Unit 3 212ECE2311 Antenna Arrays and Special Antennas PDF

Summary

This document provides an overview of antenna arrays and special antenna types. It discusses the configuration of antenna arrays, different types of array antennas, and their advantages, like high directivity and high gain. The document also touches on the applications of antenna arrays.

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212ECE2311 DR. SHASHI KANT DARGAR ASSO. PROF/ECE KARE © Kalasalingam academy of research and education Syllabus Antenna Arrays, Special Antennas Antenna array v...

212ECE2311 DR. SHASHI KANT DARGAR ASSO. PROF/ECE KARE © Kalasalingam academy of research and education Syllabus Antenna Arrays, Special Antennas Antenna array various forms, arrays of Unit 3 point sources, non-isotropic but similar point sources, multiplication of patterns, Arrays of n-isotropic sources of equal amplitude and spacing (Broad-side and End-fire array cases), array factor, Array of Outcomes n-isotropic sources of equal amplitude and Articulate an antenna system, including the structure of the spacing end-fire array with increased antenna, special characteristics, the need on the arrangement directivity, Special antennas: Broadband of the radiating elements in an array by applying the design Antennas, Yagi Uda antenna, helical principles and by selecting proper antenna type for the given antenna, Frequency independent antennas, specifications and environments Log periodic dipole array, travelling wave antennas, rhombic antenna © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Basics of antenna array What is an antenna array? ⮚ Array: a set of antennas working together to produce certain radiation pattern. ⮚ Each antenna in an array is called an element antenna (or simply an element). ⮚ The elements in an array can be the same or different. ⮚ In most practical cases, they are identical in construction (with different feedings). © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Basics of antenna array Images of antenna arrays © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Basics of antenna array Why an antenna array? ⮚ To meet long distance comm, we often need highly directive antenna which can be accomplished by Either increasing the electrical size of the antenna or by forming an assembly of radiating elements. ⮚ To provide very directive patterns, the fields from the elements interfere constructively in the desired direction and destructively in the other direction. © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Broadband Antenna an Introduction Configurations of antenna array ⮚ Linear (1D), ⮚ planar (2D), ⮚ conformal (3D). Yagi-Uda array Log –Periodic are some examples © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Basics of antenna array Why an antenna array? ⮚ In an array of identical elements the Geometrical configuration of the array Relative displacement b/w the elements Excitation amplitude of the individual element Excitation phase of the individual element Relative pattern of the individual element ⮚ Can be controlled to shape the overall pattern of the radiation. © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Array of Antenna ARRAY = Antennas are Rearranged in Required and Affective ways to Yearn Directivity. Quote: Walter Payton: “We are stronger together than we are alone.” © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 An Array of Antenna Single element is relatively wide and low directive gain (directivity) To have very high gain and long distance antenna, this can be accomplished by increasing the electrical size of the antenna. Enlarge dimensions of the antenna without increasing the size of individual elements is to form an assembly of radiating elements in an electrical and geometrical configuration. This new antenna, formed by multi-elements, is referred to as an array Most cases, the elements of an array are identical. An array is widely used as a base-station antenna for mobile communication. © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Why antenna Array ? High directivity in desired direction Point to point communication Long distance communication Narrow beam Provide diversity reception Cancel out interference from the particular set of direction Determine the direction of arrival of incoming signal This can be achieved by ◦ 1. Increase the size of antenna. ◦ (But this approach leads to mechanical problem and also difficult from a fabrication point of view) ◦ 2. Antenna array i.e. ⟹antenna