams 1 and 2 are respectively adopted as the service beams of the user, the signals received by the user can be respectively expressed as:
comparing the sizes, the invention selects the beam with the larger SINR as the service beam of the user. Due to fi,iAnd channel matrix hiIs not related, therefore, there isI.e. obeying a chi-square distribution with a degree of freedom of 2.
Throughput of the vertical domain is
<math> <mrow> <msub> <mi>R</mi> <mi>s</mi> </msub> <mo>=</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mrow> <mi>i</mi> <mo>=</mo> <msub> <mi>K</mi> <mi>s</mi> </msub> </mrow> </munderover> <mi>E</mi> <mo>[</mo> <msub> <mi>log</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <mi>S</mi> <msub> <mi>INR</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>s</mi> </mrow> </msub> <mrow> <mo>(</mo> <mi>&beta;</mi> <mo>)</mo> </mrow> <mo>)</mo> </mrow> <mo>]</mo> </mrow> </math>
Total sector throughput of
<math> <mrow> <mi>R</mi> <mo>=</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>s</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>S</mi> </munderover> <msub> <mi>R</mi> <mi>s</mi> </msub> <mo>=</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>s</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>S</mi> </munderover> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mrow> <mi>k</mi> <mo>=</mo> <msub> <mi>K</mi> <mi>s</mi> </msub> </mrow> </munderover> <mi>E</mi> <mo>[</mo> <msub> <mi>log</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <msub> <mi>SINR</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>s</mi> </mrow> </msub> <mrow> <mo>(</mo> <mi>&beta;</mi> <mo>)</mo> </mrow> <mo>)</mo> </mrow> <mo>]</mo> </mrow> </math>
The method for solving the problem is the same as the method for solving the problem of the declination based on the user. After the optimal β is solved, the sector can be naturally divided into two vertical domains according to the serving users and user positions of the two beams.
Based on the same concept, an embodiment of the present invention further provides a beamforming apparatus, which is applied to a wireless communication system as a base station, and as shown in fig. 4, the apparatus includes:
a location information obtaining module 401, configured to obtain location information of each user to be served by a next time slot service;
a function calling module 402, configured to select a combination (β) of total throughput and downward tilt angle of each user according to the location information acquired by the location information acquiring module 4011,β2) Total throughput function ofβ 1 and β 2 denote downtilt angles of the 1 st and 2 nd beams, respectively, Ri12) Throughput versus downtilt combination (β) for the ith user1,β2) I is the number of users served by the two beams.
A calculation module 403 for calculating the equation using the DIRECT algorithmMaximum down dip angle combination (beta)1M,β2M)。
A forming module 404 for correspondingly adjusting the downtilt angles of the two beams to β1M,β2M
Preferably, the function invoking module 402 is specifically configured to determine, according to the obtained location information, a combination (β) of the signal strength of each beam received by each user with respect to the downtilt angle1,β2) And take a function ofAs a combination of total throughput versus downtilt (β) for each user1,β2) Total throughput function ofWherein,
