Wave Propagation | Wave Propagation : I The ground wave or surface wave is of practica| onal important at broadcast and lower frequencies. jj is suitable for medium waves, long waves. Since the ground wave is guided along the surface of Important Points to Remember the earth es mode is eo zn eTiah pone transmitting and receiving ant are pl Ca as ee close tothe surface of the earth. In this mode FE, electromagnetic waves. waves are vertically polarized. * The orientation of the electric field vector with b) Sky wave or ionospheric wave propagation Tespect to the earth's surface is called (between 2 to 30 MHz) : Sky waves are polarization of plane EM wave. practically important for medium waves and * Types of polarization short waves. Sky wave propagation takes place 1. Linear polarization - after reflection from the ionosphere. A long distance point to point communication is possible by this mode. The signals received due to sky wave propagation scatter from fading in which signal strength varies with time. ©) Space wave propagation (above 30 MHz) : Suitable for VHF bands (between 30 MHz to 300 MHz), UHF microwaves and TV, radar communications. In this mode electromagnetic waves transmitted reach receiving antenna directly or after reflection from troposphere. Tropospheric scatter propagation or forward 5.1 : Introduction to Wave Propagation Leni tiroduction to Wave Propagation — a) Horizontal polarization b) Vertical polarization 2. Circular polarization 3. Elliptical polarization Properties of radio waves : 1. Reflection : When electromagnetic wave strikes boundary at two media, the part of the incident power gets reflected. 2 Refraction : When an electromagnetic wave is incident obliquely at the boundary of two d) media, the wave changes its direction while passing from one medium to another with change in velocity of propagation, which is scatter propagation (Above 30 MHz, UHF and microwave range) ‘The UHF and microwave signals are propagated referred as refraction. 5.2 : Different Modes of Wave Propagation Q1 Explain different modes of wave propagation Ge [INTU : Marks 5] ‘Ans. : These are different paths of wave propagation > waar are the : as types of ground waves. by which the transmitted signal can reach the a ne receiving antenna. a) Ground wave or surface wave propagation (upto | Ams.: The two types of ground waves are space __waves and surface waves. The waves which travel 2 MHz) : a ere neers eae eden na) y 6-1 beyond line of sight propagation through the forward scattering in the _ tropospheric irregularities. Hence it is also referred as tropospheric scatter propagation. 3 : Ground Wave Propagation OLAIINeU WILT Ud IIS5-2 Antennas and Propagation a 4 from the transmitter to the receiver directly without any reflection are called direct waves or space waves. ‘The waves which propagate through reflection from the earth's surface are called ground reflected waves or surface waves. Q.3 What are the factors that affect propagation of radio waves ? ESE [INTU : Dec.-16, Marks 3] Ans.: When radio wave or electromagentic (EM) wave travels from transmitter to receiver, many factors influence the propagation of wave. Some of the important factors are as follows i) Characteristics of earth such as conductivity, permittivity, permeability. Curvature of the earth, magnetic field of the earth, roughness of the earth. Frequency of operation. Height and polarization of transmitting antenna and transmitter power. v) Characteristics of ionospheric regions. vi) Distance between transmitter and receiver. vii) Refractive index and permittivity of troposphere and ionospheric regions. ii) iii) iv) Q.4 Discuss the salient features of ground wave propagation. SP [INTU : May-18, Marks 5] Ans, : 1) The ground wave propagate along the surface of the earth 2) The ground waves are produced in vertically polarized antennas which are placed very closed to earth. 3) This type of propagation is important at broadcast and lower frequencies, upto frequencies at 2 MHz. 4) Strength of ground waves varies according to characteristics of earth. 5) When ground waves propagate along the surface of the earth, the charges are induced on the surface of earth. These charges travel along the wave and current gets induced. The earth acts as leading capacitor. Q.5 Discuss the phenomenon of ground wave Propagation. Explain the polarization used. Wave Propagation Ans.:e The ground wave is a vertically polarized ‘wave that travels along the surface of the earth. For the ground wave propagation, vertical antennas are usefulThe ground wave is always a vertically polarized wave. In the ground wave, as field changes with time anc travel of the wave, varying voltage is induced ir the earth and current flows . Due to this, resistive ‘and dielectric losses take place in the ground Hence, as the ground wave travels away from th transmitting antenna, it gets attenuated. Ti minimize the losses, the transmission path must be over "ground" with high conductivity. * The important consideration for the ground wav Propagatioh is the lower frequencies. Because the ground losses increase rapidly with frequency. If the surface of the earth is rough, then th scattering of the reflected wave takes place. Witl the scattering of the wave, the amplitude of th reflected wave reduces considerably as compared t that reflected from the smooth surface. Thi “roughness” of the surface responsible for the scattering of the reflected wave can be obtained by using Rayleigh criterion. The Rayleigh criterion i given by, R = Szosing x ~@) where R= Measure of roughness ‘* Depending on the value of R, the surface can b considered as either smooth or rough. * Consider a medium with finite conductivity o anc the dielectric constant e. The Maxwell's firs equation for the fields in above medium can b written as, sae OE a= = eS to =U) * For the time varying electric field with sinusoids variation, the electric field can be written as, E- F cit eas Hence we can write, F = FeGiny2 = ja(F oie) = (aE (4 Scanned with CamSAntennas and Propagation 5-3 oE Substituting value of E in terms of © in equation (2), we can write, oE z| aE Bede sae soeal evo 3 Ux = reed een) +9] | (az) -8 Let € “(la ) © jae) * The complex dielectric constant e’ is the equivalent complex dielectric constant of a dielectric with Partial conductivity. * Let us consider that the earth's surface is partially conducting dielectric. The wave incident at the surface of the earth through the air. * Hence consider medium 1 as air whose dielectric constant is given by &> which is the dielectric constant of the free space. For the medium 2 ie. the surface of the earth, the dielectric