array is utilized to ◦ 1. to increase gain of antenna and ◦ 2. To have narrow beam © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Antenna Arrays Several individual antennas so spaced and phased that their individual contribution add in one preferred direction to give higher directivity gain and cancel in other direction This new antenna formed by multiple elements is called as antenna array An antenna array (or array antenna) is a set of multiple connected antenna which work together as a single antenna to transmit or receive radio waves. (Micro-strp Array satellite TV reception) (Crossed dipole FM radio broadcast) © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Antenna Arrays Antenna array - a configuration of multiple antennas (elements) arranged to achieve a given radiation pattern. Linear array - antenna elements arranged along a straight line. Planar array - antenna elements arranged over some planar surface. If array arranged in a plane (xy, yz or xz) then it is said to be 2-D array or Planar array Example - rectangular array, Triangular array, Spherical array, circular array, concentric array etc. © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Antenna Arrays Linear array - antenna elements arranged along a straight line. © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Linear Arrays An antenna array is a set of 2 or more antennas. The signals from the antennas are combined or processed in order to achieve improved performance over that of a single antenna. Linear Arrays ◦ Antenna elements arranged along a straight line. ◦ Elements of the array are uniformly-spaced. Classifications of Linear Array ◦ 2-Element Array with Uniform Amplitude and Spacing ◦ N-element Linear Array with Uniform Amplitude and Spacing s © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Antenna Arrays Planar array: Antenna elements arranged over some planar surface (example - rectangular array). If array arranged in a plane (xy, yz or xz) then it is said to be 2- D array or Planar array © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Antenna Arrays Planar array: ◦ Planar arrays are more versatile ◦ Antenna elements arranged over some planar surface (example rectangular array). ◦ provide more symmetrical patterns with lower side lobes much higher directivity (narrow main beam). Applications – tracking radars, remote sensing, and communications. © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Antenna Arrays © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Advantages of using antenna array Advantages of using antenna array: 1. In an array, provisions can be made to make the fields from individual elements interfere constructively in some directions and cancel in some other directions. 2. They can provide the capability of a steerable beam (radiation direction change) 3. They can provide a high gain (array gain) by using simple antenna elements 4. They provide a diversity gain in multipath signal reception. © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Advantages/ Disadvantages of antenna array Advantages of using antenna array: The following are the advantages of using antenna arrays − ◦ The signal strength increases ◦ High directivity is obtained ◦ Minor lobes are reduced much ◦ High Signal-to-noise ratio is achieved ◦ High gain is obtained ◦ Power wastage is reduced ◦ Better performance is obtained Disadvantages of using antenna array: The following are the disadvantages of array antennas − ◦ Resistive losses are increased ◦ Mounting and maintenance is difficult ◦ Huge external space is required © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Applications of antenna array Applications: The following are the applications of array antennas − ◦ Used in satellite communications ◦ Used in wireless communications ◦ Used in military radar communications ◦ Used in the astronomical study Types of Arrays: The basic types of arrays are − ◦ Collinear array ◦ Broad side array ◦ End fire array ◦ Parasitic array ◦ Yagi-Uda array ◦ Log-peroidic array ◦ Turnstile array ◦ Super-turnstile array We will discuss these arrays in the coming chapters. © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Basic principle of antenna array Array controlling factor ◦ Geometrical configurations( Linear, Rectangular, Triangular Circular, Spherical etc.) ◦ Excitation amplitude (𝐼 ) of the individual elements ◦ Excitation phase(𝛼) of each individual elements ◦ Relative spacing (𝑑) ◦ The pattern of each antenna ◦ Frequency of operation( wavelength) λ = c/ f © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Basic principle of antenna array Antenna array - 𝑬𝟏 𝑬𝟐 𝑬𝟑 𝑬𝟒 ….. 