Antenna Theory and Design (Course Code: 22LDN22) for M.Tech – VTU Syllabus.pdf
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This document outlines the course structure for Antenna Theory and Design at the University B.D.T. College of Engineering, detailing course codes, credit hours, assessment methods, and suggested learning resources. The assessment comprises continuous internal evaluation (CIE) and semester-end examination (SEE), both holding equal weight in determining final grades. Additionally, it provides an overview of antenna fundamentals, including historical milestones, electromagnetic spectrum, and radiation principles, alongside the theoretical content and practical components evaluated through various tests and assignments.
Antenna Theory and Design (Course Code: 22LDN22) for M.Tech – VTU Syllabus.pdf
- 1. Department of Studies in Electronics & Communication Engg., University B.D.T. College of Engineering Visveswaraya Technological University, Davanagere-4 Karnataka, India Dr.T.D. Shashikala 13-Jan-25
- 2. Course Code: 22LDN22 CIE Marks 50 Teaching Hours/Week (L:P:SDA) : (3:0:2) SEE Marks 50 Total Hours of Pedagogy : 40 hours Theory + 10-12 Lab; Total Marks 100 Credits 4; Exam Hours 3 ANTENNA THEORY AND DESIGN Suggested Learning Resources: Textbook: 1. ‘Antenna Theory and Design’, Stutzman and Thiele, John Wiley, 2nd Edition, 2010 2. Reference Books: 1. ‘Antenna Theory Analysis and Design’, C. A. Balanis, John Wiley, 2nd Edition, 2007 2. ‘Antennas and Wave Propagation’, J. D. Krauss, McGraw Hill TMH, 4th Edition, 2010 3. ‘Antennas and propagation’, A.R.Harish, M. Sachidanada, Pearson Education, 2015 Web links and Video Lectures (e-Resources): https://www.youtube.com/watch?v=fIbdW0NGIU0 https://nptel.ac.in/courses/117107035 13-Jan-25
- 3. Assessment Details (both CIE and SEE) The weightage of Continuous Internal Evaluation (CIE) is 50% and for Semester End Exam (SEE) is 50%. The minimum passing mark for the CIE is 50% of the maximum marks. Minimum passing marks in SEE is 40% of the maximum marks of SEE. A student shall be deemed to have satisfied the academic requirements and earned the credits allotted to each subject/ course if the student secures not less than 50% (50 marks out of 100) in the sum total of the CIE (Continuous Internal Evaluation) and SEE (Semester End Examination) taken together. CIE for the theory component of IPCC 1. Two Tests each of 20 Marks 2. Two assignments each of 10 Marks/One Skill Development Activity of 20 marks 3. Total Marks of two tests and two assignments/one Skill Development Activity added will be CIE for 60 marks, marks scored will be proportionally scaled down to 30 marks. 13-Jan-25
- 4. CIE for the practical component of IPCC 1. On completion of every experiment/program in the laboratory, the students shall be evaluated and marks shall be awarded on the same day. The15 marks are for conducting the experiment and preparation of the laboratory record, the other 05 marks shall be for the test conducted at the end of the semester. 2. The CIE marks awarded in the case of the Practical component shall be based on the continuous evaluation of the laboratory report. Each experiment report can be evaluated for 10 marks. Marks of all experiments’ write-ups are added and scaled down to 15 marks. 