Wireless Communications

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1 NETW701 Wireless Communications Dr. Wassim Alexan Winter 2018 Lecture 2

2 NETW705 Mobile Communication Networks Dr. Wassim Alexan Winter 2018 Lecture 2 Wassim Alexan 2

3 Reflection When a radio wave propagating in one medium impinges on another medium having different electrical properties, the wave is partially reflected and partially transmitted If the wave is incident on a perfect dielectric, part of the energy is transmitted into the second medium and part of the energy is reflected back into the first medium, and there is no loss of energy in absorption If the second medium is a perfect conductor, then all incident energy is reflected back into the first medium without loss of energy The electric field intensity of the reflected and transmitted waves may be related to the incident wave in the medium of origin through the Fresnel reflection coefficient Γ Wassim Alexan 3

4 Reflection From Dielectrics Fig. 1. Geometry for calculating the reflection coefficients between 2 dielectrics. (left) E field in the plane of incidence. (right) E field normal to the plane of incidence. (Rappaport, Wireless Communications: Principles and Practice, 2 nd edition) The nature of reflection varies with the direction of polarization of the E field The behavior for arbitrary directions of polarization can be studied by considering the two distinct cases shown in Fig. 1 Wassim Alexan 4

5 Reflection From Dielectrics Fig. 1. Geometry for calculating the reflection coefficients between 2 dielectrics. (left) E field in the plane of incidence. (right) E field normal to the plane of incidence. (Rappaport, Wireless Communications: Principles and Practice, 2 nd edition) The plane of incidence is defined as the plane containing the incident, reflected, and transmitted rays In Fig. 1 (left), the E field polarization is parallel with (lies within) the plane of incidence Wassim Alexan 5

6 Reflection From Dielectrics Fig. 1. Geometry for calculating the reflection coefficients between 2 dielectrics. (left) E field in the plane of incidence. (right) E field normal to the plane of incidence. (Rappaport, Wireless Communications: Principles and Practice, 2 nd edition) The plane of incidence is defined as the plane containing the incident, reflected, and transmitted rays In Fig. 1 (right), the E field polarization is orthogonal to the plane of incidence ( indicates an arrow coming out of the slide towards the reader) Wassim Alexan 6

7 Reflection From Dielectrics Fig. 1. Geometry for calculating the reflection coefficients between 2 dielectrics. (left) E field in the plane of incidence. (right) E field normal to the plane of incidence. (Rappaport, Wireless Communications: Principles and Practice, 2 nd edition) ϵ is the permittivity; μ is the permeability; σ is the conductance Using Maxwell s equations, we get θ i = θ r, E r = Γ E i, E t = (1 + Γ) E i (1) Wassim Alexan 7

8 Reflection From Dielectrics Fig. 1. Geometry for calculating the reflection coefficients between 2 dielectrics. (left) E field in the plane of incidence. (right) E field normal to the plane of incidence. (Rappaport, Wireless Communications: Principles and Practice, 2 nd edition) The reflection coefficients for the two cases of parallel and orthogonal E field polarization at the boundary of the two dielectrics are given by Γ = E r E i = η 2 sin θ t - η 1 sin θ i η 2 sin θ t + η 1 sin θ i and Γ = E r E i = η 2 sin θ i - η 1 sin θ t η 2 sin θ i + η 1 sin θ t, (2) where η i is the intrinsic impedance of the i th medium, and is given by η i = μ i / ϵ i Wassim Alexan 8

9 Reflection From Dielectrics Fig. 1. Geometry for calculating the reflection coefficients between 2 dielectrics. (left) E field in the plane of incidence. (right) E field normal to the plane of incidence. (Rappaport, Wireless Communications: Principles and Practice, 2 nd edition) In case the first medium is free space and μ 1 = μ 2, then the reflection coefficients for the two cases is given by Γ = -ϵ r sin θ i + ϵ r - cos 2 θ i and Γ = sin θ i - ϵ r - cos 2 θ i ϵ r sin θ i + ϵ r - cos 2 θ i sin θ i + ϵ r - cos 2 (3) θ i where ϵ r is the relative permittivity Wassim Alexan 9

