Presentation is loading. Please wait.

Presentation is loading. Please wait.

Wireless Communication systems

Similar presentations


Presentation on theme: "Wireless Communication systems"— Presentation transcript:

1 Wireless Communication systems
Chapter 5 Wireless Communication systems & Propagation

2 Chapter Outlines Chapter Wireless Communication Systems & propagation The Friis Transmission Equation Antenna Noise Temperature Radar Free Space Propagation Ground Reflections Ionosphere Propagation Troposphere Propagation Vegetation Propagation Urban Propagation Attenuation

3 Introduction Wireless communications involves the transfer of information between two points without direct connection  sound, infrared, optical or RF energy. Most modern wireless systems rely on RF or microwave signals, usually in the UHF to millimeter wave freq range. But why high freq?  spectrum crowding, need for higher data rates  majority of today’s wireless systems operate at freq ranging from 800MHz to few GHz. E.g. broadcast radio and TV, cellular telephones, DBS TV service, WLAN, GPS and RFID.

4 Introduction (Cont’d..)
Characterizing the wireless systems: Point to point radio systems  single transmitter with single receiver  use high gain antennas in fixed positions to max received power and minimize interference with other radios (nearby frequencies). Point to multipoint systems  connect a central station to a large number of possible receivers  commercial AM and FM radio and broadcast tv  Uses an antenna with broad beam to reach many listeners and viewers.

5 Introduction (Cont’d..)
Multipoint to multipoint systems  simultaneous communication between individual users (maybe not in fixed location)  generally not connect two users directly, but rely on a grid of base stations to provide desired interconnections between users. E.g. cellular telephone systems and WLAN. Can also be characterize in terms of directionality of communication: Simplex system  communication occurs in one direction, from tx to rx. E.g. broadcast TV, radio and paging systems.

6 Introduction (Cont’d..)
Half Duplex system  communication in two directions, but not simultaneously. E.g. early mobile radios (walkie-talkie) ..which rely on push to talk function with different intervals of transmitting and receiving. Full Duplex systems  simultaneous two-way transmission and reception. E.g. cellular telephone and point to point radio systems  require ‘duplexing’ techniques : 1. using separate freq bands for transmit and receive, 2. users to transmit and receive in certain predefined time intervals.

7 5.1 The Friis Transmission Equation
The Friis transmission equation describes how well the energy is exchanged between transmitter and receiver. Consider a pair of horn antennas with the same polarization and aligned each other.

8 The Friis Transmission Equation (Cont’d..)
The radiated power density from Horn 1 at the location of Horn 2 is : The power received by Horn 2 is product of this power density and capture area A2, written as : The power received at Horn 1 resulting from power emitted by Horn 2 :

9 The Friis Transmission Equation (Cont’d..)
The reciprocity property – the transmission pattern is the same as receive pattern, and the ratio of received power to radiated power will be the same, regardless which pair is transmitting or receiving. Therefore, or Since the directivity and area are independent each other, the ratio must be equal to constant :-

10 The Friis Transmission Equation (Cont’d..)
Generally, We find, r – receiver t – transmitter The ratio is also valid even the antennas are not in line :

11 The Friis Transmission Equation (Cont’d..)
Replace the effective area with receiving area to get : Finally consider, To get:

12 The Friis Transmission Equation (Cont’d..)
This result is known as Friis transmission equation, which addresses on how much power is received by an antenna. Practically, it can be interpreted as the max possible received power, whereby with lot of factors to reduce the received power in actual radio system: impedance mismatch at either antenna polarization mismatch between the antennas propagation effects  leads to attenuation or depolarization mutlipath effects  partial cancellation of the received field.

13 The Friis Transmission Equation (Cont’d..)
Important Notes!! The received power decreases as 1/R2 as the separation between transmitter and receiver increases. It seems large for large distance, but it is much better than the exponential decrease in power due to losses in a wired communication link (coax lines, waveguides, even fiber optic lines)  the attenuation power on Tline varies as e-2αz , with α is attenuation constant of the line  at large distance, the exp function decreases faster than an algebraic dependence like 1/R2 . For long distance communication, radio links perform better than wired links.

