The frequency scheduling improves the total throughput in the OFDM(A) system. This scheduling enables to select the best user in each frequency band so that all the frequency band are used for the best transmission in terms of downlink throughput. In FDD systems, especially the CSI (Channel State Information) feedback through the uplink channel is required to do this scheduling since channel reciprocity is not anticipated. This feedback also decreases the available data throughput of uplink channel. Thus, it seems true that we need to find out good tradeoff point between the gain of the scheduling and the penalty of the feedback. Thus, my question is whether we can find this point or not, so far? If yes, could any body tell me any reference about it! Thank you in advance. -- Best regards, James K. (txdiversity@hotmail.com) - Any remarks, proposal and/or indicator to text would be greatly respected. - Private opinions: These are not the opinions from my affiliation.
[OFDM] Frequency scheduling tradeoff
Started by ●November 23, 2003
Reply by ●November 24, 20032003-11-24
news:bpqdkq$i4j$1@news1.kornet.net...> The frequency scheduling improves the total throughput in the OFDM(A) > system.I attempt to outline the feedback methods for a frequency scheduling in FDD-OFDM(A) system to make clear what I am asking now: 1. Full (Whole) band CSI (Channel State Information) feedback 1.1 Encoding independently between each frequency band. E.g. [CSI for band1][CSI for band2]�� 1.2 Encoding dependently between all frequency bands. E.g. [CSI for all band1��L] 2. Partial (fractional) band CSI feedback 2.1 Encoding for flexible partial band by using index E.g. [index of band]&[CSI for indexed band] 2.2 Encoding for fixed partial band without using index E.g. [CSI for fixed band] Again, if you any opinions or references regarding to this issue, please let me know it. BR, -- James K. (txdiversity@hotmail.com) - Any remarks, proposal and/or indicator to text would be greatly respected. - Private opinions: These are not the opinions from my affiliation. "James K." <txdiversity@hotmail.com> wrote in message
Reply by ●November 25, 20032003-11-25
"James K." <txdiversity@hotmail.com> wrote in message news:bps7ol$57r$1@news1.kornet.net...> I attempt to outline the feedback methods for a frequency scheduling in > FDD-OFDM(A) system to make clear what I am asking now:To provide more insight into my question on the frequency scheduling for OFDM(A) system, I outline the achievable capacity of each case. In the formula, it is assumed that N is the number of frequency sub-bands, K is the # of users, SNR is the average signal-to-noise power ratio, and x(K) is 2K-order chi-square distributed random variable. Moreover, HN(K) is defined as Harmonic Number Function, which is HN(K) = 1 + 1/2 + ... + 1/K ~= log(K) + a. (a > 0)> 1. Full (Whole) band CSI (Channel State Information) feedback* Achievable capacity: C_full = N*log(1+x(K)*SNR/N) --> ~ N*log(1+HN(K)*SNR/N) (K>>1)> 1.1 Encoding independently between each frequency band > E.g. [CSI for band1][CSI for band2]--> Feedback rate: R_full <= N*r(1)> 1.2 Encoding dependently between all frequency bands > E.g. [CSI for all band1��L]> 2. Partial (fractional) band CSI feedback* Achievable capacity: C_part = N*log(1+x(K/N)*SNR/N) --> ~ N*log(1+HN(K/N)*SNR/N) (K>>1)> 2.1 Encoding for flexible partial band by using index > E.g. [index of band]&[CSI for indexed band]--> Feedback rate: R_part1 <= r(N)> 2.2 Encoding for fixed partial band without using index > E.g. [CSI for fixed band]--> Feedback rate: R_part2 <= r(1) where feedback rate r(n) == diff-entropy of 2n-order chi-sqr pdf, especially the more r(n) dec. ( the more n inc.). According to this capacity analysis, if SNR --> infinity, C_full becomes equal to C_part in probability. Therefore, it is easy to understand that in high SNR the capacity by partial feedback approaches to the capacity by full feedback.> Again, if you have any opinions or references regarding to this issue,please let> us know it to check wheather my understand is right.BR, -- James K. (txdiversity@hotmail.com) - Any remarks, proposal and/or indicator to text would be greatly respected. - Private opinions: These are not the opinions from my affiliation.