constant is given pet * Let h; and h» be the heights of the transmitting and receiving antennas respectively from the surface of the earth as shown in the Fig. Q.5.1. The waves radiated by the transmitting antenna strikes the surface of the earth at any random angle Transmitting Direct wave (space wave) ‘antenna’ Se eee ar Receiv ‘omennae Fig. Q.5.1 Schematic of direct wave and ground reflected wave Reflection Coefficient for Horizontal Polarization The horizontal polarization is also called perpendicular polarization because the electric field vector is perpendicular to the plane of incidence and parallel to the reflecting surface. For the horizontal (perpendicular) polarization the reflection coefficient is given by, iny ~ \(€s — ix) - costy R —— HO siny + (er — 2) = costy | =O) £ ecm where & 7 3, ond X= Ger Reflection Coefficient for Vertical Polarization In the case of vertical polarization, the electric fg vector is parallel to the plane of incidence while ty magnetic field vector is parallel to the reflec ‘surface. Hence the vertical polarization is cally parallel polarization. For the vertical polarization the reflection coefficient in general is given by, Rca ae Rea ® (e, -ix) sinw+ Y(€; —33) ~c0s?y € 6 where ¢, = > and x= Ge From equations (8) and (9) it is clear that the reflection coefficient Ry and Ry, both ar complex in nature. Hence we can represent these quantities in polar form as, Ry = |Ry|ZRy And Ry = |Ry|ZRy Field Strength of Ground Wave at a Distance * The field strength of the signal at a distance from the transmitting antenna due to the ground wave can be obtained by solving the Maxwell's equafin and it is given by, 120nby-hy I, E= a Vim (10) Where h, and h, = Effective heights of the transmitting and receiving antennas respectively; 1, = Antenna current, i= Wavelength, d= Distance a point from the transmitter Attenuation Characteristics for Propagation: * According to Sommerfield, for a flat earth, the Bed strength for a ground wave propagation is give" by, Ground Wa" = (tt) TECHNICAL PUBLICATIONS® - An up trust or knowedge OLdINICU WILT cainS‘Antennas and Propagation where Ey = Ground wave field strength at the earth's surface at unit distance without considering earth losses, Eg, = Ground wave field strength, ‘A = Attenuation factor accounting for earth losses + Now this unit distance field strength Eo depends on two factors mainly, such as power radiation of the transmitting antenna and the diversity in the vertical and horizontal planes. + For a vertical antenna which is non-directional in the horizontal plane, the radiation field produced is proportional to the cosine of angle of elevation Then the field at unit distance is given by, 300VP Ey cree (12) where P= Radiated power in kW + The attenuation factor A depends on i) Frequency, ii) Dielectric constant, ii) Conductivity of the earth © Thus the attenuation factor A is the function of frequency, dielectric constant and conductivity of the earth. For analysis purpose it is expressed interms of two auxillary variables such as, numerical distance p and phase constant b. A) For vertically polarized wave: For the vertically polarized wave, the two variables p and b of the attenuation factor A are given as follows. = 08x 102 )o ¥(H2) where x. O/em = B) For horizontally polarized wave : p and b are given by, Wave Propagation From Fig. Q.5.2 we can draw following conclusions. 000 1 (numerical distance) foi eet enO an T00 Fig. Q.5.2 Variation of A with numerical distance p for different values of the phase constant b i) For p < 1: The ground attenuation factor ‘A almost remains constant at unity and slowly reduces with increasing p. Then the ground losses are not significant for p <1. ii) For p> 1: As the numerical distance p becomes greater than unity, the attenuation factor decreases rapidly. iii) For p > 10 : For larger p, the ground attenuation factor is almost inversely proportional to the square of the distance. Salient Features of Ground Wave Propagation 1) The ground waves propagate along the surface of the earth. 2) When the ground waves propagate along the surface of the earth, the charges are induced on the surface of the earth. These charges travel along the wave and hence the current gets induced. 3) While carrying induced current, the earth acts as a leaky capacitor. 4) The ground waves are produced in vertically polarized antennas which are placed very close to surface of the earth. 5) The ground waves are important at broadcast and lower frequencies. These can be used upto 2 MHz. 6) According to the characteristics of the earth, the strength of ground wave varies. These waves are not affected by the changes in the atmospheric conditions. 7) The variations in surface or type of the earth affect propagation losses considerably. a2) ae ) OUdINICU WILT ULdllisAntennas and Propagation 5-5 8) The maximum range of ground wave Propagation depends on the frequency and power of the transmitter. Q6 Write short note on wave tilt. 3 [Marks 5] Ans. : ¢ In general, for a vertically polarized wave, a forward tilt is observed at surface of the earth. ‘* By how much amount the wave tilts depends on conductivity and permittivity of the earth. * Basically the electric field vector has two components; one parallel to surface of the earth while other perpendicular to surface of the earth * But due to even a slight forward tilt, these two components are not in phase and the electric field is found to be elliptically polarized. * Thus wave tilt of ground wave is defined as the change in orientation of the vertically polarized | wave at the earth's surface. © For a good conductor, over most of the frequency ranges and conductivity values, the surface impedance of the earth is given by, Pe (fates lerroat p where 1 + Permeability of the earth, o = Conductivity of the earth © = Permittivity of the earth The horizontal and vertical components of the electric field strength are respectively given by, En = 5sZ, And Ey =H-no * Then the ratio of the horizontal component to vertical component is given by, Penge eee Ey 1 L foo 1 (0? +e? )2 (2) The angle tan~!-S- gives the tilt angle as it indicates by how much amount wave is tilted. Important Points to Remember * The space wave propagation is effectively use for frequencies above 30 MHz ie. for VHF ang higher frequencies. * The ground waves above 30 MHz undergo high attenuation with considerable attenuation within very short distance. ‘© At these frequencies the sky wave propagation fails because the ionosphere cannot retract the frequencies back to the earth. Under such conditions space wave propagation is preferable, Q7 Derive and expression for the variation of field strength of space wave with antenna height and distance Involved. What happens when the distance Is large. OR Show that the field strength due to space wave is given by E = 2epsn( 2) IG [INTU : Dec.