Contents. 1.3 Array Synthesis Binomial Array Dolph-Chebyshev Array... 16

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1 Contents 1 Phased Array Antennas Introduction Array Analysis Discretization of Continuous Sources Discrete Uniformly Spaced Linear Array Discrete Uniformly Spaced Planar Array Beam Scanning using Progressive Phase Shift Grating-lobe Analysis Discrete Uniformly Spaced Linear Array Grating-lobe Analysis Discrete Uniformly Spaced Planar Array Array Synthesis Binomial Array Dolph-Chebyshev Array

2 Chapter 1 Phased Array Antennas 1.1 Introduction According to IEEE standard definition, array antenna is an antenna comprised of a number of identical 1 radiating elements in a regular arrangement and excited to obtain a prescribed radiation pattern. Why Array Antennas? Theory of array antennas is useful not only to understand discrete array antennas, but also to understand various commonly used antennas such as dipole antenna, horn antenna, etc after all, all these antennas are nothing but continuous arrays of Hertzian dipoles. For applications where narrow beam radiation patterns are required, antenna size has to be electrically large. That electrically large antenna could be a continuous source, for example, a parabolic reflector antenna. However, in the case of continuous sources, it is difficult to control amplitude and phase of the aperture distribution. So, such electrically large continuous radiating sources can be approximated by discretising 3 them. To understand the operation of such discrete arrays, once again we need to understand the theory of array antennas. Another important feature of array antennas is, the possibility of scanning the beam electrically. Unlike the conventional mechanical beam scanning antennas, array antennas can use the progressive phase shift concept to scan the beam within a fraction of second. 1 The term array antenna is sometimes applied to cases where the elements are not identical or arranged in a regular fashion. For those cases qualifiers shall be added to distinguish from the usage implied in this definition. For example, if the elements are randomly located one may use the term random array antenna. The regular arrangements possible include ones in which the elements can be made congruent by simple translation or rotation. 3 In the case of discrete arrays, one can control amplitude and phase distributions with the help of power dividing networks and phase shifters. 1

3 Analysis Versus Synthesis In the next section, definition of array factor will be provided. In the mean time, let us analyze the equation given for array factor a little bit. Array factor of a continuous linear array oriented along x axis is given as 4 AF k x = + A x exp jk x x dx, 1.1 where A x is the aperture distribution and k x = k 0 sin θ cos φ. From the above equation, it is clearly evident that the array factor and the aperture distribution form a Fourier transform pair. Here x and k x are analogous to t and ω, the parameters that we use in the signal processing and communication theory. Realizing this fact helps a lot in understanding the theory of array antennas, and the theory of antennas in general. So, ideally speaking, if array factor is known in the entire k x domain, one can obtain the corresponding aperture distribution by taking inverse Fourier transform as shown below: A x = 1 π + AF k x exp jk x x dk x 1. Here, basically 1.1 indicates the case of analysis whereas 1. indicates the synthesis case. In practice, it is possible to provide the array factor specification in the visible space region only i.e., k 0 k x k 0. So, as a part of synthesis, antenna engineer has to decide the pattern specification in the invisible region i.e., k x > k Array Analysis Consider a 3D array with all the array elements oriented along the same direction as shown in Fig Even though the scenario depicted in Fig. 1.1 is a discrete one, the concept presented in this section can be easily adapted for continuous arrays. Since all the elements are oriented along the same direction, one can apply the well known concept of pattern multiplication. So, from now onwards, let s assume that the array is simply made up of isotropic radiators. Array factor is defined as the cumulative contribution of all the isotropic radiating elements at the far-field point P r, θ, φ, and is given as AF = ] A n exp [jk 0 r r r n. 1.3 n In the above equation, the term k 0 r r r n represents the phase difference 5 between n th element and the reference point In the present case, origin is taken as the reference point, with 4 In electromagnetism, it is customary to use x, y, z and x, y, z notations for far-field and source positions, respectively. So, in this chapter the same notation will be followed. 5 Remember the equation for electrical length βl! For free space, the propagation constant β = k 0.

