Přepis řeči - TRANSCEIVER OPTIMIZATION FOR MULTI-USER MULTI-ANTENNA TWO-WAY RELAY CHANNELS

0:00:13	okay
0:00:14	a could have not everyone
0:00:16	um my talking case about a transceiver optimization for multiuser remote a on a each channel
0:00:21	and in this work
0:00:22	we can see such scenarios
0:00:24	well base station at the multiple users exchange a pink and downlink channel uh uh uh i pink and only
0:00:30	information
0:00:31	where a to will eh
0:00:33	a we're a station
0:00:34	in such scenarios
0:00:35	the
0:00:37	given the user interfere with each other
0:00:39	and that we exploit the multiple antennas at the base station the station to in to use their interfering
0:00:45	and you our work we can see that the i'm don't for the protocol
0:00:50	uh here we were talking about using much what time that's to come have to interference
0:00:55	that zero focusing is a straight for the solution
0:00:58	we have uh
0:00:59	system with and user
0:01:01	the zero forcing to game need at least a two times and and ten us at the relay station to
0:01:06	separate a with and downlink signals
0:01:08	oh in pink still
0:01:10	and the here and now the interference free game called the signal alignment is propose is proposed to by this
0:01:16	paper papers
0:01:16	uh the basic idea is that
0:01:18	through proper a base station precoding the downlink signals are projected onto the same directions
0:01:24	a pink signals
0:01:25	then the really station only needs and and a two step rate
0:01:29	given the user to imposed the signal
0:01:32	and
0:01:32	in this paper we have also proposed a of balance this P
0:01:36	and they is also a interface for scheme and it can achieve a higher by jewish bidirectional sum rate than
0:01:42	both of the zero forcing on the signal i'm i
0:01:45	but so we want to ask
0:01:46	is this game is good enough and the can we achieve a higher by just close summary
0:01:51	all that these questions we want to first all four
0:01:54	a performance benchmark uh problems but map for that concerns sister
0:01:59	a a a our system or do
0:02:01	uh
0:02:02	base station and the relay stations uh is delay you keep the with and B and and are and that's
0:02:07	each of the and you users
0:02:08	is equipped be
0:02:09	thing going on a
0:02:11	uh here H B R is the channel matrix from base station to the relay station
0:02:15	H I R is that
0:02:17	channel what or from the i-th user or to the relay station
0:02:20	and uh the that will be G and that that will be a a are respectively the
0:02:24	base station transmitter and the receiver or which a matrix
0:02:27	that we is the really station to zero with me check
0:02:30	in the first phase pose the base station and all users transmit to the really station
0:02:35	and in the second phase
0:02:36	the really stations a cost uh or its received signal
0:02:39	to the base station use user
0:02:41	we assume that the channels in the two faces
0:02:44	uh we C program
0:02:46	uh you know we introduce the interference free on screen
0:02:50	after the two face transmission
0:02:52	uh a the base station and all the users will receive that
0:02:55	or what a pink signals and all the downlink signals
0:02:58	for the base station
0:02:59	it's a knows the downing signals
0:03:01	it can remove this signal by
0:03:03	self interference cancellation
0:03:05	and uh what it has to do is to separate of that
0:03:08	a pick signals of from a different user
0:03:10	therefore we have our first the into free screen
0:03:13	which means that a
0:03:15	different
0:03:16	the a pink signals from a the users we not interfere with each other at the base station receiver
0:03:22	and for each you there
0:03:24	or or a pink and downlink signals from other users at in to user interference
0:03:28	we should be there are therefore we have a corresponding to intervenes free screen
0:03:33	which means that a pink signal and
0:03:35	oh you know
0:03:36	oh each user we will not interfere don't
0:03:38	we were not need for other user
0:03:41	well all these interference free "'cause" that come be satisfied
0:03:45	we can calculate this is system