-07, 08, 09, May-16, Marks 5] ‘Ans.: © The space wave propagation is effectively used for frequencies above 30 MHz i.e. for VHF and higher frequencies. * Because the ground waves above 30 MHz undergo high attenuation with considerable reduction in the amplitude within very short distances of the order of few 100 meters. Moreover at these frequencies the sky wave propagation fails because the ionosphere cannot refract these frequencies back to the earth. Under such conditions, the space wave propagation is the best option above 30 MHz. * The space wave propagation is through troposphere hence such propagation is limited to few hundreds of kilometer. Flold strength relation for surface wave * Consider that two antennas ie.transmitting antenna and receiving antenna are placed above the ground anywhere in the troposphere but not more than 15 km above from the surface of the earth as shown in the Fig. Q7.1. © The power radiated in the form of radio waves by the transmitter (located at point T) may be received by the receiver (located at point R) by two ways; i) By means of a direct wave traveling from transmitting to receiving. antenna directly along TECHNICAL PUBLICATIONS® - An up thrust for knowledge OUCdINICU WILT cansAntennas and Propagation Wave Propagation Indirect wave. (Ground reflected wave) Fig. Q.7.1 Space wave propagation illustrating direct and Indirect waves path TR. Such a direct wave is also called free space wave. Or ii) By means of ground reflected wave (indirect wave) along path TGR . » Thus at receiver antenna, the total field strength is equal to the vector addition of the fields of these two waves, As the propagation deals with a wave reflected from ground we must consider the coefficient of reflection of the ground which depends on factors such as conductivity of the earth, dielectric constant of the earth, frequency of the wave and polarization of the wave. ‘Also the angle made by the incident ray measured with respect to horizontal axis is also important To simplify analysis, let us assume that the distance d between two antennas along the earth's surface is very large as compared with heights h, and h, of the transmitting and receiving antenna respectively. From the concepts of optical science, a point G at which the incident wave strikes ground is located by obtaining image point R’ at distance h, below the ground level and then joining points T and R’ with a ray passing through G. Let r, be the direct distance between points T and R while r, be the distance between points T and R through G for the ground reflected wave (IR = TG + GR). ‘As 1, is approximately same as ry, the two waves received by the receiver at point R have equal amplitudes but with phase differences due to differences in path lengths. © Alongwith this, the phase reversal occurs at point G. Thus the coefficient of reflection is considered as p =12180" * From the geometry, = \d? +(b, hy)? z - aye ee © Neglecting higher order terms and simplifying, we can write, =n, cites ~() 2d Similarly we can write, | ro yd? + (hy +h)? fey - years 1 = d+ ~@) ‘© Hence subtracting equation (1) from equation (2), the path difference can be given as, ge [ Oe] [ate] (hy +h)? =(hy =h, 2d pret I TECHNICAL PUBLICATIONS® - An up thrust for knowtodge OCAINICU WILT Udillo'Wave Propagatioy Antennas and Propagation Bes ~ 3) rere thes + Hence the phase difference introduced by the path difference is given by, dn, Qhshy _ 4ndihy 4 ~@ Deere Id . + With this phase difference obtained, we can write two waves at receiver along path 1 and r, as, Ep coswt and -E, cof ors APE) respectively. Note that the minus sign in the expression for the as ground reflected wave along r, indicates phase reversed at point G. Now that resultant wave at receiver is the vector sum of two waves, hence we get, 4n ib ) By = Fyeowwr-Fyco{ ore ( ) Simplifying cosine Sone ore ge ‘s 4mhyh, ) [(4ehyh ee ce a ip 7 2Ep sin] 2 in Assuming "s+ <> hy hence R’ hi >> h7. Thus equation ( becomes, d= 2h, & d, = 28h, wal * Similarly rewriting equation (2), we get, (dg? = (R +hP (RP = 2h. +h? ( * But R’ >> he ie. R’ he >> h?. Hence equation becomes, d= 2h, TECHNICAL PUBLICATIONS® - An up tnt fr knowiedge OCdIINeU with! Udl nS‘« The maximum radio range daw is given by, one = dy + dy ‘* Substituting values of di and dz from equations (4) and (6), we get, daax ~ V2RH, + (2Rh, = @ Ble + But the radius R’ is times greater than the ideal value of the radius of the earth. ie. r-$r * Hence equation (7) becomes, danse = (feos (Sen, + 8) «The radius of the earth is given by R = 6370 km. Substituting value of R in equation (8), we get, Ogae = fx TOK I, + fEx TOK, wa (9) * We can further modify expression as, Agae = [8x 637% 108 hy + S637 108 xh, . a g 3 1 day = | fx 637%8, + [Se6a7n, pato 4.12 (iy +] (10) | *In equation (9), d is expressed in km and the heights of transmitting and receiving antennas are expressed in meter only. Tropospheric's Propagation Q.10 Describe the troposphere and explain how ducts can be used for microwave propagation. BGP [INTU : Nov.-37, Deci-16 Marks 10] ‘Ans. : Troposphere is the nearest region in the atmosphere from earth's surface around 10 km to 20 km above the earth's surface. ‘©The gas components in the troposphere remain almost constant in percentage with increase in height. * Water vapour components drastically decrease with increasing height. © The significant property of the tropospheric region is that temperature decreases with increase in the height. The troposphere is also called region of change. * At a certain height called critical height above troposphere the temperature remains constant for narrow region and then increases afterwards. This region is called tropopause. Duct propagation : The VHE, UHF and microwave frequencies are the frequencies which are neither propagated along the surface of the earth nor reflected by ionosphere. But in the troposphere region, the high frequency waves are refracted and transmission takes far beyond Line Of Sight (LOS) distance. An atmosphere where the dielectric constant is assumed to decrease uniformly with height to value equal to unity at which air density is supposed to be zero is commonly called normal atmosphere or standard atmosphere. There are different air regions or layers one above other with different temperatures and water vapour contents. In one of the regions, there is a region where “Ht is negative. In this region, the curvature along which the radio waves