4 Reference Point Figure 1.1: A typical 3D array antenna consisting of elements oriented along the same direction out loss of generality. The term A n represents the amplitude as well as phase of the n th radiating element. Equation 1.3 can be generalized for continuous arrays 6 as AF = ] A r exp [jk 0 r r r dx dy dz. 1.4 In Cartesian co-ordinate system, source position is given as r = x ˆx + y ŷ + z ẑ. Similarly, the far-field position P is given as r = r sin θ cos φˆx+r sin θ sin φŷ+r cos θẑ. From these equations, r r can be approximated as shown below: r r = r sin θ cos φˆx + r sin θ sin φŷ + r cos θẑ x ˆx y ŷ z ẑ = r sin θ cos φ x + r sin θ sin φ y + r cos θ z = r rx sin θ cos φ ry sin θ sin φ rz cos θ + x + y + z r rx sin θ cos φ ry sin θ sin φ rz cos θ = r 1 x sin θ cos φ + y sin θ sin φ + z cos θ r r 1 1 x sin θ cos φ + y sin θ sin φ + z cos θ r [r x sin θ cos φ + y sin θ sin φ + z cos θ] Some times, array factor of a continuous array is also known as space factor. 3

5 Substituting 1.5 into 1.4 gives AF = A r exp [jk 0 x sin θ cos φ + y sin θ sin φ + z cos θ] dx dy dz = A r exp [jk x x + jk y y + jk z z ] dx dy dz, 1.6 where k x = k 0 sin θ cos φ, k y = k 0 sin θ sin φ, and k z = k 0 cos θ. From the above equation, one can see that the aperture distribution A r and the array factor AF form a Fourier transform pair. Example 1. Calculate the array factor of a continuous linear array oriented along z axis and having an aperture distribution { A z 1, z L = 0, elsewhere. Also calculate a the directions in which array factor exhibits zero radiations i.e., null directions, and b null-to-null beamwidth. Solution: From 1.6, AF k z = +L/ L/ = ejkzz jk z 1 e jkzz dz +L/ L/ = ejkzl/ e jkzl/ jk z = j sin kzl = sin jk z Since k z = k 0 cos θ, AF can be written in terms of θ and φ as AF θ, φ = sin k 0 cos θl. k 0 cos θ kzl k z. From the above equation, it can be seen that array factor is not a function of φ because array is along z axis can you visualize the cylindrical symmetry along z axis?!. Now, null positions θ null can be obtained from the above equation as k 0 cos θn null L θ null n = ±nπ, where n = 1,, 3,. nπ = ± cos 1 k 0 L Note that n = 0 actually corresponds to the main beam positions. Finally null-to-null beamwidth is nothing but θ1 null, i.e., cos 1 π. k 0 L 4

6 Figure 1.: A x and AF k x corresponding to acontinuous source 1..1 Discretization of Continuous Sources From the Fourier transform relation between A and AF, it can be concluded that the beam-width of the radiation pattern is inversely proportional to the antenna s electrical length. So, in order to achieve very narrow beam widths, continuous apertures should be electrically large. However, it is very difficult to implement such electrically large continuous aperture distributions. So, antenna engineer should opt for discrete antenna arrays instead. Discrete array concept is very similar to the concept of Nyquist-Shannon sampling technique in communication theory and is explained in the following paragraph Discrete Uniformly Spaced Linear Array Let us consider a linear continuous aperture distribution oriented along x axis and the corresponding array factor shown in Fig. 1.. If this continuous source is discretized with an uniform spacing of a in between the elements, then the corresponding array factor can be obtained from 1.6 as AF k x = A m exp jk x x m = A m exp jk x ma. 1.7 m From the above equation, it can be seen that AF k x is a periodic 7 function in the k x domain 7 Array factor corresponding to a finite length continuous source oriented along x axis spreads from to +. 5