0:03:46	i pink read and the only three that
0:03:48	by this equation
0:03:50	here the P V P R and the P U are respectively the transmit power of the base station really
0:03:55	station and each user
0:03:57	uh the one or more to pretty factor is due to the two phase transmission
0:04:02	on of the people meeting on that in a minute it as the first the term is the really station
0:04:05	amplified noise
0:04:07	and the second term is the noise at the receiver a base station and you use user
0:04:12	and they say the bidirectional sum rate
0:04:15	uh
0:04:16	to at the maximum bidirectional sum rate can be achieved a
0:04:19	by talking to lay optimising the
0:04:21	street transceiver zero matrix on there these
0:04:24	can these days that base station power constraint screen and it is a is the really station power constraint
0:04:29	how one
0:04:30	this hmmm joint optimization problem is too difficult for us to
0:04:34	so they
0:04:35	therefore in this work
0:04:36	we we start to the alternating optimization
0:04:40	two all for or from a span to mark
0:04:43	here is our procedure
0:04:44	first that we was then initial is for each of the really station to you were and the base station
0:04:49	transmitter and uh
0:04:51	based there is C we C one
0:04:52	then you each that we optimize one of them by fixing the other two
0:04:56	after each side
0:04:58	we will check is the sum rate higher since that
0:05:01	a lot of the last as that
0:05:02	if yes to be go to know i mean it's the loop
0:05:05	and in the pouring
0:05:06	we will respect so with this three sample problem
0:05:10	hmmm
0:05:11	first we so the
0:05:12	we optimize might the relay station ten you work by fixing the base station to work
0:05:17	and the
0:05:17	we
0:05:18	well P mike that that we are to maximise the bidirectional sum rate
0:05:21	hi were these days that not convex summary the maximization problem
0:05:25	and we use the concept that would be to in this paper a to so to so that kind of
0:05:30	problem
0:05:31	the basic idea of the with to pull
0:05:33	it's clean here
0:05:34	the that we are maximising the summary i one plus had two
0:05:38	and the say the a two paul a region
0:05:40	and each are here corresponds to a with to pull
0:05:43	if we maximise the summary
0:05:46	uh uh with a P will be to pull we will achieve a boundary point
0:05:49	on a two on the triple be treated
0:05:52	if we can find the optimal more with me to paul we achieve
0:05:56	the optimal boundary point which will result maximum sum rate
0:06:00	and that
0:06:01	you are what system their with to paul is defined by these white or and
0:06:05	these can
0:06:06	so the original problem can be the by finding the maximum summary
0:06:11	with given
0:06:12	bits to poll
0:06:13	and then finding the optimal with to collect or
0:06:16	here
0:06:17	yeah know we use the bisection method or the to search the optimal but the bait
0:06:20	and then we have to solve this problem
0:06:24	yeah
0:06:25	and that so that problem we use the thing approach in this paper
0:06:28	and we use the bisection method that to search the maximum summary
0:06:33	which satisfies all these constraint
0:06:35	we can't be
0:06:36	some are eyes and to test a wider it satisfies all that constraint if you yes we try not a
0:06:41	one
0:06:42	you if know which files more one
0:06:44	and the full
0:06:45	for a tree
0:06:46	summary
0:06:47	we comes so this problem to see why there eight
0:06:51	satisfies is five these constraints here or not
0:06:53	if the minimize
0:06:55	mm no here is the lower than one we state
0:06:58	the people not rise is feasible
0:07:00	and
0:07:01	this problem after some mathematical derivation
0:07:04	can be
0:07:05	we write into this these form they say is the standard
0:07:08	code that collect constrained the code that the core problem
0:07:11	uh
0:07:12	which can be rewritten we which can become what had to be uh
0:07:15	semi-definite a problem with a rank one constraint
0:07:18	and so we start to the
0:07:19	why to use the semi-definite to jen to solve
0:07:22	to so wait
0:07:23	the procedure from here to here can be found in the journal paper a by provides a