pass is slightly greater than that of the earth. Due to this, the wave originally directed almost parallel to the surface of the earth gets trapped in such regions. The energy originating in this region propagates around curved surfaces in the form of series of hops with successive reflections from the earth as shown in the Fig. Q.10.1. This phenomenon is called super refraction or duct propagation. ‘Trapped wave propagating in series of hops Fig. Q.10.1 Duct propagation TECHNICAL PUBLICATIONS® - An up thus for knowledge Scanned with CamS* Two boundaries of surfaces between two air layers form a duct which guide the radio waves between walls ie. boundaries. * The concepts like line of sight and diffraction cannot be applied when the wave propagates through duct and it is found that the energy travels high distances round the earth without much attenuation. * The concept of wave trapping can be considered as @ phenomenon similar to waveguide. * But the main difference between waveguide and duct propagation is that in waveguide all the modes are confined within guide only. But in case of duct propagation, part of energy within duct May escape to the space as shown in the Fig. Q.10.2. Energy leakage Transniting ‘antenna Earth surface Fig. Q.10.2 Duct propagation as leaky waveguide There is a limit on the wavelength: of the signal of maximum value Ama to be trapped in duct. It is the maximum wavelength for which the duct Propagation holds good. If the wavelength of the signal exceeds the value Amax, then duct effect vanishes almost completely. The value of 2, given by, Amax = 25hg fA tm x 10-6) vm (I) where Aitm = Change in tq value across height of duct hg = Height of duct + In general, the duct height hg ranges from 10 to hundreds of meters. While the Au, value is typically 50 units. So considering these values, the phenomenon of duct propagation is found mostly in UHF (ultra high frequency) and microwave frequency regions. Table Q.10.1 specifies the values Of Amax and corresponding duct heights. Table 0.10.1 + Moreover the duct propagation is possible only i height of transmitting antenna is less than that of duct. height. If the transmitting antenna exists considerably above duct, there’ is comparatively less effect of presence of duct on the signal either inside or above duct. Q.11 Explain salient features of tropospheric scatter propagation. US[INTU : May-16, Marks 5] ‘Ans.:* The tropospheric propagation includes Propagation at VHF, UHF and microwave signals beyond the horizon. It is also known as forward scatter propagation or troposcatter. ‘+ Using tropospheric scatter mainly UHF and microwave signals are propagated beyond the line of sight. . + Radio waves detract or bend around curved surface of earth, The strength of the detracted field is decided by roughness of earth's surface in shadow zone. The field strength in shadow zone is very large as compared to that for the smooth earth's surface. * In addition due to the air turbulence eddies ot blobs are formed in atmosphere, * Due to such disturbances and discontinuities ther is a small irregularity in the refractive inde because of which energy gets scattered, * For commercial applications tropospheric scatter i advantageous. It provides a reliable’ multichanne communication in between areas separated b, * As compared to LOS communication tropospheri Scatter propagation requires less maintenance. * Useful for multichannel communication for militar applications for distance about 50 to 400 km. oo ts TECHNICAL PUBLICATIONS® - An up thst for knowledge OCdIINICU WILT vanSWave Propagation Shortest path, Recelving antenna Transmitting T ‘antenna Back R Fig. Q.11.1 Tropospheric scatter propagation The cost of installation is very high and signal | Due to different ionizing agents and different scatters tremendous losses than with radio link paths, | physical properties of the atmosphere at different | heights several layers are formed in ionosphere. The | most important ionizing agents are ultraviolet radiations (UV), a, Brays, cosmic rays ete. 5.5 : Sky Wave Propagation ‘© These are three principal layers during day time and are called E, F, and F, layers. * Below E layers there is D layer nearest to earth's * Sky wave propagation is effective for the surface and exists at average height of 70 km. This frequencies between 2 MHz to 30 MHz so called hayer disepponte at right short wave propagation. © The electromagnetic waves get reflected from Wave Propagatior Important Points to Remember | Characteristics of different lonospheric layers some of ionized layers of ionosphere and D-ayer : returned back to earth in single or multiple hops i) The D-layer is located about 50 to 90 km above at reflections. the surface of earth and it is nearest layer to the + This suitable for long distance communication. _ ‘earth's surface. 5.5.1 ; Structure of Ionosphere | | #9. Tt thickness: #6 about "102m. end iii) This layer is ionized by photoionization of O2 Q.12 With neat illustration explain the structure molecules. and formation of ionospheric layers and the iv) This layer is present during daytime while corresponding frequencies of Propagation. sien disappears during night time. 1: May-16, Ma | Br [ANTU + May v) It has an ionic density of about 400/em? and ‘Ans.: Ionosphere is the upper portion of the | electron density of maximum value at noon. atmosphere of earth. It absorbs large quantities of | yi) This layer reflect very low frequency (VLF) and radiant energy from the sun. Ionization is see low frequency (LF) waves. in this layer. The ionized region consists of vii) At vertical incidence, the critical frequency of the electrons, positive ions and negative ions. layer is about 100 kHz. TECHNICAL PUBLICATIONS® - An up thst for knowledge ee OUT TEU WITT UI TTO:E-Layer i) The E-layer is located about 90 to 140 km above the Surface of earth. i) Its thickness is about 25 km. Jn E-layer the ionization of all gases by X-ray radiation takes place. iv) During night time its ionization is weak. ¥) The maximum electron density is about 4x105/ cm? and is at height of 100 km, vi) It is useful for high frequency (HF) waves during day time. vii) Its critical frequency is about 3 to 5 MHz and it Provides sometimes better reception during night time. . Flay i) The F-layer is located at the height of 140 to 400 km and it is mainly combination of F,-layer (140 to 220 km) and Fy-layer (250.to 400 km), During night Fy-layer combines with F,-layer and at height of 140-300 km we get F-layer. ii) This is the only layer which is ionized during, day time as well as night time. 400) F2( June ) z 3 bn_ ZN i 3 a Fy yo) ae = 100 @ 10 12 14 16 18 20 22 24 Hours 24 6 Fig. 0.12.4 Variations of lonospheric layers ii) The maximum electron density is about 220 km approximates. iv) The critical frequency of F-layer is about 5 to 12 MHz. Daytime Height above earth's surface (km) 1o- 10? 