7 Figure 1.3: A x and AF k x corresponding to a uniformly spaced discrete array as shown in Fig Also, one can observe that the period of the array factor in the k x domain is π/a. Even though array factor is periodic and extends from to +, from application point of view k x is confined in the domain [ k 0, +k 0 ], because k 0 sin θ cos φ 1 for the entire visible domain 0 θ π and 0 φ π. This domain [ k 0, +k 0 ] is commonly known as visible-space in antenna theory and is highlighted in Fig Also seen in the figure are infinite grating-lobes placed on both sides of the main beam. If the spacing between array elements a increases, then the separation between the main-lobe and the nearest grating lobes decreases. However, in order to avoid ambiguity in applications such as radar, grating-lobes should not enter into the visible space. So, spacing should be chosen as small as possible. However, if spacing between elements is too small, then the packing density as well as the mutual coupling between the radiating elements increases. Increased mutual coupling increases the effective reflection coefficient and thus reduces the efficiency of the array. So, spacing between elements a can neither be small nor be large. So, an optimum spacing has to be chosen by antenna engineer by taking many things into consideration. This concept will be further explored in Sec Example. Calculate the array factor of a discrete uniformly spaced linear array oriented along x axis and having uniform aperture distribution. Assume that the array is having M number of elements M can be wither even or odd and the spacing between any two consecutive elements is a. Also calculate a the directions in which array factor exhibits zero radiations i.e., null directions, and b null-to-null beamwidth. So, for any discrete array, there always exists some aliasing effect as shown in Fig

8 N - Odd N - Even Figure 1.4: Even and odd linear arrays and the corresponding index values Solution: First of all one should decide the indexes of array elements. If M is odd m = 0, ±1, ±,, else m = ± 1, ± 3, as can be seen from Fig Now, using the definition of array factor given by 1.7 gives + M 1 AF k x = A m e jkxma. M 1 Remember that the above equation is valid for both even as well as odd cases. In either case, using the geometrical progression summation identity N 1 n=0 a 0r n 1 r = a N 0 and substituting 1 r A m = 1, one can prove that AF k x = = + M 1 A m e jkxma M 1 + M 1 M 1 e jkxma M 1 = e jkx a M 3 + e jkx a M e jkx a + e M 1 = e jkx [ a 1 + e jkxa + + e jkxm a + e jkxm 1a] M 1 = e jkx a [ 1 e jk xma ] M 1 = e jkx a e jkxma/ 1 e jkxa [ ] e jk xma/ e jkxma/ e jkxa/ e jkxa/ e jkxa/ [ ] e jk xma/ e jkxma/ = = sin k xma e jkxa/ e jkxa/ sin k xa. 7 M 1 jkx a

9 Similar to Ex. 1, null positions can be obtained as k null xn Ma k null xn k 0 sin θ null n = ±nπ, where n = ±1, ±, = ± nπ Ma So, null-to-null beamwidth is given as sin 1 = ± nπ nπ Ma θnull n = ± sin 1 Mak 0 π Mak 0 Example 3. Prove that array factor of any odd numbered array located symmetrically with respect to origin will be periodic in k x domain with a periodicity of π. Also prove that array factor a of any even numbered array located symmetrically with respect to origin will be periodic in k x domain with a periodicity of 4π a Solution: From 1.7, array factor is defined as.. AF k x = m A m exp jk x x m = A m exp jk x ma. Since for odd numbered arrays, m = 0, ±1, ±,, array factor for odd numbered arrays is given as AF Odd k x = + M 1 M 1 = A M 1 A m exp jk x ma [ ] M 1 exp jk x a A +1 exp +jk x a + + A + M A 1 exp jk x a + A 0 + [ ] M 1 exp jk x a. All the terms in the above summation are of the form exp jk x ma, where m is an integer. So, periodicity of such terms in k x domain will be π. So, greatest period is equal to π, which is ma a the periodicity of the array factor. In contrast to the odd numbered case, for even numbered arrays, m = 0, ± 1, ± 3. So, array factor for even numbered arrays is given as AF Odd k x = + M 1 M 1 = A M 1 A + 1 exp A m exp jk x ma [ exp M 1 jk x +jk x 1 a ] a + + A + M A 1 exp exp jk x 1 [ jk x M 1 a ] a. +