<math> <mrow> <mi>R</mi> <mrow> <mo>(</mo> <msub> <mi>&gamma;</mi> <mn>1</mn> </msub> <mo>,</mo> <msub> <mi>&gamma;</mi> <mn>2</mn> </msub> <mo>,</mo> <mi>M</mi> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mi>E</mi> <mi>X</mi> </msub> <mo>[</mo> <msub> <mi>log</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <mi>X</mi> <mo>)</mo> </mrow> <mo>]</mo> <mo>=</mo> <msub> <mi>log</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>e</mi> <mo>)</mo> </mrow> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> <mrow> <mi>M</mi> <mo>-</mo> <mn>1</mn> </mrow> </munderover> <munderover> <mi>&Sigma;</mi> <mrow> <mi>l</mi> <mo>=</mo> <mn>0</mn> </mrow> <mi>i</mi> </munderover> <mfrac> <msubsup> <mi>&gamma;</mi> <mn>1</mn> <mrow> <mi>l</mi> <mo>+</mo> <mn>1</mn> <mo>-</mo> <mi>i</mi> </mrow> </msubsup> <mrow> <msub> <mi>&gamma;</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>i</mi> <mo>-</mo> <mi>l</mi> <mo>)</mo> </mrow> <mo>!</mo> </mrow> </mfrac> <mo>&CenterDot;</mo> <msub> <mi>I</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mfrac> <mn>1</mn> <msub> <mi>&gamma;</mi> <mn>1</mn> </msub> </mfrac> <mo>,</mo> <mfrac> <msub> <mi>&gamma;</mi> <mn>1</mn> </msub> <msub> <mi>&gamma;</mi> <mn>2</mn> </msub> </mfrac> <mo>,</mo> <mi>i</mi> <mo>,</mo> <mi>l</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> </mrow> </math>
wherein I1(a, b, m, n) is an integral expression, and
<math> <mrow> <msub> <mi>I</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mi>m</mi> <mo>,</mo> <mi>n</mi> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>n</mi> </munderover> <mfrac> <msup> <mrow> <mo>(</mo> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mrow> <mi>i</mi> <mo>-</mo> <mn>1</mn> </mrow> </msup> <msup> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <mi>b</mi> <mo>)</mo> </mrow> <mi>i</mi> </msup> </mfrac> <mo>&CenterDot;</mo> <msub> <mi>I</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mi>m</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mi>i</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> <mo>+</mo> <mfrac> <mrow> <msub> <mi>I</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mn>1</mn> <mo>,</mo> <mi>m</mi> <mo>,</mo> <mn>1</mn> <mo>)</mo> </mrow> </mrow> <msup> <mrow> <mo>(</mo> <mi>b</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mi>n</mi> </msup> </mfrac> <mo>;</mo> </mrow> </math>
<math> <mrow> <msub> <mi>I</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mi>m</mi> <mo>,</mo> <mi>n</mi> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mo>&Integral;</mo> <mn>0</mn> <mo>&infin;</mo> </munderover> <mfrac> <mrow> <msup> <mi>x</mi> <mi>m</mi> </msup> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mi>ax</mi> </mrow> </msup> </mrow> <msup> <mrow> <mo>(</mo> <mi>x</mi> <mo>+</mo> <mi>b</mi> <mo>)</mo> </mrow> <mi>i</mi> </msup> </mfrac> <mi>dx</mi> <mo>=</mo> <msup> <mi>e</mi> <mi>ab</mi> </msup> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> <mi>m</mi> </munderover> <mfenced open='(' close=')'> <mtable> <mtr> <mtd> <mi>m</mi> </mtd> </mtr> <mtr> <mtd> <mi>i</mi> </mtd> </mtr> </mtable> </mfenced> <msup> <mrow> <mo>(</mo> <mo>-</mo> <mi>b</mi> <mo>)</mo> </mrow> <mrow> <mi>m</mi> <mo>-</mo> <mi>i</mi> </mrow> </msup> <msub> <mi>I</mi> <mn>3</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mi>i</mi> <mo>-</mo> <mi>n</mi> <mo>)</mo> </mrow> <mo>;</mo> </mrow> </math>