14 Example 1 Consider a pair of half wavelength dipole antennas, separated by 1 km and aligned for maximum power transfer as shown. The transmission antenna is driven with 1 kW of power at 1 GHz. Assuming antennas are 100% efficient, determine the receiving antenna’s output power.

15 Solution to Example 1 For 100% efficiency and antennas optimally aligned, For the λ/2 dipole antennas we have Dmaxt = Dmaxr = 1.64 and at 1 GHz, λ = 0.3m,

16 Solution to Example 1 (cont’d..)
In terms of decibels, So finally,

17 The Friis Transmission Equation (Cont’d..)
The Friss transmission equation can also be known as (in terms of receive and transmit) : Whereby, the product of PtGt can e interpreted equivalently as the power radiated by an isotropic antenna with input power PtGt, or effective isotropic radiated power (EIRP):

18 The Friis Transmission Equation (Cont’d..)
For a given frequency, range and receiver antenna gain, the received power is proportional to EIRP of transmitter, and can only be increased by increasing the EIRP  increase transmit power, or transmit antenna gain or both. In any RF or microwave system, impedance mismatch will reduce the power delivered from a source to a load, where the Friss formula can be multiplied by the impedance mismatch factor,

19 The Friis Transmission Equation (Cont’d..)
Max transmission between two antennas requires both antenna be polarized in the same direction. E.g. if a transmit antenna is vertically polarized, max power will be delivered to a vertically polarized receive antenna, while zero power would be delivered to a horizontally polarized received antenna. The polarization mismatch effects is measured by multiplying the Friss formula by the polarization loss factor,

20 5.2 Antenna Noise Temperature
In a receiver, noise is not only generated due to lossy components and active devices, but also by the input of a receiver by the antenna. It might received from external environment, or internally as thermal noise due to losses in the antenna itself. Noise produced within receiver is controllable (with good design and component selection), but noise from environment is uncontrollable, and may exceed the noise level of the receiver.

21 Antenna Noise Temperature (Cont’d..)
Normally, we have the simple case to measure an available output noise power N0, given by: Illustrating the concept of background temperature. (a) A resistor at temperature T. (b) An antenna in an anechoic chamber at temperature T. (c) An antenna viewing a uniform sky background at temperature T.

22 Natural and manmade sources of background noise.
Antenna Noise Temperature (Cont’d..) Natural and manmade sources of background noise. The background noise temperature, TB, is the equivalent temperature of a resistor required to produce the same noise power as the actual environment seen by the antenna

23 Antenna Noise Temperature (Cont’d..)
But when the antenna beamwidth is broad enough that different parts of the antenna pattern see different background temperatures, the temperature now is called as effective brightness temperature, Tb seen by the antenna. This antenna brightness temperature takes into account the distribution of background temperature, directivity and the power pattern function of the antenna

24 Antenna Noise Temperature (Cont’d..)
If the antenna has dissipative loss  the radiation efficiency, erad is less than 1, then the power available at the terminals of receiver is reduced by the factor of erad . This also applies to received noise power, as well as received signal power, so the noise temperature of the antenna will be reduced from the brightness temperature. Therefore, thermal noise will be generated by resistive losses in the antenna, and will increase the noise temperature of the antenna!!

25 Antenna Noise Temperature (Cont’d..)
A receiving antenna connected to a receiver through a lossy transmission line. An impedance mismatch exists between the antenna and the line. The relation between the radiation efficiency of the antenna and the attenuator loss factor is L = 1/erad

26 Antenna Noise Temperature (Cont’d..)
The resulting equivalent temperature, TA is called the antenna noise temperature , with combination of external brightness temperature seen by the antenna and thermal noise generated by the antenna. With Tp is the physical temperature. The antenna noise temperature is a useful figure for a receive antenna because it characterizes the total noise power delivered by the antenna, to the input of a receiver.