10 a b Figure 1.5: Array factors corresponding to uniform excitations with a = λ/ : a M = 10 b M = 11. All the terms in the above summation are of the form exp jk x ma, where m is always a fraction, where m { ± 1, ± 3, }. So, periodicity of such terms in k x domain will be π. So, greatest ma period is equal to π = 4π, which is the periodicity of the array 1/a a factor Discrete Uniformly Spaced Planar Array Discretization of continuous planar sources is similar to the case of linear arrays discussed in Sec The only difference is that for planar arrays one should use D Fourier transform instead of 1D. Let s consider the discrete uniformly spaced planar array shown in Fig It can be noticed that the planar array is having a very general lattice shape. If γ = 90, then it simply corresponds to the planar array with rectangular lattice. For the planar array, it can be showed qb tan γ that y m = qb and x m = pa +. Substituting these values into 1.6 gives AF k x, k y = A pq exp jk x x pq + jk y y pq p q = A pq exp [jk x pa + p q qb ] + jk y qb. 1.8 tan γ For rectangular arrays, γ = 90. So, the above equation reduces to AF k x, k y = A pq exp [jk x pa + jk y qb]. 1.9 p q Analogous to the linear case, planar discrete arrays also exhibit grating-lobe phenomena in k x k y space. It can be understood by observing 1.8. Notice that AF k x, k y given by 1.8 will have 8 In Fig. 1.5, horizontal axis indicates normalized k x values, i.e., k x /k 0 = sin θ cos φ = sin θ. So, in this domain periods will be π 4π k 0a and k 0a, for odd numbered and even numbered arrays, respectively. 9

11 VISIBLE-SPACE DISK ARBITRARY SCAN SPECIFICATION Figure 1.6: Discrete uniformly spaced planar array placed in the xy plane and the corresponding k x k y domain peaks when the following conditions are satisfied simultaneously: k x a = µπ and k xb tan γ + k yb = νπ, where µ, ν = 0, ±1, ±,. Rewriting the above set of equations gives { kx = µπ a k y = cotγk x + νπ b = tan [ 90 γ] k x + νπ b So, for discrete planar arrays, grating-lobes occur at the intersection points of the above set of straight lines as can be seen in Fig One more thing that can be noticed in Fig. 1.6 is the visible space disk. From the definitions k x = k 0 sin θ cos φ and k y = k 0 sin θ sin φ, one can derive the following properties: { k x + k y=k 0 sin θ tan 1 ky k x = φ So, the visible space i.e., 0 θ π and 0 φ π corresponds to the disk k x + k y k 0 as sin θ 1. This fact and the above properties are represented graphically in Fig The concept of visible-space will be explored further in the Sec Example 4. Calculate the array factor of a discrete uniformly spaced planar array assume rectangular lattice placed in the x y plane and having uniform aperture distribution along both x and y axes. Assume that number of elements and spacing along x axis are M and a, respectively. Similarly, Assume that number of elements and spacing along y axis are N and b, respectively. 10

12 VISIBLE-SPACE DISK Figure 1.7: Visible space representation in the k x k y domain Solution: Planar array having rectangular can be considered as a linear array of linear arrays. So, First one needs to calculate array factor along x-axis. And then using pattern multiplication principle, this array factor should be multiplied with the array factor corresponding to the distribution along y-axis. So, array factor for the given planar array is given as AF k x, k y = sin k xma sin sin k xa sin kynb kyb. 1.. Beam Scanning using Progressive Phase Shift One main type of array antennas is phased array antenna, in which case each element of the array is excited with proper progressive phase shift so that the the main-beam steers in the desired direction. In order to explain the beam scanning using progressive phase shift, the following frequency-shifting property of the Fourier transform 9 can be used: If f t F ω, then f t exp ω 0 t F ω ω In array antenna theory, x, y, z and k x, k y, k z are analogous to time and angular frequency parameters t and ω, respectively. So, analogous to the above frequency-shifting theorem, for 9 Here Fourier transform is defined as F ω = f t exp +ωt dt. 11

