Propagation Models for Wireless and Mobile Communication

Propagation Channel Modelling forWireless/Mobile Communication By Dr. Sanjay Kumar Soni (Associate Professor) Electronics and Electrical Communication Engg. Delhi Technological University Delhi-110042 • Introduction • Basic propagation model • Empirical Model • Ray Tracing model • Modeling of direction channel parameters Outline Propagation Model - Mechanisms • Reflection • Diffraction • Scattering Q1:What is propagation model ? I t is a mathematical representation of the actual physical phenomenon in the channel. General Intuition • Two main factors affecting signal at receiver – Distance (or delay) ¬ Path attenuation – Multipath ¬ Phase differences Green signal travels 1/2ì farther than Yellow to reach receiver, who sees Red. For 2.4 GHz, ì (wavelength) =12.5cm. Objective • Invent models to predict what the field looks like at the receiver. – Attenuation, absorption, reflection, diffraction... – Motion of receiver and environment… – Natural and man-made radio interference... – What does the field look like at the receiver? • Maxwell’s equations – Complex and impractical • Free space path loss model – Too simple • Ray tracing models – Requires site-specific information • Empirical Models – Don’t always generalize to other environments • Simplified power falloff models – Main characteristics: good for high-level analysis Path Loss Modeling Wireless Multipath Channel Channel varies at two spatial scales: • Large scale fading: path loss, shadowing. • Small scale fading: Multi-path fading (frequency selectivity, coherence b/w, ~500kHz), Doppler (time-selectivity, coherence time, ~2.5ms). MultiPath Interference: Constructive & Destructive Mobile Wireless Channel Multipath Three Broad classes of Channel Modeling • Empirical Modeling • Statistical Modeling • Site-specific Deterministic Modeling -Ray Tracing Algorithm - Field Integral Method Example of some real Urban Scenario for Channel Modeling Fig 3: Real Urban Scenario 1 Empirical Models for Mobile Communication 2 4 R A G P P R T T R t = 2 4 ì t R R A G = P R = Received power P T = Transmitted power G T = Gain of the Transmitting Antenna R = distance between Tx and Rx G R = gain of the Receiver Antenna ì = Wavelength Free Space Model 2 4 ) ( | . | \ | = R G G P P R T T R t ì loss path Space Free 4 2 = = | . | \ | s L R ì t L s = 92.47 + 20 log R + 20 log f R in Km dB f in GHz If antennas are included, the loss L s (modified) = L s - G T - G R dB • Note: effect of frequency f: 900 Mhz vs 5 Ghz. – Either the receiver must have greater sensitivity or the sender must pour 44W of power, even for 10m cell radius! dB (Decibel) = 10 log 10 (Pr/Pt) Log-ratio of two signal levels. Named after Alexander Graham Bell. For example, a cable has 6 dB loss or an amplifier has 15 dB of gain. System gains and losses can be added/subtracted, especially when changes are in several orders of magnitude. dBm (dB milliWatt) Relative to 1mW, i.e. 0 dBm is 1 mW (milliWatt). Small signals are -ve (e.g. -83dBm). Typical 802.11b WLAN cards have +15 dBm (32mW) of output power. They also spec a -83 dBm RX sensitivity (minimum RX signal level required for 11Mbps reception). For example, 125 mW is 21 dBm and 250 mW is 24 dBm. (commonly used numbers) dBi (dB isotropic) for EIRP (Effective Isotropic Radiated Power) The gain a given antenna has over a theoretical isotropic (point source) antenna. The gain of microwave antennas (above 1 GHz) is generally given in dBi. Decibels: dB, dBm, dBi REFLECTION 2 - Ray Model ( GROUND REFLECTION) SINGLE DIRECT PATH BETWEEN MS AND BS HARDLY EXISTS HENCE THE FREE SPACE MODEL IS HIGHLY INACCURATE. NEXT IMPROVISION : 2 RAY MODEL ÷ DIRECT PATH ÷ GROUND