hw3_2 - 3'7 8 Calls are not lost due to weak signal...

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Unformatted text preview: 3'7 8.) Calls are not lost due to weak signal condition during handoff if: distance traveled during handofl _ drm'n — dHo > mobile speed 1) ._ 4-5 seconds (2) * dmin => received power at 851 reaches Pm,"-fl P,,,,,.-,. = —2910gm(dm-,,) => elm-n = IO'P""““/29 = 1083 m (3) * dHo =9 received power at 351 reaches Pngo Pr,HO = ~2910g10(dgo) => dHO-= 10-PHo/t9 (4) Using (2), . . 1083 — lo‘PrJIQ/QQ __ > ' _ 22-22(171/5) _ 4 5 SECOIldS (5) PEG 2 —86.8 dBm (6) Thus, A = PnHO - Pam => A 2 1.2dB. (7) b) If we set A too large, several unnecessary handofi’s will be requested and performed, increasing the signaling traffic between the base stations and mobile switching center (MSG). On the other hand, if A is too small, that is, 19,, H0 is only slightly greater than Em", there will not be enough time to complete the handoff (especially for high speed mobiles), and calls may be lost due, to weak signal condition. \ S ’7 (a) zeré, r = (f?) >3/-62$ =>/v>n =>M=/Z (25) 20:2, 3?“ = N5?) >%/-62§ => ”>527“; N=7 S_ H (6) 20:1, T‘ N237) >§x~gzg => /\/>§-35 =>,v={* Fm meme! (6), We Com See Mar 6&7 535%”? (am imam 642 6'67”“? fly a fmfm of /2/§ , 07’4 For {40" Seam/E7 , mid 71mm V: may /2/7 , w ’27’4‘ . 7/41an , We Mom W 60° Seen-777. 2/7-, 2 /a) 2% My: _ H H W — 7100 CManne/s H 4 00 a I’m/76K 4 3175 = lw (”camels/cell c in} (b) 90% 070100 57/4an = 90 Er/anjs W 40“ (/44, = U/D-I) => Ll= 900 users E (a) 0&4, 90,; J 4400 => 0.05 7620,” amp/1 (,7? 3.4,) I 37, 6’05 (d) EMA seabr has 33.3 Mame/S /- 505 = 3% £0?%r:£h => 5!: 25 Erbrgs/sea/Dr l 25 = u Au $345,) H H ' => U= 250 x3 sealers H H U ‘ 750 users (a) 2500 km: ‘ H ‘ 5 km; . Hi} (4") 500 calls =~7" 5001750 uSer/wa : 375,090 users = 5170 cells a" 500 x 900 user%c//= mpaoms ‘98 7m 5] W fame MW W 572 eater/7943 3'?’ fm WW— 0&me Welt/ms “to 60" Seemed WW 7%}: W 6/1 04m; PW seem : .ng :7-5. 6mm Pr[émy]= 1;! , fmm 2%: 5% B aébz‘n‘éuzm we r/Lawe #9 total oflw/ [rec/ft 272W? fie/Seder AiH My: For #5 Mafl/Awr, fl: 2 minute/(ad, fie méerojrCa/{S Wt @206 Seat/i .56ch (M Adm/[e f6? flmr is “H _ :/25 users 9 66% 57mg? : 6X/23 = 758 0%9/5, f9???“ W’f/EDZVI. :) ”$55 7:29 first/ck”? (sigma? : /_;§% :07? :44% a 3'12 (EIRP = 32 watts, cell radius = 10 km. 005 is 5%, blocked calls cleared. H = 2 minutes, andu = 2 calls pr hour. Assume cell will be split into 4 cells.) a) What is the current capacity of the “Radio Knob" cell? Using the fimctions defined in problem 2.7 2 u. 2 . 60 Au. 0.067 Erlaugs P:=.05 Probability of blocked calls C2=57 Assume N=7 cell, AMPS Az=40 Initial guess AT(P,C):=root(GOS(A,C)-P,A) Solve iteratively for total traffic AT(P,C)=51.528 Erlangs A T (P, C)- A b) What is the radius and transmit power of the new cells? Number of users is U:= U=772.921 or 772 users II Since the 4 new cells must cover the area of the old cell, the radius of the new cells must be M, where R is the radius of the old cell. Then the area covered by the new cells is R R2 2 41!:[3] :=41t[-Z—] =nR2 which equals the area of the original cell 1? 