Reply by ●November 27, 20032003-11-27
"James K." <txdiversity@hotmail.com> wrote in message news:bpunte$654$1@news1.kornet.net... I edited some of previous typos (see lines starting with 'E>') to make this query looking prettier.> "James K." <txdiversity@hotmail.com> wrote in message> news:bps7ol$57r$1@news1.kornet.net...> > I attempt to outline the feedback methods for a frequency scheduling in> > FDD-OFDM(A) system to make clear what I am asking now:>> To provide more insight into my question on the frequency> scheduling for OFDM(A) system, I outline the achievable> capacity of each case. In the formula, it is assumed that> N is the number of frequency sub-bands, K is the> # of users, SNR is the average signal-to-noise power ratio,E> and x(K) is the maximum value selected from E> K EA 2nd-order chi-square distributed random variable. E> Moreover, HN(K) is defined as Harmonic Number Function, E> which is HN(K) = 1 + 1/2 + ... + 1/K ~= log(K) + a, E> e.g. a = 0.58 as K -> Infinity>> > 1. Full band CSI (Channel State Information) feedback> * Achievable capacity: C_full = N*log(1+x(K)*SNR/N)E> K>>1: --> ~N*log(1+HN(K)*SNR/N)> > 1.1 Encoding independently between each frequency band> > E.g. [CSI for band1][CSI for band2]E> * Feedback rate: R_full <= N*r(1)> > 1.2 Encoding dependently between all frequency bands> > E.g. [CSI for all band1��L]>> > 2. Partial (Sub) band CSI feedback> > 2.1 Encoding for flexible partial band by using index> > E.g. [index of band]&[CSI for indexed band]E> * Achievable capacity: C_part1 = ? E> * Feedback rate: R_part1 <= r(N) + log(N)> > 2.2 Encoding for fixed partial band no using index> > E.g. [CSI for fixed band]E> * Achievable capacity: C_part2 = N*log(1+x([K/N])*SNR/N) E> K>>1: --> ~N*log(1+HN([K/N])*SNR/N) E> * Feedback rate: R_part2 <= r(1)>E> where feedback rate r(n) = h(n) + d, h(n) is diff-entropy of E> the maximum value selected from N EA 2nd-order chi-sqr distributed E> r.v., d is the resolution used in channel quantization, E> [x] denotes the largest integer smaller than or equal to x, and E> ~x says approximately x, especially the more n inc, E> the more h(n) dec.>> According to this capacity analysis, if SNR --> infinity,E> C_full becomes equal to C_part in probability; thus. E> it is easy to understand that in high SNR the capacity E> by partial feedback approaches to the capacity by full E> feedback. On the other hand, if K is small, then E> the capacity of the partial band feedback system is E> less than the case K is large because of [x] effect.>> > Again, if you have any opinions or references regarding to this issue,> please let> > us know it to check wheather my understand is right.>-- Best regards, James K. (txdiversity@hotmail.com) - Any remarks, proposal and/or indicator to text would be greatly respected. - Private opinions: These are not the opinions from my affiliation.
Reply by ●November 27, 20032003-11-27
"James K." <txdiversity@hotmail.com> wrote in message
news:bpunte$654$1@news1.kornet.net...
I edited some of previous typos (see lines starting with 'E>') to make this
query looking prettier. I attempt to outline the feedback methods for a
frequency scheduling in FDD-OFDM(A) system to make clear what I am asking
now.