13 Figure 1.8: Beam scanning using progressive phase shift array antennas, if A x, y, z AF k x, k y, k z, then A x, y, z exp k x0 x k y0 y k z0 z AF k x k x0, k y k y0, k z k z From 1.13, it is evident that the array factor can be shifted/scanned in the k x, k y, k z domain by introducing proper linear progressive phase shifts along x, y, and z - directions Grating-lobe Analysis Discrete Uniformly Spaced Linear Array Now that the general theory for beam scanning is explained, in this section grating-lobe analysis will be presented for linear arrays. As mentioned earlier, for discrete arrays, spacing should neither be large nor be small. Antenna engineer has to decide an optimal spacing depending up on several factors, such as maximum scanning angle, packing density, cost of the antenna system, etc. In this section, theory is provided for deciding the optimum spacing for discrete uniformly spaced linear arrays. 1

14 It is known that by providing proper progressive phase shift, array factor can be shifted in the k x k y domain. For a given spacing a, if the array factor is shifted as shown in Fig. 1.8, then the maximum shift that can be done without grating-lobes entering into the visible space is π k a 0. So 10, π k x0 a k 0 π k 0 sin θ 0 a k 0 π θ 0 sin 1 1 ak 0 θ 0,max = sin 1 π ak However, in many applications θ 0,max is usually specified and one needs to decide the optimum spacing for the given maximum scan angle. In such cases, π k x0 a k 0 π k 0 sin θ 0,max a k 0 a 1 π k sin θ 0,max a max = 1 π k sin θ 0,max Example 5. For a given linear array oriented along x axis and having an uniform spacing of 15mm, calculate the progressive phase shift required to scan the beam to from broad-side direction i.e., θ = 0 to a θ 0 = 30 b θ 0 = 90. Assume that the operating frequency of the array is 10 GHz. Solution: From 1.13 as well as Fig. 1.8, progressive phase shift required to scan the beam to an arbitrary direction θ 0 is k x0 a = k 0 sin θ 0 a. Also, wavelength at 10 GHz in free-space is λ = c = 30mm. So, a = λ and progressive phase shifts can be calculated as shown below: f a progressive phase required to scan to θ = 30 is k 0 sin θ 0 a = π λ sin 30 λ = π b progressive phase required to scan to θ = 90 i.e., end-fire direction is k 0 sin θ 0 a = π λ sin 90 λ = π. 10 Due to cylindrical symmetry, for linear arrays along x axis, it is sufficient to know the array factor for φ = 0. So, in this section k x = k 0 sin θ cos φ = k 0 sin θ. 13

15 Example 6. For a given linear array oriented along x axis and having an uniform spacing of a = 0.75λ, calculate the maximum scanning that can be done so that grating-lobe doesn t enter into the visible-space. Solution: From 1.14, π θ 0,max = sin 1 1 ak 0 1 = sin Example 7. If the main beam corresponding to a given linear array has to be scanned to a maximum angle of θ 0 = 40, then what should be the maximum possible uniform spacing between consecutive elements. Solution: From 1.14, a max = 1 k 0 π 1 + sin θ 0,max = λ π π 1 + sin 40 = λ sin Grating-lobe Analysis Discrete Uniformly Spaced Planar Array 0.609λ. From 1.11, for a general two-dimensional lattice structure described in Fig. 1.6, grating lobes occur in the k x k y domain at k xg µ = k x0 + µπ a k yg µ,ν = k y0 + νπ b [ ] kxg µ k x0, 1.16 tan γ where k x0 and k y0 are the shifts along k x and k y axes, respectively see Sec Since by definition, k x = k 0 sin θ cos φ and k y = k 0 sin θ sin φ, radiating far fields are confined to the circular disk k x + k y 1/ k0, often known as visible space as mentioned earlier. The remaining k x k y space, described as invisible space, is related to the stored energy in the near field region, which is analogous to the phenomenon of evanescent modes in a waveguide. Usually, an antenna engineer intends to avoid the appearance of all the grating lobes except for the µ = ν = 0, main lobe case within the visible space. Fig. 1.6 shows the visible space and the grating lobe spaces placed according to 1.16 in the k x k y domain. Assuming that V 1 represents the domain of the specified main lobe scan positions, the closed loops G1 G6 represent contours of all the possible nearest grating lobe scan positions. From Fig. 1.6, it can be observed that all the grating lobe contours are just touching the visible space circle, except for G and G5. Thus, for the given scan specification V 1, the array lattice arrangement shown in Fig. 1.6 is not optimal. In this chapter, optimal array arrangement is defined as the configuration that maximizes the array s unit cell area ab. In other words, an optimal configuration minimizes the number of elements needed in a given array aperture. To achieve this optimal array lattice configuration, Fig. 1b should be modified according to the following description. 14