10° 10% 10 408 40 40? 40 40" 40% 408 | Electron density (N) fem? Electron density (N) fem? (@) © Z Fig. 0.12.2 Typical electron density variation TECHNICAL PUBLICATIONS - An up that for knowtedge OULAIINICU WILT vansvy) It is the topmost layer and highly ionized all the time compared with other 2 layers. vi) The Fy-layer reflects the high frequency (HF) waves. vil) The Fy-layer is the most important for the reflection of the high frequency radio waves. Q.13 Discuss the sallent features of sky wave Propagation. Bring out the various problems associated with this mode of propagation. US [NTU : May-19, Marks 5] ‘Ans. : Refer Q.13 of Chapter 5. *Sky wave propagation is effective for the frequencies between 2 MHz to 30 MHz so called short wave propagation. « The electromagnetic waves get reflected from some of ionized layers of ionosphere and returned back to earth in single or multiple hops at reflections. This suitable for long distance communication. Problems associated with sky wave propagation © The properties of ionosphere are highly dependent on sun, due to this there is variation in ionospheric condition. * Fading is undesirable variation in the intensity of signal received at receiver. ‘Fading is defined as fluctuations observed in received signal due to variation in height and density of the ionization in different layers. «The magnetic field of the earth affects only the radio waves propagating in ionosphere. « The radio waves incident on ionosphere is splitted into. two components namely ordinary and extraordinary wave due to earth’s magnetic field. This process is known as magneto-ionic splitting. Q.14 Explain sky wave propagation. Explain the mechanism of reflection and refraction of sky waves by lonosphere. ‘Ans.:® The sky wave propagation or ionospheric wave propagation is important as it assists global short wave communication. ‘« Due to the existence of the different ionized layers in the ionosphere, the long distance communication is possible. © Wave Propagation The D-layer exists during day time. It cannot reflect high frequency waves back to the earth. ‘Instead the intensity of the waves reflected back from the E or F layers decrease during day time due to the presence of the D-layer. ‘© The layers which exist permanently act as a radio mirror to bounce back the sky waves to the earth. ‘© The waves which return back to the earth appear to be the waves reflected by the layers of the ionosphere, But practically the ionized layers refract or bend the waves back towards the earth in much the same way as the refraction of the light waves travelling through media of different densities. The refraction mechanism can be explained in this fashion. When the wave approaches the ionized layer at an angle, the refractive index decreases as the ionization density increases. Hence the incident wave bends gradually further and further away from the normal as shown in the Fig. Q.14.1 (a). If the rate of change of the refractive index is sufficient, the refracted wave becomes parallel to the layer first, then it bends downward and then comes out of the ionized layer at an angle’ of incidence. The propagation of radio waves through ionosphere is as shown in the Fig. Q.14.1 (b). (a) Refraction mechanism ‘Waves through Ionosphere i Felayer Refracted wave Earth surface (b) Refraction by different layers, Fig. Q.14.1 Propagation through Ionosphere TECHNICAL PUBLICATIONS® - An up thrust for knowiedge OLAINICU WILT Udlllo'Antennas and Propagation 5-15 ee Wave Properetin 5.5.2 : Propagation of lonospheric Propagation Q.15 Explain in brief about the following terms with respect to wave propagation 2) Critical frequency 4) MUF ¢) Skip distance d) Virtual height SE LINTY + Nove4S, 27, May18, Maks 1) Ans. : a) Critical frequency (&) : +l is the highest frequency that can be reflected back to the earth by a particular layer at vertica incidence. It is different for different layers. * Critical frequency for a particular layer is proportional to the square root of maximum electron density in the layer. sing, BIN an At vertical incidence ; = 0, N= Nmax and £= fe = 9YNinax where Nmax : Maximum electron density per cubic meter. If the radio wave with a frequency greater than the the critical frequency is propagated through the ionosphere then the wave can also be reflected back to the earth but the angle of incidence must satisfy the expression. . SNe sind, > -—? But fy = 8INmax sin®; > ,fl- p b) Maximum usable frequency (icv) 4 Maximum usable frequency is defined as the limiting maximum frequency that éan be reflected back tothe earth by the ionospheric layer for a specific angle of incidence other than the angle of incidence for vertical incidence. + Tt denotes maximum frequency that can be used for the sky wave propagation for specific distance between two points on the earth. At £ = fur 6p 790% NNinax TECHNICAL PUBLICATIONS® - An up thst for knowedge OCAIINICU WILT vansAntennas and Propagation eae ©) Virtual height rs | oe Refracted wave 77, Earth's surface Fig. Q.15.1 Actual and virtual heights of an lonized layers: « For Jong distance communication the action of the ionized layers is to refract the sky waves back to the earth. When sky wave reaches a particular ionospheric region, the ionization density increases and refractive index reduces. According to law of refraction, as the wave enters a rarer medium from a denser medium it bends gradually further and away from normal and follow the path L-M-N as shown in Fig. Q.15.1. «The height at a point above the surface at which the wave bends down the earth is called actual height or true height. Below the ionized layer the incident and reflected waves follow the paths which are exactly the same if reflection takes place along path at a height above ‘earth's surface. It is called as virtual height. Virtual height is always greater than actual height. ) Skip distance : Earth's surface Z 3 2 g A c 8 Ground wave. i un -—ele—— Sip zine Ta Skip distanco ———————> Fig. Q.16.2 Representation of skip and effect of variation In angle ‘of Incidence of fixed frequency of transmitted wave TECHNICAL PUBLICATIONS® - An up thrust for knowiedge OCdIINICU WILT vantSAntennas and Propagation * The skip distance is the shortest distance from the transmitter measured along surface of earth, at Which a sky wave of fixed frequency will return back to earth. It also represents minimum distance for which sky wave propagation just takes place and no sky wave propagation is possible for points Nearer than this distance. © The angle of incidence for which the wave retums back to the earth at minimum distance from the transmitter ie. at skip distance is called angle of critical incidence B,. For 6; >0;, the wave escapes in space. 