REFLECTED PROPAGATION PATH • REASONABLY ACCURATE FOR PREDICTING LARGE- SCALE SIGNAL STRENGTH (MEAN SIGNAL STRENGTH FOR ARBITRARY Tx - Rx SEPERATION - USUALLY IN KM) ASSUMPTION : EARTH TO BE FLAT Path loss = 40 log d -(10 log G T + 10 log G R + 20 log h t + 20 log h r ) Classical 2-ray Ground Bounce model h b d d d r h m Direct path (line of sight) Ground reflected path d Fig 4 v/m ~ 2 2 2 0 0 d k d h h d d E E r t TOT ÷ ~ ì t and 4 2 2 d h h G G P P r t r t t r = ( ) ( ( ( ( ( ¸ ( ¸ = = = 2 2 2 2 4 120 using d G G P A E A P P r t t e e d r t ì t ( ) r t r t h h G G d log 20 log 20 log 10 log 10 log 40 Loss Path and + + + ÷ = Ex T-R separation = 5 km (mobile located 5 km away from BS). Mobile uses a vertical ì/4 monopole antenna (gain = 2.55 dB). E-field at a distance of 1 km = 10 -3 v/m, f c = 900 MHz. Find the received power at the mobile, assuming h t = 50 m, h r = 1.5 m. d >> \(h t h r ) is satisfied ( ) ( ) ( ) ( ) v/m 10 * 1 . 113 10 * 5 * 333 . 0 5 . 1 * 50 * 2 10 * 5 10 * 1 * 10 * 2 v/m ~ 2 2 ~ 6 3 3 3 3 2 0 0 ÷ ÷ = ( ¸ ( ¸ = ÷ ÷ t ì t d K d h h d d E d E r t R Power received ( ) ( ) ( ) ( ) dBm 68 . 92 w 10 * 5 4 333 . 0 8 . 1 * 377 10 * 1 . 113 4 A 120 13 2 2 6 2 e 2 ÷ = = = = = ÷ ÷ t t ì t G A d E d P e R r • The electric field flips in sign canceling the LOS field (unlike in free-space model), and hence the path loss is O(d -4 ) rather than O(d -2 ). • The frequency effect disappears! – Similar phenomenon with antenna arrays. • Near-field, far-field detail explored in next slide: – Used for cell-design 2-ray model observations 2-ray model: distance effect, critical distance • d < h t : constructive i/f • h t < d < d c : constructive and destructive i/f (multipath fading upto critical distance) • d c < d: only destructive interference “Enlist the number of scenarios where 2-Ray model is applicable” 2-ray model example, cell design • Design the cell size to be < critical distance to get O(d - 2 ) power decay in cell and O(d -4 ) outside! • Cell radii are typically much smaller than critical distance Q1:Enlist the number of scenarios where 2-Ray model is applicable. Q2: Explain how the 2-Ray Model is useful in network planning. Diffraction • Diffraction occurs when waves hit the edge of an obstacle – “Secondary” waves propagated into the shadowed region – Excess path length results in a phase shift – Fresnel zones relate phase shifts to the positions of obstacles T R 1st Fresnel zone Obstruction Fresnel Zones • Bounded by elliptical loci of constant delay • Alternate zones differ in phase by 180° – Line of sight (LOS) corresponds to 1st zone – If LOS is partially blocked, 2nd zone can destructively interfere (diffraction loss) Fresnel zones are ellipses with the T&R at the foci; L 1 = L 2 +ì Path 1 Path 2 Concentric Circles which define the boundaries of Successive Fresnel Zones Fig 10 • Radius of n’th Fresnel zone is given by: 2 1 2 1 n r d d d d n + = ì This approximation is valid for d1,d2 >> r n • Fresnel zone will have maximum radii if the plane is midway between the transmitter and receiver and the radii becomes smaller when plane is moved towards either Tx or Rx. • This explain how shadowing is sensitive to the frequency as well as the location of the obstruction with relation to the Tx or Rx. Fresnel Zone for different knife-edge diffraction scenarios Fig 11 •Phase difference between a direct LOS and diffracted path is a function of height and position of obstruction, as well as transmitter and receiver locations. •DIFFRACTION LOSS as a function of path difference around an obstruction is explained by FRESNEL ZONES. •Fresnel Zones: Successive regions where secondary waves have a path length which are nì \ 2 greater than the total path length of a LOS. •Phase difference 2 2 2 v t ì t | = A × = where A = excess path length v = Fresnel - Kirchoff diffraction parameter given as 1 2 1 2 1 2 1 2 2( ) 2 ( ) d d d d h d d d d v o ì ì + = = + The diffraction gain due to the presence of a knife edge ) ( log 20 ) ( v F dB G d = ( ) edge knife and ground both the of absence in the strength field space Free strength field diffracted edged - Knife 0 where = = E d E F v ¦ ¦ ¦ ¹ ¦ ¦ ¦ ´ ¦ > | . | \ | s s ÷ ÷ ÷ s s ÷ s s ÷ ÷ ÷ s = 4 . 