16 1. All the left-hand side grating lobe contours should be moved upward and the right-hand side contours downward.. After an optimal skew with respect to the k y axis, both left-hand side and right-hand side grating lobe contours should be moved horizontally toward each other, so that all would just touch the visible space circle. 3. If V 1 is symmetric with respect to both the k x and the k y axes, then the contours G, G3, G5 and G6 will be at the same distance from the k x axis for the optimal array lattice configuration. This optimal array element arrangement is a hexagonal array lattice γ = tan 1 b a. For further information related to grating lobe analysis, please see [1]. 1.3 Array Synthesis In the previous section, analysis of arrays with uniform excitation has been provided. However, SLR achieved with uniform aperture distributions is approximately 13 db only, which is not acceptable in many radar applications. In order to overcome this disadvantage, one has to opt for some other distributions, such as Binomial, Dolph-Chebyshev, Taylor n, etc. In this chapter theory is provided to synthesize binomial and Dolph-Chebyshev arrays only. For Taylor n and some advanced synthesis methods, please see [] Binomial Array For array synthesis cases, array factor is provided at least partially and one has to calculate the 1D aperture distribution using the formula A x = 1 π + For binomial arrays, array factor is given as AF k x exp jk x x dk x AF k x = e jkxa e jk x0a M Reasons for choosing the above array factor are: array factor should be a periodic function in k x domain with a period of π a array factor should have an order of M 1 in e jkxa domain all the zeros should present at one single location, i.e., k x = k x0, so that array factor will not have any side-lobes For synthesizing such an array, one can apply either the Fourier transform given by 1.17 or the definition of array factor AF k x = A m e jkxma itself. We will follow the second method which is easier than calculating Fourier series coefficients of The method to evaluate the array coefficients corresponding to binomial array is explained by using the following example. 15

17 a b Figure 1.9: A figure depicting the mapping process AF k x = T M 1 c cos k xa, where u = k x /k 0, a = λ/, and M is an odd number. Example 8. Design a 5-element Binomial array having all the zeros at θ 0 = 30 and having an uniform spacing of a = λ/. Solution: From 1.18 array factor is given as, F k x = e jkxa e jk x0a M 1 = e jkxa e j π λ sin 30 λ = e jkxa e j π = e jkxa j = e jkx4a + 4e jkx3a j + 6e jkxa j + 4e jkxa j 3 + j 4 = e jkx4a 4je jkx3a 6e jkxa + 4je jkxa + 1. Comparing the above equation with array factor definition AF k x = A m e jkxma gives 1.3. Dolph-Chebyshev Array A 4 = 1, A 3 = 4j, A = 6, A 1 = 4j, and A 0 = 1. Even though binomial array pattern doesn t contain any side-lobes, it suffers from broad beamwidth, which in turn reduces the directivity. So, for many practical applications binomial distribution is not an optimal distribution. An optimal distribution which exhibits the maximum possible directivity for a given SLR is Dolph-Chebyshev distribution [3]. Array factor corresponding to Dolph-Chebyshev array is given as AF k x = T M 1 c cos k xa