27 Antenna Noise Temperature (Cont’d..)
The G/T ratio is another important parameter where the signal to noise ratio (SNR) at the input of a receiver is proportional to G/TA. The SNR at the input to the receiver can be calculated as:

28 Antenna Noise Temperature (Cont’d..)
Where SNR is proportional to G/T of the receive antenna. Only Gr/TA is controllable at the receiver, and others are fixed by the transmitter design and location. G/T can be maximized by increase the gain of antenna  usually minimize reception of noise from hot sources at low elevation angles  but higher gain requires larger and more expensive antenna, and high gain may not be desirable for application of omnidirectional coverage!!

29 Example 2 The Direct Broadcast System )DBS) operates at GHz, with transmit carrier power 120W, transmit antenna gain 34dB, IF Bandwidth 20 MHz, worst case slant angle 300 of satellite to earth of 39,000km. The 18’’ receiving dish antenna has gain of 33.5dB sees an average background brightness temperature Tb = 50K, with receiver LNB noise figure of 1.1dB. Find: EIRP of the transmitter G/T for the receive antenna and LNB system. Received carrier power at the receive antenna terminal Carrier to noise ratio (CNR) at the output of LNB

30 Example 2

31 Solution to Example 2 Convert the quantities in dB to numerical values: 34 dB = 2512, dB = 1.29, dB = 2239 Take the operating frequency GHz, so wavelength m. So,

32 Solution to Example 2 (Cont’d..)
To find G/T, first find the cascaded noise temperature of the antenna and LNB, with referenced at the input of LNB: So then G/T for the antenna and LNB is:

33 Solution to Example 2 (Cont’d..)
The received carrier power is from Friis formula: The CNR at the output of the LNB is:

34 5.3 Radar The operation of monostatic radar (radio detection and ranging) system, (a) A radar antenna transmits a signal to the target. (b) The target scatters this signal, some of which is received by the radar antenna.

35 Radar (Cont’d..) The direction of antenna’s main beam determines the location of the target (azimuth and elevation). The distance or range to the target corresponds to the time between transmitting and receiving EM pulse. The speed of target, relative to antenna, can be determined by observing any frequency shift in EM energy (doppler effect). The radar equation, σs is the radar cross section

36 Radar (Cont’d..) A more popular expression in terms of an effective area of the radar antenna is : The strongest receive power occur when the antenna’s main beam is pointing at the target, D(θ,φ) = Dmax. The received power also be detectable over the noise in the system, so radar will have a minimum detectable power.

37 Example 3 A radar with minimum detectable power specified as 1 pW is 1 km distant from a target with a 1 m2 radar cross section. Operated at 1 GHz the antenna has directivity of 100. Determine how much power must be radiated to enable detection of the target.

38 Solution to Example 3 Solve the radar equation in terms of Prad1 :
At 10 GHz, we have λ = 0.3m, then we get:

39 Introduction to Propagation
The propagating wave between transmit and receive antennas in radio communication channel subjects to variety of effects (amplitude, phase or frequency) :- Reflection (from the ground or large objects) Diffraction (from edges and corners of terrain or buildings) Scattering (from foliage or other small objects) Attenuation (from rain or the atmosphere) Doppler (from moving users) This list covers the important effects for frequencies above 500 MHz.

40 Introduction to Propagation (Cont’d)
For frequencies below, about 100 MHz, other propagation effects can be important: ground surface waves atmosphere ducting ionosphere reflection Generally, propagation effects have the effect of reducing the received signal power, thus limit the usable range or maximum data rate of a wireless system.

41 5.4 Free Space Propagation
From Friis equation, the received power decreases as 1/R2 with distance from the transmitter  path loss  only applies to propagation in free space where no reflection, scattering or diffraction along the path between transmitter and receiver. Practically, the Friis equation can be used if there’s essentially a single line of sight (LOS) path between transmitter and receiver  usually implies that at least one of the link antennas has a narrow beamwidth (high gain) e.g. point to point radio links, satellite to satellite links and earth to satellite links.

42 Free Space Propagation (Cont’d..)
A point to point radio link with a single line of sight propagation path A cellular telephone channel having multiple propagation paths.

