The Cellular Concept

Transcription

1 The Cellular Concept Key problems in multi-user wireless system: spectrum is limited and expensive large # of users to accommodate high quality-of-services (QoS) is required expandable systems are needed Therefore, efficient use of the spectrum is required How? frequency re-use cellular concept Re-use the same frequencies in geographically-separated areas; use multiple BSs instead of a single one. Lecture Nov-13 1(27)

2 A single BS to serve large area: An Example F = total available bandwidth f = channel bandwidth (per user) Key relationship: F = n f n F F = f = (12.1) f n Ottawa: n 6 = 10, f = 100kHz 6 5 F = kHz = 10 MHz = 100GHz totally unrealistic! Lecture Nov-13 2(27)

3 An Example: cont. Available: F = 100MHz, F f = = 100Hz bad! n Another problem: high P t. How much Tx power is needed? Assume: 12 P r,min = 90dBm = 10 W. L P P p G = G = 1; 4 h r R r = 1m, h = 100m max = t t 20km P R 13 = 10 = 130dB 2 2 ht hr = P L = 10W/user (MU) t r p T 7 = np = 10 W =10MW! (BS) t t Pr Lp = (12.2) G G t r Lecture Nov-13 3(27)

4 Fading margin dB P t = W (MU) P = MW! (BS) How to reduce, t T P? How to improve spectral efficiency? T Lecture Nov-13 4(27)

5 Frequency Re-Use Key idea: re-use frequencies in different geographical areas, without creating interference. i f i n = 4 vs. n = 16! Cell: a small geographical area served by a single BS. Base Station: collects calls from all users within a given area and sends them to a public telephone network (add also Internet), i.e. serves as an access point for users. Adjacent cells use different frequencies (no interf.) Distant cells use the same frequencies (frequency re-use). Key system- level idea: replace single (powerful) BS with many smaller-power BSs distributed over the total coverage area. Lecture Nov-13 5(27)

6 Frequency re-use and cellular architecture Major break-through: offered high system capacity (# of users) in a limited spectrum. Expandable: new users/cells can be added, as the system grows. Mobile: users can be moving. Various services and QoS. Each cell is allocated a sub-set of frequencies (not all available). Frequency planning (allocation): insures that interference is not too high, i.e. SIR is not too low. (interference-limited system). Recent idea: cognitive radio (CR). Lecture Nov-13 6(27)

7 T.S. Rappaport, Wireless Communications, Prentice Hall, 2002 Lecture Nov-13 7(27)

8 Cell Shape Ideally, from the propagation law, circles. Cannot cover an area without gaps or overlaps. Can use triangles, squares, hexagons ( no gaps/overlaps) Important: S 2 R S = cell area R =cell radius (distance to furthest point). 2 Q:. evaluate S R for circle, square, triangle, hexagon. Which is the best? Absolute best? Lecture Nov-13 8(27)

9 Hexagon cells are used in practice (analysis/design). BS: located at the center (omni-directional antenna) or in corners (directional antennas) B.A. Black et al, Introduction to Wireless Systems, Prentice Hall, Boston, Lecture Nov-13 9(27)

10 Basic Analysis Cluster: a set of adjacent cells using all available spectrum (frequencies). Important parameters: N = # of cells in a cluster. K = total # of channels (frequencies). K k = = # of channels/cell. N M = # of clusters covering all service area. n = total # of users supported (at the same time), also system capacity or total # of channels available. Key relation: n = k N M = K M Cs = n (12.3) M = 1 K = n, as with 1 BS; S S S = cell area tot = service area cl = cluster area 1 Stot S M = = tot (12.4) S NS cl Higher M higher n higher system capacity (for fixed K ). 1 Lecture Nov-13 10(27)

11 Example: Ottawa 2 2 D = 40km S = D = 1600km, R = 1km S = 4km, 1 N = 4 S c = 16km, S M = a = 100 S c f = 100kHz, F = 100MHz, F K = = 1000 f 5 a n = M K = 10 users! enough? User activity: 1 t = 1 h over 12 h P u = n 6 n = Pu nu nu = 10 P u Lecture Nov-13 11(27)

12 Example: cont. Tx power: P t(2 km) t(20 km) 4 3 = P 10 = 10 W = 1mW/user (no fading) = mW/user (fading) P T 5 = np = 10 1mW = 100W (no F) or 1kW 10kW (F) t P BS 3 P 10 = T = = 0.25W/BS (no F.), = W (F) MN n BS (# of users/bs) = = comp. to 1 BS configuration! 5 Typical values for current cellular systems: High-Capacity Indoor Wireless Solutions: Picocell or Femtocell? by Fujitsu Lecture Nov-13 12(27)

13 # of cells in a cluster: ELG4179: Wireless Communication Fundamentals S.Loyka Not all integer N are possible. Frequency re-use factor: Hexagonal Cells 2 2 N = i + j + ij (12.5) 1 N = how often a frequency is re-used 1 1 i.e. N = 7 : N = 7 every 7 th cell is using the same frequency. Nearest co-channel cell distance: 2 2 D = R 3N = R 3( i + j + ij) (12.6) Frequency re-use ratio (distance-wise): D Q = = 3N (12.7) R Cell area: S = R (12.8) Lecture Nov-13 13(27)