To provide more insight into my question on the frequency scheduling for
OFDM(A) system, I outline the achievable capacity of each case. In the
formula, it is assumed that N is the number of frequency sub-bands, K is the
# of users, SNR is the average signal-to-noise power ratio, <E start> and
x(K) is the maximum value selected from K EA 2nd-order chi-square
distributed random variable. Moreover, HN(K) is defined as Harmonic Number
Function, which is HN(K) = 1 + 1/2 + ... + 1/K ~= log(K) + a, e.g. a = 0.58
as K -> Infinity <E end>
1. Full band CSI (Channel State Information) feedback
* Achievable capacity: C_full = N*log(1+x(K)*SNR/N)
E> K>>1: --> ~N*log(1+HN(K)*SNR/N)
1.1 Encoding independently between each frequency band
E.g. [CSI for band1][CSI for band2]
E> * Feedback rate: R_full <= N*r(1)
1.2 Encoding dependently between all frequency bands
E.g. [CSI for all band1��L]
2. Partial (Sub) band CSI feedback
2.1 Encoding for flexible partial band by using index
E.g. [index of band]&[CSI for indexed band]
E> * Achievable capacity: C_part1 = ?
E> * Feedback rate: R_part1 <= r(N) + log(N)
2.2 Encoding for fixed partial band no using index
E.g. [CSI for fixed band]
E> * Achievable capacity: C_part2 = N*log(1+x([K/N])*SNR/N)
E> K>>1: --> ~N*log(1+HN([K/N])*SNR/N)
E> * Feedback rate: R_part2 <= r(1)
<E start> where feedback rate r(n) = h(n) + d, h(n) is diff-entropy of the
maximum value selected from N EA 2nd-order chi-sqr distributed r.v., d is
the resolution used in channel quantization, [x] denotes the largest integer
smaller than or equal to x, and ~x says approximately x, especially the more
n inc, the more h(n) dec.
According to this capacity analysis, if SNR --> infinity, C_full becomes
equal to C_part in probability; thus. it is easy to understand that in high
SNR the capacity by partial feedback approaches to the capacity by full
feedback. On the other hand, if K is small, then the capacity of the partial
band feedback system is less than the case K is large because of [x] effect.
<E end>
Again, if you have any opinions or references regarding to this issue,
please let me know it to check wheather my understand is right.
Best regards,
James K. (txdiversity@hotmail.com)
- Any remarks, proposal and/or indicator to text would be greatly respected.
- Private opinions: These are not the opinions from my affiliation.
Reply by ●November 28, 20032003-11-28
"James K." <txdiversity@hotmail.com> wrote in message
news:bq55jv$e73$1@news1.kornet.net...
I revised minor point on my question to represent formulars much closer to
the exact solution. Look at text, where 'E>' denotes edited text.
I edited some of previous typos to make this query looking prettier. I
attempt to outline the feedback methods for a frequency scheduling in
FDD-OFDM(A) system to make clear what I am asking now. To provide more
insight into my question on the frequency scheduling for OFDM(A) system, I
outline the achievable capacity of each case. In the formula, it is assumed
that N is the number of frequency sub-bands, K is the # of users, SNR is the
average signal-to-noise power ratio, and x(K) is the maximum value selected
from K EA 2nd-order chi-square distributed random variable. Moreover, HN(K)
is defined as Harmonic Number Function, which is HN(K) = 1 + 1/2 + ... + 1/K
~= log(K) + a, e.g. a = 0.58 as K -> Infinity.