𝑬𝒏 d d … d Antenna’s. 𝜶𝟏 𝜶𝟐 𝜶𝟑 𝜶𝟒 𝜶𝒏 Phase Shifter’s Attenuator' s 𝑰𝟑 𝑰𝟒 ….. 𝑰𝟐 𝑰𝟏 𝑰𝒏 © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Basic Current principle of antenna supplied to different element is given. array 𝐼1 = 𝐼1 𝑒 𝑗𝛼1 , 𝐼2 = 𝐼2 𝑒 𝑗𝛼2 , 𝐼3 = 𝐼3 𝑒 𝑗𝛼3 … … , 𝐼𝑛 = 𝐼𝑛 𝑒 𝑗𝛼𝑛 Where 𝛼1 , 𝛼2 , 𝛼3 ………. 𝛼𝑛 are the phase difference of individual elements by (Phase shifter provides phase difference) and 𝐼1 , 𝐼2 , 𝐼3 ……. 𝐼𝑛 are magnitude of current of individual elements. The electric field by different element is given by: 𝐸1 = 𝐸1 𝑒 𝑗𝜓1 , 𝐸2 = 𝐸2 𝑒 𝑗𝜓2 , 𝐸3 = 𝐸3 𝑒 𝑗𝜓3 … … …. , 𝐸𝑛 = 𝐸𝑛 𝑒 𝑗𝜓𝑛 Where 𝐸1 , 𝐸2 , 𝐸3 ………. 𝐸𝑛 are the magnitude of individual elements field and 𝜓1 , 𝜓2 , 𝜓2 ……𝜓𝑛 are the phase of the of individual elements. Total electric field is the algebraic sum of all electric fields: 𝐸 = 𝐸1 + 𝐸2 + 𝐸3 + ⋯ + 𝐸𝑛 𝐸 = 𝐸1 𝑒 𝑗𝜓1 + 𝐸2 𝑒 𝑗𝜓2 + 𝐸3 𝑒 𝑗𝜓3 + … … …. +𝐸𝑛 𝑒 𝑗𝜓𝑛 © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Basic principle of antenna array The total field of antenna array is determined by the vector addition of the fields radiated by the individual elements The total phase difference (𝜓) 𝟐𝝅 𝝍 = 𝜷𝒅 + 𝜶 = 𝒅+𝜶 𝝀 Where is 𝑑 ⟶ spacing b/w two antenna, 𝛼 ⟶ initial phase (phase due to excitation) and 𝜆 ⟶ operating wavelength 𝟐𝝅 The total phase difference (𝝍) = ∗ 𝑻𝒐𝒕𝒂𝒍 𝒑𝒂𝒕𝒉 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒄𝒆 𝝀 © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Basic principle of antenna array The field from the elements of the array interfere constructively (add) in the desired directions and interfere destructively(cancel) in the remaining space But practically does not fully cancel those leading to some minor side lobes and back lobes © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Two-Element Antenna Arrays Antenna arrays : A group of several antenna elements in various configurations (straight lines, circles, triangles, and so on) with proper amplitude and phase relations to give certain desired radiation characteristics. 𝑅1 ≅ 𝑅0 − 𝑑 𝑐𝑜𝑠∅ Figure(a) Figure(b) © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Two-Element Antenna Arrays 𝑅1 ≅ 𝑅0 − 𝑑 𝑐𝑜𝑠∅ 𝒆−𝒋𝜷𝑹𝟎 𝑬𝟎 = 𝑬𝒎 𝑭 𝜽, ∅ 𝑹𝟎 Figure(a) 𝒆𝒋𝜶 𝒆−𝒋𝜷𝑹𝟏 𝑬𝟏 = 𝑬𝒎 𝑭 𝜽, ∅ 𝑹𝟏 © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Two-Element Arrays The antennas are excited with a current of the same amplitude, but the phase in antenna 1 leads that in antenna 0 by an angle 𝛼. Far-zone field phasors at point 𝑃 𝜃, 𝜙 : 𝑒 −𝑗𝛽𝑅0 𝐸0 = 𝐸𝑚 𝐹 𝜃, ∅ 𝑎𝑛𝑑 𝑅0 𝑒 𝑗𝛼 𝑒 −𝑗𝛽𝑅1 𝐸1 = 𝐸𝑚 𝐹 𝜃, ∅ 𝑅1 where 𝐹 𝜃, 𝜙 is the pattern function of the individual antennas, and 𝐸𝑚 is an amplitude function. The electric total field of the two-element array : 𝑒 −𝑗𝛽𝑅0 𝑒 𝑗𝛼 𝑒 −𝑗𝛽𝑅1 𝐸 = 𝐸0 + 𝐸1 = 𝐸𝑚 𝐹 𝜃, ∅ +. 𝑅0 𝑅1 𝑅1 ≅ 𝑅0 − 𝑑 𝑐𝑜𝑠∅ (from figure) © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Two-Element Arrays (continued) In the far-zone 𝑅0 ? 𝑑 Τ2 , 1Τ𝑅1 = 1Τ𝑅0 , 𝑎𝑛𝑑 Thus 𝑅1 ≅ 𝑅0 − 𝑑 𝑐𝑜𝑠∅ 𝑒 −𝑗𝛽𝑅0 𝑒 𝑗𝛼 𝑒 −𝑗𝛽(𝑅0−𝑑 𝑐𝑜𝑠∅) 𝐸 = 𝐸𝑚 𝐹 𝜃, ∅ +. 𝑅0 𝑅0 −𝑑 𝑐𝑜𝑠∅ Assume 𝑅0 − 𝑑 𝑐𝑜𝑠∅ ≅ 𝑅0 𝑒 −𝑗𝛽𝑅0 𝑒 𝑗𝛼 𝑒 −𝑗𝛽𝑅0 𝑒 𝑗𝛽𝑑 𝑐𝑜𝑠∅ 𝐸 = 𝐸𝑚 𝐹 𝜃, ∅ +. 𝑅0 𝑅0 𝑒 −𝑗𝛽𝑅0 𝐸 = 𝐸𝑚 𝐹 𝜃, ∅ 1 + 𝑒 𝑗(𝛽𝑑 cos ∅+𝛼). 𝑅0 𝑒 −𝑗𝛽𝑅0 𝐸 = 𝐸𝑚 𝐹 𝜃, ∅ 1 + 𝑒 𝑗𝜓. 𝑅0 Electrical phase 𝐹 𝜃,∅ −𝑗𝛽𝑅 𝐸 = 𝐸𝑚 𝑒 0 𝑒 𝑗𝜓/2 2 cos(𝜓/2). 𝑅0 Space phase © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Two-Element Arrays Magnitude of total E-field (continued) 2 𝐸𝑚 𝑬 = 𝐹 𝜃, ∅ cos(𝜓/2). 𝑅0 𝟐 𝑬𝒎 𝑬 = Element factor 𝑵𝒐𝒓𝒎𝒊𝒍𝒊𝒛𝒆𝒅 𝑨𝑭. 𝑹𝟎 The element factor : the magnitude of the pattern function of the individual radiating elements. The array factor depends on array geometry as well as on the relative amplitudes and phases of the excitations in the elements. 𝛽𝑑 𝑐𝑜𝑠𝜑 + 𝛼 𝐴𝐹 𝜙 = cos(𝜓/2) = 𝑐𝑜𝑠 2 The AF, in general, depends on: ◦ Number of elements ◦ Relative excitation (magnitudes and phases) ◦ Spacing between the elements © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Example 10-6 Plot the H-plane radiation patterns of two parallel dipoles for the following 𝜆 𝜆 𝜆 cases : 𝑎 𝑑 = 2 , 𝛼 = 0, (𝑏) 𝑑 = 4 , 𝛼 = −𝜋/2, (𝑐) 𝑑 = 4 , 𝛼 = 𝜋/2, (𝑑) 𝜆 𝑑 = , 𝛼 = 𝜋/2 2 In the H-plane (Fig. 10-6), the dipole is omnidirectional, and the normalized pattern function is equal to the normalized array factor 𝐴𝐹 𝜙 1 Thus 𝐴𝐹 𝜙 = 𝑐𝑜𝑠(𝜓/2) = 𝑐𝑜𝑠 2 𝛽𝑑 𝑐𝑜𝑠𝜑 + 𝛼 𝜆 2𝜋 𝜆 a) 𝑑 = 2 ⟹ 𝛽𝑑 = × 2 = 𝜋 , 𝛼 = 0; 𝜆 𝜋 𝐴𝐹 𝜙 = 𝑐𝑜𝑠 2 𝑐𝑜𝑠𝜑. 