Part IV: Antennas and Waveguides Communication Transmission Systems Course 1.1 An antenna-Introduction: o A metallic conductor capable of radiating and capturing EM energy. o The interface between the transmission line and the atmosphere o Converts the electrical energy travelling along the TL into an EM wave at the transmit end and vice versa. Antenna Parameters: o Radiation Pattern: . It is a polar graph representing the field strength/power densities at various angular positions relative to the antenna. There can be more than one major lobe in the radiation pattern. Major lobe propagates and receives the most energy (it called front lobe i.e. front of the antenna) . Front-to-Back/Side ratio: it is the ratio of the front lobe power to the back/side lobe power. o Antenna Types: . Isotropic pattern: The radiation pattern is the same in all directions. Antennas with isotropic radiation patterns don't exist in practice, but are sometimes discussed as a means of comparison with real antennas. Omnidirectional Antenna: the radiation pattern of an actual antenna is isotropic in a single plane such as dipole antenna and slot antenna. Directional Antenna: Do not have symmetry in the radiation pattern. These antennas typically have a single peak direction in the radiation pattern; this is the direction where the bulk of the radiated power travels. Such as dish antenna and slotted waveguide antenna. o Field Regions: Page 1 of 17 Dr. Omar R. Daoud Part IV: Antennas and Waveguides . Near Field (NF): Refers to the field pattern that is close to the antenna. Reactive NF: Region surrounding the antenna, wherein the reactive field predominates. o Angular Field distribution depends on distance from antenna. 퐷3 > 휆 푅 < 0.62√ 퐷 휆 휆 < 휆 푅 < { 2휋 퐷: Largest antenna dimension 푅: Region Radius Radiating NF: Region between reactive near-field and far-field regions (Fresnel zone). o Radiating fields predominate but angular field distribution still depends on distance from antenna. 퐷3 2퐷2 > 휆 0.62√ < 푅 < 퐷 휆 휆 휆 < 휆 < 푅 < 3휆 { 2휋 . Far Field: Refers to the field pattern that is at great distance Region where angular field distribution is essentially independent of the distance from antenna (Fraunhofer zone) 2퐷2 > 휆 푅 > 퐷 { 휆 < 휆 푅 > 3휆 Page 2 of 17 Dr. Omar R. Daoud Part IV: Antennas and Waveguides o Radiation Resistance: . Antenna resistance that equals to the ratio of the power radiated by the antenna to the square of the current at its feed input. 푃 푅 = 푟푎푑 Ω 푟 푖2 푃푟푎푑: Radiated power (Watts) 1 푃 = ∯ 푅푒[퐸⃑ × 퐻⃑⃑ ∗]. 휕푆 = ∯ 푊⃑⃑⃑⃑⃑⃑⃑ . 휕푆 푟푎푑 2 푆 푆 푎푣 푊푎푣: The average power density 휕푆: 푛̂휕푎 휕푎: Infinitesimal area of the closed surface =푟2sin (휃)휕휃휕휑 (m2) 푛̂: Unit vector normal to the surface 푖: Antenna current at the feed point (Ampere) Example 1: Find the total radiated power for an antenna has an average power sin (휃) density (푊⃑⃑⃑⃑⃑⃑⃑ ) of = 푟̂퐴 푊/푚2. 푎푣 표 푟2 Solution: 1 푃 = ∯ 푅푒[퐸⃑ × 퐻⃑⃑ ∗]. 휕푆 = ∯ 푊⃑⃑⃑⃑⃑⃑⃑ . 휕푆 푟푎푑 2 푆 푆 푎푣 2휋 휋 sin(휃) 푃 = ∯ 푊⃑⃑⃑⃑⃑⃑⃑ . 휕푆 = ∫ ∫ (푟̂퐴 ) . 