10 Exercise 1 Demonstrate that if medium 1 is free space and medium 2 is a dielectric, both Γ and Γ approach 1 as θ i approaches 0 regardless of ϵ r Wassim Alexan 10

11 Exercise 1 Solution Substituting θ i = 0 in the equations for the reflection coefficient Γ = -ϵ r sin 0 + ϵ r - cos 2 0 ϵ r sin 0 + ϵ r - cos 2 0 and Γ = ϵ r - 1 ϵ r - 1 = 1 Γ = sin 0 - ϵ r - cos 2 0 sin 0 + ϵ r - cos 2 0 Γ = - ϵ r - 1 ϵ r - 1 = -1, thus Γ = 1 Wassim Alexan 11

12 Exercise 1 Solution Γ = 1 and Γ = 1 This means that the ground may be modeled as a perfect reflector with a reflection coefficient of unit magnitude when an incident wave grazes the earth, regardless of polarization or ground dielectric properties Wassim Alexan 12

13 Ground Reflection (Two Ray) Model Fig. 2. Two ray ground reflection model. (Rappaport, Wireless Communications: Principles and Practice, 2 nd edition) A single direct path between the base station and a mobile is seldom the only means for propagation Friis equation is most likely inaccurate! Wassim Alexan 13

14 Ground Reflection (Two Ray) Model Fig. 2. Two ray ground reflection model. (Rappaport, Wireless Communications: Principles and Practice, 2 nd edition) The Two ray ground reflection model has been found to be reasonably accurate for predicting large scale signal strength over distances of several kilometers for mobile radio systems that use high towers (h >> 50 m) Wassim Alexan 14

15 Ground Reflection (Two Ray) Model Fig. 2. Two ray ground reflection model. (Rappaport, Wireless Communications: Principles and Practice, 2 nd edition) If E 0 is the free space E field at a reference distance d 0 from the transmitter, then for d > d 0, the free space propagating E field is given by E(d, t) = E 0 d 0 cos ω c t - d d c (4) where E(d, t) = E 0 d 0 / d represents the envelope of the E field at d meters from the transmitter Wassim Alexan 15

16 Ground Reflection (Two Ray) Model Two propagating waves arrive at the receiver: the direct wave travelling a distance d' and the reflected wave that travels a distance d'' E LOS (d', t) = E 0 d 0 d' E g (d'', t) = Γ E 0 d 0 d'' cos ω c t - d' c (5) cos ω c t - d'' c (6) Note that θ i = θ r, E g = Γ E i, E t = (1 + Γ) E i (7) For small values of θ i (i.e. grazing incidence), the reflected wave is equal in magnitude and 180 out of phase with the incident wave Wassim Alexan 16

17 Ground Reflection (Two Ray) Model Fig. 2. Two ray ground reflection model. (Rappaport, Wireless Communications: Principles and Practice, 2 nd edition) For small values of θ i and assuming normal incidence (i.e. Γ = -1), then E t = (1 + Γ) E i = 0 (8) Wassim Alexan 17

18 Ground Reflection (Two Ray) Model Thus, the resultant E field is the vector sum of E LOS and E g E total (d, t) = E LOS (d', t) + E g (d'', t) = E 0 d 0 d' = E 0 d 0 d' And its total envelope is given by cos ω c t - d' c + Γ E 0 d 0 d'' cos ω c t - d' c + (-1) E 0 d 0 d'' cos ω c t - d'' c (9) cos ω c t - d'' c (10) E total = E LOS + E g (11) Wassim Alexan 18

19 Ground Reflection (Two Ray) Model: The Path Difference Fig. 3. The method of images is used to find the path difference between the LOS and the ground reflected paths. (Rappaport, Wireless Communications: Principles and Practice, 2 nd edition) = d'' - d' (12) Wassim Alexan 19