3. The laboratory test at the end /after completion of all the experiments hall be conducted for 50 marks and scaled down to 05 marks. 4. Scaled-down marks of write-up evaluations and tests added will be CIE marks for the laboratory component of IPCC for 20 marks. SEE for IPCC Theory SEE will be conducted by University as per the scheduled timetable, with common question papers for the course (duration 03 hours) 1. The question paper will be set for 100 marks and marks scored will be scaled down proportionately to 50 marks. 2. The question paper will have ten questions. Each question is set for 20 marks. 3. There will be 2 questions from each module. Each of the two questions under a module (with a maximum of 3 sub- questions), should have a mix of topics under that module. 4. The students have to answer 5 full questions, selecting one full question from each module. 13-Jan-25
- 5. The theory portion of the IPCC shall be for both CIE and SEE, whereas the practical portion will have a CIE component only. Questions mentioned in the SEE paper shall include questions from the practical component). 1. The minimum marks to be secured in CIE to appear for SEE shall be the 15 (50% of maximum marks- 30) in the theory component and 10 (50% of maximum marks -20) in the practical component. The laboratory component of the IPCC shall be for CIE only. However, in SEE, the questions from the laboratory component shall be included. The maximum of 04/05 questions to be set from the practical component of IPCC, the total marks of all questions should not be more than the 20 marks. 2. SEE will be conducted for 100 marks and students shall secure 40% of the maximum marks to qualify in the SEE. Marks secured will be scaled down to 50. (Student has to secure an aggregate of 50% of maximum marks of the course(CIE+SEE) 13-Jan-25
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- 7. Module-1 Antenna Fundamentals and Definitions: Radiation Mechanisms, Overview, EM Fundamentals, Solution of Maxwell's Equations for Radiation Problems, Ideal Dipole, Radiation patterns, Directivity and Gain, Antenna impedance, Radiation efficiency, Antenna polarization. TEXT(1) 13-Jan-25
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- 9. Pre-modern civilization (up to 2 million years ago) Optical communications: Smoke signals, flags, Acoustical communications: Drums 1844 Telegraph—The beginning of electronic communication- Samuel Morse 1864 Maxwell’s equations—Principles of radio waves and the electromagnetic spectrum- James Clerk Maxwell 1866 First lasting transatlantic telegraph cable 1876 Telephone—Wireline analog communication over long distance- Alexander Bell 1887 First Antenna -Heinrich Hertz 1897 First practical wireless (radio) systems - Guglielmo Marconi 1901 First transatlantic radio - Guglielmo Marconi 1920 First broadcast radio station World War II Development of radar; horn, reflector, and array antennas 1950s Broadcast television in wide use 1960s Satellite communications and fiber optics 1980s Wireless reinvented with widespread use of cellular telephones Electromagnetics Pioneer James Clerk Maxwell Antenna Pioneer Heinrich Hertz Wireless Pioneer Guglielmo Marconi 13-Jan-25