푟 푟2 sin(휃) 휕휃휕휑 푟푎푑 푆 푎푣 0 0 표 푟2 2 푃푟푎푑 = 휋 퐴표 푊 Example 2: Find the total radiated power for an isotropic antenna has an 2 average power density (푊⃑⃑⃑⃑⃑푎푣⃑⃑ ) of = 푟̂푊표(푟) 푊/푚 . Solution: 1 푃 = ∯ 푅푒[퐸⃑ × 퐻⃑⃑ ∗]. 휕푆 = ∯ 푊⃑⃑⃑⃑⃑⃑⃑ . 휕푆 푟푎푑 2 푆 푆 푎푣 2휋 휋 푃 = 푊⃑⃑⃑⃑⃑⃑⃑ . 휕푆 = (푟̂푊 (푟)). 푟 푟2 sin(휃) 휕휃휕휑 푟푎푑 ∯푆 푎푣 ∫0 ∫0 표 2 푃푟푎푑 = 4휋푟 푊표 푊 o Radiation Intensity: . The power radiated from an antenna per unit solid angle. 푊 푈 = 푟2푊 ( 푠표푙푖푑 푎푛푔푙푒) 푟푎푑 푢푛푖푡 2 푊푟푎푑: Radiated density (W/m ) 2휋 휋 푃 = 푈(휃, 휑). 휕Ω = 푈(휃, 휑) sin(휃) 휕휃휕휑 푟푎푑 ∯Ω ∫0 ∫0 2 푟 2 1 2 푈(휃, 휑) = |퐸⃑ (푟, 휃, 휑)| ≅ [|퐸표(휃, 휑)|2 + |퐸표(휃, 휑)| ] 2휂 2휂 휃 휑 퐸⃑ (푟, 휃, 휑): far zone electric field intensity of the antenna 휂: interstice impedance (120377) Page 3 of 17 Dr. Omar R. Daoud Part IV: Antennas and Waveguides Example 1: Find the total radiated power for an antenna has a radiation 2 intensity 푈(휃, 휑) of 푈(휃, 휑) = 푟 푊푟푎푑 = 퐴표 sin(휃). Solution: 2휋 휋 푃 = 퐴 sin(휃) . sin(휃) 휕휃휕휑 푟푎푑 ∫0 ∫0 표 2 푃푟푎푑 = 휋 퐴표 푊 Example 2: Find the total radiated power for an isotropic antenna has a 2 radiation intensity 푈(휃, 휑) of 푈(휃, 휑) = 푟 푊푟푎푑 = 푈표. Solution: 2휋 휋 푃 = 푈 . sin(휃) 휕휃휕휑 푟푎푑 ∫0 ∫0 표 푃푟푎푑 = 4휋푈표 푊 o Radiation Efficiency: . It is the ratio of the power radiated by an antenna to the sum of the power radiated and the power dissipated. 푃푟푎푑 휉 = × 100% , 푃푛 = 푃푟푎푑 + 푃푑 푃푛 푃푟푎푑: Radiated power (W) 푃푑: Dissipated power (W) 푃푛: Input power (W) 2 푃푟푎푑 = 푅푟푖 . 2 푃푑 = 푅푒푖 . 푅 휉 = 푟 × 100% 푅푟 + 푅푒 푅푟: Radiation resistance 푅푒: Effective antenna resistance OR . Takes into account the physical losses of the antenna such as metal and dielectric losses, as well as mismatch between the connection of the antenna and the transmission line. 2 휉 = 푒푟푒푐푒푑 = (1 − |Γ| )푒푐푒푑 푒푟: Reflection efficiency 푒푐: Conductor efficiency 푒푑: Dielectric efficiency Γ: Reflection coefficient Page 4 of 17 Dr. Omar R. Daoud Part IV: Antennas and Waveguides o Beamwidth: . It is the angular separation between two identical points on opposite site of the pattern maximum. Typically, it is between 30o and 60o. Half-power beamwidth (HPBW): in a plane containing the direction of the maximum of a beam, the angle between the two directions in which the radiation intensity is one-half value of the beam . First-Null beamwidth (FNBW): angular separation between the first nulls of the pattern Example 1: For the normalized radiation intensity of an antenna is represented by 푈(휃) = cos2(휃), (0 ≤ 휃 ≤ 휋, 0 ≤ 휑 ≤ 2휋). Find the angle which represents the HPBW; 휃ℎ. Then find the angle which represents the FNBW; 휃푛 Solution: 1- For the HPBW 2 푈(휃)휃=휃 = cos (휃ℎ) = 0.5 ℎ 휋 휃 = cos−1 √0.5 = ℎ 4 Since the pattern is symmetric with respect to the 휋 maximum, then HPBW=2휃 = ℎ 2 2- For the FNBW ( ) 2( ) 푈 휃 휃=휃ℎ = cos 휃ℎ = 0 −1 휃ℎ = cos 0 = 휋 Since the pattern is symmetric with respect to the maximum, 휋 then HPBW=2휃 = 푛 2 o Bandwidth: . It is the frequency range over which antenna operation is satisfactory. 