1. Full band CSI (Channel State Information) feedback
* Capacity: C_full = N*log(1+x(K)*SNR/N),
K>>1: --> ~ N*log(1+HN(K)*SNR/N)
1.1 Encoding independently between each frequency band
E.g. [CSI for band1][CSI for band2]
* Feedback rate: R_full <= N*r(1)
1.2 Encoding dependently between all frequency bands
E.g. [CSI for all band1��L]
2. Partial (Sub) band CSI feedback
2.1 Encoding for flexible partial band by using index
E.g. [index of band]&[CSI for indexed band]
* Achievable capacity: C_part2 < C_part1(?) < C_full
* Feedback rate: R_part1 <= r(N) + log(N)
2.2 Encoding for fixed partial band no using index
E.g. [CSI for fixed band]
* Capacity: C_part2 = N1*log(1+x([K/N]+1)*SNR/N)
+ N2*log(1+x([K/N])*SNR/N),
K>>1: --> ~ N1*log(1+HN([K/N]+1)*SNR/N)
+ N2*log(1+HN([K/N])*SNR/N),
where N1 = K - N*[K/N], N2 = N - N1,
* Feedback rate: R_part2 <= r(1)
where feedback rate r(n) = h(n) + d, h(n) is diff-entropy of the maximum
value selected from N EA 2nd-order chi-sqr distributed r.v., d is the
resolution used in channel quantization, [x] denotes the largest integer
smaller than or equal to x, and ~x says approximately x, especially the more
n inc, the more h(n) dec.
According to this capacity analysis, if SNR --> infinity, C_full becomes
equal to C_part in probability; thus. it is easy to understand that in high
SNR the capacity by partial feedback approaches to the capacity by full
feedback. On the other hand, if K is small, then the capacity of the partial
band feedback system is less than the case K is large because of [x] effect.
Again, if you have any opinions or references regarding to this issue,
please let me know it to check wheather my understanding is correct.
--
Best regards,
James K. (txdiversity@hotmail.com, http://home.naver.com/txdiversity)
- Any remarks, proposal and/or indicator to text would be greatly respected.
- Private opinions: These are not the opinions from my affiliation.
Reply by ●November 29, 20032003-11-29
"James K." <txdiversity@hotmail.com> wrote in message
news:bq7o29$ise$1@news1.kornet.net...
[2003.11.30 01a:28]
I add one possible formulation for C_part1.
[2003.11.29 12a:30]
I revised minor point on my question to represent formulars much closer to
the exact solution.
[2003.11.28 01a:18]
I edited some of previous typos to make this query looking prettier. I
attempt to outline the feedback methods for a frequency scheduling in
FDD-OFDM(A) system to make clear what I am asking now.
[2003.11.25 02p:57]
To provide more insight into my question on the frequency scheduling for
OFDM(A) system, I outline the achievable capacity of each case.
[2003.11.23 06p:27]
The frequency scheduling improves the total throughput in the OFDM(A)
system. This scheduling enables to select the best user in each frequency
band so that all the frequency band are used for the best transmission in
terms of downlink throughput. In FDD systems, especially the CSI (Channel
State Information) feedback through the uplink channel is required to do
this scheduling since channel reciprocity is not anticipated. This feedback
also decreases the available data throughput of uplink channel. Thus, it
seems true that we need to find out good tradeoff point between the gain of
the scheduling and the penalty of the feedback. Thus, my question is whether
we can find this point or not, so far? If yes, could any body tell me any
reference about it! Thank you in advance.
In the formula, it is assumed that N is the number of frequency sub-bands, K
is the # of users, SNR is the average signal-to-noise power ratio, and x(K)
is r.v. having distribution of the maximum value selected from K EA
2nd-order chi-square distributed random variable. Moreover, HN(K) is defined
as Harmonic Number Function, which is HN(K) = 1 + 1/2 + ... + 1/K ~= log(K)
+ a, e.g. a = 0.58 as K -> Infinity.