<math> <mrow> <msub> <mi>I</mi> <mn>3</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mi>m</mi> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mo>&Integral;</mo> <mi>b</mi> <mo>&infin;</mo> </munderover> <msup> <mi>x</mi> <mi>m</mi> </msup> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mi>ax</mi> </mrow> </msup> <mi>dx</mi> <mo>=</mo> <mfenced open='{' close=''> <mtable> <mtr> <mtd> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mi>ab</mi> </mrow> </msup> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> <mi>m</mi> </munderover> <mfrac> <mrow> <mi>m</mi> <mo>!</mo> </mrow> <mrow> <mi>i</mi> <mo>!</mo> </mrow> </mfrac> <mfrac> <msup> <mi>b</mi> <mi>i</mi> </msup> <msup> <mi>a</mi> <mrow> <mi>m</mi> <mo>-</mo> <mi>i</mi> <mo>+</mo> <mn>1</mn> </mrow> </msup> </mfrac> </mtd> <mtd> <mi>m</mi> <mo>&GreaterEqual;</mo> <mn>0</mn> </mtd> </mtr> <mtr> <mtd> <msub> <mi>E</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>ab</mi> <mo>)</mo> </mrow> </mtd> <mtd> <mi>m</mi> <mo>=</mo> <mo>-</mo> <mn>1</mn> </mtd> </mtr> <mtr> <mtd> <mfrac> <mrow> <msub> <mi>E</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>ab</mi> <mo>)</mo> </mrow> </mrow> <mrow> <msup> <mrow> <mo>(</mo> <mo>-</mo> <mi>a</mi> <mo>)</mo> </mrow> <mrow> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mn>1</mn> </mrow> </msup> <mrow> <mo>(</mo> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mo>!</mo> </mrow> </mfrac> <mo>+</mo> <mfrac> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mi>ab</mi> </mrow> </msup> <msup> <mi>b</mi> <mrow> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mn>1</mn> </mrow> </msup> </mfrac> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> <mrow> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mn>2</mn> </mrow> </munderover> <mfrac> <mrow> <msup> <mrow> <mo>(</mo> <mo>-</mo> <mi>ab</mi> <mo>)</mo> </mrow> <mi>i</mi> </msup> <mrow> <mo>(</mo> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mi>i</mi> <mo>-</mo> <mn>2</mn> <mo>)</mo> </mrow> <mo>!</mo> </mrow> <mrow> <mrow> <mo>(</mo> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mo>!</mo> </mrow> </mfrac> </mtd> <mtd> <mi>m</mi> <mo>&le;</mo> <mo>-</mo> <mn>2</mn> </mtd> </mtr> </mtable> </mfenced> </mrow> </math>
wherein,received power of s beam for i user, Gsis) Represents the downtilt angle beta of the reception gain of the ith user to the s beam relative to the s beamsAs a function of (c).
Preferably, the function call module calculates the receiving gain G of the ith user to the ith beam by the following formulasis):
Wherein,and thetaiAzimuth angle and pitch angle, SLL, corresponding to the ith user respectivelyazAnd SLLelSide lobe levels, SLL, of the horizontal and vertical patterns, respectivelytotIs the total side-lobe level of the signal,for the received power of the ith user on the s-th beam,is the angle between the aiming axis of the antenna array of the base station where the ith user is located and the x axis.
The foregoing is only a preferred embodiment of the present invention, and it should be noted that, for those skilled in the art, various modifications and decorations can be made without departing from the technical principle of the present invention, and these modifications and decorations should also be regarded as the protection scope of the present invention.

Claims (6)

1. A beamforming method applied to a dual-beam communication system, the method comprising:
acquiring the position information of each user to be served by the current time slot;
selecting a combination (beta) of the total throughput of each user with respect to the downtilt angle according to the acquired location information1,β2) Total throughput function ofBeta 1 and beta 2 minutesRespectively, the downtilt angles, R, of the 1 st and 2 nd beamsi12) Throughput versus downtilt combination (β) for the ith user1,β2) I is the number of users served by the two beams;