2 225 . 0 log 20 4 . 2 1 ) 1 . 0 38 . 0 ( 1184 . 0 4 . 0 log( 20 1 0 )) 95 . 0 exp( 5 . 0 log( 20 0 1 ) 62 . 0 5 . 0 log( 20 1 0 ) ( 2 v v v v v v v v v dB G d Fig 12 Example: Compute diffraction loss for the three cases shown in Fig( 11). Assume λ=1/3 m, d1=1Km, d2=1Km. (a) h =25 m (b) h =0 (c) h=-25 m. To compare the answers using values from Fig ( 12) as well as approximate solution given by G d (B) formulation. Solution: (a) h =25m Fresnel diffraction parameter using this Fresnel value and from Fig ( ), the diffraction loss is 22 dB. Using numerical approximation, the diffraction loss is equal to 21.7 dB. 74 . 2 1000 * 1000 * ) 3 / 1 ( ) 1000 1000 ( 2 25 ] d d / ) d [2(d h 2 1 2 1 = + = + = ì v The path length difference between the direct and diffracted rays is given as : To find Fresnel zone in which the tip of the obstruction lies, we need to compute n which satisfies Hence, the tip of the obstruction completely blocks the first three Fresnel zone. (b) h=0 it implies . Hence from Fig , diffraction loss is obtained as 6 dB. Using numerical approximation, the diffraction loss is 6 dB. For h =0, we have . Hence, tip of the obstruction lies in the middle of the first Fresnel zone. m 625 . 0 1000 * 1000 ) 1000 1000 ( * 2 ) 25 ( ) ( 2 h 2 2 1 2 1 2 = + = ( ¸ ( ¸ + = A d d d d 2 / *ì n = A 75 . 3 / 2 n , m 0.625 , 3 / 1 = A = ¬ = A = ì ì 0 = A (c ) h =-25 m it implies from Fig , the diffraction loss is approximately 1 dB, using numerical approximation, diffraction loss is 0 dB. Since absolute value of h is same as in part (a), the excess path length and hence n will also be same. Hence though the tip of the obstruction completely blocks the first three Fresnel zones, the diffraction loss are negligible since obstruction is below the line-of-sight (h is negative). Free Space Path Loss reconsidered • Path Loss is a measure of attenuation based only on the distance to the transmitter • Free space model only valid in far-field; – Path loss models typically define a “close-in” point d 0 and reference other points from there: 2 0 0 ) ( ) ( | . | \ | = d d d P d P r r | | | | | | | | 0 0 ( ) ( ) ( ) 2* 10*log( / ) r dB dB dB PL d P d PL d d d = = + Shadowing 43 Average received Signal power decreases logarithmically with distance.The path loss where ¸ is the path loss exponent which indicates the rate at which the path loss increases with distance , d o = close -in reference distance d = T X - R X separation NOTE: when exponent is taken to be free space path loss exponent 2,then, this model converts into free space path loss model | | | | 0 0 0 PL(d) - : ensemble average i.e. PL(d) PL(d ) 10 log dB dB d d d d ¸ o ¸ | | | \ . | | = + | \ . LOG - DISTANCE PATHLOSS MODEL 44 Path loss exponent for different environment 45 LOG -NORMAL SHADOWING Surrounding environment clutter image be vastly different at different locations having same T X -R X separation (This has not been considered in previous model). If so, measured signals would vary from the average value given above. Path loss PL(d) at a particular location is a log - normal distributed r.v. (normal is dB) about the mean distance dependent value. ( ) ( ) PL d PL d X o = + 0 0 ( ) 10 log d PL d X d o ¸ | | = + + | \ . 