9; = 6, the wave retumns back to the earth at point B. 8; <0c, the waves are received beyond point B. For a given frequency f = fyaup z to =f) Dyup = 2h, fue ip * 2 A) 1 Multi-hop Propagation Q.16 Write a short note on multi-hop propagation. ‘ES [INTU : Nov-17, Marks 5] Ans.:* The wave originating from transmitter reaches receiver without to ground anywhere in between transmitter and receiver is single hop transmission. ‘But single hop transmission is sometimes not possible due to : i) Curvature of earth i) Distance between transmitter and receiver is greater as compared with skip distance allowed. * Under such conditions multi hop propagation takes place. In practice the minimum angle at which a radio wave leaves the transmitter is usually 35° above TECHNICAL PUBLICATIONS® 5-17 Wave Propepatig, horizon. For E layer and Fz layer the maxim, ship distances are about 1700 kam and 3509 |= approximately. When the distance betyg transmitter and receiver is greater than the skp distance, the wave takes two or more hops reaching the receiver as shown in Fig. Q161 guy communication is called multi-hop propagation one ce “VV. jas [| a a ~ = Fig. Q.16.1 Multl-hop propagation Q.17 Derive Friss free space equation. | OR Derive fundamental equation for free space propagation. Ans.:+ In general, when the power is radiated by the antenna in the free space, the radio energy may be absorbed or radiated by the objects in. the region. So it becomes essential to calculate the loss during radio transmission. This is nothing but radio transmission loss, * The basic definition of radio transmission loss is the ratio of the radiated power to the received power. This loss is based on the concept of the inverse ‘square law in optics applied to radio transmission. * For the isotropic radiator, the radiation is uniform in all the directions. Hence the power density is same everywhere at all the points on the surface of a sphere with radius r. * Thus the average power can be expressed inter! of radiated power as, Bs ) 4nr? * The maximum directive gain or directivity of th test antenna is given by, (Com), Pavg W/m? Pamax P, Peg Ani? + A Up toast fr knowledge 5 OLAIHINCU WILT Udo‘Antennas and Propagation Pamax * Gp max w= (2) + Now the receiving antenna is placed such that ideally it receives total power from the radio waves. 4nr? + Let Pee be the maximum power delivered by the receiving antenna to the receiver load under matched load conditions. Let (A.), be the effective aperture of receiving antenna. ee we can write im "Fanaa (Ae) * (Sons), Fir Ae = @) * But in general, the roa and effective aperture area for any antenna are related as, 4 Somax = 37 (Ac) » * Let (Gp max), be the directivity of the receiving antenna, then we can write, (Goma), * $F Asde (4. = = (Gone), ~@) Substituting value of (A,), in ees @), we get, ~ (Com), Zh [Fe (Comm )e] . BS ~ (Com): (Com). (amz) ~ (0 ©The equation (6) called fundamental equation for free space propagation. This is also called Friss free space equation. Q.18 The transmitter and receiver are placed 45 lan away from each other. Calculate minimum height of transmitting and receiving antennas to ensure line of sight communication assuming same heights for both antennas above ground level. Ans. : Given : dmax= 45 km, (km) hy = h, for line of sight communication By formula, dmax = 4.12 (Vir + Jr) h, = hy, hence we get, = 412 imax Rr 45 = 8.24 J, But Wave Propagation hy = hy = 29.824 km Thus minimum height of transmitting and receiving antennas must be 29.824 km above the earth surface to ensure line of sight communication. Q.19 Two alrcrafts are flying at altitudes of 3000 m and 5000 m respectively. What Is the minimum possible distance along the surface of the earth over which they can have effective point to point microwave communication? Radius of earth is 637x106m. 0a [INTU : May-09, Marks 8] Ans. : Let, hi = Altitude of one aircraft = 3000 m h, = Altitude of other aircraft = 5000 m The maximum possible distance along surface of earth is given by, a= BFL Vin + Jie] = 263710" [ 3000 + 15000 ] = 447.8879 x 10° = 447.8879 km Q.20 VHF communication Is to be established with 50 W transmitter at 100 MHz. Calculate the LOS distance if the heights of transmitting and recelving antennas are respectively 50 m and 10 m. Assuming the capture area of transmitting antenna Is 25m’, calculate the field strength at the receiving end neglecting ground reflected wave. TE [INTU : Nov.-07, Marks 8 ] Ans. : Power transmitted = 50 W Height of transmitting antenna = 50 m Height of receiving antenna = 10 m ne " = Frequency = 100 MHz = 100x106 Hz a Capture area = 25 m? The line of sight (LOS) distance is given by, @ = 412[ fh; + Yh; ] = 412[ 50+ V0] = 42.1613 km The field strength at the receiving end is given by, Ro 88VP hy h, oS h, — eee TECHNICAL PUBLICATIONS" An up thrust fr knowledge OLAINICU WILT Udlllo'ground wave has an antenna current of 8 A. What Antennas and Propagation : 5-19 Wes Property Bs 88.50 (50) (10) | LOS = 4.12 1h + Rel ( 3x10? ] (421613x10? | where hn ind hy in meter ete | 38 = 4.12 [hy +V24] = 58.3429 uV/m j . (= Q.21 A 150 m antenna transmitting at 1.2 MHz by | 412 i voltage Is received by the receiving antenna 40 km away, with a helght of 2m ? EGE [INTU : Nov.-08, 09, Marks 8] Ans.: hy = Transmitting antenna height = 150 m f = Tansmitting signal frequency = 12x 10° Hz h, = Receiving antenna height = 2 m I= Antenna current = 8 A d = Distance between transmitter and receiver = 40 km = 40x 10° m The expression for field strength at a distance from the tramsmitting antenna is given by, p= 120m, hy T Ment 3 x108 F 2 x108 = (1207) (150) 2) (8) 250(40x103) But Ae = 0.09048 Vim ‘Thus the voltage received by antenna, V = E (hy) = (0.09048) (2) = 0.18096 volts Q.22 Determine the height of the transmitting antenna to obtain a maximum distance of transmission upto 38 km from a 24 meter high receiving antenna ? US [INTU : Nov.