computing using a DIRECT algorithmMaximum down dip angle combination (beta)1M,β2M);
Correspondingly adjusting the downtilt angles of the two beams to beta1M,β2M
2. The method of claim 1, wherein selecting a combination of total throughput versus downtilt (β) for each user based on the obtained location information1,β2) Total throughput function ofThe method specifically comprises the following steps:
for each user, the received power of each beam is calculated, and the receiving gain of each beam is combined with the declination angle (beta)1,β2) A function of (a);
getAs a combination of total throughput versus downtilt (β) for each user1,β2) Total throughput function ofWherein,
<math> <mrow> <mi>R</mi> <mrow> <mo>(</mo> <msub> <mi>&gamma;</mi> <mn>1</mn> </msub> <mo>,</mo> <msub> <mi>&gamma;</mi> <mn>2</mn> </msub> <mo>,</mo> <mi>M</mi> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mi>E</mi> <mi>X</mi> </msub> <mo>[</mo> <msub> <mi>log</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <mi>X</mi> <mo>)</mo> </mrow> <mo>]</mo> <mo>=</mo> <msub> <mi>log</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>e</mi> <mo>)</mo> </mrow> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> <mrow> <mi>M</mi> <mo>-</mo> <mn>1</mn> </mrow> </munderover> <munderover> <mi>&Sigma;</mi> <mrow> <mi>l</mi> <mo>=</mo> <mn>0</mn> </mrow> <mi>i</mi> </munderover> <mfrac> <msubsup> <mi>&gamma;</mi> <mn>1</mn> <mrow> <mi>l</mi> <mo>+</mo> <mn>1</mn> <mo>-</mo> <mi>i</mi> </mrow> </msubsup> <mrow> <msub> <mi>&gamma;</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>i</mi> <mo>-</mo> <mi>l</mi> <mo>)</mo> </mrow> <mo>!</mo> </mrow> </mfrac> <mo>&CenterDot;</mo> <msub> <mi>I</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mfrac> <mn>1</mn> <msub> <mi>&gamma;</mi> <mn>1</mn> </msub> </mfrac> <mo>,</mo> <mfrac> <msub> <mi>&gamma;</mi> <mn>1</mn> </msub> <msub> <mi>&gamma;</mi> <mn>2</mn> </msub> </mfrac> <mo>,</mo> <mi>i</mi> <mo>,</mo> <mi>l</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> </mrow> </math>
wherein I1(a, b, m, n) is an integral expression, and
<math> <mrow> <msub> <mi>I</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mi>m</mi> <mo>,</mo> <mi>n</mi> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>n</mi> </munderover> <mfrac> <msup> <mrow> <mo>(</mo> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mrow> <mi>i</mi> <mo>-</mo> <mn>1</mn> </mrow> </msup> <msup> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <mi>b</mi> <mo>)</mo> </mrow> <mi>i</mi> </msup> </mfrac> <mo>&CenterDot;</mo> <msub> <mi>I</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mi>m</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mi>i</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> <mo>+</mo> <mfrac> <mrow> <msub> <mi>I</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mn>1</mn> <mo>,</mo> <mi>m</mi> <mo>,</mo> <mn>1</mn> <mo>)</mo> </mrow> </mrow> <msup> <mrow> <mo>(</mo> <mi>b</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mi>n</mi> </msup> </mfrac> <mo>;</mo> </mrow> </math>
<math> <mrow> <msub> <mi>I</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mi>m</mi> <mo>,</mo> <mi>n</mi> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mo>&Integral;</mo> <mn>0</mn> <mo>&infin;</mo> </munderover> <mfrac> <mrow> <msup> <mi>x</mi> <mi>m</mi> </msup> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mi>ax</mi> </mrow> </msup> </mrow> <msup> <mrow> <mo>(</mo> <mi>x</mi> <mo>+</mo> <mi>b</mi> <mo>)</mo> </mrow> <mi>i</mi> </msup> </mfrac> <mi>dx</mi> <mo>=</mo> <msup> <mi>e</mi> <mi>ab</mi> </msup> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> <mi>m</mi> </munderover> <mfenced open='(' close=')'> <mtable> <mtr> <mtd> <mi>m</mi> </mtd> </mtr> <mtr> <mtd> <mi>i</mi> </mtd> </mtr> </mtable> </mfenced> <msup> <mrow> <mo>(</mo> <mo>-</mo> <mi>b</mi> <mo>)</mo> </mrow> <mrow> <mi>m</mi> <mo>-</mo> <mi>i</mi> </mrow> </msup> <msub> <mi>I</mi> <mn>3</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mi>i</mi> <mo>-</mo> <mi>n</mi> <mo>)</mo> </mrow> <mo>;</mo> </mrow> </math>