- 10. Table 1-2 The Electromagnetic Spectrum Band Designation Frequency Wavelength Example Uses ELF 3 to 30 Hz 100 to 10 Mm SLF 30 to 300 Hz 10 to 1 Mm Power lines ULF 300 to 3 kHz 1 Mm to 100 km VLF 3 to 30 kHz 100 to 10 km Submarine comm. LF 30 to 300 kHz 10 to 1 km RFID MF 300 kHz to 3 MHz 1 km to 100 m AM broadcast HF 3 to 30 MHz 100 to 10 m Shortwave broadcast VHF 30 to 300 MHz 10 to 1 m FM and TV broadcast UHF 300 MHz to 3 GHz 1 m to 10 cm TV, WLAN, GPS, Microwave ovens SHF 3 to 30 GHz 10 to 1 cm Radar, WLAN, Satellite comm. EHF 30 to 300 GHz 10 to 1 mm Radar, Radio astronomy, Point to point high rate data links, Satellite comm. Microwaves 1 to 300 GHz 30 cm to 1 mm Millimeter waves 30 to 300 GHz 10 to 1 mm Submm waves .300 GHz ,1 mm 13-Jan-25
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- 39. The formulations antenna problems are in vector form and expressed in spherical coordinates. This is because electromagnetic fields have polarization Polarization is the orientation of the electric field. Spherical coordinates are required because antennas radiate in all directions (i.e.3 dimensions) The fields are expressed as a function of the spherical coordinate angles θ and φ around the antenna ANTENNA FUNDAMENTALS The fundamental electromagnetic equations in the time domain E = Re(Eejωt), H= Re(H ejωt) ∇ x E = - 𝟃 𝟃𝒕 B ∇ x H = 𝟃 𝟃𝒕 D + JT ∇ .D = ρT ∇. B = 0 ∇ . JT = 𝟃 𝟃𝒕 ρT (t) D= ε E B= µ H ∇ x E = -jwH ∇ x H = 𝒋𝒘ε E+ J ∇ x E = -jwB ∇ x H = 𝒋𝒘D + JT ∇ .D = ρT ∇. B = 0 ∇ . JT = 𝒋𝒘ρT ∇ .E = ρ ε ∇. H = 0 ∇ . J = -𝒋𝒘ρ P = Re( 𝑺. 𝒅𝒔)= 𝟏 𝟐 Re( 𝑬𝒙𝑯∗. 𝒅𝒔) 13-Jan-25
- 40. • E and H with a given J, by simplifying maxwells equations by defining the scalar and vector potential functions Φ and A. • H= 𝟏 µ ∇ x E ; E= - jwA-∇ Φ • ∇ x H = 𝟏 µ ∇ x ∇ x A = 𝒋𝒘ε E+ J • ∇ x ∇ x A = ∇ ( ∇. A) - ∇ 𝟐 A ; vector identity • ∇. A = jw µ ε Φ ; Lorentz condition • ∇ 𝟐 A= 𝒘𝟐 µ ε A= - µ J • E= jwA - j ∇ ( ∇. A) 𝑤µ ε ; vector wave equation SOLUTION OF MAXWELL’S EQUATIONS FOR RADIATION PROBLEMS β = ω με = 2π/λ ; c= 𝟏 με = f λ Maxwell found this result, concluding correctly that the velocity is a finite constant and that this result also applies to light because light is an electromagnetic wave 13-Jan-25
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- 42. THE IDEAL DIPOLE Hertzian electric dipole, electric dipole, infinitesimal dipole, and doublet 13-Jan-25
- 43. • Consider an element of current of length Δz ≪1 along the z-axis centered on the coordinate origin. It is of constant amplitude I. • The volume integral of for vector potential reduces to the one-dimensional integral A = 𝑣 𝜇𝐽 𝑒 − 𝑗𝛽𝑅 4𝜋𝑅 dv = Ƹ 𝑧 𝜇I − ∆𝑧 2 ∆𝑧 2 𝑒 − 𝑗𝛽𝑅 4𝜋𝑅 dz’ ; R≈ 𝑟 A= 𝜇𝐼 𝑒 − 𝑗𝛽𝑟 4𝜋𝑟 ∆𝑧ෝ 𝑧 ; Δz ≪ λ & Δz≪ R H= 𝟏 µ ∇ x A = 𝟏 µ ∇ x ( Az ෝ 𝑧) = ∇ ( 𝑰∆𝑧𝑒 − 𝑗𝛽𝑟 4𝜋𝑟 ) x ෝ 𝑧 → Since curl of constant vector is zero Applying the gradient in spherical coordinates H= 𝑰∆𝑧 4𝜋 −𝑗𝛽𝑒 − 𝑗𝛽𝑟 𝑟 − 𝑒 − 𝑗𝛽𝑟 𝑟2 ෝ 𝑟xෝ 𝑧 The electromagnetic fields created by the ideal dipole. 13-Jan-25
- 44. A is existing only in Z direction hence , ෝ 𝑟xෝ 𝑧 = ෝ 𝑟x (ෝ 𝑟 cosθ − θ sinθ = − ො 𝜑 sinθ H= 𝑰∆𝑧 4𝜋 𝑗𝛽 𝑟 + 1 𝑟2 𝑒 − 𝑗𝛽𝑟 sinθ ො 𝜑 -----(1a) The electric field can be obtained from E= 1 𝒋𝒘ε ∇ x H ; E= 𝑰∆𝑧 4𝜋 𝑗𝜔𝜇 𝑟 + 𝜇 𝜀 1 𝑟2 + 1 𝒋𝒘ε𝑟3 𝑒 − 𝑗𝛽𝑟 sinθ θ + 𝑰∆𝑧 2𝜋 𝜇 𝜀 1 𝑟2 + 1 𝒋𝒘ε𝑟3 𝑒 − 𝑗𝛽𝑟 𝑐𝑜𝑠θ ෝ 𝑟 --(2a) H= j𝛽 𝑰∆𝑧 4𝜋𝑟 1 − 1 𝑗𝛽𝑟 𝑒 − 𝑗𝛽𝑟 sinθ ො 𝜑 -----(1b) E= 𝑗𝜔𝜇 𝑰∆𝑧 4𝜋𝑟 1 + 1 𝑗𝛽𝑟 − 1 (𝛽𝑟)2 𝑒 − 𝑗𝛽𝑟 sinθ θ + 𝑰∆𝑧 2𝜋𝑟 1 𝑟 − 𝑗 𝛽𝑟3 𝑒 − 𝑗𝛽𝑟 𝑐𝑜𝑠θ ෝ 𝑟 --(2b) The Fields in Eqns 1 & 2 are complex and exists everywhere near antenna 13-Jan-25