푓 − 푓 퐵푊 = 퐻 퐿 × 100% 푓표 푓퐻: Highest frequency of operation 푓퐿: Lowest frequency of operation 푓표: Optimum frequency of operation Example 1: Determine the percent bandwidth for an antenna which has an optimum frequency of 600MHz, and a -3dB frequencies of 570MHz and 630MHz. Page 5 of 17 Dr. Omar R. Daoud Part IV: Antennas and Waveguides Solution: 푓 − 푓 630 − 570 퐵푊 = 퐻 퐿 × 100% = × 100% = 10% 푓표 600 o Directivity: . Ratio of radiation intensity in a given direction from the antenna to the average radiation intensity 4휋푈(휃, 휑) 4휋푈(휃, 휑) 퐷(휃, 휑) = = 푃푟푎푑 푃 = 푈(휃, 휑). 휕Ω 푟푎푑 ∯Ω 4휋 퐷(휃, 휑)푚푎푥 = Ω퐴 Ω퐴: Beam Solid angle (all the power of the antenna would flow if its radiation intensity were constant and equals to 푈푚푎푥) . The average radiation intensity equals to the radiation intensity of an isotropic source. If the direction is not specified, it implies the directivity of maximum radiation intensity (maximum directivity) 4휋푈푚푎푥 퐷푚푎푥 = 퐷표 = 푃푟푎푑 Example 1: If the radial component of the radiated power density of an infinitesimal linear dipole is given by 푠푖푛2휃 푊 = 푟̂퐴 푊/푚2 푎푣 표 푟2 Then calculate the maximum directivity. Solution: 2 2 - The radiation intensity is 푈 = 푟 푊푟 = 퐴표푠푖푛 휃 휋 The maximum radiation is directed along 휃 = 2 푈푚푎푥 = 퐴표 - The total radiated power 푃 = 푈(휃, 휑). 휕Ω 푟푎푑 ∯Ω 2휋 휋 2 8휋 푃푟푎푑 = 퐴표 ∫ ∫ 푠푖푛 휃. 푠푖푛휃 휕휃휕휑 = 퐴표 0 0 3 4휋푈푚푎푥 - The maximum directivity 퐷푚푎푥 = 퐷표 = 푃푟푎푑 3 퐷 = 표 2 o Gain: . The ratio of radiation intensity in a given direction to the average radiation intensity that would be obtained if all the power input to the antenna were radiated isotropic. It takes into account the efficiency of the antenna as well as its directional properties. Page 6 of 17 Dr. Omar R. Daoud Part IV: Antennas and Waveguides 4휋푈(휃, 휑) 4휋푈(휃, 휑) 퐺(휃, 휑) = = 푒푐푒푑 = 푒푐푒푑퐷(휃, 휑) 푃푛 푃푟푎푑 2 푃푛: Input power to antenna = (1 − |Γ| )푃표 = 푃푟푎푑 + 푃푙표푠푠 . Directive/Relative Gain: Ratio of the power density radiated in a particular direction to the power density radiated to the same point by a reference antenna, assuming both antenna are radiating the same amount of power. Power Gain (Ap): it is the same as directive gain except that the total power fed to the antenna is used. It is assumed that the given antenna and the reference antenna have the same input power and the reference antenna is lossless. (휉 = 100%, 푃푟푎푑 = 푃푛, 푃푙표푠푠푒푠 = 0) 퐴푝 = 휉퐷 Example 1: A lossless resonant half-wave dipole with an impedance of 73Ωis connected to a transmission line with characteristic impedance of 50Ω. If the radiation pattern is given by 3 푈 = 퐵표푠푖푛 휃 Then calculate the maximum realized gain of the antenna. Solution: 푈푚푎푥 = 퐵표 - The total radiated power 2휋 휋 2 3 3휋 푃푟푎푑 = 퐵표 ∫ ∫ 푠푖푛 휃. 푠푖푛휃 휕휃휕휑 = 퐵표 ( ) 0 0 4 4휋푈푚푎푥 - The maximum directivity 퐷푚푎푥 = 퐷표 = 푃푟푎푑 퐷표 = 1.697 - Since the antenna is lossless, then 푒푐푒푑 = 1 - 퐺표 = 푒푐푒푑퐷표 = 1.697 - 퐺표(푑퐵) = 10 log(1.697) = 2.297 - With the realized gain, we can take into account the mismatch losses between the TL and the antenna impedance as: 73−50 2 푒 = (1 − |Γ|2) = (1 − | | ) = 0.965 푟 73+50 푒푟(푑퐵) = 10 log(0.965) = −0.155 푒표 = 푒푐푒푑푒푟 = 0.965 = −0.155 푑퐵 퐺표푟푒푎푙푧푒푑 = 푒푟푒푐푒푑퐷표 = 1.6376 = 2.142 푑퐵 Page 7 of 17 Dr.