1. Full band CSI (Channel State Information) feedback
* Capacity: C_full = N*Fc(x(K)),
K>>1: --> ~ N*Fc(HN(K))
1.1 Encoding independently between each frequency band
E.g. [CSI for band1][CSI for band2]
* Feedback rate: R_full1 <= N*r(1)
1.2 Encoding dependently between all frequency bands
E.g. [CSI for all band1��L]
* Feedback rate: R_full2 <= R_full1
2. Partial (Sub) band CSI feedback
2.1 Encoding for flexible partial band by using index
E.g. [index of band]&[CSI for indexed band]
* Achievable capacity: C_part2 < C_part1(?) < C_full
(Prop1) C_part1 (N=2 case) -->
Sum(Comb(K,k)*(1/2)^K*(Fc(x(k))+Fc(x(K-k))),k=0..K)
* Feedback rate: R_part1 <= r(N) + log(N)
2.2 Encoding for fixed partial band no using index
E.g. [CSI for fixed band]
* Capacity: C_part2 = N1*Fc(x([K/N]+1)*SNR/N)
+ N2*Fc(x([K/N])*SNR/N),
K>>1: --> ~ N1*Fc(HN([K/N]+1)*SNR/N)
+ N2*Fc(HN([K/N])*SNR/N),
where N1 = K - N*[K/N], N2 = N - N1,
* Feedback rate: R_part2 <= r(1)
where Fc(x) = log(1+x*SNR/N), feedback rate r(n) = h(n) + d, h(n) is
diff-entropy of the maximum value selected from N EA 2nd-order chi-sqr
distributed r.v., d is the resolution used in channel quantization, [x]
denotes the largest integer smaller than or equal to x, and ~x says
approximately x, especially the more n inc, the more h(n) dec.
According to this capacity analysis, if SNR --> infinity, C_full becomes
equal to C_part in probability; thus. it is easy to understand that in high
SNR the capacity by partial feedback approaches to the capacity by full
feedback. On the other hand, if K is small, then the capacity of the partial
band feedback system is less than the case K is large because of [x] effect.
Again, if you have any opinions or references regarding to this issue,
please let me know it to check wheather my understanding is correct.
--
Best regards,
James K. (txdiversity@hotmail.com, http://home.naver.com/txdiversity)
- Any remarks, proposal and/or indicator to text would be greatly respected.
- Private opinions: These are not the opinions from my affiliation.
Reply by ●November 29, 20032003-11-29
"James K." <txdiversity@hotmail.com> wrote in message news:bqahq6$6am$1@news1.kornet.net... [2003.11.30 03a:02] Additionally, I correct small part but important on version of [2003.11.30 01a:28].> "James K." <txdiversity@hotmail.com> wrote in message > news:bq7o29$ise$1@news1.kornet.net... > > [2003.11.30 01a:28] > I add one possible formulation for C_part1. > > [2003.11.29 12a:30] > I revised minor point on my question to represent formulars much closer to > the exact solution. > > [2003.11.28 01a:18] > I edited some of previous typos to make this query looking prettier. I > attempt to outline the feedback methods for a frequency scheduling in > FDD-OFDM(A) system to make clear what I am asking now. > > [2003.11.25 02p:57] > To provide more insight into my question on the frequency scheduling for > OFDM(A) system, I outline the achievable capacity of each case. > > [2003.11.23 06p:27] > The frequency scheduling improves the total throughput in the OFDM(A) > system. This scheduling enables to select the best user in each frequency > band so that all the frequency band are used for the best transmission in > terms of downlink throughput. In FDD systems, especially the CSI (Channel > State Information) feedback through the uplink channel is required to do > this scheduling since channel reciprocity is not anticipated. Thisfeedback> also decreases the available data throughput of uplink channel. Thus, it > seems true that we need to find out good tradeoff point between the gainof> the scheduling and the penalty of the feedback. Thus, my question iswhether> we can find this point or not, so far? If yes, could any body tell me any > reference about it! Thank you in advance. > > In the formula, it is assumed that N is the number of frequency sub-bands,K> is the # of users, SNR is the average signal-to-noise power ratio, andx(K)> is r.v. having distribution of the maximum value selected from K EA > 2nd-order chi-square distributed random variable. Moreover, HN(K) isdefined> as Harmonic Number Function, which is HN(K) = 1 + 1/2 + ... + 1/K ~=log(K)> + a, e.g. a = 0.58 as K -> Infinity. > > 1. Full band CSI (Channel State Information) feedback > * Capacity: C_full = N*Fc(x(K)), > K>>1: --> ~ N*Fc(HN(K)) > 1.1 Encoding independently between each frequency band > E.g. [CSI for band1][CSI for band2] > * Feedback rate: R_full1 <= N*r(1) > 1.2 Encoding dependently between all frequency bands > E.g. [CSI for all band1��L] > * Feedback rate: R_full2 <= R_full1 > > 2. Partial (Sub) band CSI feedback > 2.1 Encoding for flexible partial band by using index > E.g. [index of band]&[CSI for indexed band] > * Achievable capacity: C_part2 < C_part1(?) < C_full > (Prop1) C_part1 (N=2 case) -->*E* Sum(Comb(K,k)*(1/2)^K*(Fc(x(2*k))+Fc(x(2*(K-k)))),k=0..K)> * Feedback rate: R_part1 <= r(N) + log(N) > 2.2 Encoding for fixed partial band no using index > E.g. [CSI for fixed band] > * Capacity: C_part2 = N1*Fc(x([K/N]+1)*SNR/N) > + N2*Fc(x([K/N])*SNR/N), > K>>1: --> ~ N1*Fc(HN([K/N]+1)*SNR/N) > + N2*Fc(HN([K/N])*SNR/N), > where N1 = K - N*[K/N], N2 = N - N1, > * Feedback rate: R_part2 <= r(1) > > where Fc(x) = log(1+x*SNR/N), feedback rate r(n) = h(n) + d, h(n) is > diff-entropy of the maximum value selected from N EA 2nd-order chi-sqr > distributed r.v., d is the resolution used in channel quantization, [x] > denotes the largest integer smaller than or equal to x, and ~x says > approximately x, especially the more n inc, the more h(n) dec. > > According to this capacity analysis, if SNR --> infinity, C_full becomes > equal to C_part in probability; thus. it is easy to understand that inhigh> SNR the capacity by partial feedback approaches to the capacity by full > feedback. On the other hand, if K is small, then the capacity of thepartial> band feedback system is less than the case K is large because of [x]effect.> Again, if you have any opinions or references regarding to this issue, > please let me know it to check wheather my understanding is correct. > > -- > Best regards, > James K. (txdiversity@hotmail.com, http://home.naver.com/txdiversity) > - Any remarks, proposal and/or indicator to text would be greatlyrespected.> - Private opinions: These are not the opinions from my affiliation.-- Best regards, James K. (txdiversity@hotmail.com, http://home.naver.com/txdiversity) - Any remarks, proposal and/or indicator to text would be greatly respected. - Private opinions: These are not the opinions from my affiliation.