3,12 Cont '61 To maintain the same SNR, the power at the edge of the new cells must equal the power at the edge of the original cell or R -4 ngz=PW P,R‘:=P2[7] and P2-= 1% where P, and P2 are the powers of the base station in the old and new cells respectively. If P1‘32 watts, then P2=2 watts. c) How many channels are needed in the new cells to maintain frequency reuse stability in the system? C:=57 Each new cell gets the number of channels of the original cell once '" the cell splitting process is complete. d) It trafiic is uniformly distributed, what is the new traffic carried by each new cell? Will the probability of blocking in these new cells be below 0.1% after the split? 772 FT U-l93 users per new cell Az=U-Au A= 12.867 Erlangs GOS(12.87,57)=0 The probability of blocking is less than .1% BfSince users are uniformly distributed over the area, each cell in the cluster is assigned the same number of channels: M NC = —, (8) where N NC = number of channels per cell M = number of channels available in the system (300 channels) N = cluster size ' (9) Given the number of channels per cell and the designed blocking probability Pb = 1%, we can compute the maximum carried traflic per cell in Erlang (Cc) using the Erlang B formula Cc = Erlang(Nc, Pb), (10) and the maximum carried traffic in the system C: C 2 Cc x 84 (11) Since each user offers a traffic of 0.04 Erlangs, the maximum number of users supported by the system is C 6.0—4 Table 2: Number of channels per cell (NC), carried trafiic per cell (Cc), total carried traffic in the system (C), and maximum number of users in the system (NU), for cluster sizes N= 4, 7 and 12. Blocking probability 1% Cluster channels per carried traffic total carried number of size N cell 7(NC) per cell (NC) traffic 0 users NU 60. 73 Erl 5101. 09 Erl 127527 2584 81 Erl 64620 1354. 49 E11 33862 Nu = (12) j H1 345 Cent’a’ ,9? c : a» mug , AW : ML 5sz , we Ame Fri—M >0] :lo‘og ‘ ‘ 2 Prbtaéj > awed = Pideéj >0} 9:71; (C ‘AtMD‘Q‘SeC/H] =‘ M“ ex? [— (fig/4)X&ofl05]50~°/7 M AW 3’ 54175! M W- PYMcész] =‘ 0-07 => PJdéjzma] = MFA exfl'Mo-i/Whflfikofi/S (d) Fm (c) we (M See 5461‘ M2 fwéaéx‘é‘tj W a 644 1 MI! ée delayed! jg, mm; 920% 257 a (vital! Maya” {firm as {es-s Zrflm 01/1 fir m “E aéfi‘e’enz 0mm Am}. 74% a Zm‘ cad “37;; 7mm fajem War Man a system 744: $7795 #0ch Ms. M73 W W WW 5%! 1 1131 i] 5 few from Mafia [mwm‘flgr LS P2: @173“: PM “57'0"“ do $.01;st Z? 749 (War off/{e WM (WW (6665‘ ‘ 2’ ks 27/»? my‘w ran/1M, I2; m3 we , pgmw) ”:7 1 0/0 =IM, IVE, MM we flat/e / 1m <-/ood,gm 2) r>4704 m For 4 6%! W miter/1, #:4, Wu? Ame fligfi mm of a cell (Won): 1 r23: — (047003‘ 0574M numfiero of ms 2?; «(Bill/{z am ofacell KW @1552} = 0524x7000 :‘5/5/ mars Q A = M419: mix 3’27! = 854 541,75 GlVen [= 70 fa». 