<math> <mrow> <msub> <mi>I</mi> <mn>3</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mi>m</mi> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mo>&Integral;</mo> <mi>b</mi> <mo>&infin;</mo> </munderover> <msup> <mi>x</mi> <mi>m</mi> </msup> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mi>ax</mi> </mrow> </msup> <mi>dx</mi> <mo>=</mo> <mfenced open='{' close=''> <mtable> <mtr> <mtd> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mi>ab</mi> </mrow> </msup> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> <mi>m</mi> </munderover> <mfrac> <mrow> <mi>m</mi> <mo>!</mo> </mrow> <mrow> <mi>i</mi> <mo>!</mo> </mrow> </mfrac> <mfrac> <msup> <mi>b</mi> <mi>i</mi> </msup> <msup> <mi>a</mi> <mrow> <mi>m</mi> <mo>-</mo> <mi>i</mi> <mo>+</mo> <mn>1</mn> </mrow> </msup> </mfrac> </mtd> <mtd> <mi>m</mi> <mo>&GreaterEqual;</mo> <mn>0</mn> </mtd> </mtr> <mtr> <mtd> <msub> <mi>E</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>ab</mi> <mo>)</mo> </mrow> </mtd> <mtd> <mi>m</mi> <mo>=</mo> <mo>-</mo> <mn>1</mn> </mtd> </mtr> <mtr> <mtd> <mfrac> <mrow> <msub> <mi>E</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>ab</mi> <mo>)</mo> </mrow> </mrow> <mrow> <msup> <mrow> <mo>(</mo> <mo>-</mo> <mi>a</mi> <mo>)</mo> </mrow> <mrow> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mn>1</mn> </mrow> </msup> <mrow> <mo>(</mo> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mo>!</mo> </mrow> </mfrac> <mo>+</mo> <mfrac> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mi>ab</mi> </mrow> </msup> <msup> <mi>b</mi> <mrow> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mn>1</mn> </mrow> </msup> </mfrac> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> <mrow> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mn>2</mn> </mrow> </munderover> <mfrac> <mrow> <msup> <mrow> <mo>(</mo> <mo>-</mo> <mi>ab</mi> <mo>)</mo> </mrow> <mi>i</mi> </msup> <mrow> <mo>(</mo> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mi>i</mi> <mo>-</mo> <mn>2</mn> <mo>)</mo> </mrow> <mo>!</mo> </mrow> <mrow> <mrow> <mo>(</mo> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mo>!</mo> </mrow> </mfrac> </mtd> <mtd> <mi>m</mi> <mo>&le;</mo> <mo>-</mo> <mn>2</mn> </mtd> </mtr> </mtable> </mfenced> </mrow> </math>
wherein,received power of s beam for i user, Gsis) Represents the downtilt angle beta of the reception gain of the ith user to the s beam relative to the s beamsAs a function of (c).
3. The method of claim 2, wherein for each user, its receive gain versus downtilt combination (β) for each beam is calculated separately1,β2) Comprises:
calculating the receiving gain G of the ith user to the s beam by the following formulasis):
Wherein,and thetaiAzimuth angle and pitch angle corresponding to the ith user, SLLaz and SLLel are side lobe levels of horizontal and vertical directional diagrams, SLLtotIs the total side-lobe level of the signal,for the received power of the ith user on the s-th beam,is the angle between the aiming axis of the antenna array of the base station where the ith user is located and the x axis.