Reply by ●December 11, 20032003-12-11
This is a multi-part message in MIME format. ------=_NextPart_000_002F_01C3C024.C4221730 Content-Type: text/plain; charset="ks_c_5601-1987" Content-Transfer-Encoding: quoted-printable "James K." <txdiversity@hotmail.com> wrote in message = news:bqakic$73r$1@news1.kornet.net... So far, we have discussed two different cases: one is the full band CSI = feedback, and the other is the partial band CSI feedback.=20> > "James K." <txdiversity@hotmail.com> wrote in message > > news:bq7o29$ise$1@news1.kornet.net... > > > > 1. Full band CSI (Channel State Information) feedback > > > > 2. Partial (Sub) band CSI feedbackNow let's discuss about, say, the conventional method. This case is = added to give a reference for other two cases. In this new scheme, it = is provided that the scheduling is done only in time but not in = frequency. Thus, whole frequencies are allocated for one user who has = best capacity average over all frequencies as C[k] =3D Fc[x[1]] + Fc[x[2]] + ... Fc[x[N]] ,where x[n] is the 2nd order chi-squre distributed r.v for n-th user. = If there is more than 2 frequency bands, we can approximate the = distribution of C as a Gaussian with extreamly high accuracy[1].=20 Then, we need to find maximum user who provides best capacity at the = moment. Let us assume that there are only two users, then the = distribution of Max[C[1], C[2]] is easily founded by the simple = manupulation of Gaussian pdf. In practice, if C[k] is i.i.d and = ~N[m,s^2], applying [2] yields Max[C[1],C[2]] ~ N[m+s/Sqrt(Pi),(-1+Pi)*s^2/Pi]=20 In breief, the new scheme has been shown that it is considering the = scheduling only in time domain but not in frequency domain for the = referece of other two cases. References [1] ... <By using the central limit theorem> [2] miller (amiller@bigpond.net.au), "[Re-post] Pdf of larger one in two = Gaussian r.v?", sci.stat.math in newsgroup, Date: 2003-12-02 22:51:09 = PST=20 --=20 Best regards, James K. (txdiversity@hotmail.com) - Any remarks, proposal and/or indicator to text would be greatly = respected. - Private opinions: These are not the opinions from my affiliation. [HOME] http://home.naver.com/txdiversity ------=_NextPart_000_002F_01C3C024.C4221730 Content-Type: text/html; charset="ks_c_5601-1987" Content-Transfer-Encoding: quoted-printable <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <HTML><HEAD> <META http-equiv=3DContent-Type content=3D"text/html; = charset=3Dks_c_5601-1987"> <META content=3D"MSHTML 6.00.2800.1276" name=3DGENERATOR> <STYLE></STYLE> </HEAD> <BODY> <DIV><FONT face=3DArial size=3D2>"James K." <</FONT><A=20 href=3D"mailto:txdiversity@hotmail.com"><FONT face=3DArial=20 size=3D2>txdiversity@hotmail.com</FONT></A><FONT face=3DArial = size=3D2>> wrote in=20 message </FONT><A href=3D"news:bqakic$73r$1@news1.kornet.net"><FONT = face=3DArial=20 size=3D2>news:bqakic$73r$1@news1.kornet.net</FONT></A><FONT face=3DArial = size=3D2>...</FONT></DIV> <DIV><FONT face=3DArial size=3D2></FONT> </DIV> <DIV> <DIV><FONT face=3DArial size=3D2>So far, we have discussed two different = cases: one=20 is the full band CSI feedback, and the other is the partial band CSI = feedback.=20 </FONT></DIV> <DIV><FONT face=3DArial size=3D2></FONT> </DIV></DIV> <DIV><FONT face=3DArial size=3D2>> > "James K." <</FONT><A=20 href=3D"mailto:txdiversity@hotmail.com"><FONT face=3DArial=20 size=3D2>txdiversity@hotmail.com</FONT></A><FONT face=3DArial = size=3D2>> wrote in=20 message<BR>> > </FONT><A = href=3D"news:bq7o29$ise$1@news1.kornet.net"><FONT=20 face=3DArial size=3D2>news:bq7o29$ise$1@news1.kornet.net</FONT></A><FONT = face=3DArial=20 size=3D2>...