Radio Wave Propagation Prepared by Dr. Abbou Fouad Mohammed, Multimedia University, 1 Radio Wave Propagation Terminology Free space (propagation medium) u Electromagnetic waves u Transmitting and receiving antenna u Prepared by Dr. Abbou Fouad Mohammed, Multimedia University, 2 Transverse Electromagnetic Waves Dir e n ctio of P ag rop atio n z y Magnetic Field Electric Field x u The waves propagate as transverse electromagnetic waves (TEM) - i.e. the electric field, the magnetic field, and the direction of travel of the waves are all mutually perpendicular. 3 Prepared by Dr. Abbou Fouad Mohammed, Multimedia University, Speed & Wavelength of em Waves u The speed of propagation (ν) and the wavelength (λ) of an electromagnetic wave are given, respectively, by: v= c v and λ = f ∈r where c = 3x108 m/s, ∈ r = medium’s relative permittivity or dielectric constant, and f = frequency of wave in Hz. Prepared by Dr. Abbou Fouad Mohammed, Multimedia University, 4 Characteristic Impedance u u The characteristic impedance of a medium is the ratio of the electric field intensity and the magnetic field intensity, i.e., η = E/H. For free space, η = 120π = 377 Ω. For other media: (η η is also called, the intrinsic impedance) η= µ 377 or ∈ ∈r where µ = medium’s permeability, in H/m and ∈ = m e d i u m ’s permittivity in F/m Prepared by Dr. Abbou Fouad Mohammed, Multimedia University, 5 Electromagnetic spectrum Radio waves, infrared, visible light, ultraviolet, X rays, and gamma rays are all different forms of electromagnetic radiation. Prepared by Dr. Abbou Fouad Mohammed, Multimedia University, 6 Relation between power density and transmitted power u u Let P t be the transmitted power into free space. The power density at any point at distance R is SR = Pt 4πR 2 Watts / m 2 Prepared by Dr. Abbou Fouad Mohammed, Multimedia University, 7 Example 1: u The power density from a point source in free space at 20 km is 200 micro-watts/m 2. Find the power density at 25 km away from the source Prepared by Dr. Abbou Fouad Mohammed, Multimedia University, 8 Relation between power density and electric field u u u For an uniform plan wave the E and H are related by: η = E R/HR ER is the rms value of the electric field The average poynting vector at distance R from the transmitter SR = Hence ER = Pt E 2R E 2R = E xH = = R R 4πR 2 η 120π 30Pt R Volts / m 9 Prepared by Dr. Abbou Fouad Mohammed, Multimedia University, The Effective Isotropic Radiated Power Since a transmitting antenna focuses energy in a specific way, it has “gain” over an isotropic radiator in a particular direction. One can speak of the effective isotropic radiated power: EIRP = G t Pt where Pt = total transmitter power, and G t = gain of transmitter antenna If a transmitting antenna has a gain Gt , then ER = Prepared by Dr. Abbou Fouad Mohammed, Multimedia University, 30Pt G t R 10 Transmission between two antennas 11 Prepared by Dr. Abbou Fouad Mohammed, Multimedia University, Friis transmission formula u The average power density at the receiving antenna is Sr = G t Whereby u Pt ξDP ξAP = t t 2t = t 2 t 2 t 2 4 πR 4 πR λR Dt = 4π λ2 At Has been used The power intercepted by the receiving antenna with effective area A r is Pint = S r A r = Prepared by Dr. Abbou Fouad Mohammed, Multimedia University, ξ t Pt A t A r λ2 R 2 12 Friis transmission formula u The receiving power Pr delivered to the receiver is ξA PA λ Pr = ξ r t 2t t2 r = G t G r Pt λR 4πR 2 Or λ Pr = G t G r Pt 4πR 2 Friis transmission formula 13 Prepared by Dr. Abbou Fouad Mohammed, Multimedia University, Path Loss u As the signal propagates there will be loss due to the spreading of the wave outward from the source Pr = G t G r Pt 4 πR λ 2 P 4πR 1 = t λ G t G r Pr 2 hence P c 4 πdf Path Loss = 10 Log t = 10 log − G t (dBi ) − G r(dBi ) ; R = d and λ = f c Pr 2 Or Path Loss = Pt ( dB) − Pr (dB) = 33 .44 + 20 log( d km ) + 20 log( f MHz ) − G t ( dBi) − G r ( dBi) Prepared by Dr. Abbou Fouad Mohammed, Multimedia University, 14 Example 2: In a satellite communication system, the free-space conditions may be assumed. The satellite is at a height of 36,000 km above the earth, the frequency used is 4000 MHz, the transmitting antenna gain is 15 dB, and the receiving antenna gain is 45 dB. Calculate (a) the free space transmission loss (b) the received power when the transmitted power is 200W. Prepared by Dr. Abbou Fouad Mohammed, Multimedia University, 15 Radio Propagation Mechanisms: Effects of environment During the propagation of electromagnetic waves the waves experience: Different environmental characteristics: land, sea, space, etc. Reflection u Refraction u Diffraction u Interference u Prepared by Dr. Abbou Fouad Mohammed, Multimedia University, 16 Reflection Radio waves behave like light waves: u These are reflected by a surface where the angle of incidence, θi = the angle of reflection, θr. Incident Ray Normal θi θr Reflected Ray Conductor The reflection coefficient is defined as:. ρ = E reflected E incident ρ is unity for a perfect conductor and will be less than 1 in case of practical situations 17 Prepared by Dr. Abbou Fouad Mohammed, Multimedia University, Refraction u Radio waves will be bend or refracted when they travel from one medium with refractive index, n 1 to another with refractive index, n 2. The angles involved are given by : n1 θi n1<n2 sinθ t v 2 ε = = 1 sinθ i v1 ε2 ε1and ε 2 n2 θt Prepared by Dr. Abbou Fouad Mohammed, Multimedia University, are dielectric constants of medium 1 and 2 respectively. 