43 Free Space Propagation (Cont’d..)
Multipath propagation is particularly likely when the antennas have broad beams (low gain) and in close proximity to the ground or other large reflecting structures i.e. buildings, vehicles or foliage. May be no LOS path at all!!  common situation for cellular phone located in a building or vehicle. Communication still possible in multipath or even in the absence of LOS path  but the total signal voltage received will have varying degrees of destructive or constructive interference due to the variable phase delays that occur at different paths  Friis can not be used!

44 5.5 Ground Reflections Consider an LOS path with a single reflected signal  it’s useful for ground reflections, which frequently occur in practice + reflections from building, vehicles etc. By Snells Law, the incident wave is specularly reflected from the ground, so that angle of incident = angle of reflection.

45 Ground Reflections (Cont’d..)
The electric field of an arbitrary antenna can be expresses as: And with the well known Friis equation, we can write the received voltage due to direct wave as: Rd is path length of direct ray, Z0 is receiver load impedance

46 Ground Reflections (Cont’d..)
The constant C : Assume d>>h1 and d>>h2, so the Rd can be approximated using Taylor expansion to get: Similarly for received reflected voltage,

47 Ground Reflections (Cont’d..)
With reflection coefficient close to -1 due to angles of incidence close to grazing (small θ ), and then combine the direct and reflected voltages, give: Assume Rr ≈ Rd ≈ d with negligible error because d>>h1 and d>>h2, it reduces to:

48 Ground Reflections (Cont’d..)
Where the magnitude of last factor  path gain factor , F Observe 0 ≤ F ≤ 2, so that the received voltage maybe doubled (power quadrupled) when 2 signals in phase, or reduced to zero if complete destructive interference occurred.

49 Ground Reflections (Cont’d..)
Define the angle ψ as the elevation angle of the receive antenna as seen at the transmitter So then the path gain factor can be rewritten as

50 Example 4 The height of cellular telephone transmit antenna operating at 1800 MHz is 8.33m. If the distance to the receiver is 1 km, find the smallest receiver antenna height that will maximize the receive signal voltage

51 Solution to Example 4 At 1800 MHz, wavelength = 0.1667m. So,
The path gain factor has a maximum when the argument of sin function is π/2, 3π/2.. for n=0,1,2,.. So the min height for max path gain is,

52 Path Loss for Ground Reflections
The received signal in the presence of a ground reflections varies according to the path gain factor and is not simply a function of separation distance. Applying Taylor series gives the received voltage as:

53 Path Loss for Ground Reflections
Since the signal voltage decreases as 1/R2, the received signal power will decrease as 1/R4 This result applies when k0h1h2/d < 0.3 or for

54 Typical Path Loss The resulting path loss can be expressed as 1/Rn , where the exponent may vary depends on the environment. Environment Path Loss Exponent Free Space 2 Urban 2.7 – 3.5 Shadowed Urban 3 – 5 In Building LOS 1.6 – 1.8 In Building Shadowed 4 – 6 Factory Shadowed 2 – 3 Retail Store 2.2 Office – Soft Partitions 2.4

55 5.6 Ionosphere Propagation
UV light, gamma light and cosmic particles such as electron and proton will ionized gas to generate equal layer labeled as layer D, E and F in ionosphere layer. F1 Virtual height F2 h E D EARTH

56 Ionosphere Propagation (Cont’d..)
The frequency range that can propagates in the ionosphere layer is within 50 kHz to 30 MHz. Frequencies above 30 MHz can’t get reflected by ionosphere layer where it could get through it. Meanwhile, frequency range between 50 kHz to 30 MHz can get through the lowest layer up to F layer. Frequencies lower that 50 kHz can propagate lower than the ionosphere layer.

57 Ionosphere Propagation (Cont’d..)
The highest critical wave frequency that incident vertically and the ionosphere able to reflect with is called critical frequency, fc. N = ion density The maximum frequency that can propagates in the ionosphere is called maximum usable frequency, fMUF, determined by critical frequency of F layer and incident angle.

58 Ionosphere Propagation (Cont’d..)
Φi is the incident angle of waves in ionosphere, d is the distance between transmitter and receiver, and h is the virtual height of F layer. The minimum frequency that can propagates in the ionosphere is called least usable frequency, fLUF, determined by critical frequency of D layer and incident angle.