<BR>> ><BR>> > 1. Full band CSI (Channel State=20 Information) feedback<BR>> ><BR>> > 2. Partial (Sub) band = CSI=20 feedback<BR></FONT></DIV> <DIV> <DIV><FONT face=3DArial size=3D2>Now let's discuss about, say, the=20 conventional method. This case is added to give a=20 reference for other two cases. </FONT><FONT face=3DArial = size=3D2>In=20 this new scheme, it is provided that the scheduling is = done only=20 in time but not in frequency. Thus, whole frequencies are=20 allocated for one user who has best capacity average over = all=20 frequencies as</FONT></DIV> <DIV><FONT face=3DArial size=3D2></FONT> </DIV> <DIV><FONT face=3DArial size=3D2> C[k] =3D Fc[x[1]] + = Fc[x[2]] + ...=20 Fc[x[N]]</FONT></DIV> <DIV><FONT face=3DArial size=3D2></FONT> </DIV> <DIV><FONT face=3DArial size=3D2>,where x[n] is the 2nd order = chi-squre=20 distributed r.v for n-th user. If there is more than 2 frequency = bands, we=20 can approximate the distribution of C as a Gaussian = with extreamly=20 high accuracy[1]. </FONT></DIV> <DIV><FONT face=3DArial size=3D2></FONT> </DIV> <DIV><FONT face=3DArial size=3D2>Then, we need to find maximum user who = provides=20 best capacity at the moment. Let us assume that there are only two = users, then=20 the distribution of Max[C[1], C[2]] is easily founded by the simple = manupulation=20 of Gaussian pdf. In practice, if C[k] is i.i.d and=20 ~N[m,s^2], applying [2] yields</FONT></DIV> <DIV><FONT face=3DArial size=3D2></FONT> </DIV> <DIV><FONT face=3DArial size=3D2> Max[C[1],C[2]] ~=20 N[m+s/Sqrt(Pi),(-1+Pi)*s^2/Pi] </FONT></DIV> <DIV><FONT face=3DArial size=3D2></FONT> </DIV> <DIV><FONT face=3DArial size=3D2>In breief, the new scheme has = been shown=20 that it is considering the scheduling only in time domain but not = in=20 frequency domain for the referece of other two cases.</FONT></DIV> <DIV><FONT face=3DArial size=3D2></FONT> </DIV> <DIV><FONT face=3DArial size=3D2>References</FONT></DIV> <DIV><FONT face=3DArial size=3D2>[1] ... <By using the central = limit=20 theorem></FONT></DIV> <DIV><FONT face=3DArial size=3D2>[2] </FONT><A=20 href=3D"http://groups.google.com/groups?hl=3Den&lr=3D&ie=3DUTF-8&= amp;oe=3DUTF-8&edition=3Dus&q=3Dauthor:amiller%40bigpond.net.au+"= =20 target=3D_top><FONT face=3DArial size=3D2>miller</FONT></A><FONT = face=3DArial size=3D2>=20 (</FONT><A href=3D"mailto:amiller@bigpond.net.au"><FONT face=3DArial=20 size=3D2>amiller@bigpond.net.au</FONT></A><FONT face=3DArial size=3D2>), = "[Re-post]=20 Pdf of larger one in two Gaussian r.v?", </FONT><A=20 href=3D"http://groups.google.com/groups?hl=3Den&lr=3D&ie=3DUTF-8&= amp;oe=3DUTF-8&edition=3Dus&group=3Dsci.stat.math"=20 target=3D_top><FONT face=3DArial size=3D2>sci.stat.math</FONT></A><FONT = face=3DArial=20 size=3D2> in newsgroup, </FONT><FONT face=3DArial size=3D2>Date: = 2003-12-02=20 22:51:09 PST <BR></FONT></DIV></DIV> <DIV><FONT face=3DArial size=3D2>-- <BR>Best regards,<BR>James K. = (</FONT><A=20 href=3D"mailto:txdiversity@hotmail.com"><FONT face=3DArial=20 size=3D2>txdiversity@hotmail.com</FONT></A><FONT face=3DArial = size=3D2>)<BR>- Any=20 remarks, proposal and/or indicator to text would be greatly = respected.<BR>-=20 Private opinions: These are not the opinions from my=20 affiliation.<BR><STRONG><FONT=20 color=3D#ff0000>[HOME]</FONT></STRONG> </FONT><A=20 href=3D"http://home.naver.com/txdiversity"><FONT face=3DArial=20 size=3D2>http://home.naver.com/txdiversity</FONT></A></DIV></BODY></HTML>= ------=_NextPart_000_002F_01C3C024.C4221730--