46 L t r P P d P ÷ = ) ( and dBm dBm dB X o = zero-mean Gaussian r.v. with standard deviation o(in dB) The log- normal distribution describes the random SHADOWING effects( due to different levels of clutter on the propagation path). • Okumura model – Empirically based (site/freq specific) – Awkward (uses graphs) • Hata model – Analytical approximation to Okumura model • Cost 136 Model: – Extends Hata model to higher frequency (2 GHz) • Walfish/Bertoni: – Cost 136 extension to include diffraction from rooftops Empirical Models 48 Okumura Model :(1968) Curve fitting experimental procedure, Okumura model: 150 -1920 MHz range 1 Km -100 Km Tx -Rx distance 30 m -1000 m base station heights. 49 Empirical curves of variation of median attenuation (relative to free space) in an urban area over a quasi-smooth terrain vs. frequency is given. Curves obtained from extensive measurements using vertical omni - directional antennas at both the BS and MS. L (dB) = L F + A mu (f,d) - G(h te ) - G(h re ) -G AREA L : path loss value L F : Free space propagation loss A mu : median attenuation relative to free space 50 G(h te ) : Base station antenna height gain factor G(h re ): MS Antenna height gain factor G AREA : Gain due to the type of environment m h G m h G m h G re re te 3 h 10m 3 h log 20 ) ( 3 h 3 h log 10 ) ( 30 h m 1000 200 h log 20 ) ( re re re re te te > > | . | \ | = s | . | \ | = > > | . | \ | = Okumura model : Simple and fairly accurate Major disadvantage : Slow response to rapid change in terrain. (not good for rural areas) 51 MEAN ATTENUATION RELATIVE TO FREE SPACE 52 CORRECTION FACTOR 53 Ex Find the path loss using Okumura’s model for d = 50 km, h te = 100 m, h re = 10 m in a suburban environment. If Base Station transmitter radiates an EIRP of 1 kw at ƒ c = 900 MHz, find P R (assuming unity gain receiving antenna) Free space path loss ( ) ( ) ( ) ( ) dB 5 . 125 10 * 50 * 4 10 * 900 / 10 * 3 log 10 4 log 10 2 3 2 2 6 8 2 2 2 = ( ( ¸ ( ¸ = ( ¸ ( ¸ = t t ì d L F 54 Okumura’s curve for 50 km distance at 900 MHz gives A mu = 43 dB and G AREA = 9 dB for suburban area ( ) ( ) dB 46 . 10 3 10 log 20 3 log 20 dB 6 200 100 log 20 200 log 20 = | . | \ | = | . | \ | = ÷ = | . | \ | = | . | \ | = re re te te h h G h h G L(dB) = L F + A mu (ƒ, d) - G(h te )- G(h re ) - G AREA = 125.5 + 43 - (-6) - 10.46 - 9 dB = 155.04 dB 55 Therefore, the received power P r (d) = EIRP(dBm) - L(dB) + G r (dB) = 60 dBm - 155.04 + 0 = - 95.04 dBm. Using two-ray model, the received power is given by: 4 2 r 2 t r t t R d h h G G P P = P t G t =EIRP=1kW=1*10 6 mW=60dBm, G r =1, h t =100m, h r =10m, d=50Km Hence, received power(dB) is: P r (dB)=60dBm+0dB+40dB+20dB-187.6dB = -67.6dBm Hence, two-ray model predicts the received power to be far greater than that predicted by Okumura model. 57 HATA MODEL :150 – 1500MHz (1980) Urban area propagation loss model L(urban,dB) = 69.55 + 26.16 log f c - 13.82 log h te -a(h re ) + (44.9-6.55 log h te )log d f c = frequency in MHz; 150 s f s 1500 MHz ; h te = effective base station antenna height ; m in range 30 m to 200 m h re = effective mobile receiver antenna height ; m in range 1 m to 10m d = T x - R x separate distance ; Km a(h re ) = correction factor for effective mobile antenna height ; a function of coverage area size. 58 ( ) ( ) ( ) ( ) ¦ ¹ ¦ ´ ¦ ¦ ) ¦ ` ¹ > ÷ s ÷ ÷ ÷ = city large a for 300MHz c f 4.97dB 2 re log11.75h 3.2 300MHz c f 1.1dB 2 re log1.54h 8.29 city sized medium for dB 0.8 c 1.56logf re h 0.7 - c 1.1logf ) ( re h a For Suburban area, Hata Model for path loss is modified as L(dB) = L(urban)-2 [log (f c /28)] 2 - 5.4 and for path loss in rural area, it is given as L(dB) = L(urban)-4.78 (log f c ) 2 + 18.33 log (f c ) - 40.94 Hata model compares well (very closely) with the Okumura model for d>1 km. Well suited for large cell areas, but not for PCS ( cell area of the order of 1 Km.) 59 Extension to Hata Model for Personal Communication service (PCS) (COST-231):The European Cooperative for Scientific and Technical research (EURO-COST) formed the COST-231 working committee to develope an extended version of Hata model to make it suitable for PCS cell (which is of the order of 1 Km)Cost-231 proposed following formula to extend Hata model to 2GHz. L 50 (urban) = 46.3+33.9 log f c -13.82 log h te -a(h re ) + (44.9 - 6.55 log h te )log d + C M centres an metropolit for dB 3 areas suburban and city sized medium for dB 0 ¹ ´ ¦ = M C For PCS, this model is extended to frequency upto 2 GHz, i.e. f : 1500 MHz to 2000 MHz ,h te : 30 m to 200 m ,h re : 1 m to 10 m d : 1 km. to 20 km. Deterministic Model Main Component of Deterministic Channel Model • Reflection From the Wall • Diffraction from the Building Corners and edge • Ray tracing •For the large cell, the received signal mostly depends upon the LOS path but for small cell such as urban scenario, the energy spreading around the building also becomes very significant. • For urban scenario, Rizk et. Al.[1] predicted the signal using only reflection component fig (2) and found that whatever electrical parameters taken, it never matches with measurement. • Then, they took reflection and diffraction both in prediction programme and found it to be very close to measurement thus verifying the importance of diffraction fig(3). Why diffracted signal energy is important component for Modeling urban scenario [1] Rizk K, Wagen JF, Gardiol F.,“ Two-dimensional ray-tracing modeling for propagation prediction in microcellular environments,” IEEE Trnas Veh Technol 1997; 46:508-18 Fig 2: Prediction result using only Reflection Fig 3: Prediction result using Reflection and Diffraction both Ray Tracing approach is most widely used approach for deterministic channel modeling Why Ray Tracing ? • Highly accurate path loss prediction • Gives clear insight into physical phenomenon taking place in channel • Best for characterizing channel parameters such as impulse response for MIMO channel, delay spread, Doppler spread, angular spread and distribution function of short-term fading [4] [4] Thomas Fugen, Jurgen Maurer, “Capability of 3-D Ray tracing for Defining parameter sets for the specification of future mobile communication systems,” IEEE Trans. on Antennas and Propagation, vol. 54, No. 11, Nov. 2006. Two Dimensional description of Urban environment Image Method I2 2 1 I3 I1 Tx 3 Rx Tx I1 Rx Tx I2 I1 Solving Ray tracing path Lit region visibility test[1]:(1999) Here, it is just checked if the mobile is in the lit region (visibility region) of the facet or not. If it comes in lit region, then, its corresponding field at Mobile is computed Rx Tx I1 2 1 First Layer Tx Rx I1 3 1 1 2 3 Tx Tree for Ray tracing Second layer Development of 2D Ray Tracing tool in MATLAB Flowchart for the ray tracing tool start Check for LOS power. If LOS exists calculate LOS power 1 Find the facet illuminated by the source, i.e. find the facets of the given terrain which falls under the lit region of the source and are unblocked Find the image source/diffraction source and corresponding lit region of each illuminated facet 2 3 If the receiver lies unobstructed in the lit region of these image or diffraction sources find the corresponding reflected and diffracted fields . Obtain the terrain geometry Tx, Rx, linear piecewise data points of the terrain as obtained through plot digitizer,N =0; Check for shadow of path If path shadowed, P N =0, else compute the total power and store it 2 1 N=N+1. This completes