6%? C Mar-f we Awe fig jaméaé/ff W a Cad Willéed’Z/ajea/ ’ Pr[de(aj>°] :‘05 19 PrIa/e/aj >0IOSEC] 3/)r[5{e/437r >0] Pr[a[aéj>aa/de(zy] — 0-5 X 97> [— (?0- ~85!) w/o’v] ._ "40/35 9/31 See seam» 3.7-1. Iffl‘i Pz‘ZIfZZ=—:fl 3.3 5E ‘ A A ‘4. ,_ ___fl,, 7, r am a) Mininum SIR In order to compute the minimun SIR at the mobile, we need to determine the number of interfering base stations in each possible configuration, which can be ' done by inspecting Figuresl and Z. Table 1 shows the number of interfering base stations in the first tier, when 3'sectors (BW = 120°) and 6 sectors (BW = 60°) \are used, for cluster sizes N = 3 and 4. ; Using expression (1), we determine the minimum SIR (approximation) in each l configuration (path loss exponent n = 4). Results are shown in Table 2. ‘ Therefore, cluster size N = 3 cannot be used, since the minimum SIR achieved 5‘ is below SIR = 18.7 dB. On the other hand, both configurations using cluster l size N = 4 are feasible, regarding co-channel interference (assuming that a l. difference of 0.1 dB is negligible). b) Maximum carried trafiic per cell Let us now computer the carried traffic per cell, when sectoring is used. As we ‘ * know, each sector is assigned a subset of the set of channels assigned to the cell. ; l 2‘ For example, for cluster size N = 3, each cell is assigned 300/3 = 100 channels. If six sectors are employed, each sector is assigned 100/6 z 16 channels. Using Erlang B formula, we find that each sector carries a maximum traffic of 9.83 Erlangs at a blocking probability of 0.02. Therefore, the maximum trafic carried by each cell is 9.83 x 6 = 58.97 Erlangs. Repeating this procedure, we can ‘ ‘ compute the maximum carried traffic per cell for other beamwidths and cluster ” sizes Table 3 presents the results. ~ Table 1: Number -of interfering base stations in the first tier (in) when 3 sectors (BW = 120°) and 6 sectors (BW = 60°)are used. BW =60" BW: 120° N ”I I I 2 i‘ l Table 2: Minimum SIR achieved when sectoring is used, for cluster size: N = 3 and 4. N BW = 60° BW = 120° i 3 16.1 dB 14.3 dB ‘ J( 4 21.6 dB 18.6 dB l i 1 Table 3: Maximum carried traffic per cell (in Erlangs) when sectoring is used, for i “ cluster sizes N = 3 and 4. 300 channels available in the system, Pb = 2% BW = 60° BW = 120° 58.97 73.88 39.69 52.51 l ll 3 4 : . 42. . 3.257 Conic! Cluster size N = 3 mobile station mobile station (worst case) (worst case) 65659 a a base station 2 intc ering BS’s BW = 60° 3 interfering BS’s BW = 120° six sectors three sectors sziwe .7 : Clusv‘er 5:53 IV=3 , +hrec «1" Si)! sca-fvrS Cluster sizeN s 4 mobile station 43 Mama's/1‘ . Cad: $5X/o5, we I’m/e, ‘ 1%? WW of Mi éam 5121:4773 We are “5’55 ‘59 ”553% N: W — 5W” ’5’” 3105. c 3 ‘ Cb: FXIOE p“ (B GMw Ah 8cdu, Zmdchgyeww.flm=fiflfi. “ ‘ _ _ A?“ _ [5,”? z ‘ ‘w :9 mflfi‘ @m% r [ Avd“77—-*=/Z5km i 1 . ””76 0f «9f 8 \ 3m Ame = 2-6 K M We ‘1‘“ : _ ,5 J . ““‘:‘ IQ éefifi - "£277; : 076/071 (6) Far and, year, em, cmm M4 fag P=gox/2:$6vo, /455ua1. fifig IMAméwr 3/: CoaQQDnCrS 077 542 ifgféf oeg/ ch G {M+2M+4M+3M) /D=/5/"I-/D ‘3‘ We [Le—2d G; $/0X/0‘ I) /5MIDZ /0X/06 _ flax/06 ‘ _> M >/oX/05 _ “‘ [5 P /5 X600 :5 /,,,./ ‘4‘ /§éh¢e ZZE inaufimanw naavwiw— ef' caéfWWWW’ an? ZWQP jfinfir‘iéf ‘ ‘ 5v; §wwvai $5 $02 ___——— _—_——— ‘ ' ‘ (‘0 MM” "f M" f” Sim km :fiy /. ...
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  • Spring '16
  • hamza rehman
  • new cells, base station, History of mobile phones

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