4. A beamforming apparatus for use in a wireless communication system as a base station, the apparatus comprising:
the system comprises a position information acquisition module, a position information acquisition module and a time slot processing module, wherein the position information acquisition module is used for acquiring the position information of each user to be served by the next time slot service;
a function calling module for selecting the total throughput of each user relative to the downward inclination angle combination (beta) according to the position information acquired by the position information acquisition module1,β2) Total throughput function ofβ 1 and β 2 denote downtilt angles of the 1 st and 2 nd beams, respectively, Ri12) Throughput versus downtilt combination (β) for the ith user1,β2) I is the number of users served by the two beams;
a calculation module for calculating the equation using the DIRECT algorithmMaximum down dip angle combination (beta)1M,β2M);
A shaping module for correspondingly adjusting the downtilt angles of the two beams to beta1M,β2M
5. The device of claim 4, wherein the function call module is specifically configured to determine, according to the obtained location information, a combination (β) of signal strength of each user receiving each beam with respect to downtilt angle1,β2) Is taken as a function ofAs a combination of total throughput versus downtilt (β) for each user1,β2) Total throughput function of <math> <mrow> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>I</mi> </munderover> <msub> <mi>R</mi> <mi>i</mi> </msub> <mrow> <mo>(</mo> <msub> <mi>&beta;</mi> <mn>1</mn> </msub> <mo>,</mo> <msub> <mi>&beta;</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> <mo>;</mo> </mrow> </math> Wherein,
<math> <mrow> <mi>R</mi> <mrow> <mo>(</mo> <msub> <mi>&gamma;</mi> <mn>1</mn> </msub> <mo>,</mo> <msub> <mi>&gamma;</mi> <mn>2</mn> </msub> <mo>,</mo> <mi>M</mi> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mi>E</mi> <mi>X</mi> </msub> <mo>[</mo> <msub> <mi>log</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <mi>X</mi> <mo>)</mo> </mrow> <mo>]</mo> <mo>=</mo> <msub> <mi>log</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>e</mi> <mo>)</mo> </mrow> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> <mrow> <mi>M</mi> <mo>-</mo> <mn>1</mn> </mrow> </munderover> <munderover> <mi>&Sigma;</mi> <mrow> <mi>l</mi> <mo>=</mo> <mn>0</mn> </mrow> <mi>i</mi> </munderover> <mfrac> <msubsup> <mi>&gamma;</mi> <mn>1</mn> <mrow> <mi>l</mi> <mo>+</mo> <mn>1</mn> <mo>-</mo> <mi>i</mi> </mrow> </msubsup> <mrow> <msub> <mi>&gamma;</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>i</mi> <mo>-</mo> <mi>l</mi> <mo>)</mo> </mrow> <mo>!</mo> </mrow> </mfrac> <mo>&CenterDot;</mo> <msub> <mi>I</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mfrac> <mn>1</mn> <msub> <mi>&gamma;</mi> <mn>1</mn> </msub> </mfrac> <mo>,</mo> <mfrac> <msub> <mi>&gamma;</mi> <mn>1</mn> </msub> <msub> <mi>&gamma;</mi> <mn>2</mn> </msub> </mfrac> <mo>,</mo> <mi>i</mi> <mo>,</mo> <mi>l</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> </mrow> </math>
wherein I1(a, b, m, n) is an integral expression, and
<math> <mrow> <msub> <mi>I</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mi>m</mi> <mo>,</mo> <mi>n</mi> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>n</mi> </munderover> <mfrac> <msup> <mrow> <mo>(</mo> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mrow> <mi>i</mi> <mo>-</mo> <mn>1</mn> </mrow> </msup> <msup> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <mi>b</mi> <mo>)</mo> </mrow> <mi>i</mi> </msup> </mfrac> <mo>&CenterDot;</mo> <msub> <mi>I</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mi>m</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mi>i</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> <mo>+</mo> <mfrac> <mrow> <msub> <mi>I</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mn>1</mn> <mo>,</mo> <mi>m</mi> <mo>,</mo> <mn>1</mn> <mo>)</mo> </mrow> </mrow> <msup> <mrow> <mo>(</mo> <mi>b</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mi>n</mi> </msup> </mfrac> <mo>;</mo> </mrow> </math>