𝜙 0 𝜋/4 𝜋/2 𝜋/3 𝐴𝐹 𝜙 0 1 1/ 2 © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Example 10-6 (continued) 𝜋 Broadside array (Fig. 10-7(a)) : the pattern has its maximum at 𝜙0 = ± 2 Their electric fields add in the broadside direction and cancel each other at 𝜙 = 0 𝑎𝑛𝑑 𝜙 = 𝜋 𝜆 2𝜋 𝜆 𝜋 b) 𝑑 = 4 ⟹ 𝛽𝑑 = ×4= , 𝛼 = −𝜋/2; 𝜆 2 𝜋 𝐴𝐹 𝜙 = 𝑐𝑜𝑠 4 𝑐𝑜𝑠𝜑 − 1. which has maximum at 𝜙0 = 0 and vanishes at 𝜙 = 𝜋 The pattern maximum (Fig. 10-7(b)) : in a direction along the line of array (end 0 fire array).𝜙 𝜋/4 𝜋/2 𝜋/3 𝐴𝐹 𝜙 0 © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Example 10-6 (continued) (c) 𝑑 = 𝜆Τ4, 𝛼 = 𝜋/2; 𝑑 = 𝜆Τ4, 𝛼 = 0; 𝑑 = 𝜆Τ4, 𝛼 = −𝜋/2; © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Example 10-6 (continued) 𝜆 (𝑑) 𝑑 = , 𝛼 = 𝜋/2 2 © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Example 10-7 Discuss radiation pattern of a linear array of the three isotropic sources spaces 𝜆/2 apart. The excitations in the sources are in-phase and have amplitude ratios 1:2:1. This three-source array : two two-element arrays displaced 𝜆/2 (Fig. 10-8). © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Example 10-7 (continued) By the principle of pattern multiplication we obtain The radiation pattern is sketched in Fig. 10-9 (sharper). © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 * P(𝜃, ∅) Two-Element Arrays (Two Isotropic Point Sources of Same Amplitude and Phase) Figure(b) © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 1. Two-Element Arrays (Two Isotropic Point Sources of Same Amplitude and Phase) The pathE difference -field by source 1 w.r.t origin b/w source 1 and 2 will be: 𝑑 𝑐𝑜𝑠∅ 𝐸 = 𝐸 𝑒 −𝑗𝜓/2 1 1 Point source (1) lead Path Difference by (-𝜓/2) phase Point source (2) lead by (+𝜓/2) phase Array Axis Figure(b) E -field by source 2 w.r.t origin 𝐸2 = 𝐸2 𝑒 +𝑗𝜓/2 © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Two-Element Arrays (Two Isotropic Point Sources of Same Amplitude and Phase) Total E-field: 𝐸 = 𝐸1 + 𝐸2 = 𝐸1 𝑒 −𝑗𝜓/2 + 𝐸2 𝑒 +𝑗𝜓/2 Two Isotropic Point Sources of Same Amplitude ⟹ 𝐸1 = 𝐸2 = 𝐸0 2 𝐸 = 𝐸0 𝑒 −𝑗𝜓/2 + 𝑒 +𝑗𝜓/2 × 2 𝐸 = 2𝐸0 cos 𝜓/2 To normalize field 𝐸𝑛 = 2𝐸0 𝛽𝑑 𝑐𝑜𝑠𝜑 + 𝛼 𝐸 = 𝐸𝑛 cos 𝜓/2 = 𝐸𝑛 cos 𝜓/2 = 𝐸𝑛 𝑐𝑜𝑠 2 2𝜋 Where 𝛽 = 𝜆 © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Radiation pattern of Two-Element Antenna Arrays Cases I : 𝒅 = 𝟐𝝀 , 𝜶 = 𝟎 ; 2𝜋 𝜆 ⟹ 𝛽𝑑 = × =𝜋 𝜆 2 𝜋 𝐸 = 𝐸𝑛 𝑐𝑜𝑠 𝑐𝑜𝑠𝜑 2 Direction of Maxima: ⟹ 𝐸 = ±1 𝜋 ⟹ 𝐸𝑛 𝑐𝑜𝑠 𝑐𝑜𝑠𝜑 = ±1 2 𝜋 ⟹ cos ∅𝑚𝑎𝑥 = ± 𝑛𝜋; 𝑛 = 0, 1, 2, … 2 𝜋 ⟹ cos ∅𝑚𝑎𝑥 = ±0; 𝑓𝑜𝑟 𝑛 = 0 2 𝜋 3𝜋 ⟹ ∅𝑚𝑎𝑥 = 𝑐𝑜𝑠 −1 ±0 = and 2 2 ∅𝑚𝑎𝑥 = 900 &2700 © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Radiation pattern Direction of Minima: ⟹ 𝐸 = ±0 𝜋 ⟹ 𝐸𝑛 𝑐𝑜𝑠 𝑐𝑜𝑠𝜑 = ±0 2 𝜋 ⟹ cos ∅𝑚𝑖𝑛 = ± 2𝑛 + 1 𝜋/2; 𝑛 = 0, 1, 2, … 2 𝜋 ⟹ cos ∅𝑚𝑖𝑛 = ±1 ∗ 𝜋/2; 𝑓𝑜𝑟 𝑛 = 0 2 ⟹ ∅𝑚𝑖𝑛 = 𝑐𝑜𝑠 −1 ±1 = 0 and 𝜋 ∅𝑚𝑖𝑛 = 00 &1800 1 Direction of HPBW: ⟹ 𝐸 = ± 2 𝜋 1 ⟹ 𝐸𝑛 𝑐𝑜𝑠 𝑐𝑜𝑠𝜑 =± 2 2 𝜋 ⟹ cos ∅𝐻𝑃 = ± 2𝑛 + 1 𝜋/4; 𝑛 = 0, 1, 2, … 2 𝜋 𝜋 ⟹ cos ∅𝐻𝑃 = ± ; 𝑓𝑜𝑟 𝑛 = 0 2 4 1 ⟹ ∅𝐻𝑃 = 𝑐𝑜𝑠 −1 ± = 600 and1200 2 © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Radiation pattern Cases II : 𝒅 = 𝝀, 𝜶 = 𝟎 2𝜋 ⟹ 𝛽𝑑 = 𝜆 × 𝜆 = 2𝜋 𝐸 = 𝐸𝑛 𝑐𝑜𝑠 𝜋 𝑐𝑜𝑠𝜑 Direction of Maxima: ⟹ ∅𝑚𝑎𝑥 = 00 , 900 , 1800 , 2700 Direction of Minima: ⟹ ∅𝑚𝑖𝑛 = 600 , 1200 , 2400 , 3000 Cases III : 𝒅 = 𝟐𝝀, 𝜶 = 𝟎 2𝜋 ⟹ 𝛽𝑑 = 𝜆 × 2𝜆 = 4𝜋 𝐸 = 𝐸𝑛 𝑐𝑜𝑠 2𝜋 𝑐𝑜𝑠𝜑 Direction of Maxima: ⟹ ∅𝑚𝑎𝑥 = 00 , 600 , 900 , 1200 , 1800 , 2400 , 2700. Direction of Minima: ⟹ ∅𝑚𝑖𝑛 = 48, 760 , 1040 , 3000 © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 2. Two-Element Arrays (Two Isotropic Point Sources of Same Amplitude and opposite Phase current) Total E-field: 𝐸 = 𝐸1 + 𝐸2 = 𝐸1 𝑒 −𝑗𝜓/2 + 𝐸2 𝑒 +𝑗𝜓/2 Two Isotropic Point Sources of Same Amplitude ⟹ 𝐸2 =-𝐸1 = −𝐸0 2𝑗 𝐸 = 𝐸0 𝑒 +𝑗𝜓/2 − 𝑒 −𝑗𝜓/2 × 2𝑗 𝐸 = 2𝑗 𝐸0 sin 𝜓/2 To normalize field 𝐸𝑛 = 𝑗2𝐸0 𝛽𝑑 𝑐𝑜𝑠𝜑 + 𝛼 𝐸 = 𝐸𝑛 sin 𝜓/2 = 𝐸𝑛 𝑠𝑖𝑛 2 2𝜋 Where 𝛽 = 𝜆 © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Radiation pattern of Two-Element Antenna Arrays 𝝀 Cases :𝒅 = ,𝜶 = 𝟎 𝟐 2𝜋 𝜆 ⟹ 𝛽𝑑 = × =𝜋 𝜆 2 𝜋 𝐸 = 𝐸𝑛 𝑠𝑖𝑛 𝑐𝑜𝑠𝜑 2 Direction of Maxima: ⟹ ∅𝑚𝑎𝑥 = 00 &1800 Direction of Minima: ⟹ ∅𝑛𝑢𝑙𝑙𝑠 = 900 &2700 © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Pattern Multiplication Principle of pattern multiplication: the pattern function of an array of identical elements is described by the product of the element factor and the array factor. 2 𝐸𝑚 𝑬 = 𝐹 𝜃, ∅ cos(𝜓/2). 𝑅0 𝟐 𝑬𝒎 𝑬 = Element factor 𝑵𝒐𝒓𝒎𝒊𝒍𝒊𝒛𝒆𝒅 𝑨𝑭. 