20 Ground Reflection (Two Ray) Model: The Path Difference Fig. 3. The method of images is used to find the path difference between the LOS and the ground reflected paths. (Rappaport, Wireless Communications: Principles and Practice, 2 nd edition) d' = (h t - h r ) 2 + d 2 (13) Wassim Alexan 20

21 Ground Reflection (Two Ray) Model: The Path Difference Fig. 3. The method of images is used to find the path difference between the LOS and the ground reflected paths. (Rappaport, Wireless Communications: Principles and Practice, 2 nd edition) d'' = (h t + h r ) 2 + d 2 (14) Wassim Alexan 21

22 Ground Reflection (Two Ray) Model: The Path Difference Usually, the T R separation is much larger than the heights h t and h r, thus the following approximation can be safely used and d' = (h t - h r ) 2 + d 2 = d 1 + (h t - h r ) 2 d'' = (h t + h r ) 2 + d 2 = d 1 + (h t + h r ) 2 d 2 d 1 + (h t - h r ) 2 2 d 2 (15) d 2 d 1 + (h t + h r ) 2 2 d 2 (16) where 1 + x 1 + x 2 for x << 1 Wassim Alexan 22

23 Ground Reflection (Two Ray) Model: The Path Difference The path difference is given by = d'' - d' 2 h t h r d The phase difference is given by (17) θ = 2 π λ = ω c c (18) The time delay between the arrival of the two components is given by τ d = c = θ 2 π f c (19) Wassim Alexan 23

24 Ground Reflection (Two Ray) Model Assuming a very large d, the path difference becomes very small and thus E LOS and E g are identical in magnitude and differ only in phase Remember that E 0 d 0 d E 0 d 0 d' E 0 d 0 d'' (20) E tot (d, t) = E 0 d 0 d' cos ω c t - d' c + (-1) E 0 d 0 d'' cos ω c t - d'' c (21) Evaluating (21) at t = d'' / c, we get E tot d, t = d'' c = E 0 d 0 d' cos ω c d'' c - d' c + (-1) E 0 d 0 d'' cos ω c d'' c - d'' c (22) Wassim Alexan 24

25 Ground Reflection (Two Ray) Model Simplifying (22), we get E tot d, t = d'' c = E 0 d 0 d' Keep in mind that = E 0 d 0 d' = E 0 d 0 d' cos ω c d'' c - d' c + (-1) E 0 d 0 d'' cos ω c d'' c - d' c - E 0 d 0 d'' cos ω c d'' c - d' c - E 0 d 0 d'' cos 0 cos ω c d'' c - d'' c (23) Thus, θ = E tot d, t = d'' c = E 0 d 0 d' E 0 d 0 d d'' - d' c ω c = ω c c cos θ - E 0 d 0 d'' (cos θ - 1) (24) (25) Wassim Alexan 25

26 Resultant E tot E LOS = E 0 d 0 d' θ E g = E 0 d 0 d'' Fig. 4. The E fields from the two rays. Wassim Alexan 26

27 Resultant E tot E 0 d 0 sin θ d' θ E g = E 0 d 0 d'' E 0 d 0 cos θ d' Fig. 5. The E fields from the two rays and the resolution of E LOS. Wassim Alexan 27

28 Resultant E tot E tot E LOS E LOS θ Fig. 6. The E fields from the two rays and the resultant E tot. Wassim Alexan 28

29 Resultant E tot E tot E LOS E LOS E 0 d 0 d sin θ θ E g E 0 d 0 d - E 0 d 0 d cos θ Fig. 7. The E fields from the two rays, their i and j components and the resultant E tot. The resultant envelope of the E field is then expressed as E tot (d) = E 0 d 0 d 2 (1 - cos θ ) 2 + E 0 d 0 d 2 sin 2 θ (26) Wassim Alexan 29