<math> <mrow> <msub> <mi>I</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mi>m</mi> <mo>,</mo> <mi>n</mi> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mo>&Integral;</mo> <mn>0</mn> <mo>&infin;</mo> </munderover> <mfrac> <mrow> <msup> <mi>x</mi> <mi>m</mi> </msup> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mi>ax</mi> </mrow> </msup> </mrow> <msup> <mrow> <mo>(</mo> <mi>x</mi> <mo>+</mo> <mi>b</mi> <mo>)</mo> </mrow> <mi>i</mi> </msup> </mfrac> <mi>dx</mi> <mo>=</mo> <msup> <mi>e</mi> <mi>ab</mi> </msup> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> <mi>m</mi> </munderover> <mfenced open='(' close=')'> <mtable> <mtr> <mtd> <mi>m</mi> </mtd> </mtr> <mtr> <mtd> <mi>i</mi> </mtd> </mtr> </mtable> </mfenced> <msup> <mrow> <mo>(</mo> <mo>-</mo> <mi>b</mi> <mo>)</mo> </mrow> <mrow> <mi>m</mi> <mo>-</mo> <mi>i</mi> </mrow> </msup> <msub> <mi>I</mi> <mn>3</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mi>i</mi> <mo>-</mo> <mi>n</mi> <mo>)</mo> </mrow> <mo>;</mo> </mrow> </math>
<math> <mrow> <msub> <mi>I</mi> <mn>3</mn> </msub> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mi>m</mi> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mo>&Integral;</mo> <mi>b</mi> <mo>&infin;</mo> </munderover> <msup> <mi>x</mi> <mi>m</mi> </msup> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mi>ax</mi> </mrow> </msup> <mi>dx</mi> <mo>=</mo> <mfenced open='{' close=''> <mtable> <mtr> <mtd> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mi>ab</mi> </mrow> </msup> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> <mi>m</mi> </munderover> <mfrac> <mrow> <mi>m</mi> <mo>!</mo> </mrow> <mrow> <mi>i</mi> <mo>!</mo> </mrow> </mfrac> <mfrac> <msup> <mi>b</mi> <mi>i</mi> </msup> <msup> <mi>a</mi> <mrow> <mi>m</mi> <mo>-</mo> <mi>i</mi> <mo>+</mo> <mn>1</mn> </mrow> </msup> </mfrac> </mtd> <mtd> <mi>m</mi> <mo>&GreaterEqual;</mo> <mn>0</mn> </mtd> </mtr> <mtr> <mtd> <msub> <mi>E</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>ab</mi> <mo>)</mo> </mrow> </mtd> <mtd> <mi>m</mi> <mo>=</mo> <mo>-</mo> <mn>1</mn> </mtd> </mtr> <mtr> <mtd> <mfrac> <mrow> <msub> <mi>E</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>ab</mi> <mo>)</mo> </mrow> </mrow> <mrow> <msup> <mrow> <mo>(</mo> <mo>-</mo> <mi>a</mi> <mo>)</mo> </mrow> <mrow> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mn>1</mn> </mrow> </msup> <mrow> <mo>(</mo> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mo>!</mo> </mrow> </mfrac> <mo>+</mo> <mfrac> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mi>ab</mi> </mrow> </msup> <msup> <mi>b</mi> <mrow> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mn>1</mn> </mrow> </msup> </mfrac> <munderover> <mi>&Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> <mrow> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mn>2</mn> </mrow> </munderover> <mfrac> <mrow> <msup> <mrow> <mo>(</mo> <mo>-</mo> <mi>ab</mi> <mo>)</mo> </mrow> <mi>i</mi> </msup> <mrow> <mo>(</mo> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mi>i</mi> <mo>-</mo> <mn>2</mn> <mo>)</mo> </mrow> <mo>!</mo> </mrow> <mrow> <mrow> <mo>(</mo> <mo>-</mo> <mi>m</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mo>!</mo> </mrow> </mfrac> </mtd> <mtd> <mi>m</mi> <mo>&le;</mo> <mo>-</mo> <mn>2</mn> </mtd> </mtr> </mtable> </mfenced> </mrow> </math>
wherein,received power of s beam for i user, Gsis) Represents the downtilt angle beta of the reception gain of the ith user to the s beam relative to the s beamsAs a function of (c).
6. The device of claim 5, wherein the function call module calculates the reception gain G for the ith user for the ith beam by specifically using the following formulasis):
Wherein,and thetaiAzimuth angle and pitch angle, SLL, corresponding to the ith user respectivelyazAnd SLLelSide lobe levels, SLL, of the horizontal and vertical patterns, respectivelytotIs the total side-lobe level of the signal,for the received power of the ith user on the s-th beam,is the angle between the aiming axis of the antenna array of the base station where the ith user is located and the x axis.
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