𝑹𝟎 © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Pattern Multiplication Array Factor (AF) Pattern at different spacing figure1: 𝑑 = 𝜆2 , 𝛼 = 0 figure2: 𝑑 = 𝜆, 𝛼 = 0 figure3: 𝑑 = 2𝜆, 𝛼 = 0 © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Two Same Dipoles and Pattern Multiplication Two Same Dipoles and Pattern Multiplication Dipole AF Final Pattern For 𝜶 = 0, Array Factor (AF) will give max. radiation in Broadside Direction © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Two Same Dipoles and Pattern Multiplication Two Same Dipoles and Pattern Multiplication Dipole Element Pattern AF Final Pattern Pattern For 𝜶 = 0, Array Factor (AF) will give max. radiation in Broadside Direction © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Pattern multiplication of 4- isotropic sources: Pattern multiplication of 4- isotropic sources: © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Pattern multiplication of 8- isotropic sources: Pattern multiplication of 8- isotropic sources: © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Exercise Q-1: Obtain the pattern of two isotropic point sources with identical amplitude and phase currents, and spaced one-half wavelength apart. Q-2: Obtain the pattern of two isotropic point sources with identical amplitude and opposite phase currents, and spaced one-half wavelength apart. Q-3: Obtain the pattern of two isotropic point sources with identical amplitude and in phase quadrate currents, and spaced (i) one-half wavelength, (ii) quarter wavelength apart. Q-4: An array consists of two horizontal dipoles located along z-axis at one-half wavelength apart. If the excitation magnitudes and phases are same, then obtain the shape of the radiation pattern. © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Exercise Q-5: For two point source antenna array, find the nulls of the total fields when 𝑑 = 𝜆Τ4 and. (i) 𝛼 = 0; (ii) 𝛼 = +𝜋/2; (iii) 𝛼 = −𝜋/2; Q-6: Prove that the directivity of a Broadside array of two identical isotropic in phase point source spaced at ‘d’ apart is given by: 2 𝐷 𝜃, ∅ = sin 𝛽𝑑 1+ 𝛽𝑑 © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Exercise Q-7: Prove that the directivity of a End-fire array of two identical isotropic in phase point source spaced at ‘d’ apart is given by: 4 𝐷 𝜃, ∅ = sin 𝛽𝑑 1+ 𝛽𝑑 © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 N- Elements Uniform Antenna Array N-element Linear Array with Uniform Amplitude and Spacing ◦ An array of identical elements with identical magnitudes and with a progressive phase is called a uniform array. In a uniform array the antennas are equi-spaced and are excited with uniform current with constant progressive phase shift (phase shift between adjacent antenna elements) as shown in Fig. © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Length of array N- Elements Uniform Antenna Array Let the array have N elements and let the antennas be isotropic (this condition will be relaxed later). All the antennas are excited with equal amplitude currents. Let us define the following for the array. Inter-element spacing d, : This the spacing between any two adjacent elements of the array. Progressive phase-shift 𝛼, : This is the phase shift between currents on any two adjacent antenna elements of the array. The field due an antenna is proportional to its current. Also for a far away point, the fields due to individual antennas have equal amplitude but different phases. The phase of the field has two components: ◦ (1) The phase due to the phase of the excitation current. ◦ (2) The phase due to propagation. If the observation point P is in a direction which makes an angle φ with the array axis, the propagation phase difference between radiation from two adjacent elements is 𝛽𝑑 𝑐𝑜𝑠𝜑. Total phase difference is written as: 𝜓 = 𝛽𝑑 𝑐𝑜𝑠∅ + 𝛼 © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 N- Elements Uniform Total Electric fields at observation pointAntenna Array P (Apply the superposition of E-fields) 𝐸𝑡𝑜𝑡 = 𝐸1 + 𝐸2 + 𝐸3 + 𝐸4 … … + 𝐸𝑁 = 𝐸0 𝑒 𝑗0 + 𝐸0 𝑒 𝑗𝜓 + 𝐸0 𝑒 𝑗2𝜓 … … + 𝐸0 𝑒 𝑗 𝑁−1 𝜓 = 𝐸0 [1 + 𝑒 𝑗𝜓 + 𝑒 𝑗2𝜓 … … + 𝑒 𝑗 𝑁−1 𝜓 ].GP series 1−𝑒 𝑗 𝑁−1 +1 𝜓 1−𝑒 𝑗𝑁𝜓 𝑒 −𝑗𝑁𝜓/2 −𝑒 +𝑗𝑁𝜓/2 𝑒 +𝑗𝑁𝜓/2 = 𝐸0 = 𝐸0 = 𝐸0 1−𝑒 𝑗𝜓 1−𝑒 𝑗𝜓 𝑒 −𝑗𝜓/2 −𝑒 +𝑗𝜓/2 𝑒 +𝑗𝜓/2 𝑁 2𝑗 sin 𝜓 𝑒 +𝑗𝑁𝜓/2 2 = 𝐸0 1 2𝑗 sin 𝜓 𝑒 +𝑗𝜓/2 2 𝑁 2𝑗 sin 𝜓 𝐸𝑡𝑜𝑡 = 𝐸0 𝑒 𝑗(𝑁−1) 𝜓/2 2 1 2𝑗 sin 𝜓 2 𝑁 𝐸𝑡𝑜𝑡 sin 𝜓 𝐴𝐹 = = 2 𝐸0 1 sin 𝜓 2 𝑵 𝑨𝑭 𝟏 𝐬𝐢𝐧 𝟐 𝝍 𝑨𝑭 𝒏 = = (Normalized AF) 𝑨𝑭𝒎𝒂𝒙 𝑵 𝒔𝒊𝒏 𝟏𝝍 𝟐 © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 N- Elements Uniform Antenna Array Total Electric fields at observation point P 𝐸𝑡𝑜𝑡 = 𝐸1 + 𝐸2 + 𝐸3 + 𝐸4 … … + 𝐸𝑁 = 𝐴[𝐼1 𝑒 𝑗𝛽𝑟1 + 𝐼2 𝑒 𝑗𝛽𝑟2 + 𝐼3 𝑒 𝑗𝛽𝑟3 … … + 𝐼𝑁 𝑒 𝑗𝛽𝑟𝑁 ] 𝑗𝜂𝛽𝐿 𝑤ℎ𝑒𝑟𝑒 𝐴 = 4𝜋𝑟 𝑠𝑖𝑛𝜃 for infinitesimal dipole 𝐼2 = 𝐼1 𝑒 𝑗𝛼 𝑟2 = 𝑟1 − 𝑑 𝑐𝑜𝑠𝜃 𝐸𝑡𝑜𝑡 = 𝐴 𝐼1 𝑒 𝑗𝛽𝑟1 [ 1 + 𝐼2 𝑒 𝑗𝛼 𝑒 𝑗𝛽𝑑 𝑐𝑜𝑠𝜃 + 𝐼3 𝑒 𝑗2𝛼 𝑒 𝑗2𝛽𝑑 𝑐𝑜𝑠𝜃 + … … + 𝐼𝑁 𝑒 𝑗(𝑁−1)𝛼 𝑒 𝑗(𝑁−1)𝛽𝑑 𝑐𝑜𝑠𝜃 ] 𝐸𝑡𝑜𝑡 = 𝐸0 [ 1 + 𝐼2 𝑒 𝑗𝛼 𝑒 𝑗𝛽𝑑 𝑐𝑜𝑠𝜃 + 𝐼3 𝑒 𝑗2𝛼 𝑒 𝑗2𝛽𝑑 𝑐𝑜𝑠𝜃 + … … + 𝐼𝑁 𝑒 𝑗(𝑁−1)𝛼 𝑒 𝑗(𝑁−1)𝛽𝑑 𝑐𝑜𝑠𝜃 ] © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 N- Elements Uniform Antenna Array𝐸𝑡𝑜𝑡 𝐴𝐹 = = 1 + 𝑒 𝑗(𝛽𝑑 𝑐𝑜𝑠∅+𝛼) +𝑒 𝑗2(𝛽𝑑 𝑐𝑜𝑠∅+𝛼) 𝐸0 + ⋯ + 𝑒 𝑗(𝑁−1)(𝛽𝑑 𝑐𝑜𝑠∅+𝛼) 𝐴𝐹 = 1 + 𝑒 𝑗𝜓 +𝑒 𝑗2𝜓 + 𝑒 𝑗3𝜓 + ⋯ + 𝑒 𝑗(𝑁−1)𝜓 Where, 𝜓 = 𝛽𝑑 𝑐𝑜𝑠∅ + 𝛼 𝐴𝐹 = 1 + 𝑒 𝑗𝜓 +𝑒 𝑗2𝜓 + 𝑒 𝑗3𝜓 + ⋯ + 𝑒 𝑗(𝑁−1)𝜓............................ (1) 𝐴𝐹𝑒 𝑗𝜓 = 𝑒 𝑗𝜓 + 𝑒 𝑗2𝜓 +𝑒 𝑗3𝜓 + ⋯ + 𝑒 𝑗 𝑁−1 𝜓 + 𝑒 𝑗𝑁𝜓................. (2) Subtract (1) from (2) 𝐴𝐹(𝑒 𝑗𝜓 −1) = (−1 + 𝑒 𝑗𝑁𝜓 ) 𝑒 𝑗𝑁𝜓 − 1 𝑒 +𝑗𝑁𝜓/2 − 𝑒 −𝑗𝑁𝜓/2 𝑒 +𝑗𝑁𝜓/2 𝐴𝐹 = = 𝑒 𝑗𝜓 − 1 𝑒 +𝑗𝜓/2 − 𝑒 −𝑗𝜓/2 𝑒 +𝑗𝜓/2 𝑁 sin 𝜓 𝐴𝐹 = 𝑒 𝑗(𝑁−1) 𝜓/2 2 1 sin 𝜓 2 © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 N- Elements Uniform Antenna Array 0 Max occurred at AF = which occurred at ψ = ± mπ (for m = 0,1,2,….) 0 To get Max value⟹ differentiate num and denum w.r.t. ψ (and substitute ψ=0) 𝐴𝐹𝑚𝑎𝑥 = 𝑁 hence 𝑁 𝐴𝐹 1 sin 2 𝜓 𝐴𝐹 𝑛 = = 𝐴𝐹𝑚𝑎𝑥 𝑁 sin 1 𝜓 2 The radiation pattern is generally normalized with respect to the maximum value N to get the ‘ Array Factor ' as Observations: (1) Main lobe is in the direction so that 𝜓 = 𝛽𝑑 𝑐𝑜𝑠∅ + 𝛼 = 0 (2) The main lobe narrows as N increases. © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 N- Elements Uniform Antenna Array Approximated normalized 𝐴𝐹 𝜓 1 𝜓 If 𝑖𝑠 𝑣𝑒𝑟𝑦 𝑠𝑚𝑎𝑙𝑙 ⟹ sin 𝜓 ≈ 2 2 2 𝑁 𝐴𝐹 1 sin 2 𝜓 𝐴𝐹 𝑛 = ≈ 1 Approximated 𝐴𝐹𝑚𝑎𝑥 𝑁 𝜓 2 © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Exercise Q.1 Show that the peaks of the array factor of an N-element uniform array are given by the solution of the equation: 𝜓 𝜓 𝑁 𝑡𝑎𝑛 = 𝑡𝑎𝑛 𝑁 2 2 Q.2 For a uniform 7-elements array with a=0, calculate the exact location of the peak of the 𝜓 𝜓 first side lobe by the equation: 𝑁 𝑡𝑎𝑛 = 𝑡𝑎𝑛 𝑁 and calculate its level in dB w.r.t the 2 2 main lobe peak. © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Radiation pattern of N- elements Array A typical radiation pattern is shown in Fig. The range of the angle is from 0 to π , and the 3-D radiation pattern is the figure of revolution of the Array Factor around the axis of the array. © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Radiation pattern of N- elements Array © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Direction of Maximum Radiation The direction of maximum radiation (also called the direction of the main beam) is one of the important features of the array. The maximum radiation is obtained when 𝝍 = 𝟎 or multiple of 2𝝅. If the direction of maximum radiation is denoted by we have ∅𝑚𝑎𝑥 𝛼 𝜓 = 𝛽𝑑 𝑐𝑜𝑠∅𝑚𝑎𝑥 + 𝛼 = 0 ⟹ 𝑐𝑜𝑠∅𝑚𝑎𝑥 = − 𝛽𝑑 𝛼 ⟹ ∅𝑚𝑎𝑥 = 𝑐𝑜𝑠 −1 − 𝛽𝑑 𝛼𝜆 ⟹ ∅𝑚𝑎𝑥 = 𝑐𝑜𝑠 −1 − 2𝜋𝑑 𝛼 = −𝛽𝑑 𝑐𝑜𝑠∅𝑚𝑎𝑥 Total phase difference (The array phase in terms of the direction of main beam) is written as 𝜓 = 𝛽𝑑 𝑐𝑜𝑠∅ + 𝛼 ⟹ 𝛽𝑑 𝑐𝑜𝑠∅ − 𝛽𝑑 𝑐𝑜𝑠∅𝑚𝑎𝑥 𝝍 = 𝜷𝒅 (𝒄𝒐𝒔∅ − 𝒄𝒐𝒔∅𝒎𝒂𝒙 ); ∅𝑚𝑎𝑥 = 0, 𝜋 © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Direction of Maximum Radiation Two things can be noted from the equation. ◦ 1. The direction of the maximum radiation is independent of the number of elements in the array. ◦ 2. The direction of the main beam can be changed from 0 to 𝜋 by changing the progressive phase shift 𝛼 from −𝛽𝑑 to +𝛽𝑑. ◦ ∅𝑚𝑎𝑥 ⇒ (0 ⟶ 𝜋) ⟹ 𝛼 ⇒ (−𝛽𝑑 ⟶ +𝛽𝑑) For uniform array, there are two cases ◦ 1. End fire array (symmetry from axis ⟶ radiation goes along the axis) ⟹ ∅𝑚𝑎𝑥 = 0, 𝜋 𝜋 ◦ 2. Broad side array ( radiation goes perpendicular to the axis) ⟹ ∅𝑚𝑎𝑥 = ± the 2 broadside is a plane perpendicular to the array axis (see Fig below) © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Direction of Maximum (A). End fire array: (∅ Radiation = 0, 𝜋) ⟹ An array is said to be an end fire array if the main 𝑚𝑎𝑥 beam is along the axis of array. −𝛽𝑑; ∅𝑚𝑎𝑥 = 0 𝛼 = −𝛽𝑑 𝑐𝑜𝑠∅𝑚𝑎𝑥 ⟹ ቊ +𝛽𝑑; ∅𝑚𝑎𝑥 = 𝜋 Total phase difference 𝜓 = 𝛽𝑑 𝑐𝑜𝑠∅ + 𝛼 ⟹ 𝜓 = 𝛽𝑑𝑐𝑜𝑠∅ ± 𝛽𝑑 𝝍𝑬𝒏𝒅 = 𝜷𝒅 𝒄𝒐𝒔∅ ± 𝟏 𝜋 (B). Broadside side array: (∅𝑚𝑎𝑥 = ± ) ⟹ An array is said to be an Broadside array if 2 the main beam is goes perpendicular to the axis of array. 