30 Exact Expression for E tot E tot (d) = E 0 d 0 d 2 (1 - cos θ ) 2 + E 0 d 0 d 2 sin 2 θ (27) = E 0 d 0 d 2-2 cos θ Using the following trigonometric identity We can simplify (27), getting sin 2 θ 2 = 1 2 (1 - cos θ ) (28) E tot (d) = 2 E 0 d 0 d sin θ 2 (29) The significance of the path difference appears in the phase difference θ between the two waves (LOS and reflected) Wassim Alexan 30

31 Exact Expression for E tot at a Large d The expression for E tot (d) in (29) can be further simplified if d is very large d being very large allows for sin θ 2 θ 2, which implies that Remember from (17) and (18) that θ 2 < 0.3 rad (30) 2 h t h r d and θ = 2 π λ = ω c c, (31) Then θ 2 = 2 π h t h r λ d < 0.3 rad (32) Wassim Alexan 31

32 Which means that the expression for E tot (d, t) can be simplified whenever d > 2 π h t h r 0.3 λ d > 20 π h t h r 3 λ 20 h t h r λ (33) Thus, as long as d satisfies this condition, E tot (d, t) can be expressed as E tot (d) = 2 E 0 d 0 d sin θ 2 2 E 0 d 0 d 2 π h t h r λ d [V / m] (34) Keep in mind that the power flux density can is expressed as Φ r = E tot(d) 2 η = 2 E 0 d 0 d 2 π h t h r λ d π = E 0 d 0 d π 4 π h t h r λ d 2 (35) From the LOS analysis, we have E 0 d 0 d π = P t G t 4 π d 2 (36) Wassim Alexan 32

33 This allows us to express the receive antenna power flux density as Φ r = E tot(d) 2 η = P t G t 4 π d 2 4 π h t h r 2 (37) λ d Also, since (A e ) iso = λ2 4 π (38) Then P r (d) = Φ r (A e ) dir = Φ r G r (A e ) iso = P t G t 4 π d 2 4 π h t h r 2 λ 2 G λ r d 4 π (39) Expanding and simplifying, we get P r (d) = P t G t G r h t 2 h r 2 d 4 (40) P r (d) falls off with distance raised to the fourth power (a rate of 40 db/decade) Wassim Alexan 33

34 This is a much more rapid path loss than is experienced in free space For large values of d, the received power and the path loss are frequency dependent Obtaining 10 log (P t / P r ) in db: PL = 40 log d - (10 log G t + 10 log G r + 20 log h t + 20 log h r ) (41) Wassim Alexan 34

35 Exercise 2 A mobile is located 5 km away from a base station and uses a vertical λ/4 monopole antenna with a gain of 2.55 db to receive cellular radio signals. The E field at 1 km from the transmitter is measured to be 10-3 V/m. The carrier frequency used for this system is 900 MHz a) Find the length and the effective aperture of the receiving antenna b) Find the received power at the mobile using the two way ground reflection model assuming the height of the transmitting antenna is 50 m and the receiving antenna is 1.5 m above ground Wassim Alexan 35

36 Exercise 2 Solution Given: T R separation distance = 5 km E field at a distance of 1 km = 10-3 V/m Frequency of operation, f = 900 MHz λ = c f = = m a) Length of the antenna, L = λ / 4 = / 4 = m = 8.33 cm Effective aperture of the antenna A e = G λ2 4 π = (0.3333) 2 4 π = m 2 Wassim Alexan 36

37 Exercise 2 Solution b) Since d >> h t h r, the electric field is then given by E r (d) 2 E 0 d 0 d 2 π h t h r λ d k d π V / m The received power at a distance d can then be obtained as follows P r (d) = E r π A e = π = = db = dbm Wassim Alexan 37

Reflection/Refraction

$Reflection/Refraction$ Reflection/Refraction Page Reflection/Refraction Boundary Conditions Interfaces between different media imposed special boundary conditions on Maxwell s equations. It is important to understand what restrictions