-10, R-07, Marks 8] Ans. : Given : ‘Height of receiving antenna = hr = 24 m Maximum distance of transmission = LOS = 38 km The Line Of Sight (LOS) or maximum distance of transmission is given by, | | | | hy = 18.6997 m = 18.7 m Q.23 For a flat earth assume that at 400 km reflection takes place. The maximum density of Ionosphere corresponds to a refractive Index of 0.9 at 10 MHz. Calculate range for which fyur = 10 MHz. EG [INTU : March-06, Marks 4) ‘Ans.:n = Refractive index = 0.9 fur = 10 MHz = 10x 106Hz h = 400 kn ‘The refractive index is given by, 81 Nmax ae 81 Ninax 09 = tee Ninax = 4.84322 x 105/m?3 But fo, = (81 Ninax = 4.3588x 10°Hz = 4.3588 MHz ‘The range for which fyqur = 10 MHz is given by, = 1651.8 km Q.24 The critical frequency for reflection at vertical Incidence of an ionospheric wave is 10 MHz. Calculate the maximum value of the electron density. ‘1 [INTU = May-09, Marks 8) ‘Ans. : Given : f£, = 10 MHz = 10x106 Hz. For vertical incidence, the refractive index is 2210. Thus the expression for critical frequency for becomes, fa 7 9VNeax TECHNICAL PUBLICATIONS® - An up thus for krowedge > OLAIHINCU WILIT UdIIIOwhere N,,,¢ = maximum electron density ss | 10x10 = 9./Niax 2 } 1.2345x 10? /m?> Q.25 Assume that reflection takes place at a height of 400 km and that the maximum electron density in the ionosphere corresponds to a 0.9 refractive index at 10 MHz. What will be the range for which MUF Is 10 MHz ? UGE [INTU : March-06, Marks 6] Ans. : Given : h = height = 400 km = 400x103 m n= refractive index = 0.9 fyqur = 10 MHz = 10x10° Hz. By definition, INnax ne fe 81(Nmax) = 09 = fie (1ox108)? hax = 23456 x 10! mn f= ¥8INmax [s1(23456x10"") = 4.3588 x10° Hz ~ 4.3588 MHz 4) For Flat Earth : The skip distance for flat earth is, p 72h (f= 6 = 2x 400%10? Fees 1 Hence D, z 43588%108 = 1651.84 lan ii) For Curved Earth : The skip distance for curved: earth is, E 7 D2) If fur ) _ Data * fn+Pa| (“et uy 32 s = 2[ soox108 16517610" tox10®_) _, 8(6370x10°)? |\{ 43588106 | | | | | | i Daa p 7 1872.95 kn Q.26 A radio link has to be established between two earth stations placed at a distance of 25000 kam between them. If the height of the Ionosphere Is 200 km and its critical frequency is 5 MHz. Calculate the MUF for given path. Also calculate the electron density in the Ionosphere layer. 1 [INTU : Nov.-07, Feb.-08, Marks 8] Ans. : Given : D = Distance between two stations = 25000 km = 25000 x10%m h = Height of ionosphere = 200 km = 200 x 10° m fq = 5 MHz = 5x 106 Hz The maximum usable frequency, fyyyp for a given path is, < D bar * (8) 2 25000105 * faye =, 5x108 +1 aU, (an ] 312.5399 x 106 Hz =312.54 MHz At critical frequency, 1 ~ refractive index = 0. Hence we can write, fg = 9Nimaw 5x10® = 9./Ninax @.27 Communication by fonospheric propagation Is required for a distance of 200 km. Height of the layer is 220 km and critical frequency is 5 MHz. Find MUP. 0 [JNTU : Feb.:08, Marks 8) yins. : Given : D = Distance = 200 km = 200x109 m 62 pad = 3.0864 x 107 m=? h = Height pf ionosphere = 220x10° m fq ~ 5 MHz TECHNICAL PUBLICATIONS® - An up thus for krowtedye OLdINICU WILT vadThe maximum usable frequency for a path is given by, : 200x103 5x106 }« 2x220x10 5.4923 10° Hz faup ~ 5.4923 MHz Q.28 It Is required to establish a short wave communication between two” points on earth's surface separated by 1200 km. Calculate fycyp and angle of take off of the transmitted wave from the following data. Highest signal frequency returned to earth after vertically upward propagation is 7.2 MHz and virtual height of ionized layer is 200 km. Assume surface of the earth to be flat. TP [INTU : Nov.-07, Marks 6] Ans. : Given : D = Distance between two points on the earth = 1200 km = 1200x103 m fq ~ Highest frequency = Critical frequency = 7.2 MHz = 7.2 x10 Hz h = Height of ionized layer = 200 km = 200x108 m i) Now fur = = 22.7684 x 10° Hz fyup = 22.7684 MHz ii) The angle of take off is assumed to be fas shown in the Fig. Q281. Hence refering Fig. Q28.1, ney Fig. 0.28.1 9.29 Calculate the value of frequency at which at EM wave must propagate through the D-regio: | with an index of refraction of 0.5 and an electro, density of 3.24 x 10” etectrons/m°, 12F [INT : Dec.-09, Marks 8] Given : n = Refractive index = 0.5 Ans. N =3.24 x 104 electrons/m? | The refractive index is given by, | [, SIN 2 7 05 = 8162410") aa ‘Squaring both the sides, n= (os? = 18162410") gee ‘ 025-1 = =81626«108) 2 @ = 81(8.24x104) | 075 tn B= DD = Db | ft @ eee t= [81(8.24x 10°) (2h O75 7 fé = tan-}( 24 B = Angle of take off = tan’ (3) | Pawar | TECHNICAL PUBLICATIONS® An up tht for Inoutedge OcdINICU WILT vansAntennas and Propagation 2.30 Determine the maximum usable frequency for a critical frequency of 20 MHz and an angle of Incidence of 35°. EGF [INTU : Nov.-10, Set-a, R-07, Marks 8] ” Ans. : Given : Critical frequency = f= 20 MHz = 20x 106 Hz Angle of incidence = 6,. By definition, the maximum usable frequency fyqur is given by, fer __ 20x 106 fyuE (S€C6;) fey =—lt_ = 20% 10° Mur = (S004) fy = Ge = AT = 24.4155x 106 Hz=24.4155MHz Q.31 Find the maximum range of tropospheric transmission for which the height of transmitting antenna Is 100 ft and that of receiving antenna is 50 ft. ‘ES [INTU : May-18, Marks 5] Ans. d= 14M2[Jhy + Fr] d= 1.4142[V/100 + 50] = 24.1418 miles Q.32 Calculate the wave tilt in degrees of the surface wave over an earth of 6 mm conductivity and relative permitivity of 12 at 2 MHz. [NTU = May-18, Marks 5] Ans, : tant Wave tilt = tan’ oe € = €9 €, = 12x8.854x10- 2 @ = 2nx2x108 rad/sec Q.33 Discuss about the atmospheric effects in space wave propagation. [&[JNTU : Dec.-19, Marks 5] Ans. : Atmospheric Effects on Space Wave Propagation *The space wave when propagates through atmosphere, it is affected by the atmospheric conditions. * Significantly one of the reasons behind this, is the presence of gas molecules of water vapours mainly. In general, dielectric constant of water is very high (about 80). Moreover the dielectric constant of moist air further varies over a higher range. 5-2 Wave Propagation Hence summarizing all these facts, we can say that the refractive index of the air with such water vapour is slightly greater than unity. * The density of air varies with height. Thus the dielectric constant and the refractive index of air also depends on height. In general, the refractive index decreases with height. Modified Refractive Index of Alr (1.) © The refractive index of the air is defined as the root of the dielectric constant and is denoted by p. ‘* For study of radio propagation, the actual refractive index () is expressed interms of a modified refractive index (1 ,) which is given by, Bm = [nag }xaoe a () where 1 = Refractive index, h = Height above ground, Re = Radius of earth = 6.37x 10° m «Practically it is found that the effects produced by the atmospheric refraction do not depend on modified refractive index (1) but depend on the rate of change of modified refractive index (i) Sika te, dtm with height (b) ie. S, * At high altitudes, in the troposphere, the dielectric constant and hence refractive index of air do not vary with