14 B.A. Black et al, Introduction to Wireless Systems, Prentice Hall, Boston, Lecture Nov-13 14(27)

15 B.A. Black et al, Introduction to Wireless Systems, Prentice Hall, Boston, Lecture Nov-13 15(27)

16 Co-Channel Interference Frequency re-use co-channel interference. Careful analysis/design is required. Recall the threshold effect: SNIR S γ = = γ I + N th for good performance (12.9) Interference-limited system noise is negligible: S I N γ = SIR γ I From the path loss (av. power): R = user - BS distance R = interferer - BS i distance th (12.10) avpt a, vp S = I t v i = (12.11) R R v i Lecture Nov-13 16(27)

17 Co-Channel Interference: cont. 1 st tier interfering cells: the smallest R i dominates SIR. For hexagonal cells, Ri = D. N I = # of 1 st tier interferers. v v v S R ( D R) Q γ SIR = = = I v N D N N i i I I I (12.12) assuming P t is the same for all users. Hexagonal cells: Q = 3N and 2 (3 N) v γ = (12.13) N I Increasing N decreasing I increasing SIR but decreasing n = C! s Acceptable QoS: γ γ th (= 18dB for AMPS) 2 v 2 v ( γth N I ) ( thni ) N N γ = 3 3 where x = ceiling (smallest integer x). (12.14) Lecture Nov-13 17(27)

18 γ = 18dB ( 63), v = 4. N th An Example (3 N) = 7 NI = 6 γ = 74 > 63 O.K. N I v 2 But γ = 20dB would require larger N. th Recall: 2 2 N = i + j + ij, so that: i = 1, j = 2 : N = 7; i = 1, j = 1: N = 3; i = 1, j = 0 : N = 1; i = 2, j = 2 : N = 12. Lecture Nov-13 18(27)

19 Parameters: ELG4179: Wireless Communication Fundamentals S.Loyka System Design Trade-offs Cell radius R, cluster size N, # of clusters M, # of users n ( C s ) K = total # of channels Key equations: cl 1 =, Stot S M = = tot n = Cs = KM (12.15) S NS K = fixed: M n γ = (3 N) v N I 2 How to increase M? S tot = fixed: Scl M S : via N, fixed S or N fixed, S cl N γ : keep γ γth! n S n : γ fixed, but: 1 S 1 1 trade-off 1. n tot BS = (complexity cost) S 1 trade-off 2. Lecture Nov-13 19(27)

20 Cell Splitting System expansion: in the area (add more cells/clusters) in user density (splitting/sectoring) Cell Splitting: split a big (macro) cell into # of smaller (micro) cells. Can accommodate: growing demand (# of users) non-uniform user density Congested cells (downtown): split into micro-cells. Microcells: lower P t and BS antenna height maintains the same Q = D R (samesir) preserves frequency re-use plan Tx power scaling P R P = P v t 2 Pt 1 2 v ( R 2) t1 t2 v i.e. P 2 P1 = 1 16 for v = 4. t t BS antennas: center vs. corners. = (12.16) Lecture Nov-13 20(27)

21 B.A. Black et al, Introduction to Wireless Systems, Prentice Hall, Boston, Lecture Nov-13 21(27)

22 Sectoring Key idea: using directional antennas concentrates radiation and thus decreases interference. No sectoring: sectors: 3 x sectors: 6 x 60 # of 1 st tier interfering cells decreases: 360 = 6 N I 3 x 120 = 2 N I 6 x 60 = 1 Recall that: N I v 2 fixed γ (3 N) th γ = NI γ N Cs (12.17) N I Smart antennas: much better improvement is possible, via SDMA. MIMO: much longer link capacity (Mb/s). Lecture Nov-13 22(27)

23 B.A. Black et al, Introduction to Wireless Systems, Prentice Hall, Boston, Lecture Nov-13 23(27)

24 B.A. Black et al, Introduction to Wireless Systems, Prentice Hall, Boston, Lecture Nov-13 24(27)

25 Traffic Engineering Grade of service(gos) = quality of service (QoS) = P b (blockage prob.) = outage probability. Given # of channels K, how many users can be supported? Key observation: not all users are active at the same time, na n. n K, n = # of active users a n a = total # of users (12.18) Average # of active users: n = p n, p = prob. of a user being active (12.19) a a a Using na = K, Blocking probability: n = na K p = p (12.20) a a P K λn K! na K! = = i! n i! i= 0 i= 0 K b K K i i n a λ (12.21) Lecture Nov-13 25(27)

26 λ = nλ = n λ = av. # of calls/unit time/user where: n a n λ = Tλ = traffic intensity = T = n av. holding time/call Q.: assuming that b 1 that n a hint: assume that P = ε is required, prove using (12.21) K K K n = 1 ε (1 ε) p p a p K i K K 1 n a n a n a i= 0 i! K! +, ( K 1)! justify this assumption. a a (12.22) Q.: how many users can be supported if ε = 1 2, 0.1? 0.01? Assume K = 100. Lecture Nov-13 26(27)

27 Summary The cellular concept. Frequency re-use. Cell shape. Hexagonal cells. Basic analysis. Co-channel interference. System design trade-offs. Cell splitting. Sectoring. Traffic engineering. Reading: Rappaport, Ch. 3. Other books (see the reference list). Note: Do not forget to do end-of-chapter problems. Remember the learning efficiency pyramid! Lecture Nov-13 27(27)