18 Diffraction u Diffraction is the phenomenon which results in radio waves that normally travel in a straight line to bend around an obstacle. Direction of wave propagation T R Obstacle Obstruction u Diffraction generates unwanted side lobes and prevents the desired narrow pencil beam of radiation. 19 Prepared by Dr. Abbou Fouad Mohammed, Multimedia University, Interference u When two em-waves that left the source, travel different length paths and arrive at a point at the same instant of time, l there will be interference due to the phase shift of the two waves.. T R ht hr Phase shift! u When the antenna is located near ground the direct signal and ground reflection will reach the receiving antenna. Prepared by Dr. Abbou Fouad Mohammed, Multimedia University, 20 Interference Lobes Null Null Ground surface u If the path difference corresponds to half wavelength then there will be phase opposition between the two, resulting in cancellation at the receiver. u If the path difference is one wavelength, the two waves reinforce each other resulting in lobe and null antenna pattern. If the target lies in null direction, then it cannot be detected. u Prepared by Dr. Abbou Fouad Mohammed, Multimedia University, 21 Radio Wave Propagation Modes u depends on several factors such as u u u frequency of operation, distance between transmitting and receiving antennas Radio wave propagation can be : u u u Ground wave or surface wave propagation Sky wave or ionospheric propagation Space wave propagation Prepared by Dr. Abbou Fouad Mohammed, Multimedia University, 22 Prepared by Dr. Abbou Fouad Mohammed, Multimedia University, 23 Ground Wave or Surface Wave Propagation u In ground wave propagation the electromagnetic wave is guided along the surface of the earth. u The ground wave is of practical importance at broadcast and lower frequencies i.e. for medium waves, long waves and very long waves. l l It is used in local broadcasting. All the broadcast signals received during daytime are due to ground wave propagation. Prepared by Dr. Abbou Fouad Mohammed, Multimedia University, 24 Ground Wave or Surface Wave Propagation u The ground wave, therefore, suffers attenuation while propagating along the curvature of the earth, depending on: l u frequency, surface irregularities, permittivity and conductivity. Earth's attenuation increases as the frequency increases and hence the mode of propagation is suitable for low and medium frequencies up to 2 MHz only. 25 Prepared by Dr. Abbou Fouad Mohammed, Multimedia University, surface wave attenuation due to wave diffraction and tilting u As the wave progress over the curvature of the earth, the wave fronts starts gradually tilting more and more. u The tilt of wave causes short circuit of the electric field component and hence the field strength goes on reducing. Wave tilting Earth surface Prepared by Dr. Abbou Fouad Mohammed, Multimedia University, 26 surface wave attenuation due to wave diffraction and tilting The surface wave dies because of the losses mentioned above. It may be noted that maximum range of surface wave propagation depends u u not only on the frequency but on the power as well. The range of transmission can be increased by increasing the power of the transmitter in the VLF band but this method can not be effective at the MF band where the tilting due to diffraction is more effective. u 27 Prepared by Dr. Abbou Fouad Mohammed, Multimedia University, u The field strength of ground wave at a distance “d” from transmitter is given by E= u 120πh t I λd The voltage induced in receiving antenna placed at this point is given by V = Exhr = 120πh thr I λd where ht and hr are the effective heights of transmitting and receiving antennas respectively. I = antenna current, λ = wavelength, d = distance from transmitter. Prepared by Dr. Abbou Fouad Mohammed, Multimedia University, 28 Example 3: A 150 m antenna transmitting at 1.2 M Hz has an antenna current of 8A. Find the voltage received by a receiving antenna of height 2 m and 40 km away from transmitting antenna Prepared by Dr. Abbou Fouad Mohammed, Multimedia University, 29