59 Ionosphere Propagation (Cont’d..)
Refraction index, n in the ionosphere can be determined as: Where its relationship with incident and critical frequency: f = incident frequency

60 5.7 Troposphere Propagation
Troposphere layer is the bottom layer in atmosphere from ground to 10km at poles and up to 18km at equatorial. Several mechanisms in troposphere propagation, i.e. direct propagation, refraction, ducting or scattering. Refraction Three types of refraction: Normal refraction Superfraction Trapping

61 Troposphere Propagation (Cont’d..)
Normal Superfraction Trapping

62 Troposphere Propagation (Cont’d..)
Ducting When trapping occurs, wave propagation is concentrated on small area, i.e. same as in waveguide  gives optimum signal received, even at far distance away  centimeter wave at hot area and wide ocean area, e.g. Mediterranean and Carribean. Earth Troposphere layer

63 Troposphere Propagation (Cont’d..)
Troposphere Scattering It occurs due to inhomogeneous troposphere layer structure at antenna’s cross section path between the transmitter and the receiver. Earth Antenna path

64 Troposphere Propagation (Cont’d..)
This happen because due to discontinuity of dielectric constant in troposphere layer. The propagation is useful for frequency between 400 MHz to 5 GHz with distance between 300 km to 600 km. However, the received power will be too small. Therefore, high power is needed (1 to 10kW) for transmission. Scattering is also happen at the bottom of ionosphere with frequency range between 30 MHz to 60 MHz and distance between 1000 km to 2000km with power of 50kW.

65 5.8 Vegetation Propagation
Few models being used for estimation of radio wave loss in vegetation area. The first model, where it considers the received wave comes from direct rays and ground reflections. The E field in free space can be described as : Assume that h<<d and H<<d

66 Vegetation Propagation (Cont’d..)
From previous chapter, The received power after propagated in vegetation area, φ = flux power at distance d Ae = antenna effective area Af = vegetation loss As = earth unevenness loss For flat ground surface, As = 1.

67 Vegetation Propagation (Cont’d..)
Here, So that,

68 Vegetation Propagation (Cont’d..)
For effective area of aperture antenna, So then substitute into previous equation, Where in terms of decibels, Where Lf= 10 log Af

69 Vegetation Propagation (Cont’d..)
Therefore, the radio wave loss in vegetation area : Or if d in km,

70 Vegetation Propagation (Cont’d..)
The second model, from ‘Jansky and Bailey’, considers the frequency of radio wave. Consider, Replace with λ=c/f and As = 1 for flat surface,

71 Vegetation Propagation (Cont’d..)
Where in terms of decibels, Therefore, the radio wave loss in vegetation area :

72 5.9 Urban Propagation Urban area can be described as rough surface area, where it increases the interference of direct waves and reflected waves. The Lf variable in vegetation loss equation can be replace with urban loss factor, b. Hence, the loss is: Where, f = frequency in MHz, U is soil usage factor and Hb is the height difference between the transmitter and receiver.

73 Example 5 A moving communication system station is operating at 900 MHz. The transmitting antenna’s gain and height are 3 dB and 5m respectively. The soil usage factor at that area is 30%. A moving car as receiving side is at 1 km distance from the station with SNR = 20 dB and 10 kHz bandwidth. The antenna’s height is 2m with 2 dB gain. Determine the received power so that the receiver can get a signal from the station.

74 Solution to Example 5 Substitute into the urban loss factor equation,
So that, The received power must meet this :

75 Solution to Example 5 (Cont’d..)
Where, NR = kTB = 10 log (1.38x10–23 x 290 x 10 x 103) = –164 Finally, Or,

76 Attenuation Attenuation decrease in signal power due to losses in the propagation path. Material Frequency Loss, dB Concrete Block Wall 1300 MHz 13 Sheetrock 2 x 3/8” 9.6 GHz 2 Plywood 2 x 3/4” 4 Concrete Wall 8-15 Chain Link Fence 5-12 Loss Between Floors 20-30 Corner in Corridor 10-15

77 Wireless Communication Systems and Propagation
End


Download ppt "Wireless Communication systems"

Similar presentations


Ads by Google