0:07:30	and then we and the base station transmitter
0:07:32	oh when the really station
0:07:34	just he is fixed
0:07:36	the base station trans
0:07:37	the base station transmitter meter are only affects the town read
0:07:40	therefore we
0:07:41	maximise the that only grid here
0:07:43	a a it is also not comics summary max
0:07:47	maximization problem
0:07:48	we can use the with to pull here
0:07:51	original problem can be
0:07:53	can be so by
0:07:55	finding the maximum downlink three
0:07:57	uh with the people will be to people and since search open more with to pull or
0:08:03	and
0:08:04	uh a to find is the maximum
0:08:06	well to find the maximal maximum only agreed with that be it to pull we can of this problem
0:08:11	and we also use the bisection method is two
0:08:14	find is a mark or more D which satisfies all these constraint
0:08:18	and it these screen
0:08:19	um forms a second order cone people read in
0:08:23	there for each feed but it the problem can be reading into a second order cone problem the costs
0:08:29	a a late we optimize the base station receiver was based a C what only a pink read of the
0:08:35	of all we only might the having lead here
0:08:38	and the from the a three expression here we found that
0:08:41	each user ping
0:08:43	read is only a function
0:08:45	of the
0:08:46	i score along with a of this matrix
0:08:48	therefore this problem can be decoupled into
0:08:51	the and use up problem
0:08:53	each problem optimize one column of this matrix
0:08:57	and this problem can be easily be right in into the form of a
0:09:00	really racial maximization problem
0:09:02	that so
0:09:04	hmmm finally i want to talk about the convergence of the alternating optimization
0:09:10	uh a to me
0:09:11	in that summary increase ease
0:09:13	by iterations
0:09:14	so the alternating optimization will surely come work
0:09:17	but
0:09:18	since the original problem is not common
0:09:21	therefore of the come to the result depends on the initial value
0:09:25	hmmm
0:09:26	we cannot guarantee a uh
0:09:28	baltimore optimal out but
0:09:29	we can perform alternating optimization with mode all different initial is and is and choose the best one which can
0:09:36	words to the
0:09:36	i is the maximum
0:09:38	uh uh which which can to the high the summary
0:09:41	and the by doing so we can increase the probability to achieve optimal
0:09:44	so
0:09:46	here is the some some simulation results first i want to show the
0:09:50	a bidirectional sum rate was as uh
0:09:53	each region number
0:09:54	uh the in this speaker the release a number at a number is that has full the base station and
0:10:00	the number and the use number instead
0:10:02	a set that's two
0:10:03	the
0:10:04	in the power of each user is no that's one and the
0:10:07	and those of the base station the it's it has set has two
0:10:11	this three blue curve
0:10:13	uh the convergence performance of you
0:10:16	by using the you need to let was as the uh
0:10:19	and is this again uh
0:10:21	zero for the a game and the signal i gay
0:10:23	and the
0:10:24	to read curve
0:10:25	uh the convergence performance by using a me use initial value
0:10:29	the lower lower use these only one initial values and
0:10:32	a problem right curve use
0:10:34	use it time a random initial values and the choose is the best one
0:10:38	we can see that by using different in usual is the come were to the result out E
0:10:43	and a by using multiple initial values and the choose the best one we can increase the performance
0:10:49	and is a single galatians we found that
0:10:51	by using more and time take initial values
0:10:54	the
0:10:55	performance gain or were these right curve
0:10:57	is my to or four we can take the
0:11:00	right curve as a new optimal solution and sorrow as a bit performance but file
0:11:06	and in this curve
0:11:07	in this speaker are we compare as a sum rate performance of different at time zero
0:11:11	it's
0:11:12	uh okay as the base station the you base station and a number and the use a number as that
0:11:17	has to
0:11:18	uh the how uh
0:11:20	note chance or a set has is
