Homework 4
Problem Capacty wth CSI only at Recever: C = log det I + E )) s HH H N M T R SS = I) SVD of the Channel Matrx: H = UΣV H = [U 1 U ] [ Σr ] [V 1 V ] H Capacty wth CSI at both transmtter and Recever: ) r C = log 1 + σ E s γ opt N M T Waterfllng: γ opt = X) + = r γ opt = M T µ M ) T N E S σ + { X f X f X < Start wth teraton counter p=1 Calculate the constant µ µ = M T r p + 1 [ 1 + N E S r p+1 ] 1 Usng µ calculate the power n the th sub-channel from, γ = µ M ) T N E S σ = 1, r p + 1) σ If the energy n the channel wth lowest gan s negatve, dscard t by settng γ opt r p+1 = Repeat the procedure by settng, p = p + 1 The optmal power allocaton strategy s found when all the allocated power s non-negatve 1. ρ = 1 db = E s N = 1 H= [ 1 1 1 1 ] = [ 1 1 1 1 ] [ ] [ 1 1 ] H 3
σ 1 =, σ = Snce both the sngular values are equal, we can allocate equal power to the two modes We can see ths even from the waterfllng procedure Start wth teraton counter p=1 Calculate the constant µ µ = M T r p + 1 [ 1 + N E S r p+1 ] [ 1 = 1 + 1 1 σ ] 1 σ Usng µ calculate the power n the t h sub-channel from, γ = µ M ) T N E S σ = 1, r p + 1) 11 γ 1 = 1 ) = 1 = γ Snce both γ 1, γ 1 are not negatve,γ opt 1 = 1, γ opt = 1 = 11 1 a) Capacty wth CSI at both transmtter and Recever: ) ) r C = log 1 + σ E s γ opt = log 1 + σ 1γ opt = log N M T 11) = 6.9189 b) Capacty wth CSI only at Recever: C = log det I + E )) s HH H = 6.9189 N M T Comment: Snce our optmum power allocaton strategy wth CSI at transmtter gave R ss = I whch s what we use for CSI unknown case), channel knowledge at the transmtter doesn t help n ths case.. H= [ 1 1 1 1 ] = σ 1 = 4, σ = [ 1 1 1 1 ] [ ] [ 1 1 1 1 H s rank defcent and so t has only one sngular value. We should use only one mode and put all power nto t. Waterfllng procedure s not requred n ths case. a) Capacty wth CSI at both transmtter and Recever: ) ) r C = log 1 + σ E s γ opt = log 1 + σ 1γ opt = log N M T 41) = 5.3576 ] H 4
b) Capacty wth CSI only at Recever: C = log det I + E )) s HH H = 4.393 N M T Comments: Capacty values wth or wthout CSI at transmtter) are lower for ths channel compared to the prevous channel, ths can be attrbuted to the rank defcent nature of ths channel. The capacty values wth and wthout CSI are dfferent n ths case, For the prevous channel they are same). Ths s due to the fact that we avoded puttng energy nto the mode that s n the nullspace, whch we can t do f there s no CSI. As a consequence, the capacty wthout CSI s less than wth CSI. Problem 3 15 Capacty Vs. SNR 1 Capacty b/s/hz) 5 5 1 15 5 Fgure 1: Capacty wth dfferent antenna confguratons At hgher SNR s, Capacty ncreases lnearly wth the rank of channel matrx. The performance wth s better than. Ths s due to the array gan seen n case. Better performance of can be explaned by the addtonal degrees of freedom avalable. All these plots assume that transmtter doesn t have CSI. 5
Capacty wth CSI only at Recever: C = log det I + ρ )) HH H M T let, r = mn {M T, M R } Usng a low SNR approxmaton, outage formulaton s gven by, { } ρ P H F C = x M T { H P F C } = x M T M R ρm R Correspondngly for, { P h C } = x ρ Based on the tals of the dstrbutons of H F M T M R and h we can see that to have the same outage x), we need to ntegrate across more regon n case compared to. Ths leads to the followng nequalty, C C M R The above nequalty s notced to be vald from smulatons also. A smlar analyss can be carred out for hgh SNR regme, usng the followng approxmaton. C = log det I + ρ )) HH H M T Some observatons from plots: = r log ρ + r log λ For small ɛ s, and at hgh SNR s, the outage capacty rato between and s more than the rato at hgher outage values. Whch means for smaller outage values havng more antennas at the transmtter or recever) s more meanngful than at hgher ɛ s. At moderate to hgh ɛ s, outage capacty follows the pattern of ergodc capacty, whch s a lnear growth n mn {M T, M R } At very hgh ɛ s 99%), outage capacty has nterestng behavor. At moderate to hgh SNR s lnear growth n mn {M T, M R } s possble. But at very low SNR s, performs better than all other schemes. has the worst performance of all, performs slghtly better than. Ths can be explaned n terms of power dstrbuton at the transmtter. At low SNR s t s better to use only one channel, snce we don t have channel knowledge f we put energy n both modes the chance of t beng n outage s pretty hgh. Ths could be one reason for better performance of compared to. performs slghtly better than because of the recevng antennas. P. 187-189 Tse and Vshwanath) have an analytcal treatment of outage capacty for, and cases. 6
Ergodc Capacty: MATLAB code - Slght modfcaton for outage code % mt, Number of transmttng antennas % mr, Number of transmttng antennas % ITER, number of trals % SNRdB, Range of % C, varable for capacty of a system % C, varable for capacty of a system % C, varable for capacty of a system % C, varable for capacty of a system % h, random channel for wth zero mean unt varance) % h, random channel for % h, random channel for % h, random channel for clc; close all; clear all; mt = ; mr = ; ITER = 1; SNRdB = [:5]; SNR = 1.ˆSNRdB/1); C = zeros1,lengthsnr)); C = zeros1,lengthsnr)); C = zeros1,lengthsnr)); C = zeros1,lengthsnr)); for te = 1:ITER h = randn +j*randn)/sqrt); h = randnmr,1)+j*randnmr,1))/sqrt); h = randn1,mt)+j*randn1,mt))/sqrt); h = randnmr,mt)+j*randnmr,mt))/sqrt); for K = 1:lengthSNR) C K) = C K) + log1+ SNRK)*normh )ˆ); C K) = C K) + log1+ SNRK)*normh )ˆ); C K) = C K) + log1+ SNRK)*normh )ˆ/mT); C K) = C K) + logabsdeteyemr)+snrk)*h *h /mt))); end end C = C /ITER; C = C /ITER; C = C /ITER; C = C /ITER; plotsnrdb,c, r -.,SNRdB,C, b - o,snrdb,c, m,snrdb,c, k - * ) legend,,,,) xlabel ) ylabel Capacty b/s/hz) ) ttle Capacty Vs. SNR ) grd; 7
1 9 8 7 1% Outage Capacty Capacty b/s/hz) 6 5 4 3 1 5 1 15 5 1 1 1 % Outage Capacty Capacty b/s/hz) 8 6 4 5 1 15 5 18 16 99 % Outage Capacty Capacty b/s/hz) 14 1 1 8 6 4 5 1 15 5 Fgure : Outage Capacty: Increases lnearly wth the rank of channel matrx at hgh SNR s 8
.6.5 1% Outage Capacty at low SNR s Capacty b/s/hz).4.3..1 18 16 14 1 1.14.1.1 1 % Outage Capacty at Low SNR s Capacty b/s/hz).8.6.4. 18 16 14 1 1.8.7.6 99 % Outage Capacty at low SNR s Capacty b/s/hz).5.4.3..1 18 16 14 1 1 Fgure 3: Outage Capacty at low SNR s 9