height. But even though is constant, the modified refractive index jt, defined by equation (1) varies with height Its value increases at a constant rate of 0.1575 units per meter. lt is observed that near the earth's surface, the dielectric constant and so the refractive index of the atmosphere decreases with height for standard atmospheric conditions. © Thus for standard atmospheric conditions, near the earth surface aa value is less than the standard value and it is equal to 0.1181 units per meter (less than 0.1575 units per meter). The variations in the modified refractive index values with height are as shown in the Fig. Q.33.1 TECHNICAL PUBLICATIONS® - An up trust for knowledge OLdINICU WILT UdoHm (b) Refraction at lower heights ‘atmosphere (c) Simple surface ‘duct duct (ground based) (4) Elevated Fig. Q.33.1 Variation of modified refractive Index (iim) with height (h) (@) to (d). All the variations are for the conditions near surface of the earth. * Note that at greater heights, value of the modified refractive index varies with height at a constant rate of 0.1575 units per meter. Refraction of Waves ‘+ As the refractive index 1 decreases with height in standard atmosphere, the wave gets refracted away from the normal similar to optical rays getting refracted when travel from denser to rarer medium. * Thus atmosphere causes radio wave to bend towards the earth and then travel to reach point located away than the optical horizon as shown in the Fig. 0.332. = Constant “e515 permetr Transmit A ‘antenna "| Sandard amncsphere. “i= 0.1181 per meter veoh Fig. Q.33.2 Paths of radio ways showing refraction of waves from denser to rarer medium * As the refractive index varies with height, the wave travelling in the atmosphere bends away from low dielectric constant region to high dielectric constany region. « The effect of variation in refractive index value is ag shown in the Fig. Q.33.2. . active index do not vary with he as shown in the Fig. Q:33.2. For constant For u varying with h_ refractive index Standard ‘atmosphere Le Fig, Q.33.3 Effect of variation of 1. with height on the radio waves: ‘* In standard atmospheric condition, the radio waves are refracted slightly (a5 [1m increases with height) shown by path b in the Fig. Q.333. + When the rate of jim with height becomes zero at lower heights, the waves follow path as if curvature of the earth shown by path c in the Fig. Q333, When the dielectric constant of atmosphere increases with height, the waves move away from the earth represented by path d as shown in the Fig. Q.33.3. Short Answered Questions Q.1 Define wave tilt of ground wave. ES [INTU ; Nov.-15, Marks 2] Ans. : Refer Q.6. Q.2 Write the expression for relation between MUF and skip distance. C&P [JNTU : March-16, Marks 2] Ans. : Refer Q.14 (d) Q.3 What are the types of ground waves ? EGP [INTU : Dec.-16, Marks 2] Ans, : Refer Q2. TECHNICAL PUBLICATIONS® «An up tus fr knowledge OcdINICU WILT Udo‘Antennas and Propagation Q.4 What are factor that affect the propagation of radio waves. BGP [INTU : Dec.-16, Marks 3] Ans. : Refer Q3. Q.5 Write briefly about duct propagation. USP [INTU = July-t7, Marks 2] ‘Ans, : Refer Q.10. Q6 Briefly explain about D region. EGP [ANTU : May-18, Marks 3] Ans. : Refer Q.12. Q7 Derive the expression for refractive Index for ionosphere. B@[INTU : Dec.-17, Marks 2] Ans. : Refer Q.12. Q.8 Write short note on super refraction. SGP [INTU : Nov-i5, Dec.-17, Marks 3] ‘Ans. : Refer duct propagation Q.9 Write the expression for relation between MUF and skip distance. ES [JNTU : March-16, Marks 2] Refer Q.32. Ans. Q.10 Define optimum usable frequency of an ionospheric layer. ES INTU : May-19, Marks 2] ‘Ans.: The frequency used for the ionospheric transmission ‘should be maximum usable frequency MUF. But due to. continuous changes and irregularities in the ionosphere, the MUF varies about 15 % of value of maximum value. ‘The optimum usable frequency is defined as the frequency in the range between 50 % to 85 % of predicted MUF between transmission and reception points. Q.11 Find the maximum range of tropospheric for which the transmitting antenna height is 100 ft and receiving antenna height Is 50 ft. 0 [May-18, Marks 2] Ans. > d= 14142[ Yh, + hy] miles = 1.4142[v100 + /50] = 24,1418 miles | | Wave Propagation. = 24.1418 (1.609) kn = 38.844 km Q.12 Obtain the roughness factor at 3 MHz for an earth having o = 0.5 with @ = 30°. Calculate the ratio of roughness factors for the same earth and same 0 It frequency Is doubled. SG [INTU : March-16, Marks 3] Ans. : Roughness factor R = 485580 4n(05)(05) = joo ~ 100 Frequency foubled, 2 will be half, R will be doubled. Ri J Riles Rh Fill in the Blanks for Mid Term Exam The orientation of electric field vector with respect to the earth's surface is called of plane EM wave. Charge in the direction of EM wave incident obliquely of the boundary of two media is known as- of wave. qa Q3 If the total retained power is constant, then power density is proportional to square of distance. Using mainly VHF, UHF and microwave signals are propagated beyond the line of sight. Sky wave propagation takes place due to the reflection by The highest frequency that can be reflected back to the earth by a particular layer for vertical incidence is known as ‘The denotes the . maximum frequency that can be used for the sky wave propagation for specific distance. The ____ is the shortest distance from the transmitter measured along surface of earth at which a sky wave of fixed frequency will return back to earth. as as as a7 TECHNICAL PUBLICATIONS® - An up tris for knowledge OLAINICU WILT Udlllo'Se a i | QS The critical frequency Tage 09 Sip distance is related to fygp_ and for by denoly i 10" per cable metering 40 Line of sight distance d in km is related to. [a] 2.846 MHz B) 2246 oir heights of transmitting and receiving antennas @] 2.846 KHz [a] 1.423 Mite (in meters) by expression | - Multiple Choice Questions for Mid Term i Qt The critical frequency of radio wave for reflection at vertical incidence if the maximum. value of electron density is 1.26x10® cm”? is, ‘Answer Keys for Fill in the Blanks [a) 10.1024 MHz (B) s Mune [e] 2 Miz (a) tmz 2.2 The transmitting and receiving antennas of same height are 45 km away from each other. ‘The minimum height of antennas to line of sight communication is, [al 20 km [B) 29.824 bm [Ee] 35 km [a] 15 im \3° The sky wave propagation takes place in the range of. [a] 30 k#iz.to 3 MHz b) 3 MHz to 30 MHz [e] 30 MHz to 300 MHz [a 300 Mriz to 3 GHz 4 The roughness of earth surface responsible for the scattering of reflected wave is given by pa 2tosing [a] R-*S nosing [R= OCAIIICU WILT vain