- 45. At large distances from the antenna, the terms 1/r2 and 1/r3 are negligible H= j𝛽 𝑰∆𝑧 4𝜋𝑟 𝑒 − 𝑗𝛽𝑟 sinθ ො 𝜑 → H𝜑 --(1c) E= 𝑗𝜔𝜇 𝑰∆𝑧 4𝜋𝑟 𝑒 − 𝑗𝛽𝑟 sinθ θ → Eθ --(2c) Eqns 1c & 2c represent the radiations fields of the ideal short dipole The ratio of these electric and magnetic field components is Eθ H𝜑 = 𝜔𝜇 𝛽 = 𝜇 𝜀 = η ; intrinsic impedance of the medium S= 𝟏 𝟐 E x H = 𝟏 𝟐 ( 𝑰∆𝑧 4𝜋 ) 2𝜔𝜇 𝛽 𝑠𝑖𝑛θ2 𝑟2 ෝ 𝑟 ; Pf = 𝜔𝜇 𝛽 12𝜋 (𝑰∆𝑧)2 ------- (A) 13-Jan-25
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- 52. ANTENNA IMPEDANCE The primary function of a transmitting antenna is to convert a bound wave to an unbound (i.e., radiated) wave, and vice versa for a receiving antenna. Whereas the transmission line connected to an antenna binds the wave and prevents it from radiating, the antenna itself should instead enable radio waves to leave the structure. The antenna is an interface between wave phenomena on and beyond the antenna to the connecting circuit hardware. The antenna input terminals form the interface point and the circuit parameter of impedance is used to characterize the input to the antenna. The input impedance of an antenna (or simply antenna impedance) will be affected by other antennas or objects that are nearby, but the discussion here assumes an isolated antenna. 13-Jan-25
- 53. As with conventional circuits, antenna impedance is composed of real and imaginary parts. ZA = RA + j XA → input impedance. Fig. shows the general antenna model And its equivalent model for a transmitting antenna. As a consequence of reciprocity, the impedance of an antenna is identical for receiving and transmitting operations. The input resistance, RA, represents dissipation which occurs in two ways. 1. Power that leaves the antenna and never returns (i.e., radiation), Rr 2. And ohmic losses just as in a lumped resistor, R0 RA= Rr+R0 13-Jan-25
- 54. Electrically small antennas can have significant ohmic losses but other antennas usually have ohmic losses that are small compared to their radiation dissipation. The input reactance, XA, represents power stored in the near fields of the antenna. The average power dissipated in an antenna is Pin = 1 2 Ra IA 2 Where IA Peak current at the input terminals Separating the dissipated power into radiative and ohmic losses gives, Pin = P + P0 1 2 Ra IA 2 = 1 2 Rr IA 2 + 1 2 R0 IA 2 The Radiation & Loss resistance of an antenna referred to the input terminals is defined as Rr= 2𝑃 IA 2 & R0= 2𝑃0 IA 2 The radiated power is P= 1 2 𝑠𝑓𝑓 (𝐸𝑥𝐻 ∗ ).ds = 1 2 𝐸𝑥𝐻 ∗ as P in far field is real valued RADIATION RESISTANCE 13-Jan-25
- 55. The power radiated from an ideal dipole of length Δz≪ λ and input current IA = I From Eqn (A), Pf = 𝜔𝜇 𝛽 12𝜋 (𝑰∆𝑧)2 Rr= 2𝑃 IA 2 = 2 IA 2 𝜔𝜇 𝛽 12𝜋 (𝑰∆𝑧)2 = 𝜔𝜇 𝜇 𝝐 𝛽(∆𝑧)2 𝝐 6 𝜋 The power radiated from an ideal dipole of length Δz≪ λ and input current IA = I Rr = η (𝛽∆𝑧)2 6 𝜋 = 80 𝜋2 ∆𝑧 λ 2 Ω -----Ideal Dipole For ideal dipoles, Rr is very small since Δz≪ λ. 13-Jan-25
- 56. RADIATION EFFICIENCY The efficiency factor is the ratio of wanted power to total power supplied. The radiation efficiency er of an antenna defined as the ratio of radiated power (which is the wanted power) to the net power accepted by the antenna. Er = 𝑃 𝑃𝑖𝑛 = 𝑃 𝑃+𝑃0 where P= power radiated Po= power dissipated in ohmic losses on the antenna Pin= P+Po = input power = power accepted by the antenna Er = 1 2 Rr IA 2 1 2 Rr IA 2 + 1 2 R0 IA 2 = 𝑅𝑟 𝑅𝑟+𝑅0 = 𝑅𝑟 𝑅𝐴 13-Jan-25