0:11:22	uh you know a the bidirectional sum rate words as the relay station on a number
0:11:26	because see that the
0:11:27	traditional zero forcing to game
0:11:29	performs performs bad out when the at than and when the
0:11:33	and than a number at the relay station
0:11:35	is
0:11:36	is a low
0:11:36	and it out performs the signal i'm and
0:11:40	when the when when the relay station and the number is a lot
0:11:43	and are well form our proposed to balance this again outperformed performs ball
0:11:48	and as the
0:11:48	but curve here
0:11:50	is the alternating an eighteen optimisation
0:11:52	we can see that the performance gap between the
0:11:55	alternating optimization and a lot from our proposed the balance gain is quite small
0:12:00	therefore by four by proposing this performance but of we can see that our form proposed a scheme is the
0:12:06	new optimal solution
0:12:08	yeah i'd like to conclude my talk
0:12:10	uh uh this work we employed of alternating optimization
0:12:13	these to design a base station and a station transceiver us
0:12:16	you know a two we will a system
0:12:19	the in at to we aim at maximizing the bidirectional sum rate on the interference free tree
0:12:25	and we use a multiple initial values to perform alternating my station and select like the best one
0:12:30	by doing so we can treat the probability to achieve the global optimal solution
0:12:35	and the and the of the a the formants can be taken as a performance a by back for the
0:12:39	can them
0:12:41	and that we found that performance gap between the bit to our form a proposed the it that's again and
0:12:46	alternating optimization is small
0:12:48	it indicates that the balance this game is then the optimal solution
0:12:52	these uh some reference you use the in these talk
0:12:55	uh in this paper as you can find of the signal and scheme
0:12:59	and that they is our uh and of the scheme
0:13:02	and they
0:13:03	uh uh you this paper as you can send is uh
0:13:06	the concept of a it to pull
0:13:08	and they they say a in this it provides a lot paper you can fans them
0:13:12	techniques the a convex optimization
0:13:14	that's source and Q
0:13:28	sorry um
0:13:29	so you you mentioned a your that the is for to really a your systems
0:13:34	but made it is really thing with a so
0:13:36	just or and can you comment on
0:13:38	just just to about so interference "'cause" it seems like it's going to
0:13:42	um um
0:13:43	drastically in the to roots so you gonna achieve
0:13:46	uh yeah no i you insist can we only consider a single cell scenario
0:13:51	and is the intra sell you and the so so there is no inter self interference
0:13:56	and i was interest out interference a handled by
0:13:59	these
0:14:00	by these
0:14:01	no interference a constraint
0:14:03	the you
0:14:04	a three constraint
0:14:06	i is that the we will
0:14:07	we we don't have the rip we there will be no interest that interference
0:14:12	but uh and they
0:14:13	that's a
0:14:14	yeah
0:14:21	what do you to just three constraints
0:14:24	so this just a a a a uh a
0:14:27	the shows use six
0:14:29	sorry what we have been to students G constraint well don't to just
0:14:33	the max most the rate
0:14:34	oh yeah why don't use you can do as you have two reasons for was
0:14:38	for was it for the constraint
0:14:39	the first one we can well oh
0:14:41	because there's that if the
0:14:43	i so nice highly nap
0:14:44	the therefore for
0:14:46	by doing by doing so the sum the we will be
0:14:49	uh the same as to what you does the same we be to is don't to call you pick can
0:14:53	see that the just to maximise the sum rate
0:14:56	right
0:14:56	and the otherwise that the back can see didn't that can screen
0:14:59	and uh
0:15:00	we will achieve them
0:15:02	good mathematical result
0:15:05	because the
0:15:07	summary here are we'll be where simple
0:15:22	yeah

TRANSCEIVER OPTIMIZATION FOR MULTI-USER MULTI-ANTENNA TWO-WAY RELAY CHANNELS

Multiuser and Network MIMO

Přednášející: Can Sun, Autoři: Can Sun, Chenyang Yang, Beihang University, China; Yonghui Li, Branka Vucetic, University of Sydney, Australia