𝜋 𝛼 = −𝛽𝑑 𝑐𝑜𝑠 2 ⟹ 𝟎 Total phase difference 𝜓 = 𝛽𝑑 𝑐𝑜𝑠∅ + 0 ⟹ 𝜓 = 𝛽𝑑𝑐𝑜𝑠∅ 𝝍𝑩𝒔 = 𝜷𝒅𝒄𝒐𝒔∅ © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 MAXIMUM: 𝜓 Maximum occurred at ⟹ = ±𝑛𝜋; 𝑤ℎ𝑒𝑟𝑒 𝑚𝑛 = 1, 2, 3, … (for (AF)n=0/0 ) 2 ⟹ 𝜓 = 𝛽𝑑 𝑐𝑜𝑠∅ + 𝛼 = ±2𝑛𝜋; 𝑤ℎ𝑒𝑟𝑒 𝑛 = 1, 2, 3, … 1 𝜆 ∅𝑚𝑎𝑥 = 𝑐𝑜𝑠 −1 −𝛼 ± 2𝑛𝜋 = 𝑐𝑜𝑠 −1 −𝛼 ± 2𝑛𝜋 𝛽𝑑 2𝜋𝑑 For sin(Nx)/Nsin(x)maximum occurred at x=0 or ψ/2 = 0 i.e m=0 𝑛𝜆 𝑛𝜆 ∅𝑚𝑎𝑥 = ± 𝑐𝑜𝑠 −1 ∅𝑚𝑎𝑥 = 𝑐𝑜𝑠 −1 1− 𝑑 𝑑 n=1, 2,in3, Broadside Array (sources … 𝜶 = 0) phase End fire Array n=1, (𝜶 = - 2, βd)3, … 𝑛 ≠ 𝑁, 2𝑁, 3𝑁, … 𝑛 ≠ 𝑁, 2𝑁, 3𝑁, … © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Direction of Maximum Radiation An array is said to be End-fire array if the main beam is along the axis of the array. An array is said to be Broadside array if the main beam is perpendicular to the axis of the array. Fig: Broadside and end-fire direction of an antenna array © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Directions of Nulls The nulls of the radiation patterns can be obtained by equating the array factor to zero. ⟹ 𝐴𝐹 𝑛 = 0 or 𝐸𝑛 = 0 𝑁 𝐸0 sin 2 𝜓 𝐸𝑛 = 𝑁 sin 1𝜓 2 The directions of the nulls ∅𝑁𝑢𝑙𝑙𝑠 , is given as 𝑁 𝐸𝑛 = 0 ⟹ sin 𝜓 =0 2 𝑁 ⟹ 𝜓 = ±𝑚𝜋; 𝑤ℎ𝑒𝑟𝑒 𝑚 = 1, 2, 3, … 2 2𝑚𝜋 ⟹𝜓=± ; 𝑤ℎ𝑒𝑟𝑒 𝑚 = 1, 2, 3, … 𝑁 𝟐𝒎𝝅 Which can be simplified as 𝝍 = 𝜷𝒅 𝒄𝒐𝒔∅𝑵𝒖𝒍𝒍𝒔 − 𝒄𝒐𝒔∅𝒎𝒂𝒙 = ± 𝑵 𝟐𝒎𝝅 ⟹ 𝒄𝒐𝒔∅𝑵𝒖𝒍𝒍𝒔 = 𝒄𝒐𝒔∅𝒎𝒂𝒙 ± 𝜷𝒅𝑵 © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 NULLS: 𝑁 Nulls occurred at ⟹ sin 𝜓 =0 2 𝑁 ⟹ 𝜓 = ±𝑚𝜋; 𝑤ℎ𝑒𝑟𝑒 𝑚 = 1, 2, 3, … 2 2𝑚𝜋 ⟹𝜓=± ; 𝑤ℎ𝑒𝑟𝑒 𝑚 = 1, 2, 3, … 𝑁 2𝑚𝜋 ⟹ 𝛽𝑑 𝑐𝑜𝑠∅ + 𝛼 = ± ; 𝑤ℎ𝑒𝑟𝑒 𝑚 = 1, 2, 3, …(again n≠ 0 or N or 𝑁 2N…….this make (AF)n= 0/0 which is max condition) −1 1 2𝑚𝜋 −1 𝜆 2𝑚𝜋 ∅𝑁𝑢𝑙𝑙𝑠 = 𝑐𝑜𝑠 −𝛼 ± = 𝑐𝑜𝑠 −𝛼 ± 𝛽𝑑 𝑁 2𝜋𝑑 𝑁 −1 ± 𝑚𝜆 𝑚𝜆 Broadside ∅𝑁𝑢𝑙𝑙𝑠Array = 𝑐𝑜𝑠(sources in phase 𝜶 =0) End 𝑐𝑜𝑠 −1 fire=Array ∅𝑁𝑢𝑙𝑙𝑠 (𝜶 =- 1− β d) 𝑁𝑑 𝑁𝑑 m=1, 2, 3, … m=1, 2, 3, … 𝑚 ≠ 𝑁, 2𝑁, 3𝑁, … 𝑚 ≠ 𝑁, 2𝑁, 3𝑁, … © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Directions of Nulls Fig: Nulls of a linear array in three dimensional space © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Directions of Side-lobes Local maximum in the radiation pattern is called the side lobe. There is one side lobe between two adjacent nulls except the main beam. Whenever the numerator of the AF is maximum, there is a side-lobe in the radiation pattern. Generally, the directions of the side-lobes are taken approximately half way between the two adjacent nulls. 𝑁 Side-lobes occur: when ( 2 𝜓) = odd multiple of 𝜋/2 𝑁 1 ⟹ 𝜓 =± 𝑚+ 𝜋; 𝑤ℎ𝑒𝑟𝑒 𝑚 = 1, 2, 3, … 2 2 2𝑚+1 ⟹𝜓=± 𝜋; 𝑤ℎ𝑒𝑟𝑒 𝑚 = 1, 2, 3, … 𝑁 2𝑚+1 ⟹ 𝜓 = 𝛽𝑑 𝑐𝑜𝑠∅𝑆𝐿 − 𝑐𝑜𝑠∅𝑚𝑎𝑥 = ± 𝜋; 𝑁 𝟐𝒎 + 𝟏 𝝅 ⟹ 𝒄𝒐𝒔∅𝑺𝑳 = 𝒄𝒐𝒔∅𝒎𝒂𝒙 ± 𝑵 𝜷𝒅 m=1 ⟹direction of 1st SL, m=2 ⟹direction of 2nd SL, ……. © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Amplitudes of Side-lobes (Secondary Maxima for AF) The amplitude of the side lobe is obtained by substituting the value of 𝜓 in the AF. 𝑁 1 sin 2 𝜓 ⟹ 𝐴𝐹 𝑛 = 𝑁 sin 1 𝜓 2 Maximum amplitude of main beam =1 ⟹ m=0; Represents the main lobe: ⟹ only one maxima occurs Amplitude of 1st Side Lobe ⟹ m=1 𝑁 3𝜋 1 3𝜋 ⟹ 𝜓 =± ; and 𝜓 =± 2 2 2 2𝑁 3𝜋 1 sin ± 2 1 1 ⟹ 𝐴𝐹 𝑛 = = 𝑁 sin ±3𝜋 𝑁 sin ±3𝜋 2𝑁 2𝑁 3𝜋 3𝜋 3𝜋 For large array 𝑁 ≫ 1 ⟹ ± 2𝑁 is very small ⟹ sin ± 2𝑁 ≈ ± 2𝑁 1 1 2 ⟹ 𝐴𝐹 𝑛 ≈𝑁 3𝜋 ≈ 3𝜋 ≈ 0.2123 = −13 𝑑𝐵 (20log() as it is amplitude) ± 2𝑁 © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Amplitudes of Side-lobes (Secondary Maximandfor AF) 2 Amplitude of 2 Side Lobe ⟹ m=2 ⟹ 𝐴𝐹 𝑛 ≈ 5𝜋 ≈ 0.1273 2 Amplitude of 3rd Side Lobe ⟹ m=3 ⟹ 𝐴𝐹 𝑛 ≈ 7𝜋 ≈ 0.090.. 1 1 Amplitude of Nth Side Lobe ⟹ m=N ⟹ 𝐴𝐹 𝑛 = 𝑁 sin ± 𝑚+1 𝜋 2 𝑁 The first, second, third side lobe amplitudes are 2/3𝜋, 2/5𝜋, 2/7𝜋 respectively. The important thing to note: ◦ 1. The side lobe amplitudes are independent of the array size and the direction of the main beam. ◦ 2.Nor of SL are also independent of the no. of elements. 𝐼 ◦ 3. Level of SL is not depends on the array parameter (like 𝑑, 𝛼, or current ratio 𝐼2) 1 © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Half Power Beam Width (HPBW) HPBW of radiation pattern © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Radiation pattern of N- elements Array © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Half Power Beam Width (HPBW) ∅𝑛1 + − ∅𝑚𝑎𝑥 ≈ ∅𝑚𝑎𝑥 − ∅𝑛1 − © Kalasalingam academy of research and education ANTENNA AND Radiation PROPAGATION-ECE18R372 pattern showing direction of maximum radiations and nulls Half Power Beam Width (HPBW) 1 For HPBW: 𝐴𝐹 𝑛 = 2 𝑁 1 sin 2 𝜓 1 ⟹ = 𝑁 sin 1𝜓 2 2 (This equation can’t solved by analytically so solve by numerically) For large arrays (𝑁 ≫ 1), 1 ⟹ HPBW ≅ 𝐵𝑊𝐹𝑁. 2 1 ⟹ ∅2 − ∅1 ≅ ∅𝑛1 + − ∅𝑛1 −. 