18 a b Figure 1.10: A figure depicting the mapping process AF k x = T M 1 c cos k xa, where u = k x /k 0, a = λ/, and M is an even number. where Chebyshev polynomial T N x is defined as T N x = { cos N cos 1 x, x 1 cosh N cosh 1 x, x The parameter c is decided by the given SLR. If the given SLR is R in linear scale, then c is decided such that T M 1 c = R cosh [ M 1 cosh 1 c ] = R c = cosh cosh 1 R. 1.1 M 1 Both Chebyshev polynomials as well as the mapping given by 1.19 are given in Fig. 1.9 and 1.10 for odd and even arrays, respectively. Reasons for choosing array factor given by 1.19 are as given below: array factor should be a periodic function in k x domain with a period of π a even numbered arrays array factor should have an order of M 1 in e jkxa domain ] region of kx do- [ c, +c] region of Chebyshev polynomial should get mapped to [ 0, π a main as shown in Fig π a for odd

19 In order to synthesize Dolph-Chebyshev array for a given set of SLR and M values, all we need to know is M 1 zeros of array factor These zeros can be obtained as shown below: T M 1 c cos knull x a = 0 ] cos [M 1 cos c 1 cos knull x a = 0 M 1 cos 1 c cos knull x,n a = ± n 1 π, where n = ±1, ±, ±3, k null x,n = a cos 1 { 1 c cos [ n 1 M 1 ]} π. 1. Once array factor zeros are obtained, these zeros should be converted into zeros in Z transform domain. Till now, it has been emphasized that AF is nothing but Fourier transform of aperture distribution A x, y, z. This Fourier transform can be re-written in terms of Z transforms as shown below: AF k x = A m e jkxma AF Z = A m Z m, where Z = e jkxa. 1.3 So, converting array factor zeros obtained using 1. into Z domain gives Z null n = e jknull x,n a. 1.4 Finally, one can obtain array coefficients by expanding array factor obtained by AF Z = Z Z null n. 1.5 An example describing the method given in this section is given below.e. Example 9. Design a 5-element Dolph-Chebyshev array having SLR of 30 db and spacing a = λ/. Solution: First of all, we need to calculate the parameter c using 1.1 and the given SLR in db as shown below: So, R = 10 SLR/0 = = c = cosh cosh 1 R = M 1 We need to knowm 1 zeros i.e., 4 zeros of the array factor to calculate A m values. However, due to symmetry, it is sufficient to calculate of zeros. Because remaining two zeros 18

20 are nothing but mirror images of the first two with respect to k x = 0. So, from 1. { [ ]} 1 n 1 π kx,n null a = cos 1 c cos, n = 1, M 1 = cos 1 [ cos π 8 = and.654. So, first two zeros in the Z domain are given as Due to symmetry, other two zeros are given as So, from 1.5, array factor is given as AF Z = Z Z null n ] and cos 1 [ cos Z null 1, = e j1.899, and e j.654. Z null 1, = e j1.899, and e j.654. = Z e j1.899 Z e j1.899 Z e j.654 Z e j.654 = Z + 1 Z cos Z + 1 Z cos.654 ] 3π 8 = Z 4 Z 3 cos cos Z Z Z cos cos = Z Z Z +.41Z + 1. Comparing the above equation with the definition of array factor AF Z = A m Z m gives A 4 = 1, A 3 =.41, A = 3.14, A 1 =.41, A 0 = 1. 19

21 0

22 Bibliography [1] S. R. Zinka, I. B. Jeong, J. H. Chun, and J. P. Kim, A novel geometrical technique for determining optimal array antenna lattice configuration, IEEE Transactions on Antennas and Propagation, vol. 58, no., pp , [] A. K. Bhattacharyya, Phased Array Antennas, Floquet analysis, Synthesis, BFNs, and Active Array Systems. Hoboken, NJ: John Willey, [3] C. L. Dolph, A current distribution for broadside arrays which optimizes the relationship between beam width and side-lobe level, Proceedings of the IRE, vol. 34, no. 6, pp ,