one level of child sources. Keep track of the parent source and parent propagation phenomena N=Nmax? Each source in the above level becomes the new transmitting source with their lit region Add all powers, LOS, reflected, diffracted, multiple reflected, multiple diffracted, reflected- diffracted combination sto p 3 Yes No 1 2 3 4 5 2 Rx 3 Rx 3 Tx Rx Dominant Path Tree Conversion of 2D path to 3D path Determining Lateral rays Two-dimension path can be converted to 3D ray using 2D-3D ray path conversion. This is explained for single reflected ray as follows: Here, AB=D2, BD=D1, BC=X2, CD=X1 In the above figure , suppose that we are given 2D ray E-F-G with reflection at point F. In 3D this may represent two 3D rays with one as E-H-I and another one E-C-I (ground reflected at C). Since F is the reflection point hence we know θ Φ Φ G F K J A B D C I E Reflectio n point H the distances AB and BD. The height of Tx and Rx are ht and hr respectively. 1 2 1 D X X = + 2 2 1 2 2 1 tan X D h X h X D h X h r t r + = ¬ + = = | X1 and X2can be solved as other parameters are known. To calculate the height of reflection point KI KE JI JH = = u tan Hence, except JH, rest parameters are known, and hence JH can be calculated. Thus, with the knowledge of height of both transmitting and receiver antenna, the 3D ray can be obtained from 2D ray. Determining Rooftop Rays Path loss using 2.5D Ray tracing Work to be done Verification of Important channel parameters such as delay spread, Doppler spread, angular spread, and distribution function of short-term fading using 2.5 D Ray tracing tool. Why 2.5 D Ray tracing ? • 2.5 D Ray tracing is much faster than full 3D ray tracing Base Station (BS) Mobile Station (MS) multi-path propagation Path Delay P o w e r path-2 path-2 path-3 path-3 path-1 path-1 Source #2: Multipaths: Power-Delay Profile Channel Impulse Response: Channel amplitude |h| correlated at delays t. Each “tap” value @ kTs Rayleigh distributed (actually the sum of several sub-paths) Doppler spread (Mathematical formulation) Doppler: Non-Stationary Impulse Response. Impulse response Narro Band Channel Characterization Complex low pass frequency response at single frequency can be described as: Window length T w to obtain average of long term fading is related to speed of vehicle [25] 2 ( ) ( ) ( ) ( ) ( ) LP LP LP D D D D HH D t HH H t H f H f S f r t ÷ ¬ = A Coherence time: Time interval Δt that fulfills the following condition: ( ) 1/ (0) t HH t HH r t e r A = Average Doppler shift and Doppler spread are given as: 2 ( ) ( ) ( ) 2 ( ) D D HH D D D HH D D D HH D D f D HH D D f S f df f S f df f S f df f S f df o +· ÷· +· ÷· +· ÷· = +· ÷· × = × ÷ } } } } . D f coherent T const o × = Dispersion-Selectivity Duality Delay Spread: Wideband channel characteristic Mean Arrival Time: Delay Spread Measurement Plan Digital Map Channel impulse response Measurement Objective: To obtain Power-delay profile (PDP) Method: 1 ( ) ( ) ( ) ( ) ( ) ( ) y t p t h t Y f H f P f = © ¬ = Method2: Spread Spectrum Channel Impulse response measurement system Method 3: Frequency Domain Impulse Response Measurement Requirement for this measurement: • Suitable building locations for which 3D database is available • Channel sounder to record impulse response of channel • Vehicle mounted with antenna to move the Tx Once impulse response found, delay spread is calculated as: 0 1 2 5 -30dB -10dB -20dB 0dB τ (μs) P r (τ) Power-delay Profile → Long-fading, short-fading, Path Loss • We need a channel sounder that can record the complex channel transfer function H(t) with sampling time Ts • Do the averaging of these samples for a given window ( ) LP H t t Long-term fading signal (Local Mean area) W T Thank you