2 1 ⟹ ∅2 − ∅1 ≅ 2 ∅𝑛1 + − ∅𝑚𝑎𝑥 + ∅𝑚𝑎𝑥 − ∅𝑛1 −. ∅𝑛1 + − ∅𝑚𝑎𝑥 ≈ ∅𝑚𝑎𝑥 − ∅𝑛1 − (from above figure) 1 ⟹ ∅𝐻𝑃𝐵𝑊 ≅ 2 2(∅𝑛1 + − ∅𝑚𝑎𝑥 ) ≅ (∅𝑛1 + − ∅𝑚𝑎𝑥 ) © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 𝜆 Direction of nulls: 𝑐𝑜𝑠∅𝑁𝑢𝑙𝑙𝑠 = 𝑐𝑜𝑠∅𝑚𝑎𝑥 ± (Already computed) 𝑑𝑁 𝜆 cos∅𝑛1 + = 𝑐𝑜𝑠∅𝑚𝑎𝑥 ± (visible nulls for positive angle) 𝑑𝑁 𝜆 ⟹ cos∅𝑛1 + − 𝑐𝑜𝑠∅𝑚𝑎𝑥 = 𝑑𝑁 𝐴−𝐵 𝐴+𝐵 Using: 𝑐𝑜𝑠𝐴 − 𝑐𝑜𝑠𝐵 = 2 𝑠𝑖𝑛 𝑠𝑖𝑛 2 2 ∅𝑛1 + − ∅𝑚𝑎𝑥 ∅𝑛1 + + ∅𝑚𝑎𝑥 𝜆 ⟹ 2 𝑠𝑖𝑛 𝑠𝑖𝑛 = 2 2 𝑑𝑁 ∅𝑛1 + − ∅𝑚𝑎𝑥 2∅𝑚𝑎𝑥 + ∅𝑛1 + − ∅𝑚𝑎𝑥 𝜆 ⟹ 2 𝑠𝑖𝑛 𝑠𝑖𝑛 = 2 2 𝑑𝑁 ∅𝐻𝑃𝐵𝑊 2∅𝑚𝑎𝑥 +∅𝐻𝑃𝐵𝑊 𝜆 ⟹ 2 𝑠𝑖𝑛 𝑠𝑖𝑛 = 2 2 𝑑𝑁 ∅𝐻𝑃𝐵𝑊 ∅𝐻𝑃𝐵𝑊 𝜆 ⟹ 2 𝑠𝑖𝑛 𝑠𝑖𝑛 ∅𝑚𝑎𝑥 + = 2 2 𝑑𝑁 Using: 𝑠𝑖𝑛 𝐴 + 𝐵 = 𝑠𝑖𝑛A cosB + cosA sinB ∅𝐻𝑃𝐵𝑊 ∅𝐻𝑃𝐵𝑊 ∅𝐻𝑃𝐵𝑊 𝜆 ⟹ 2 𝑠𝑖𝑛 𝑠𝑖𝑛∅𝑚𝑎𝑥 𝑐𝑜𝑠 + 𝑐𝑜𝑠∅𝑚𝑎𝑥 𝑠𝑖𝑛 = 2 2 2 𝑑𝑁 © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Half Power Beam Width (HPBW) 𝑓𝑜𝑟 𝑁 ≫ 1 ⟹ ∅𝐻𝑃𝐵𝑊 ≪ 1 (𝑖𝑓 𝑥 𝑖𝑠 𝑣𝑒𝑟𝑦 𝑠𝑚𝑎𝑙𝑙 ⟹ 𝑠𝑖𝑛𝑥 ≈ 𝑥 𝑎𝑛𝑑 cos 𝑥 ≈ 1) ∅𝐻𝑃𝐵𝑊 ∅𝐻𝑃𝐵𝑊 𝜆 ⟹2 2 𝑠𝑖𝑛∅𝑚𝑎𝑥 ∗ 1 + 𝑐𝑜𝑠∅𝑚𝑎𝑥 ∗ 2 = 𝑑𝑁 𝟐𝝀 𝟐∅𝑯𝑷𝑩𝑾 𝒔𝒊𝒏∅𝒎𝒂𝒙 + ∅𝟐 𝑯𝑷𝑩𝑾 𝒄𝒐𝒔∅𝒎𝒂𝒙 = 𝒅𝑵 For a given array the HPBW is a function of the direction of the main beam. The HPBW is minimum for a broadside direction and maximum for the end-fire direction. The beam width monotonically increases as the main beam tilts towards the axis of the array. © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Half Power Beam Width (HPBW) A. For broad side array: (∅𝑚𝑎𝑥 = 𝜋/2) 2𝜆 ⟹ 2∅𝐻𝑃𝐵𝑊 sin 𝜋/2 + ∅2 𝐻𝑃𝐵𝑊 cos 𝜋/2 = 𝑑𝑁 𝝀 ⟹ ∅𝑯𝑷𝑩𝑾 = 𝒅𝑵 radian The HPBW for the broad-side array is approximately given as 𝜆 𝜆 ∅𝐻𝑃 𝐵𝑆 = ≈ 𝑑𝑁 𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝐴𝑟𝑟𝑎𝑦 𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝐴𝑟𝑟𝑎𝑦 (L) = 𝑁 − 1 𝑑 ≈ 𝑁𝑑; (𝑓𝑜𝑟 𝑁 ≫ 1) 𝜆 𝜆 𝜆 180 57.3 𝜆 ∅𝐻𝑃 𝐵𝑆 = ≈ 𝑟𝑎𝑑𝑖𝑎𝑛 ≈ × 𝑑𝑒𝑔𝑟𝑒𝑒 ≈ 𝑑𝑒𝑔𝑟𝑒𝑒 𝑑𝑁 𝐿 𝐿 𝜋 𝐿 © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Half Power Beam Width (HPBW) Note: 1 ◦ 1. ∅𝐻𝑃 𝐵𝑆 ∝ 𝑁; N = 𝑛𝑜. 𝑜𝑓 𝑎𝑟𝑟𝑎𝑦 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 1 ◦ 2. ∅𝐻𝑃 𝐵𝑆 ∝ 𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝐴𝑟𝑟𝑎𝑦 IN BS array, the no. of elements essentially play the role in deciding the angular zone in which effectively energy goes. The HPBW is inversely related to the array length. Larger the array narrower is the beam i.e., smaller HPBW. 𝑅𝑎𝑑𝑖𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑛𝑎𝑟𝑟𝑜𝑤𝑣𝑒𝑟 𝑀𝑜𝑟𝑒 𝑓𝑜𝑐𝑢𝑠 𝑖𝑛 𝑎 𝑁 ↑ ⟹ ∅𝐻𝑃 𝐵𝑆 ↓ ⟹ ⟹ 𝑖𝑛 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑧𝑜𝑛𝑒 𝑔𝑖𝑣𝑒𝑛 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 ⟹ 𝐷 ↑ ⟹ 𝐴𝑒 ↑ © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Half Power Beam Width (HPBW) B. For End Fire array (∅𝒎𝒂𝒙 = 𝟎, 𝝅) 2𝜆 ⟹ 2∅𝐻𝑃𝐵𝑊 𝑠𝑖𝑛0 + ∅2 𝐻𝑃𝐵𝑊 𝑐𝑜𝑠0 = 𝑑𝑁 2𝜆 2𝜆 ⟹ ∅2 𝐻𝑃𝐵𝑊 = ⟹ ∅ 𝐻𝑃𝐵𝑊 = radian 𝑑𝑁 𝑑𝑁 The HPBW for the end-fire array are approximately given as 2𝜆 2𝜆 ∅𝐻𝑃 𝐸𝐹 = = 𝑑𝑁 𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝐴𝑟𝑟𝑎𝑦 2𝜆 2𝜆 180 2𝜆 ∅𝐻𝑃 𝐸𝐹 ≈ 𝑟𝑎𝑑𝑖𝑎𝑛 ≈ × 𝑑𝑒𝑔𝑟𝑒𝑒 ≈ 57.3 × 𝑑𝑒𝑔𝑟𝑒𝑒 𝐿 𝐿 𝜋 𝐿 © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 1 First Null Beam Width (FNBW) Important Observation: The HPBW of the broadside array is less than that of the end-fire array but the directivity of the end-fire array is larger than the broadside array. ⟹ FNBW ≅ 2 ∗ 𝐻𝑃𝐵𝑊 ∅𝐹𝑁𝐵𝑊 ≈ 2∅𝐻𝑃𝐵𝑊 𝜆 114.6 𝜆 ±2 ∗ 𝑟𝑎𝑑𝑖𝑎𝑛 ≈ 𝑑𝑒𝑔𝑟𝑒𝑒 ⟹ 𝑓𝑜𝑟 𝑏𝑟𝑜𝑎𝑑 𝑠𝑖𝑑𝑒 𝑎𝑟𝑟𝑎𝑦 𝑑𝑁 𝐿 =. 2𝜆 2𝜆 ±2 ∗ 𝑟𝑎𝑑𝑖𝑎𝑛 ≈ 114.6 × 𝑑𝑒𝑔𝑟𝑒𝑒 ⟹ 𝑓𝑜𝑟 𝑒𝑛𝑑 𝑓𝑖𝑟𝑒 𝑎𝑟𝑟𝑎𝑦 𝑑𝑁 𝐿 © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Grating Lobe A grating lobe or grating beam is a beam identical to the main beam but in undesired direction. The power radiated by the array gets divided between the main beam and the grating lobe. The power efficiency in the direction of the main beam is consequently reduced. A grating lobe should be avoided in the radiation pattern. A grating lobe appears when 𝜓 = 2𝑚𝜋; Where 𝑚 is an integer. For broadside array the grating lobe appears when 𝑑 ≥ 𝜆, and for the end-fire array it appears when 𝑑 ≥ 𝜆/2. We can conclude that to avoid grating lobe in the radiation pattern for any array the inter- element spacing should be < 𝜆/2. 𝜆 𝑑 < 2 ⟹ 𝐺𝑟𝑎𝑡𝑖𝑛𝑔 ; 𝑙𝑜𝑏𝑒 𝑐𝑎𝑛 𝑏𝑒 𝑎𝑣𝑜𝑖𝑑𝑒𝑑 © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 Grating Lobe Figure 1: Radiation pattern and visible range of broad side array Figure2: Radiation pattern and visible range of end fire array © Kalasalingam academy of research and education ANTENNA AND PROPAGATION-ECE18R372 N- Elements Uniform Antenna Array Max occurred at ψ= 0 = 𝜓 = 𝛽𝑑 𝑐𝑜𝑠∅ + 𝛼 (for AF pattern ) Broad Side Array End fire Array To detect the first maximum toward ∅ = 𝟎 then 𝜓 = 𝛽𝑑 𝑐𝑜𝑠∅ + 𝛼 ⟹ 𝛼 = Since it is desired to have the first maximum detected toward ∅ = 𝜋 /2 − 𝛽𝑑 then Array Setting For 𝜓 = 𝛽𝑑 𝑐𝑜𝑠∅ + 𝛼 ⟹ 𝛼 = 0 broad side AF If theFor Setting firstEnd maxim is detected toward ∅ = 𝜋 then Fire © Kalasalingam academy of research and education pattern ANTENNA AND PROPAGATION-ECE18R372 AF pattern 𝜓 = 𝛽𝑑 𝑐𝑜𝑠∅ + 𝛼 ⟹ 𝛼 = It is required to study (AF)n 𝜓 Nulls: 𝑁 = ±𝑛𝜋, 𝑛 = 1, 2, 3,.. ≠ 0, 𝑁, 2𝑁, …. 2 𝜓 Maximum: =±m𝜋, m=0, 1, 2,..(0 for main lobe) 2 (Grating lobe condition (at m=1,2,3,…) 𝜓 3-dB point: 𝑁 = 1.39 2 Secondary Maximum for minor lobes: 𝜓 2𝑠+1 𝑁 = 𝜋, 𝑠 = 1, 2, 3,.. 2 2 𝜓 3π Maximum of first minor lobe occurred at 𝑁 =

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